Mathematics Analysis and Approaches for the IB Diploma Higher Level Worked Solutions

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WORKED SOLUTIONS

Exercise 1.1 1.

In each of the following, make the indicated letter the subject of the left side of the equality: (a) (b) (c) (d) (e)

if r is a length, then

f g

h k

fk gh

k

gh f

(f) (g) (h)

2.

Factor out k at the denominator of the fraction:

In parts (a) to (d), find m, the slope of the line, using

, then find the y-intercept c,

by choosing one of the given points and substituting its coordinates and the value of the slope into the slope-intercept form of the equation of a line, . (a)

(b)

EITHER:

OR:

The given points have the same y-coordinate, so the line is horizontal, the equation of the line is

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(c)

(d)

By inspection, the given points have the same x-coordinate, so the line is vertical, and its equation is

(e) Parallel lines have the same slope, consequently the slope of the required line is

(f) The slope of a line perpendicular to another line with slope is

3.

In each of these, substitute the coordinates of the given points in the relevant formulae: (a)

(b)

(c)

(d)

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4.

Substitute the coordinates of the given points in the distance formula and set it equal to 5: (a)

(b)

5.

Use the distance formula to show that properties regarding the length of sides for the indicated shapes hold true. (a)

Let , and The lengths of the sides are:

be the vertices of triangle

If the

.

theorem

, then

is a right-angled triangle.

is a right-angled triangle. (b)

Let

,

and

be the vertices of triangle

.

is isosceles, because two of its sides have the same length. (c)

Let If

,

, and

and , then

be the vertices of quadrilateral is a parallelogram,

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6.

(a)

Subtract the two equations to eliminate x:

Substitute

into either of the original equations to find x:

The solution is (b)

Multiply the second equation by 3 and add it to the first equation to eliminate y:

Substitute

into either of the original equations to find x:

The solution is (c)

Multiply the first equation by 4 and the second equation by 3, then subtract the two equations, to eliminate y:

Substitute

into either of the original equations to find x:

The solution is (d)

Multiply the second equation by 4 and add it to the first equation to eliminate x:

This is not true, so the system of equations has no solution.

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(e)

Multiply the first equation by 5 and the second equation by 7, then add the two equations, to eliminate y:

Substitute

into either of the original equations to find y:

The solution is 7.

(a)

Make y the subject in the first equation:

Substitute the expression of y in the second equation and solve for x:

Substitute

into either of the original equations to find y:

The solution is (b)

Make y the subject in the second equation:

Substitute the expression of y in the first equation and solve for x:

Substitute

into the expression of y:

The solution is (c)

Divide the first equation by 2:

Make x the subject in this equation:

Substitute the expression of x in the second equation and solve for y: true for all This means that there are an infinite number of solutions, due to the fact that the two equations are multiples of each other (the lines representing the two equations are coincident). It follows that the solution of this system is the set of all points on the line with equation (or , or ) © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

(d)

Make y the subject in the second equation:

Substitute the expression of y in the first equation and solve for x:

Substitute

into the expression of y:

The solution is (e)

Make y the subject in the second equation:

Substitute the expression of y in the first equation and solve for x:

Substitute

into the expression of y:

The solution is (f)

Multiply the second equation by 10:

Make y the subject in this equation:

Substitute the expression of y in the first equation and solve for x:

Substitute

into the expression of y:

The solution is

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8.

9.

(a)

The solution is

(b)

The solution is

(c)

The solution is

(a)

The solution is

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(b)

The solution is (c)

infinite solutions (d)

The solution is

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10.

(a)

The solution is

(b)

The system of equations has no solution.

(c)

The solution is

(d)

Infinite solutions.

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11.

The last equation is:

In order for the system to have no solutions, the coefficient of z must be 0, and the right side of the equation should not be equal to 0.

When

, the right-hand side of the equation is: the required value is

Exercise 1.2 1.

(a)

(i)

Graph G

(ii)

represents a straight line with a positive slope , passing through the origin

(b)

(i)

Graph L

(ii)

(c)

(i)

Graph H

(ii)

represents a horizontal line with positive slope y-intercept

represents a straight line , and with a negative

(d)

(i)

Graph K

(ii)

represents a circle centred at the origin and with radius 2 (this is not a function)

(e)

(i)

Graph J

(ii)

represents a straight line with a negative slope

(f)

(i)

Graph C

(ii)

(g)

(i)

Graph A

(ii)

(h)

(i)

Graph I

(ii)

, and with a positive y-intercept

represents a parabola opening upwards, with vertex at

represents a rational function with asymptotes and

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(i)

(i)

Graph F

(ii)

represents a parabola opening downwards, with vertex at

2.

3.

The area of a triangle is

, where b and h are the base and the height of the triangle,

respectively. Let l be the side of the equilateral triangle. When drawing one of the heights, a right-angled triangle is formed with sides h, l and . Us

l 2

4.

h:

The area of the pavement can be calculated as the difference between the areas of two rectangles:

5.

6.

(a)

Substitute the given values to find k:

(b) 7.

(a) (b) (c)

8.

(a)

The domain is the set of all x-values:

(b)

The radius is a length, so

(c)

f is a linear function, so

(d)

h is a quadratic function, so

(e)

The radicand must be greater or equal to zero, so

(f)

All real numbers can be cube rooted, so

(g)

f is a rational function; its denominator cannot be equal to 0.

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(h)

There are two conditions: x cannot be 0, as it is at the denominator of a fraction, and , consequently the domain is ,

9.

No, because a line with equation , where c is a constant, fails the requirement of a function to map one x-value to only one y-value.

10.

(a)

Substitute the given x-value into the expression of the function: (i) (ii) (iii)

11.

(b)

The radicand must be negative for the function to be undefined, so

(c)

The domain of h is given by , the range is (the square root of any positive number is positive, the square root of 0 is 0).

(a)

(i)

is a rational function, its denominator cannot be equal to 0, so as

(ii)

the domain is . The range of f is , as the function can take all real values, except 0, .

To find the x-intercept, set y (or

) equal to 0, and solve for x:

no solution, so no x-intercept. To find the y-intercept, set x equal to 0, and calculate y: the y-intercept is the equation of the vertical asymptote is

.

the equation of the horizontal asymptote is The graph of the function is:

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(b)

(i)

is a rational function with a square root at the denominator, so the radicand has to be strictly positive, so meaning the domain is The values given by function g will be strictly positive, because the square root at the denominator will always result in a positive number, so the range of g is

(ii)

To find the x-intercept, set y (or

) equal to 0, and solve for x:

no solution, so no x-intercept. The y-intercept is found when , but this value for x is not in the domain of the function, so there is no y-intercept. there are two vertical asymptotes: the equation of the horizontal asymptote is The graph of the function is:

(c)

(i)

is a rational function, its denominator cannot be equal to 0, so

the domain is

.

, this means that the range of h is (ii)

To find the x-intercept, set y (or

) equal to 0, and solve for x: so the x-intercept is

To find the y-intercept, set x equal to 0, and calculate y: the y-intercept is the equation of the vertical asymptote is

.

the equation of the horizontal asymptote is The graph of the function is:

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.

(d)

(i)

is a square root function, so the radicand must be positive or 0:

the domain is The values given by function g will be positive, because the square root at the denominator will always result in 0 or a positive number, but, because the radicand is a quadratic expression, the range of must also be considered. The x-coordinate of the vertex of the downwards parabola representing is consequently the y-coordinate is meaning the y-values given by g are always less than 5, This means that the range of p is (ii)

To find the x-intercept, set y (or

the x-intercepts are

) equal to 0, and solve for x:

and

.

To find the y-intercept, set x equal to 0, and calculate y: the y-intercept is This function does not have any asymptotes. The graph of the function is:

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(e)

(i)

is a rational function, its denominator cannot be equal to 0, so

the domain is

.

, this means that the range of f is (ii)

To find the x-intercept, set y (or

) equal to 0, and solve for x:

the x-intercept is

.

The y-intercept is found when , but this value for x is not in the domain of the function, so there is no y-intercept. The equation of the vertical asymptote is asymptote is ( ).

, the equation of the horizontal

The graph of the function is:

Exercise 1.3 1.

(a) (b) (c) (d)

2.

(a) (b) (c) (d) (e) (f)

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(g) (h)

(i) (j)

3.

In each question, when finding the domain of

, check the following two conditions:

the input x is in the domain of , because the first rule to be applied to x is g, the inside function of . the output is in the domain of f (the range of g must be either equal to or a subset of the domain of f ). The same applies when finding the domain of : x must be in the domain of , because the first rule to be applied to x is f, the inside function of , and is in the domain of g. (a)

f

g

Domain Range

of

is also

, this set of values is the same as the domain for f, so the domain .

of

is also

, this set of values is the same as the domain for g, so the domain .

(b) f

g

Domain Range

, this set of values is the same set of values as the domain for f, so the domain of is also . © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

of

, this set of values is a subset of the domain of g, so the domain .

is also

(c) f

g

Domain Range

of

, this set of values is a subset of the domain of f, so the domain .

is also

domain of

, this set of values is a subset of the domain of g, so the is also .

(d)

f

g

Domain Range

, but this set of values is not a subset of the domain of f. To be able to compose the two functions, the x-value, which is the input of g resulting in an output of , must be excluded from the domain of g. , so the domain of

is

.

, this set of values is a subset of the domain of g, so the domain of

is also

.

(e) f

g

Domain Range

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, this set of values is the same as the domain for f, so the domain of is also

of

.

, this set of values is the same as the domain for g, so the domain .

is also

(f) f

g

Domain Range

, this set of values is a subset of the domain of f, so the domain of

of

is also

.

is also

, this set of values is the same as the domain for g, so the domain .

(g) f

g

Domain Range

, but this set of values is not a subset of the domain of f. To be able to compose the two functions, the x-value, which is the input of g resulting in an output of must be excluded from the domain of g. , so the domain of

is:

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.

, but this is not a subset of the domain of g. To be able to compose the two functions, the x-value, which is the input of f resulting in an output of must be excluded from the domain of f. , so the domain of

(h)

f

is:

.

g

Domain Range

, this set of values is the same as the domain for f, so the domain of

is also

.

, this set of values is the same as the domain for g, so the domain of

is also

.

(i) f

g

Domain Range

, but this set of values is not a subset of the domain of f. To be able to compose the two functions, the x-value, which is the input of g resulting in an output of must be excluded from the domain of g. , so the domain of

is

.

, this set of values is a subset of the domain of g, so the domain of

is also

.

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4.

(a)

g

h

Domain Range

, this is not a subset of the domain of f. To be able to compose the two functions, only the x-values which are common to both the domain of g (real numbers greater than or equal to 1) and the range of h (real numbers less than or equal to 10) are acceptable, so only real numbers between 1 and 10, including 1 and 10, can be inputs for g. This means that the domain of will not contain the x-values that will lead to outputs which are less than 1. , so the domain of The outputs of

is:

, corresponding to inputs taking values from the set , will be elements of the set , so the range of

. is

. (b) , this is a subset of the domain of h, so the domain of is also

.

The range of is the range of the quadratic 5.

, as the restriction on the domain of h does not impact .

(a) f

g

Domain Range

, this set of values is not a subset of the domain of f. To be able to compose the two functions, the x-value, which is the input of g resulting in an output of must be excluded from the domain of g. , so the domain of The outputs of will be elements of the set cannot be 0, so the range of is:

is: , as the input of g

.

(b)

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of

, this set of values is a subset of the domain of g, so the domain .

is also

The outputs of for 6.

, corresponding to inputs taking values from the set , will be elements of the set ,( , which is not in the domain), so the range of is:

is obtained .

In each of the following, try to identify the rules transforming the input into the expression given by , and take into consideration the order in which the two functions are combined. (a)

3 is added to x, and the result is squared

(b)

5 is subtracted from x, and the result is square rooted

(c)

x is square rooted, and the result is subtracted from 7

(d)

3 is added to x, and the reciprocal of the result is computed

(e)

1 is added to x, this result is then the power to which 10 is raised

(f)

9 is subtracted from x, and the result is cube rooted

(g)

9 is subtracted from the square of x, and the absolute value of result is taken

(h)

the square root of the difference between x and 5, and the reciprocal of the result is computed

7.

(a)

f

g

Domain Range

, this is a subset of the domain of f, so the domain of is also

.

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(b)

f

g

Domain Range

, but this is not a subset of the domain of f. To be able to compose the two functions, the x-value which is the input of g resulting in an output of must be excluded from the domain of g. , so the domain of

is

(c) f

g

Domain Range

, but this set of values is not a subset of the domain of f. To be able to compose the two functions, the x-values, which are the inputs of g resulting in the outputs must be excluded from the domain of f. , so the domain of

is

(d)

f

g

Domain Range

, this set of values is the same set of values as the domain for f, so the domain of

is also

.

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Exercise 1.4 1.

2.

3.

(a)

For

, so for

(b)

For

, so for

(a)

For

(b)

For

If

, this means , this means

, so for

, this means

, so for then

so

, this means is an output for the original function f.

4.

or

, but

so 5.

(a)

(i)

(ii)

The two graphs are reflections of each other in the line (b)

, so f and g are inverse functions.

(i)

(ii)

The two graphs are reflections of each other in the line

, so f and g are inverse functions.

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(c)

(i)

(ii)

The two graphs are reflections of each other in the line (d)

, so f and g are inverse functions.

(i) (ii)

The graph of

is symmetrical about the line y = x, so the reflected graph

overlaps with the original graph. (e)

(i)

(ii)

The two graphs are reflections of each other in the line

, so f and g are inverse functions.

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(f)

(i)

(ii)

The two graphs are reflections of each other in the line (g)

(i)

The two graphs are reflections of each other in the line (h)

, so f and g are inverse functions.

, so f and g are inverse functions.

(i)

(ii)

The two graphs are reflections of each other in the line

, so f and g are inverse functions.

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(i)

(i)

(ii)

The two graphs are reflections of each other in the line (j)

, so f and g are inverse functions.

(i)

(ii)

6.

The two graphs are reflections of each other in the line

, so f and g are inverse functions.

The fact that the domain of following questions.

is used when answering the

is the same as the range of

(a)

The domain of

is

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(b)

The domain of

is

The domain of

is

The domain of

is

The domain of

is

(c)

(d)

(e)

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(f)

The domain of

is

The domain of

is

The domain of

is

(g)

(h)

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(i)

The domain of

is

The domain of

is

(j)

7.

(a)

The graph of

shows that f is a decreasing function on its

domain (no need for restricting the domain). This means that f is a one-to-one function, consequently exists. Next, find the expression of

:

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The domain of

is the range of f , namely

The two graphs are shown below:

(b)

The graph of domain. In order for .

shows that f is not a one-to-one function on its to exist, the domain of f must be restricted to either or

Next, find the expression of

for

:

The domain of is the range of f , namely below for a domain for f restricted to :

(c)

The graph of domain. In order for or .

The graphs are shown

shows that f is not a one-to-one function on its to exist, the domain of f must be restricted to either

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Next, find the expression of

for

:

The domain of is the range of f , namely below for a domain for f restricted to :

(d)

The graph of domain. In order for or .

The graphs are shown

shows that f is not a one-to-one function on its to exist, the domain of f must be restricted to either

Next, find the expression of

for

:

The domain of is the range of f , namely below for a domain for f restricted to :

The graphs are shown

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8.

The graph of

has a minimum point at

and a maximum point at

. Only a one-to-one function has an inverse, so the three intervals on which f satisfies this requirement are: (f is decreasing), (f is increasing), and (f is decreasing). 9.

First, the inverse functions of g and h need to be found.

(a) (b) (c) (d) (e) (f)

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(g)

(h)

10.

Exercise 1.5 1.

(a)

The graph of is a vertical translation of the parabola representing the basic function . The transformed graph is obtained by shifting all points on the original parabola 6 units down, so the new vertex will be at , and the new x-intercepts are at and

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(b)

The graph of is a horizontal translation of the basic function All points on the original parabola will shift 6 units right, including the original vertex, , which will now be located at .

(c)

The graph of is a vertical translation of the graph representing the basic function . The transformed graph is obtained by shifting all points on the original graph 4 units up, including the original x-intercept, , now at .

(d)

The graph of is a horizontal translation of the graph representing the basic function . The transformed graph is obtained by shifting all points on the original graph 4 units left, including the original x-intercept, , now at .

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.

(e)

The graph of is obtained by translating all points on the graph of the basic function by 2 units right and 5 units up. The original x-intercept, , will now be located at .

(f)

The graph of

is a horizontal translation of the basic function

.

All points on the original graph will shift 3 units right, including the original vertical asymptote, , which will now have equation . The horizontal asymptote is the same,

(g)

The graph of

is a translation of the graph of the basic function

. All points on the original graph will shift 5 units left and 2 units up, including the original asymptotes: the equations of the new vertical and horizontal asymptotes are and , respectively.

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(h)

(i)

The graph of is obtained by transforming the graph of the basic function . First, the original graph is reflected in the x-axis, then all points shift 4 units down, including the original x-intercept, which is now at .

The graph of is obtained by transforming the graph representing the basic function . First, shift all points on the original graph 1 unit right, then reflect the graph in the x-axis, and shift it 6 units up. The original x-intercept, , will first shift to , then to The reflection in the x-axis has no effect on point , as it is on the line of reflection.

(j)

The graph of is obtained by transforming the graph of the basic function as follows: first, shift all points on the graph 3 units left, then reflect it in the x-axis. The transformation can also be performed in a different order if we consider the equivalent form of the equation of f , : first, reflect the graph in the x-axis and then shift it right 3 units, you will obtain the same graph as before. In both cases, the original x-intercept, , will end up at .

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(k)

The graph of is obtained by transforming the graph of the basic function by stretching it vertically using a scale factor of 3 (the y-coordinates of all points on the original graph are multiplied by 3). The x-intercept, , is not impacted by this transformation, it stays at .

(l)

The graph of function

is obtained by transforming the graph of the basic quadratic by stretching it vertically using a scale factor of (the y-coordinates

of all points on the original graph are multiplied by ). The x-intercept, impacted by this transformation, it stays at

(m)

The graph of

, is not

.

is obtained by transforming the graph of the basic

quadratic function by stretching it horizontally using a scale factor of 2 (the x-coordinates of all points on the original graph are multiplied by 2). The transformation can also be performed in a different way if we consider the equivalent form of the equation of f , , this can be done by vertically stretching the graph of

with a scale factor of (the y-coordinates of all

points on the original graph are multiplied by ). In both cases, the original xintercept,

, will not be impacted by the transformation, it stays at

.

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(n)

2.

(a)

The graph of is obtained by transforming the graph of the basic function by reflecting the original graph in the y-axis. Alternatively, if we consider the equivalent form of the equation of f, , then the graph of should be reflected in the x-axis. In both cases, the original x-intercept, will not be impacted by the transformation; it stays at .

,

The given graph represents a parabola, so the transformations are applied to the graph of the basic quadratic function . Based on the given graph, the transformations are: reflection in the x-axis, followed by a vertical shift of 5 units up. the equation of the graph is

(b)

The shape of the given graph suggests that the basic function to work with is . The graph is a reflection in the y-axis, as checked when considering the point . This is the image of point , so the required equation is .

(c)

The basic graph in this case is , the applied transformations are: reflection in the x-axis, followed by a horizontal shift of 1 unit left. the equation of the graph is

(d)

The graph to be transformed is

, the applied transformations are: a horizontal

translation of 2 units right, followed by a vertical shift of 3 units down (this is easily deduced when looking at the original vertical and horizontal asymptotes, the lines and have changed to and , respectively). the equation of the graph is 3.

In the following questions, first recognise the transformations to be applied by analysing the given equation, then transform the important points of the original graph: the two ends, and , the maximum point, , and the minimum point, , then plot and join these images to obtain the final graph. (a)

The transformation to be applied is a vertical shift of 3 units down. ,

,

,

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(b)

The transformation to be applied is a horizontal shift of 3 units to the right. ,

,

,

(c)

The transformation to be applied is a vertical stretch of scale factor 2. ,

,

,

(d)

The transformation to be applied is a horizontal stretch of scale factor . ,

,

,

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(e)

The transformation to be applied is a reflection in the x-axis. ,

(f)

,

The transformation to be applied is a reflection in the y-axis. ,

(g)

,

,

,

The transformations to be applied are: a vertical stretch of scale factor 2, followed by a vertical translation of 4 units up. ,

,

,

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4.

(a)

the transformations to be applied are a horizontal shift of 3 units right, followed by a vertical translation of 5 units up (the order of the translations does not matter).

(b)

the transformations to be applied are a reflection in the x-axis,, followed by a vertical translation of 2 units up (the order of the transformations does not matter).

(c)

the transformations to be applied are a horizontal translation of 4 units left, followed by a vertical stretch with scale factor (the order of the transformations does not matter).

(d)

the transformations to be applied are: a horizontal stretch with scale factor , followed by a horizontal translation of 1 unit right and a vertical translation of 6 units down.

5.

(a)

(i)

The function

is not defined where

This means that the domain of

:

will be

will have a vertical asymptote, should be investigated:

and its graph

The behaviour of the graph about

(because

for

(because

for

) )

As x becomes infinitely large, either positively or negatively, the values of will decrease and will approach 0, this means that the graph of will have a horizontal asymptote, The y-intercept can easily be calculated, Consequently, the graph of

is:

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(b)

(ii)

To obtain the graph of , the part of the graph of which is below the x-axis will be reflected above the x-axis. The resulting graph is:

(iii)

To obtain the graph of , the part of the graph of to the left of the y-axis is eliminated and the remaining part is reflected in the y-axis. The resulting graph is will be symmetrical about the y-axis:

(i)

The function

is not defined where

This means that the domain of

:

will be

and its

graph will have two vertical asymptotes, and The behaviour of the graph of about and should be investigated: (because

)

for

(because

for

)

, (because

for

)

(because

for

)

As x becomes infinitely large, either positively or negatively, the values of will decrease and will approach 0, this means that the graph of will have a horizontal asymptote, The y-intercept can easily be calculated, Consequently, the graph of

is:

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(c)

(ii)

To obtain the graph of , the part of the graph of which is below the x-axis will be reflected above the x-axis. The resulting graph is:

(iii)

To obtain the graph of , the part of the graph of to the left of the y-axis is eliminated and the remaining part is reflected in the y-axis. The resulting graph is will be symmetrical about the y-axis:

(i)

The function

is not defined where

This means that the domain of will have a vertical asymptote, should be investigated: (because (because

:

will be

and its graph

The behaviour of the graph about for for

) )

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As x becomes infinitely large, either positively or negatively, the values of will decrease and will approach 0, this means that the graph of will have a horizontal asymptote, Consequently, the graph of

is:

(ii)

To obtain the graph of , the part of the graph of which is below the x-axis will be reflected above the x-axis. The resulting graph is:

(iii)

To obtain the graph of , the part of the graph of to the left of the y-axis is eliminated and the remaining part is reflected in the y-axis. The resulting graph is will be symmetrical about the y-axis:

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Chapter 1 practice questions 1.

(a)

f

g

Domain Range

, this is not a subset of the domain of f. To be able to compose the two functions, only the x-values which are common to both the domain of f (real numbers greater than or equal to 3) and the range of g (real numbers greater than or equal to ) are acceptable. This means only real numbers greater than or equal to 3 can be inputs for f. Consequently, the domain of will not contain the x-values that will lead to outputs for g which are less than 3.

It follows that the domain of (b) 2.

The range of

is

is

, so

and

, as the outputs are square roots of positive numbers or 0.

(a)

(b) 3.

(a)

(b)

4.

(a) (b)

First, find the expressions of the required inverse functions, substitute the given x-values, then calculate the value of the left side of the given conjecture, to see if it equals 24.

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5.

infinite solutions, as the equation holds true for any

6.

The solution is

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.

7.

(a)

The transformation

represents a horizontal translation of 1 unit

left followed by a vertical shift of unit down.

(b)

Minimum point: Maximum point:

8.

Reflection in the line Vertical stretch with scale factor Horizontal shift of p units left, Vertical shift of q units up, When comparing

to

, it follows that:

(a) (b) (c) 9.

(a)

is a rational function with a square root at the denominator, it follows that the radicand has to be strictly positive (it cannot be equal to 0), so This set of values coincides with the given domain. To find the x-intercept, set y (or

) equal to 0, and solve for x:

no solution, so no x-intercept. The y-intercept is found when there are two vertical asymptotes:

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(b)

10.

(c)

The values given by function f will be strictly positive and greater or equal to 1, as seen in the graph, so the range of g is

(a)

This is the basic rational function, it has asymptotes intercepts.

(b)

Translation of 4 units left:

and no axes

Translation of 2 units down: The expression of function h is (c)

(i)

To find the x-intercept, set y (or

) equal to 0, and solve for x:

The x-intercept is The y-intercept is found by substituting the function:

into the expression of

The y-intercept is (ii)

The equation of the vertical asymptote is found by setting the denominator equal to 0 and solving for x: the equation of the vertical asymptote is The equation of the horizontal asymptote is found after evaluating

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.

This means that as , the values of horizontal asymptote and its equation is (iii)

11.

(a)

approach

, so there is a

Plot the axes intercepts and draw the asymptotes, then sketch the graph of h.

(i) (ii) (iii)

(b)

f is undefined when the radicand is negative: for .

, so f is undefined

(c) 12.

(a)

First, find the expression of

, the inverse function of g:

(b)

(c) 13.

(a)

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(b) (c)

(d) 14.

(a)

The equation of the vertical line is (from graph). This means that the denominator of the function takes value 0 when .

If is on the graph it means that its coordinates satisfy the equation of the curve:

(i) (ii)

15.

(b)

represents a reflection in the x-axis, as the only change in the coordinates is that the original y-value changes to its opposite.

(a)

The first transformation represents a vertical stretch with scale factor 2 (the y-coordinates of all points on the graph are multiplied by 2).

The second transformation represents a horizontal translation of 4 units right (all x-coordinates are increased by 5).

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(b)

represents a reflection in the x-axis. represents a vertical translation of 1 unit down.

The coordinates of 16.

The transformation to the right. As x-intercept of .

The first x-intercept of The second x-intercept of

are represents a horizontal translation of k units , the first x-intercept of will not be greater than the second

is the image of the first x-intercept of is the image of the second x-intercept of

17.

18.

(a) (b)

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19.

(a)

In order for f to have real values the radicand has to be positive or 0:

In order for f to take finite values the denominator of the function should not take the value 0:

The required set of values is: (b)

f is a square root function, so its outputs will be either positive or 0, so the range of f is:

20.

The domain of asymptote, 21.

is the same as the range of f, so , so 2 will not be an output for f ).

(a)

(f has a horizontal

f has a horizontal asymptote, will not include the value

, so the range of f

.

The graph of the function is shown below:

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Based on the graph, if than .

, the values of f will not exceed 2, but they will be greater the range of is:

(b)

22.

(a)

(b)

g must have a rule that will transform into root to obtain , then 1 should be added to

23.

This means it first has to cube to obtain

(a)

In order for f to have real values the radicand has to be positive, but not 0, as the square root is at the denominator of the function:

(b)

The graph of the function is shown below:

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Based on the shape of the graph, the y-values will be greater than or equal to

.

the range of is: 24.

f must have a rule that will transform divide by 2 to obtain , then 1 to be added to

25.

into

This means it first has to add 1, then

and the result divided by 2 to obtain

Points A, B, C and D are (from graph): (a)

The transformation to be applied to the graph of shift of 4 units right.

is a horizontal

The graphs of both the original and the transformed function are shown below:

(b)

The transformations to be applied to the graph of are, in any order, a reflection in the x-axis and a horizontal stretch with scale factor .

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The graphs of both the original and the transformed function are shown below:

OR: The transformations to be applied to the graph of are, in any order, a reflection in the y-axis and a horizontal stretch with scale factor .

The graphs of both the original and the transformed function are shown below:

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Exercise 2.2 1. (a) f : x

x 2 10 x 32

Converting to vertex form: h

b2 c 4a

k

f x

10 4

32 x 5

2

b 2a

10 2

5,

2

32 25 7

7

(i) axis of symmetry: x 5 , coordinates of the vertex: 5, 7 (ii) horizontal translation 5 units right, vertical translation 7 units up (iii) since the coefficient of the leading term is positive, the function f has minimum value 7 for x = 5. (b) f : x

x2 6x 8

Converting to vertex form: h

f x

x 3

2

b 2a

6 2

3, k

b2 c 4a

62 8 4

8 9

1

1

(i) axis of symmetry: x 3 , coordinates of the vertex: 3, 1 (ii) horizontal translation 3 units left, vertical translation 1 unit down (iii) since the coefficient of the leading term is positive, the function f has minimum value 1 for x = 3. (c) f : x

2 x 2 4 x 10

Converting to vertex form: h

f x (i) (ii) (iii)

2 x 1

2

b 2a

4 4

1, k

c

b2 10 4a

4 8

2

10 2 12

12

1 , coordinates of the vertex: 1, 12 axis of symmetry: x horizontal translation 1 unit left, reflection in the x-axis, vertical stretch by factor 2, vertical translation 12 units up since the coefficient of the leading term is negative, the function f has maximum value 12 for x = 1.

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(d) f : x

4 x2 4 x 9

Converting to vertex form: h

f x

1 4 x 2

b 2a

4 8

1 ,k 2

b2 c 4a

9

4 16

2

9 1 8

2

8

1 1 ,8 , coordinates of the vertex: 2 2 1 (ii) horizontal translation unit right, vertical stretch by factor 4, vertical translation 2 8 units up (iii) since the coefficient of the leading term is positive, the function f has minimum 1 value 8 for x . 2

(i) axis of symmetry: x

(e) f : x

1 2 x 7 x 26 2

Converting to vertex form: h f x

1 x 7 2

2

b 2a

7 1

7, k

c

b2 4a

26

72 2

26

49 2

3 2

3 2

3 2 1 (ii) horizontal translation 7 units left, vertical shrink by factor , vertical translation 2 3 units up 2 (iii) since the coefficient of the leading term is positive, the function f has minimum 3 value for x 7. 2

(i) axis of symmetry: x

7 , coordinates of the vertex:

7,

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2. (a) x 2 2 x 8 0 (b)

x 4 x 2

x 4 0 or x 2 0

x 2 3x 10 x 2 3x 10 0 x 5 x 2 x 5 0 or x 2 0 x 5 or x 2

(c) 6 x 2 9 x

0

3x 2 x 3

(d) 3 x 2 11x 4 0 (e) 3x 2 18 15 x

9 x 2 4 x2

0

3x 1 x 4

0

x

1 or x 4

3. (a) x 2 4 x 3 0 2 x 2 7 0

7 x 2

x 2 x

7 2

x

x2 2x 1 2 0 x 1

2

3 0

No real solution

2

3 x 2 5x 6

3 2

0 or x

3 x 1 0 or x 4 0

1 or x 3

x

0

x 3 or x 2

4x 1 x 2

0

2

7

2

0 7 7

x 2 0

2

9 0

x 2 3 x 2 3

0

x 5 0 or x 1 0 x 5 or x

(c) x 2 2 x 3 0

4 or x

(b) x 2 4 x 5 0 x2 4 x 4 9 0

0 or x 2 7 or x

0 or 2 x 3 0 0

x

0

x 3 0 or x 2 0

4 x2 9 x 2 0

4 x 1 0 or x 2 0

x 2

x

3 x 2 15 x 18 0

3 x 3 x 2 (f)

0

1

(d) 2 x2 16x 6 0 Divide both sides by 2: x 2 8x 3 0 x 2 8 x 16 13 0 x 4

x 4 x 4 x

2

13 0

13 x 4 13

4

13

0 or x 4

13 or x

4

0 13

0

13

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4

(e) x 2 2 x 8 0 x2

(f) 2 x 2 4 x 9 0 Divide both sides by 1:

2x 1 9 0

x 1

2

2x2 4x 9 0

9 0

x 1 3 x 1 3

2 x2 2x

0

0

x 4 0 or x 2 0

2 x2 2 x 1 1

x

2 x 1

4 or x

2

x

f x

4

2

41

Quadratic formula: x x1

9 0

11 0

x 1

11 2

x 1

11 2

x 1

22 2

x 1

22 2

2

22

x

2 2

22 2

2

22 2

2

11 2

0

22

2

0 22

2

2

0

0

2

0 or x or x

2

0

22 2

x2 4 x 1

b2 4ac

(a)

1

9 0

x 1

x

x

2

2

2 x 1

4.

9

4 2 5 2

2

1

b

5, x2

2a

20,

20

2 5

. Therefore the zeros are: 4 2 5 2

2

5

(b) Axis of symmetry passes through the vertex. The x-coordinate of the vertex: x1 x2 2 5 2 5 x 2 . Therefore, equation of the axis of symmetry: x = 2 2 2 (c) Since the coefficient of the leading term is positive, the function f has minimum value f 2 22 4 2 1 5

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5. (a) x 2 3x 2 0 b2 4ac 32 4 1 2 (b) 2 x2 3x 2 0 b2 4ac 3

2

0 , therefore two real and distinct solutions.

9 8 1.

4 2 2

9 16

0 , therefore no real solutions.

7.

(c) x 2 1 0 b 2 4ac

(d) 2 x 2

02 4 1

9 4

6. The equation 2 x 2 b 0 p

0 , therefore two distinct solutions.

4.

9 x 1 0 4

b 2 4ac

2

1

4ac

p

4 2 1

81 8 16

47 . 16

0 , therefore no real solutions.

px 1 0 has one real solution if its discriminant is equal to 0.

2

4 2 1

p2 8 0 2 2 or p

2

p2 8

p 2 2

p 2 2

0

p 2 2

0 or p 2 2

0

2 2

7. The equation x 2 4 x k 0 has two distinct real solutions if its discriminant is greater than 0. b2 4ac 42 4 1 k 16 4k

0

16 4k 0

k 4

8. If the equation x2 4kx 4 0 has two distinct real solutions, then its discriminant is greater than 0. 2 b2 4ac 4k 4 1 4 16k 2 16

0

16k 2 16 0

k2 1 0

k

, 1

1,

as seen on the graph:

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mx 2 6 x m will have no common point with the x-axis

9. The graph of the function g : x if m 0 0. b2 4ac 62 4m2

36 4m2

0

36 4m2

9 m2

0

0

m

, 3

as seen on the graph:

3,

10. The inequality 3x2 12 x k 0 is true for all real values of x if the quadratic function f x 3 x 2 12 x k has no real zeros. Since the coefficient of the leading term is positive, the discriminant of the function must be negative. 2 b2 4ac 12 4 3 k 144 12k

0

144 12k 0

k 12

11. The expression can be rewritten in the form f x 2

x2

x 2 . The discriminant of the

2

b 4ac 1 4 1 2 1 8 9 0 . The coefficient of the quadratic function leading term is negative. Consequently, the graph of the function opens downwards and the function has no real zeros. Therefore f x 0 for all real values of x.

12. (a) Since two zeros of the function are given, then y function passes through the point 0, 8 , so

8 a 0 1 0 4 Now, y

4a 8

a

a x 1 x 4 . The graph of the

2

2 x2 3x 4 , or y

2 x 1 x 4

(b) Since two zeros of the function are given, then y function passes through the point 4

a

Now, y

1

1 2

1 3

2 1 x 3 2

6a x 3

4 2 2 x 3

2x2 a x

6x 8 1 2

x 3 . The graph of the

1, 4 , so

a

2 3

7 x 2

2 , or y 3

2 2 x 3

7 x 1 3

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13. The equation 2 x 2 b2 4ac

2

3 k

4 2 k 3

0.

9 6k k 2 8k 24 k 2 14k 15

k 2 14k 15 0

0

k 2 14k 15 k1

3 k x k 3 0 has two imaginary solutions

k 1 k 15

0

1, k2 15

Therefore k

1, 15

5 x 2 kx 2 has two distinct real zeros if

14. The function f x 2

k

2

b

4ac

0

k 2 40 0

4 5 2

k

2

0.

40

k 2 10 k 2 10

0

As can be seen on the graph, k

,

2 10

2 10,

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15.

If the graph of a quadratic function f x ax 2 bx c, a 0 passes through the 17 55 7 55 , , 4, 4 , , points , their coordinates have to satisfy the formula of 4 8 4 8 the function. Therefore: 2

17 4 7 4

a

2

a

17 b c 4

7 b c 4

55 8

55 8

2

4 a 4b c

4

289 17 55 a b c 16 4 8 289a 68b 16c 110 49 7 55 a b c Simplifying: , or 49a 28b 16c 110 16 4 8 16a 4b c 4 16a 4b c 4 Using a GDC we get a

2, b 5, c

8 as can be seen on following pictures:

2 x2 5x 8

Therefore f x

3 1 1 (by symmetry of the 2 2 graph). The function can be now expressed in the form f x a x 1 10 .

16. If f 3

f

Since f 3

f x

1

2 , then the x-coordinate of the vertex x

2 , then f 3

2 x 1

2

a 3 1

10 . Now f 2

2

10 2 2 2 1

4a 2

8

a

2 and

10 8 .

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17. 4 x 1 x 2 4 x2 4x 3 0

x 3 x 1

0

x 3 or x 1

Therefore x

,1

3,

18. The discriminant of the quadratic function f x

2x2

2 t x t 2 3 is:

2

b2 4ac 2 t 4 2 t 2 3 4 4t t 2 8t 2 24 Since the discriminant is a quadratic itself, then: 2 b2 4ac 4 4 7 20 16 560 544 1

Since

7t 2 4t 20 .

0 and the coefficient of the leading term in the expression for

1

is negative,

the expression is negative for all values of t. It follows that there is no value of t for which the equation 2 x 2 2 t x t 2 3 0 has real roots. 19. The equation ax2 bx a 0 must have two distinct roots. In a quadratic equation ax2 bx c 0 the product of the roots is a a

b

1. Alternatively, the quadratic formula: x1

b2 4a2 2a

b

x2

x1 x2

b2 4a 2 2a

b

x1 x2

b2

(or, x1 x2

b2 4a 2 4a2 a a

b 2 4a 2 2a

b2 4a2 2a

b b

b

b2 4a 2 and 2a

f the roots:

b 2 4a 2 2a

b2 b2 4a 2 4a 2

c , which in this case is a

4a 2 4a2

b

b 2 4a 2 b 4a 2

b 2 4a 2

1

1)

Therefore the two roots of the equation are reciprocals of each other.

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20. (a) 2 x 2 6 x 5 0 x1

c a

x1 x2

(b) x 2

b a

x2

6 2

5 2

3 5 2

1 3x

x 2 3x 1 0

b a

x1 x2 c a

x1 x2

3 1

3

1 1

1

b a

0 4

(c) 4 x 2 6 0 x1

x2 c a

x1 x2

6 4

0 3 2

(d) x2 ax 2a 0 x1

b a

x2 c a

x1 x2

(e) m m 2

a 1 2a 1

a 2a

4 m 1

m2 6m 4 0 b a

x1 x2 c a

x1 x2

(f) 3 x 3x 2 x1 x1 x2

2 x

6 1

4 1 1, x

6

4 0

x 2 0 b a

x2 c a

2 3

1 3 2 3

1 3

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21. Let the new equation be ax 2 bx c 2

2

2

0, a

0 . If

and

are its roots, then:

2

1 3 and 2

are roots of the equation 2 x2 3x 6 0 , then

Since and

3 2

Now

2

9 6 4 3

2 3 3

c 1 a

c

5 4

b a

5 4

b

6 2

3.

5 a and 4 5 ax a 4

a . Substituting in the new equation we get ax 2

0

and finally 4 x2 5x 4 0 . 22. If

and are roots of the equation 3x2 5x 4 0 , then: 2

(a)

2

2

2

b2 a2

2

2

2

(b) 3

(c)

3

2

b2 a2

b a

x2 (b) k

x2

2 , x1 x1x2

2

b2

2ac a

1 9 4 3

2

3c a

52 2 3 4 32

1 9

1 12

2

23. (a) Given the equation x 2 8x k and x1 x2 k . Now: 3 x2 4 x2 8

2c a

2

b b 2 3ac a a2 0 and that x1

2

2

5 52 3 3 4 3 32

3x2 we can write x1

55 27

x2

8

8

6 2

6

12

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3

and are roots of the equation x 2 b 1 1 b 1 a (a) c c 4 a 24. If

x 4 0 then:

(b) If the roots of a quadratic equation ax 2 bx c b a

1

1

1 4

1 c a and 4 a

b

1

1

1

1 4

1

Let the new equation be ax 2 bx c b a

1

1

2

c a

2

1

1

2

2

2 2

2

1 5

Therefore ax 2 19ax 25a 2

b a

2

3

3 5

2

2

0 and finally 4 x 2 3 and 5

25

2

1 5

2 1 5

1 2

c

2

19

2

Therefore ax 2

2

2

72 1 ax a 25 5

19a

25a

0 and finally x 2 19 x 25 0

3

2

x 1 0 1 . 5

b

3 5

3

3 5

2

3 1 5

c a

then

0.

2

1

2

0, a

1

and

1 a 4

c

and are roots of the equation 5x 2 3x 1 0 then

25. If

(b)

0 are

1 1 ax a 4 4

Substituting in the above equation we get ax 2

(a)

0, a

1 5

c

1 5

72 25

1 a 5

0 and finally 25 x 2

72 x 5 0

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b

72 a 25

Exercise 2.3 1. (a) x 3 3 x

2

3x 4 5x 5

3x 2 9 x 4x 5 4 x 12 7

Then 3x 2 5 x 5 (b)

x 3 3x 4

7

3x3 2 x 2 4 x 1 x 2 3x 4 8 x3 9 x 5 3x 4 6 x 3 2 x3 9 x 2 x3 4 x 2 4x2 9x 4 x2 8x x 5 x 2 7

(c)

Then 3x 4 8 x3 9 x 5 x2 x 1 3 2 x 4 x 5 x 3x 7

x 2 3x3 2 x 2 4 x 1

7

x3 4 x 2 x 2 3x x2 4 x x 7 x 4 11 Then x 3 5 x 2 3x 7 x 4 x 2 3x 2 5 x (d) 3x 1 9 x 3 12 x 2 5 x 1

x 1

11

9 x3 3x 2 15 x 2 5 x 15 x 2 5 x 1 Then 9 x3 12 x 2 5 x 1

3x 1 3x 2 5 x

1

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2 x 3 17 x 2 22 x 7 . Divide f x by x 1 :

2. The binomial x 1 is a factor of f x x 1 2 x3

2 x 2 15 x 7 17 x 2 22 x 7

2 x3 2 x2 15 x 2 22 x 15 x 2 15 x 7x 7 7x 7 0 The expression 2 x 2 15 x 7 can be further factored out to give 2 x 2 15 x 7 2 x 1 x 7 . Therefore f x

2 x 3 17 x 2

22 x 7

x 7 x 1 2x 1

3. The binomial 2 x 1 is a factor of f x 2 x 1 6 x3

6 x3 5 x 2 12 x 4 . Divide f x by 2 x 1 : 3x 2 4 x 4 5 x 2 12 x 4

6 x3 3x 2 8 x 2 12 x 8x2

4x 8x 4 8x 4 0

The expression 3x 2 4 x 4 can be further factored out to give 3x 2 4 x 4 3x 2 x 2 . Therefore f x

6 x 3 5 x 2 12 x 4

x 2 2 x 1 3x 2

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4. The binomial x x

2 3x 4 2 x 3 3 3x 4 2 x 3

2 is a factor of f x 3 3x3 36 x 48 36 x 2 24 x 32

3x 4

2 x3 36 x 2

24 x 32 . Divide by x

2 : 3

36 x 2 24 x 36 x 2 24 x 48 x 32 48 x 32 0 Now, 3x3 36 x 48 3 x3 12 x 16 3 x x 2 x 2

3 x3 4 x 8x 16 8 x 2

x 2 x 2 x 4

3 x x2 4

3 x 2 x

x 2

2

2

8 x 2

2x 8

x 4

Therefore f x

5.

3x 4

2 x 3 36 x 2

(a) x 2 x

2

24 x 32 3 x 2

x 2 5x 4

x 2 3x

x 4

x

2 3

x 2

2

x 4 3x 2

x2 2 2x 1

(b) x 2 x 3 2 x 2 x3 2 x 2

2x 4 2x 6 2

Quotient = x 2 Remainder = 2

2

2x 1 2x 4 3 Quotient = x 2 2 Remainder = 3

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(c) 3 x

2

7x

9x

2

9x2

3 x 5

x 4 x3 4 x 2 4 x 4 (d) x 1 x 5 3 x 3 6

21x

x5

20 x 5

x4 x 4 3x3

Quotient = 3 Remainder = 20x 5

x4

x3

4 x3 6 4 x3 4 x 2 4x2 6 4x2 4x 4x 6 4x 4 2 Quotient = x 4 x 3 4 x 2 Remainder = 2

4x 4

6. (a) By the remainder theorem, the required remainder is equal P 2 3 2 P 2 2 2 3 2 4 2 7 5 (b) By the remainder theorem, the required remainder is equal P 5 4 2 P 1 1 2 1 3 1 20 1 3 23

1

(c) By the remainder theorem, the required remainder is equal P

7

P

7

5 7

4

30

7

3

40

7

2

36

7

14

483

(d) By the remainder theorem, the required remainder is equal P 1 P 4

1 4

3

1 4

1 49 1 4 64

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7. If x 2 is a double root of the polynomial P x x 4 5 x3 7 x 2 4 then the polynomial 2 P x is exactly divisible by x 2 x2 4 x 4 . x2 x 1 x2 4 x 4 x4 5x3 7 x2 4 x 4 4 x3 4 x 2 x3 3x 2 4 x3 4 x 2

4x

x

2

4x 4

x

2

4x 4 0

The remaining roots are the roots of f x f x

x2

0

b2 4ac

x2

x 1 0 1

2

41

1

5

x

8. If x 3 is a zero of the polynomial f x Therefore 33 32 3k 1 0

3k

37

1

conditions must hold: f 1

f 1

21

f 4

2 4

51 4

3

5 4

14 1 3

0 and f 4 2

14 4

a1 2

a 4

5 2

or x

x3

1

5 2

x 2 kx 1 , then f 3 37 k 3

9. Since 1 and 4 are zeros of the polynomial f x 4

x 1

0

2 x 4 5 x3 14 x 2 ax b , the following

0

b a b 17 0 b 4a b 32 0

We need to solve the system of equations:

a b 17 4a b 32 Subtracting the first equation from the second one we get 3a 15 or a 5 Then b 17 a 17 5 12 10. (a) The polynomial can be written in the product form as f x a x 2 x 1 x 4 . We can assume a 1 . Then f x x2 x 2 x 4 x3 4x2 x 2 4x 2x 8 x3 3x2 6x 8

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(b) The polynomial can be written in the product form as f x We can assume a 1 . Then f x

x

2

3x 2 x

2

6x 9

a x 1 x 3 4

x

(c) The polynomial can be written in the product form as f x We can assume a 1 . Then f x

x 2

3

3

3x

7x

a x 2

3

2

2

x 2

15x 18

x3 6 x2 12 x 8

11. (a) The polynomial P x 6 x3 7 x 2 ax b satisfies the following conditions: P 2 72 and P 1 0 (by the remainder theorem and the factor theorem respectively). Then: 3 2 3 2 6 2 7 2 2a b 2a b 76 72 and 6 1 7 1 a b a b 1 0 We need to solve the system of equations: 2a b 4 a b 1 Subtracting second equation from the first one we get 3a 3 a 1 . Then b 1 a 2 (b) The polynomial can be written as P x 1 P x then P must be equal to 0. 2 1 P 2

1 6 2

3

1 7 2

2

1 2 2

6 8

7 4

6 x3 7 x 2

x 2 . If 2 x 1 is a factor of

1 2 0 . Therefore 2 x 1 is a factor. 2

To find the third factor we can divide P x by 2 x 1 x 1 2x

2

x 1 6x

3

6x

7x 3

2

2x2

x 1

3x 2 x 2

3x 2 3 x 4x2 4x

2

2x 2 2x 2 0

Since the remainder is equal to 0, then the remaining factor of P x is 3x 2 12. By the remainder theorem: 3 p 1 a( 1) b a b

3

1 and p 2

a(2) b

3

2a b

3

27

Now we need to solve the following system of equations: a b

3

1 and 2a b

3

27

Taking the third root of both sides we get

a b

1 and 2a b 3

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Subtracting the second equation from the first one gives 3a b

4 , so a

4 and 3

4 1 1 3 3

a 1

13. Let P x x3 ax 2 bx c . By the factor theorem and by the remainder theorem the following conditions must be satisfied simultaneously: P 2 0 P

2

P 3

0 10

Therefore:

P

2

23 a 2

P 2 2

2

3

a

33 a 3

P 3

b 2 2

2

c

2

b

b 3

4a 2b c 8 0 2

c

4a 2b c 8 0

c 9a 3b c 27 10

To determine values of a, b and c we need to solve the following system of equations:

4a 2b c

8

4a 2b c 8 9a 3b c

17

16 b 4 and Subtracting the second equation from the first one we get 4b c 4a . Substituting these values into the third equation we have 1, b 4, c 4 . 9a 12 4a 17 5a 5 a 1 . Therefore a 5 x3 3x 2 ax 7 and Q x

14. Let P x P

2

5

2

3

3

2

Q

2

4

2

3

a

2

2

2

4 x 3 ax 2

a

2

7

2 a 45

7

2

4 4a 50 2 R

7 x 4 . By the remainder theorem:

R

2a 45 into the second equation we have Substituting R 4a 50 2 2a 45 4a 50 4a 90 8a 40 a 15. Let f x x3 19 x 2 bx 216 . If the roots of the equation f x terms of a geometric sequence, we can write

a ar ar 2

19

a ar ar 2

216

5 0 are consecutive

where a is the first term of a geometric sequence and r is the common ratio.

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Simplifying the system of equations, we get a ar ar 2 19

a ar ar 2 19

a3 r 3

ar

216

6

a 6 ar 2 a

19

6 r

a ar 2 a

13

6 r

6 into the first equation we have r 6 6 2 6 r 13 6r 13 6r 2 13r 6 0 . Factoring out gives r r r 2 3 . Then a 9 or a 4 6r 2 13r 6 0 3r 2 2 r 3 0 r or r 3 2 respectively. Therefore, the roots of the given equation are 9, 6, 4 or 4, 6, 9 . Since these are the same roots, we can write

Substituting a

x3 19 x 2 bx 216

x 4 x 6 x 9

x 3 19 x 2 bx 216

x 2 10 x 24 x 9

x3 19x 2 bx 216

x3 19x 2 114 x 216

from which it follows that b 114 16. (a) Since a polynomial P x is divided by first degree binomial ax b , the remainder can only be a constant. We can write P x

ax b Q x b ,a a

Let x

R

0 . Then P

b a

a

b a

b Q

b a

R

b b Q

b a

R 0 R

Therefore, when a polynomial P x is divided by ax b , the remainder is b R P . a (b) The remainder when P x

P

2 3

9

2 3

3

2 3

9 x3 x 5 is divided by 3x 2 is equal P

2 . 3

5 3

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R

17. (a) By the formula, the sum of the roots of the equation x 4 2 3 1

an 1 an

2 and the product of the roots is 3

(b) Expanding the equation x 2 x3 6 x 2 12 x 8 x4

3

n

a0 an

1

4

5 1

5

x4 1 we have

x4 1

x 3 6 x 2 12 x 7 0

and by the formula: the sum of the roots is

1

roots is

n

a0 an

1

(c) Convert the equation 3 2 x5 1

6 x5 2 x 4

7 1

4

x

2

2

2x2 x , x 2 x5 1

n

1 1 and the product of the 1

1 5

2

into an algebraic sum.

x

x3 4 x 2 2 x

x3 4 x 2

2x 3 0

and by the formula: the sum of the roots is

1

an 1 an

7

3

x2 2 2x2

6 x5 3 2 x 4

is

1

2 3 x 3x 2 2 x 5 0 is 3

a0 an

1

5

3 6

an 1 an

x

a x2

x

1 , the product of the roots 3

1 2

18. The given equation can be written as ax3 bx2 cx d Expand the right-hand side of the equation: a x

2 6

a x3

x2

a x3

x2

ax3 a

x2 a

x x

a x

x

x

.

x

x x

x a

Comparing coefficients of x in both forms of the given function we have: a

b

b a

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,2 ,

be the three roots of the equation x3 63 x 162 0 . Then:

2

0

19. Let 2

3 162

Now, 2

2

3

2

2 162

3

Then

0 162 3

6

3 3

3 2

2 3

216

162 27

3

9 and the zeros are 3, 6, and 9

20. Let the zeros of the equation x3 6 x 2 24 x 64 0 be represented by

r

,

,

r

for some constant r. Then:

r 6

r

r

r and

r 6

r

r 6

r

3

64

r 6

r 4

64

4 4 4r r

4 4r 10 0 r

6

2r 2 5r 2 0

Factorising the last equation, we have 2r 2 5r 2 0

2r 1 r 2

0

1 or r 2 and 2 4 2 or 2, 4, r r 2 r

4 1 8, 4, r 4 4 2 1 r 2 2 Therefore, the three zeros of the equation x3 6 x 2 24 x 64 0 are 2, 4, 8. 21. Zeros of the equation x3 6 x 2 kx 10 0 can be represented as some constant d. Then

d d and

2d 2d

2 d 2 2d

6 10

d

10

2 d 2d

Now, 2 d 2 d 2d Therefore d

5, 1,

3d

3

5

3 or d 3 and

6

d 2d

3

10

2 d

2d for

2d

. 5

2 d 2 d

2

d,

,

8

5 or

5

4 d2

2 3

5

d2 9

1 respectively. Then

d 5 3 2, 2d 5 6 1 or d 1 3 2, 2d 1 6 5

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10

and the zeros of the given equation are 5, 2, and 1. The given polynomial can be converted to product form: x3 6 x 2 kx 10 x 5 x 2 x 1 Then: x 3 6 x 2 kx 10 x3 6 x 2

kx 10

It follows that k

x 2 7 x 10 x 1 x3 6x2 3x 10

3 d, ,

Note: If we had ignored the given hint and replaced the roots by solution would have been more straightforward! 22. Let the zeros of the equation x3

x2 2x k

d , the

, , r for some r constant r. Then, using result of question 18 and key facts from this section: 1 1 r 1 r 1 r 1

0 be represented by

r

r

r r

r r

r

r

r k

2

2

2

r 3

k

2

r 2

2

1

1 r

r k

2

3

Dividing the second equation by the first one side by side we get 2 1 1 r 3 2 r 3 2 8 2 . Therefore, k 1 1 1 r

r

Note: This question is an investigation that goes beyond the syllabus coverage. Consider it a challenge! The roots are not all real numbers.

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Exercise 2.4 1

1. (a) f x

x 2 Domain:

2 , so there is a vertical

0, asymptote x , f x 2 . As x so there is a horizontal asymptote y 0 . f x

0 on its domain, so the graph of

f x has no x-intercept. 1 1 As x 0, f 0 0 2 2

The y-intercept has coordinates 0,

(b) g x

1 2

3 x 2

Domain:

2 , so there is a vertical

0, asymptote x 2 . As x , g x so there is a horizontal asymptote y 0 . g x

0 on its domain, so the graph of

g x has no x-intercept. 3 3 As x 0, g 0 0 2 2

The y-intercept has coordinates 0,

3 2

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(c) h x

1 4x 1 x

Domain:

1 , so there is a vertical

asymptote x 1 . The function h x can be 1 4 x expressed in the form h x . As 1 1 x 1 4 , so there is a x , 0 , so h x x horizontal asymptote y 4 . 1 , so the graph h x 0 1 4x 0 x 4 1 of h x has x-intercept , 0 . As x 0, h 0 4 The y-intercept has coordinates 0, 1 . (d) R x

1 4 0 1 0

1.

x x

2

9

Domain:

3, 3 , so vertical asymptotes are

3 and x 3 .The function R x can be 1 x . For x written in the form R x 9 1 2 x x

,

1 9 0 and 2 0 , so the graph of R x has a x x horizontal asymptote y 0 . Also,

R x

0

x 0 , so the graph of R x has x-intercept 0, 0 .

For x

0, R 0

0 , so the y-intercept has coordinates 0, 0 .

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(e) p x

x

2 2x 3

2

Domain:

x

2 x 3 x 1

3, 1 , so vertical asymptotes are

3 and x 1. As x

x

3 , p x

x

3 , p x

, as

. As

1, p x

, as x

1, p x

.

Since the function y x 2 2 x 3 for x assumes the minimum value ymin

x

1

2

1

4 , then p x for

3

1 assumes its maximum value

pmax x

2

1

2

p

1

2 2x 3

2 4

1 . For x 2

,

0 , so the graph of p x has horizontal asymptote y

p x has no x-intercept. For x 0,

0 . The graph of

2 , so the y-intercept has coordinates 3

0, p 0

2 . 3 x2 1 x

(f) M x Domain:

0 , so vertical asymptote has

equation x 0 . As x

0 ,M x

, as

x 0 , p x . Since the degree of the numerator is one more than the degree of the denominator, the graph of M x has oblique x2 1 1 x x x M x has no x-intercept.

asymptote:

For x

y

x .The graph of

0, M 0 is indeterminate so there is no y-intercept.

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(g) f x

x

x 4x 4

2

Domain:

x x 2

2

2 , so vertical asymptote has

equation x

2 . As x

2 , f x

, as

x 2 , f x . The degree of the numerator is one less than the degree of the denominator, so the horizontal asymptote has equation y 0 . If x 0 , f x 0 , so the x-intercept of the graph of f x is 0, 0 . For x

0, f 0

0 so the y-intercept of the

graph of f x is 0, 0 . x2 2x x 1

(h) h x

Domain:

1 , so vertical asymptote has

equation x 1 . As x

1,h x

, as

x 1,h x . The degree of the numerator is one more than the degree of the denominator, so the graph of h x has an oblique asymptote y ax b . Dividing x 1 into x 2 x 1x

2 x we get: 2

x2

x 3 2x x 3x 3x 3 3

Therefore, the oblique asymptote has equation y so the x-intercepts of the graph of h x are

x 3. h x

0 if x

2, 0 and 0, 0 . For x

2 or x 0 , 0, h 0

the y-intercept of the graph of h x is 0, 0 .

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0 so

(i) g x

2x 8 x x 12 2

Domain:

3, 4 , so vertical asymptotes

have equations x x 3 ,g x x

2x 8 x 4 x 3

3 and x 4 . As , as

3 ,g x

. As

x 4 ,g x , as x 4 , g x . The degree of the numerator is one less than the degree of the denominator, so the graph of g x has horizontal asymptote y 0 . If x g x (j) C x

2 , so the y-intercept has coordinates 0, 3

0, g 0

0

2x 8 0

x 2 x2 4x

Domain:

x

4 . The x-intercept is

4, 0 .

x 2 x x 4

0, 4 , so vertical asymptotes

have equations x 0 and x 4 . As x 0 ,C x , as x 0 , C x As x

2 . 3

4 ,C x

.

, as

x 4 ,C x . The degree of the numerator is one less than the degree of the denominator, so the graph of C x has horizontal asymptote y has no y-intercept. C x

0 . The graph of C x

0

x 2 0

x 2 . The x-intercept is 2, 0 .

2. (a)

Domain:

2, 2 , Range:

,

5 4

2,

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(b)

x 4 x 4 can be written in the form f x . x 4 x 1 x 3x 4 1 1 If x and it would assume the value of if x could be 4 , then f x x 1 5 equal to 4. Therefore: 1 1 Domain: 4, 1 , Range: , ,0 0, 5 5 The function f x

2

(c)

Domain:

, Range: 0, 1

Domain:

1 , Range:

(d)

0

3. (a)

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The x-intercept:

x

6 and x

5 5 , 0 , the y-intercept: 0, , vertical asymptotes: 18 2 3 , horizontal asymptote: y 2

0.

(b)

The x-intercept: none, the y-intercept: 0, 1 , vertical asymptote: x 1 , oblique asymptote: y x 2 (c)

The x-intercept and the y-intercept: 0, 0 , horizontal asymptote: y 3 (d)

The x-intercept: none, the y-intercept: 0, and x

1 , vertical asymptotes: x 4

2, x 1

2 , horizontal asymptote: y 0

x a : x b x c (a) for a b c (say, a 1, b 4, c

4. The function y

6)

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(b) for b a c (say, a

2, b 1, c

4)

(c) for b c a (say, a 5, b 2, c

4)

5. (a) (b)

The highest concentration occurs at t

2 minutes and is equal to 6.25 mg/L.

(c)

eventually decreases to zero.

(d) C t

25t t 4

0.5

2

25t

0.5

0.5 t 2

4

0.5t 2 25t 2 0 and by the quadratic formula: t

25

252 4 0.5 2 2 0.5

0.080 or t

25

252 4 0.5 2 2 0.5

49.9

bloodstream to drop below 0.5 mg/L.

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Exercise 2.5 x 6 2x 9

1. (a)

x 6

9 2x

Squaring both sides of the equation we get x 6 81 36 x 4 x 2

4 x 2 37 x 75 0

2

37

37 4 4 75 2 4

When x 3,

3 6 2 3

x

When x 6.25,

x 3 or x

6.25

3 6 9

6.25 6 2 6.25

3.5 12.5 16 9

The only solution is x 3 (b)

x 7

x 5

Squaring both sides of the equation we get x 7

x 2 10 x 25

x2 11x 18 0

Factorising: x 2 x 9

0

x

2 or x 9

When x 2,

2 7 5 3 5 8 2

When x 9,

9 7 5 9

The only solution is x 9 (c)

7 x 14

x 2

Squaring both sides of the equation we get 7 x 14

x2 4 x 4

x 2 3x 10 0

Factorising: x 5 x 2 When x

2,

When x 5,

7 7 5

The solutions are x

2

0

14 2

x 5 or x

2

2

14 2 7 2 5

2 or x 5

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(d)

2x 3

x 2

2x 3

2

2

x 2

Squaring both sides of the equation we get 2x 3 4 4 x 2

x 2

4 x 2

x 1

Squaring again we get 16 x 32

x2 2x 1

Factorising: x 11 x 3

0

x 11 or x 3

When x 11,

11 2

2 11

When x 3,

2 3

3 3

3 2

x 2 14 x 33 0

5 3 2

3 1 2

The solutions are x 3 or x 11 (e)

5 x 4

4 x

21 5 x 20

5 x 4

4 x

21 5 x 4

Multiplying by the least common denominator 5 x x 4 we get

25 x 20 x 4 Therefore, x Check:

21x

x 1 2x 3

16 x 80

5

5 5 4

4 5

The solution is x (f)

25x 20 x 80 21x

5 1

4 5

21 and 5 5

21 5 20

21 5

21 5

5

5x 1 7x 3

Multiplying by the least common denominator 2 x 3 7 x 3 we get

x 1 7x 3

Factorising we get 3 x 2 When x When x 1,

2,

7 x 2 10 x 3 10 x 2 13x 3

5x 1 2 x 3

2 1 2 2 3

1 1 21 3

The solutions are x

x 2

0

x 2 x 1

1 5 1 and 1 7

2 51 and 5 7 1

1 3

2 2 4 10

1 3

0

x

3 x 2 3x 6 0

2 or x 1

11 1 11

2 5

2 or x 1

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(g)

1 x

1

1

x 1

x 4

Multiplying by the least common denominator x x 1 x 4 we get

x 1 x 4

x x 4

Therefore, x 2 x 2 When x

1 2

2,

When x

2,

1 2

(h)

2x 1 x2

1

0

x

1 2 1 1

1 2

3 6

2 1

The solutions are x

x 2 5x 4 x 2 4x

x x 1

x2 4 0

x

2 or x 2 1

2 6

x2

1 and 2

1 1 and 6 2 4

1 2 4

1 2

1 6

2 or x 2

2

x 1

Multiplying by the least common denominator 1 x 2 we get 2x 1 x

2 2 x2

2 x2

x 1 0

Factorising, 2 x 1 x 1

When x

When x

1 , 2

2 1

1 2 1 2

x

1 2

1 2

1

1 or x 2

1 3 4

1 3 2

4 3

1

2 3

2

1 both denominators are equal to 0 so x

The only solution is x (i) x 4 2x 2 15 0 Let t

0

x2 , t

x2

1 is not a solution.

1 2 2

2x 2 15 0

0 . Then 2t 2 2t 15 0

Factorising, t 3 t 5 The solutions are x

0 5 or x

t

3 0 or t

5 . Now, x 2

5

x

5

5

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(j) 2 x

2 3

x

1 3

15 0

2 x

1 2 3

1

x 3 15 0

1 3

x . Then 2t 2 t 15 0

Let t

Factorising, 2t 5 t 3 Now, x

1 3

0

5 or t 2

t

3

1

5 2

125 or x 3 8

x

3

125 or x 8

The solutions are x (k) x6 35x3 256 0

2

x3

x

27

27

35x3 256 0

x3 . Then t 2 35t 216 0

Let t

Factorising, t 8 t 27 Now, x

3

8

x

2, x 3

When x

When x 3, x 3

2 or x

3

0

t 8 or t

27

x 3

8 and 82 35 8

27

216 64 280 216 0

27 and 27 2 35 27

216 729 945 216 0

The solutions are x 2 or x 3 (l) 5x

2

x

1

2 0

x 1, t

Let t

5 x

1

x

1

2 0

0 . Then 5t 2 t 2 0

The discriminant, Now, t

1 2

1

41 or t 10

1

2

4 5

2

1 40 41 .

41 . 10

When x t

1

x t

1

10 1 41 10 1 41

10 1 1

41

41 1 10 1

1

10 1

41

10 1 41 1 41

10 1

41

10 1 41 1 41

41

41 1

41

1

40

4 41

40

41

1

41 4

Since x can be any real number except 0, the solutions are x

1

41 4

or x

1

41 4

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or

(m) 3x 4

8

3x 4

8 or 3x 4 8

Now:

3x x

12 or3x 4 4 3

4 or x

(n) x 6

3 x 24

Since both sides of the equation are non-negative: 2

x 6

3x 24

2

x 2 12 x 36 9 x 2 144 x 576

The discriminant,

39

Now, x

39

441 2 2

4 2 135

39 21 18 4 4

2 x 2 39 x 135 0

1521 1080 441

9 or x 2

39

441 2 2

39 21 15 4

1

1 3

9 or x 15 2

The solutions are: x (o) 5 x 1

2

8 x 2 156 x 540 0

2x

When 5 x 1 0

x

1 , 5x 1 2x 5

When 5 x 1 0

x

1 , 5

5x 1

3x 2x

x

5x 1 2x

1 5 7x 1

x

Therefore, there is no solution. (p) x 1

x

3

x 1

3

x

Squaring both sides: x 1

2

9 6x

When x 0 : 3x

x2

6x

x 4

When x 0 : 3x The solutions are x

x 4

2x 8

2x 4 4x 4

3x

x 4

x 2 x

1

1 or x 2

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1 7

1 5

(q)

x 1 x 1

x 1 x 1

3

x 1 3 and x x 1

3 or

x 1

3 x 1 or x 1 3 x 1

x 1

3x 3 or x 1 3x 3

x

1 or x 2

2

When x

1 1 1 2 , 2 1 1 2

When x

2,

2 1 2 1

3 2 1 2 3 1

The solutions are: x

3

3

3

1 or x 2

6 x

(r) In the equation x

2

1 clearly x

0

Multiplying both sides of the equation by x 6

x

Let t

x, t

0

x

x we get

x 6 0 0 . Then t 2 t 6 0 .

Factorising, t 2 t 3

0

t

2 0 or t

3 . Now,

x

3

x 9.

The solution is x 9 . (s)

4 x

6 x

4 x

14 2 x

14 2 x 6 x

Squaring both sides of the equation we get

4 x 14 2 x 2 14 2 x 6 x 2 14 2 x 6 x

16 4 x

14 2 x 6 x

Squaring again we get 84 26x 2 x2 64 32 x 4x2 Factorising: x 5 x 2

0

6 x

2 x2 6x 20 0

x

8 2x x2 3x 10 0

5 or x 2

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When x

5,

4

When x 2,

5

4 2

9

1 2 and

2 2 2

2 and

14 2 14 2 2

2

1

1 x2

10 x

2

4

, x

x2 1 x 4

4

Therefore, 6 x4 24 x2 Let t

x2, t

4 5

x 10

4 we get

5x4 9x2 4 0

0 . Then 5t 2 9t 4 0 .

0

0

2 5

x

4 or t 1 5

t

2 5 or x 2 1 5

2 5 or x 5

The solutions are x

(u) x

18 3 2

10 x 2 x 2 1

x 4 5x 2 4 10 x4 10 x2

Factorising, 5t 4 t 1 Now, x 2

2

0

Multiplying by the least common denominator x 2 x 2 1 x 2 6x2 x2

5

5 6

x

5

6 2

The solution is x (t) In the equation

6

x

x

1

1

x 10

Squaring both sides of the equation we get x2

x2

x 10

x 10 0

The discriminant Now, x

Since x

1

41 2

1

41 2

1

2

0 or x

0 and

the only solution is x

41

1

10

41 2

x 10

1

1 40 41

1

41 2

10

0,

41 2

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(v) In the equation 6 x 37 x 56 0 clearly x 0 Let t

x, t

2 0 . Then 6t 37t 56 0

The discriminant, Now, t

37

37 25 2 6

32 12 8 3

Therefore, x t 2

2

4 6 56

8 or t 3

2

37 25 2 6

64 or x t 2 9

64 or x 9

The solutions are x

1369 1344 25

7 2

42 12 2

7 2

49 4

49 4

2. (a) 3x2 4 4 x 3x 2 4 x 4 0

3x 2 x 2

0 2 3

3x 2 x 2 3x 2 x 2 Answer: x (b)

0 +

2 +

+ 0 0

0

+ + +

2 ,2 3

2x 1 1 x 2 2x 1 1 0 x 2

2x 1 x 2 x 2

2 0

x 2 x 3 x 3 x 2

Answer: x

+

,

X

2

x 3 x 2

0

+

3 + 0

+ +

0

+

0

3,

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(c) 2 x 2 8 x 120

x 2 4 x 60 0 10 0

x 10 x 6 x 10 x 6 Answer: x (d) 1 4 x

+

x 10 x 6 6 + 0 0

+

0

0

+ + +

10, 6

7

We can write 1 4 x 7 or 1 4 x 7 . Then 4 x 8 or 3 3 . The answer is x x 2 or x , 2, 2 2 (e) x 3

4 x 6 and

x 14

Since the expressions on both sides of the inequality are non-negative, x 3

2

x 14

2

x 3

2

x 14

2

0

x 3 x 14 x 3 x 14 11 2 x 17

0

x

x2 4 x x2 4 x x2

4 x

3

x2 4 3 and x 0 x

3

3 and

x2 4 x

3 0 and

x 2 3x 4 x

x 4 x 1 x

17 2

17 , 2

The answer is x (f)

0

0 and

x2

3 4

x

3 0

x 2 3x 4 x

0 and

0

x 4 x 1 x

0

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4

x 4

0

0

+

+

+

x 1

X

+

+

+

x

0

+

x 4

0

+

x 1 x 4 x 1

0

+

4, 0

1,

x

x

X

1

x 1

x

1

x 2

x 1

x 2

x 1

x2 x x 2 x 2 x 1

0

1

4, 1

x

x x 1 x 2 x 2 x 1

0

x 4 x 1

0

x

x

4

(g)

1

+

+

x

1

, 1

0 +

+

+

+

+

+

X

+

+

+

0

+

0

+

X

0, 4

4

1, 4

0

0

x2 2 x 2 x 1

0

0

Since x 2 2 0 for all real values of x, then the inequality holds when x 2 x 1 0 1 X

x 1 x 2 x 1 x 2 x (h)

+

, 1

4x 1 x 2x 3 2

+

X

2 + X X

+ + +

2, 3

4x 1 3 0 x 2x 3

4 x 1 3x 2 6 x 9 x2 2 x 3

2

0

3x2 10 x 8 x 1 x 3

Factorising the numerator, we get

0

3x 2 x 4 x 1 x 3

3x 2 10 x 8 x 1 x 3

4

0

0

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2 3 + 0

+ +

+ + X

X

0

+

X

, 1

2 ,3 3

1

x 3x x x

1 2 3 4

X

3x 2 x 4

+

x 1 x 3

x

+

3

4 + + +

+ + + 0

+ + + +

0

+

4,

3. The function f x 3kx 2 k 3 x k 2 has no real zeros if k 0 (note that for k 0 the function has one real zero). 2

k 3

4 3k k 2

Now 11k 2 30k 9 0

11k 2 30k 9 0 3 11 0 +

11k 3 k 3 11k 3 k 3 + k

k 2 6k 9 12k 2 24k

3 11

11k 2 30k 9

11k 3 k 1

+ 0 0

+ + +

3,

4. (a) The equation px 2 3 x 1 0 has one real solution when p discriminant 0.

3

2

Now 9 4 p

4p 1 0

0

p

p

9 . Therefore, one real solution when p 4

9 but also p 4

0

p

0 and the

9 . Therefore, p 4

0 or p

0 and the discriminant

0 . Therefore, p

(c) The equation has no solution when p 9 4p

0 or p

9 4p

(b) The equation has two real solutions when p 9 4p

0

3

0 ,

0 and the discriminant

,0

0 and the discriminant

0,

9 4

0.

9 4

0.

9 , 4

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5. The function f x k 1

2

4k 2

x2

k2

x k 1

k 2 2k 1 4k 2

Now 3k 2 2k 1 0

3k 2 2k 1

3k 2 2k 1 0

0 +

k

3k 1 k 1

1 3 + 0

1

k 1 3k 1 k 1 3k 1

0.

0 for all real values of x when

+

0

0

+ + +

0 1 , 3

, 1

1 2. n The inequality is equivalent to mn 2n 1 0 .

6. (a) Assume m n 0 , prove m

For m n we have mn n2 and mn 2n 1 n2 2n 1 which is true for all real values of n.

1 m

(b) Notice that m n

1 n

m n

m n mn

m n mn

n 1

2

2

. We need to show that

2

m n 4 for all real numbers m and n such that m 0 and n 0 . mn Since m 0 and n 0 : 2

m n 4mn m2 2mn n2 4mn m2 2mn n2 which is true for all real positive values of m and n.

0

m n

2

0

Alternative solution: m n

2

0

m2 m n

Now,

2mn n 2 2

m n m n mn

4mn 4

m2

0 m n mn m n

2mn n 2

4mn

4mn

2

4

m n mn

4

m n

1 m

1 n

4

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7.

x2

x

x2

x

x2

x

2

5 x 2 5x 6

2

5 x2

x

6

2

5 x2

x

6 0

Let x2

x t . We have:

t 2 5t 6 0 Now, x 2 x2 x

x

t 3 t 2

x 3 or x 2

x

1 41 21

1

13

3

or

Show a b c If a, b, c

3 a

0 and a

2

b

2

x 2 x 1

0

2 or x 1

8. Assumption: a, b, c 0 and a 2

3 or t

x 2 0

or x

2

t

2

x 3 0 or x 2 1

0

b

2

b c

c

2

c , then a b

2

a c

a2 2ab b2 a2 2ac c2 b2 2bc c 2

2

b c

2

0 . We have:

0 , so 2ab 2bc 2ac

2a 2

2b 2 2c 2

Adding a 2 b2 c 2 to both sides of the last inequality we get a2 b2 c2 2ab 2bc 2ac 3a 2 3b2 3c2

which is equivalent to a b c 9. (a)

2

3 a 2 b2 c 2

2x 3 1, x 0 . This inequality is equivalent to 2 x 3 x Now, squaring both sides of the inequality we get: 2x 3

2

x.

x2

4 x 2 12 x 9

x2

3x 2 12 x 9 0

x2 4x 3 0

x 3 x 1

0

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1 0

x 1 x 3 x 3 x 1 x (b)

+

0

+ + +

1, 3

3

2

x 1

x 1

1, where x

3x 3 2 x 2 1 x 1 x 1 x 5 x2 1 x 1 x 1

1 and x 1

x 5 x 1 x 1 x2 x 6 x 1 x 1

0

x 3 x 2 x 1 x 1

x x x x

3 + 0 0

+

x 5 x 1 x 1

1 0

x2 x 6 x 1 x 1

0

1 0

0

0

2 0

2 1 1 3

x 2 x 3 x 1 x 1

+

x

+

0

,

2

1 + X

+ +

1 + + X

X

+

X

1, 1

+ + +

3 + + + 0

+ + + +

0

+

3,

10. Since both sides of the inequality a b a b are nonnegative, we can square 2ab 2 ab ab ab . both sides to get a 2 2ab b2 a 2 2 ab b 2 If a and b have the same sign, then ab ab and the inequality above is satisfied. If a and b have opposite signs, then ab

ab

ab 0 and ab 0

ab for any real values of a and b which means that a b

ab

ab . Therefore,

a

b for any real

values of a and b.

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Exercise 2.6 1. (a)

5x 1 x x 2

5x 1 x 1 x 2

2

A

B

x 1

x 2 5x 1 B x 1 A x 2 x 2 6 A 2 3

Multiply both sides of the identity by x 1 . We get 51 1 1 2

Now, let x 1 . Then

A

B 1 1 1 2

Multiply both sides of the identity by x 2 . We get Now, let x

5x 1 x x 2

Therefore (b)

x 4 x2 2x

5

2 . Then

2

x 4 x x 2

2 1 2 1

A

2

3

x 1

x 2

A x

2 2 2 1

B

0 4 0 2

A

Therefore (c)

x 2 x 4x 3 2

x 4 x2 2x

A

4 2

x 4 x 2

3 x 2

Bx . Now, let x x 2

B

B

6 2

x  4 A x  2   B. x x

3

2 x

x 2 x 1 x 3

1 . Then

3

2

A 2 2 2

2 4 2

A

A

B

x 1

x 3 x 2 x 3 1 A 2

Multiply both sides of the identity by x 1 . We get Now, let x

B

x 2

B 0 0 2

2 . Then

9 3

B

Multiply both sides of the identity by x 2 . We get Now, let x

A x 2 x 1

B

Multiply both sides of the identity by x . We get Then

5x 1 x 1

1 2 1 3

A

B

1 1 1 3

Multiply both sides of the identity by x 3 . We get

x 2 x 1

A

B x 1 . x 3

A x 3 x 1

B.

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0.

Now, let x Therefore

(d)

3 2 3 1

3 . Then x 2 x 4x 3

1 2x 2

2

5 x 2 20 x 6 x3 2 x 2 x

A

3 3 3 1

1 2x 6

5 x 2 20 x 6 x x2 2x 1

5x2

B 1 2 x 1

20 x 6

1 2

1 2 x 3

A x

2

x x 1

1 2

B

B

C

x 1

x 1

2

2

Multiply both sides of the identity by x x 1 : 5x2 20x 6

Now, let x

A x 1

Let x 1 . Then 5 1 24 2 B 9 31

(e)

Bx x 1

Cx 2

1 . Then 5x2 20x 6 A x 1 2

Let x 0 . Then 5 0

Therefore

2

2

20 0

20 1

A 0 1

2

6 x

B

B 0 0 1

B 1 1 1

Cx

C

9

C 0

A 6

C

91

1

1

9

x 1

x 1

2 x 2 x 12 x x 2 5x 6

2

2

6 61 1

2B

5 x 2 20 x 6 x3 2 x 2 x

2 x2 x 12 x3 5x 2 6 x

6

Bx x 1

2

2 x 2 x 12 x x 2 x 3

A x

B

C

x 2

x 3

Multiply both sides of the identity by x x 2 x 3 :

2 x2

x 12

A x 2 x 3

Now, let x 0 . Then 2 0 6A 12 A 2 Let x 2

2 2 12 2B 6

3

2

Cx x 2

0 12

A 0 2 0 3

B 0 0 3

A 2 2 B 3

2 3

B

2

2 3

C

2

2 2

A 3 2

3 3

B

3

3 3

C

3

3 2

1

3

x 3

x 2

2 x

C 0 0 2

2 . Then 2

Let x 2

Bx x 3

3 . Then 2

3C

3 12 3

Therefore,

C 1 2 x 2 x 12 x3 5 x 2 6 x

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9

(f)

4 x2 2 x 1 x3 x 2

4x2 2x 1 x2 x 1

A x

B x2

C x 1

Multiply both sides of the identity by x 2 x 1 :

4 x 2 2 x 1 Ax x 1

Now, let x 0 . Then 4 0 Let x

1 . Then 4

Cx 2

B x 1 2

2

1

2 0

2

1

1 A 0 0 1

1 A

1

1 1

2

21 1 A1 1 1 Let x 1 . Then 4 1 2A 1 5 2A 6 A 3 Therefore, (g)

x

4x2 2x 1 x3 x 2

3 x 2

2

3 x

1 x2

x 1

3

A

B

x 2 x 1

x 2

x 1

B 0 1

B

1 1 1

C 0

1 1 11

C

2

B

1

1

2

C 1

A

2

2

1

Multiply both sides of the identity by x 2 x 1 :

3

A x 1

Then 3

B x 2 . Now, let x A 2 1 B 2 2 3

Now, let x 1 . Then 3 Therefore, (h)

5 x 2x x 1 2

x

3 x 2

2

A1 1

A

B1 2

1

1

x 1

x 2

5 x 2x 1 x 1

2 3A

1

3 3B

B 1

A

B x 1 2x 1

Multiply both sides of the identity by 2 x 1 x 1 : 1 5 x A 2 x 1 B x 1 . Now, let x . Then 2 1 1 1 9 3 B 3 5 A 2 1 B 1 B 2 2 2 2 2 Now, let x Therefore,

1 . Then 5 5 x 2x x 1 2

1 3 2x 1

A 2

1

1

B

1 1

6

3A

2 x 1

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(i)

3x 4 x 2

2

A

B

x 2

x 2

2

2

Multiply both sides of the identity by x 2 :

3x 4

A x 2

B . Now, let x

Now, let x 0 . Then 3 0 3x 4

Therefore, (j)

x

x 2

12 x3 2 x2

4

2

4

A 0 2

3

2

x 2

x 2

12 x

2

x

2 . Then 3 2

2

x

x 2

2

2

4

A

2A 6

2 2

B

B

2

A 3

2

12 x 2 x 1

A x

B x2

C

D

x 2

x 1

Multiply both sides of the identity by x2 x 2 x 1 :

12

Ax x 2 x 1

Cx2 x 1

B x 2 x 1

Dx2 x 2

Now, let x 0 . Then 12

A 0 0 2 0 1

B

2

0 1

D 0

2

0 2

2 B 12

1 . Then A

1

D

Let x 12

C 0

6

Let x 12

B 0 2 0 1

1 2

1 1

B

1 2

1 1

C

1

2

1 1

D

1

2

1 2

4

2 . Then A 2 2 2 2 1

B 2 2 2 1

C 2

2

2 1

D 2

2

2 2

12C 12

C 1

(k)

Let x 1 . Then

12

Therefore,

12 x3 2 x 2

x

4

2 x

3

12

2 x

x x

2

A1 1 2 1 1

1

A x

61 2 1 1

2 A 12 2 4 12 1 x 2

4

3 x 1 x

2A

1 1

2

6

A 3

1 1

41

2

1 2

6 x2

Bx C x2 1 x x2 1 :

2

A x2 1

Bx C x

Now, A 2 and C

A B x 2 Cx A

0 and A B 0

B

2

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3D 12

Therefore,

(l)

x 2 x3 3x

2 x

3

x

x 2 x x2 3

2 x

2x x

A x

2

1

Bx C x2 3

Multiply both sides of the identity by x x 2 3 : x 2

A x2 3

A B x 2 Cx 3 A

2 and C 1 and A B 3

Now, A Therefore,

x 2 x3 3x

3x 2 x3 6 x

3x 2 x x2 6

(m)

Bx C x

2 3x A x

0

B

2 3

3 2x 3 x2 3 Bx C x2 6

Multiply both sides of the identity by x x 2 6 : 3x 2

A x2 6 1 and C 3

Now, A Therefore, (n)

2x 3 x3 8x

A B x 2 Cx 6 A

Bx C x 3 and A B

3x 2 x3 6 x

1 3x

2x 3 x x2 8

A x

0

B

1 3

9 x 3 x2 6 Bx C x2 8

Multiply both sides of the identity by x x 2 8 : 2x 3

A x2 8

Now, A Therefore,

3 and C 8 2x 3 x3 8 x

Bx C x

A B x 2 Cx 8 A

2 and A B 3 8x

0

B

3 8

16 3 x 8 x2 8

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(o)

x

3

x 5 5 x2 4 x

x 5 x x 5x 4 2

x 5 x x 1 x 4

A x

B

C

x 1 x 4

by x x 1 x 4 :

x 5

A x 1 x 4

Bx x 4

Cx x 1

Now, let x 0 . Then 0 5

A 0 1 0 4

Let x 1 . Then 1 5

A1 1 1 4

Let x

B 0 0 4

B1 1 4

C 0 0 1

C 1 1 1

5 4A

3B 6

5 4

A

B

2

4.

Then 4 5 Therefore,

A 4 1 4 4

x

x 5 5x2 4x

3

B 4 4 4

5 4x

C 4 4 1

12C 9

3 4

C

2

3 x 1 4 x 4

Chapter 2 practice questions 1. x 2 a 3b x 3ab 0 This appears to be a factorable equation.

x 3b x a

0

Therefore, the solutions are x 3b or x a 2.

3x 2 4x 1 . Multiplying both sides of the inequality by 15 we get: 3 5 3 3 3 x 2 3(15) 5 4 x 1 9 x 6 45 20 x 5 11x 44 x 4

3 x 2 8 x c is

3. If the vertex of the parabola y Therefore,

1 3

16 9

3

32 c 3

So, f x

a x

a x 4

2

p

16 3

32 c 3

1 , then 3 c

1 3

4 3 3

2

8

4 3

5.

ax 2 bx c can be expressed in the turning

4. The quadratic function f x point form f x

1 3

4 , 3

2

q , where p and q are coordinates of the vertex.

6.

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c.

Since the graph of the function passes through the point 2, 4 then 4 a 2 4 1 It follows that 4 4a 6 and a . Therefore, 2 1 1 2 1 2 2 f x x 4 6 x 8 x 16 6 x 4x 2 . 2 2 2 We need to verify that x 4 2 3 is a zero of the quadratic function: 2 1 1 f 4 2 3 4 2 3 4 6 12 6 0 , 2 2 so x

4 2 3 is the zero of the function. Therefore, a

5. If the roots of the polynomial f x b 2 3 4 3 5 a

2 and

If f

2

f 1

2

x3 5x2

px q are

1 , b 2

4, c

, 2

and

2

6.

2

3 then:

2

3 1 are roots of the equation, then 3

5

2

1 5 p q

6

2

2 p q 12 2 p q p q

0

0

If we subtract the two equations, we get p = 2, and consequently q

8.

6. (a) The equation mx 2 2 m 2 x m 2 0 has two real roots if and only if m 0 and 0 (there can be a double root, which satisfies requirements of the problem). 4 m 2

0

2

4m m 2

8m 18 0

m

4m2 16m 16 4m2 8m 8m 16

2 but also m 0 . Therefore, m

2, 0

0,

(b) The equation mx 2 2 m 2 x m 2 0 has two real roots with opposite signs if m 2 m 0 and 0 , additionally, x1 x2 0. m 2 0

m 2 m m 2 m

m

+

0

+

0 + 0

+ +

X

+

2, 0

Now, the three conditions, m 0 , m satisfied simultaneously. Therefore, m

2, 0

0,

and m

2, 0 must be

2, 0

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7. Since x 1 and x 1 are factors of the polynomial f x x3 ax 2 bx c 0 then by the factor theorem: f 1 0 and f 1 0 . Additionally, by the remainder theorem

12 . We have:

f 2

2

13 a 1 3

1

b 1

a

23 a 2

1 2

2

c

0

b

b 2

It follows, that b

1

a b c 1 a b c 1 4a 2b c 4

c 0

c 12 1 and

a c

0

4a c

6

c a 4a a 6

Solving the system of equations, we get Therefore a

2, b

1, c

c a 3a 6

c a

2 2

2

8. Squaring both sides of the inequality x x 2 25 x 2 12 x 36 x2

2b 2 a b c 1 4a 2b c 4

5 x 6 we get

25 x 2 300 x 900

24 x 2 300 x 900 0

2 x 2 25 x 75 0

2 x 15 x 5 5

x 5 2 x 15 2 x 15 x 5 + x

,5

9. The equation 2 x 2 3 k

0

2

k

+

0

15 2 + 0 0

+ + +

15 , 2

3 k x k 3 0 has two imaginary roots when

4 2 k 3

k 2 14k 15

k 1 k 15 k 1 k 15

0

0

9 6k k

k 1 k 15 1 0

+

2

0

+

15 + 0 0

8k 24 k

2

0.

14k 15

0 + + +

1, 15

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2 x2 8x 7 2 x2 8x 10 3 x2 4 x 5 x2 4 x 5 3 2 2 x 2 1

10. (a) f x

f x

(b) (i) lim f x

x

3

and

x 2

2

x

2

2 since in both cases as x

2 , (ii) lim f x

x

2 x2 4 x 5 x2 4x 5

3 4x 5

, x 2

2

1

0

1

3

(c) f x has a minimum when the expression possible. This happens when y of the parabola y

x 2

2

2

x 2

x 2

2

assumes the maximum value

1

1 has the minimum value. The minimum value

1 occurs at the vertex, whose coordinates are

2, 1 .

11. The equation k 2 x 2 4 x 2k 1 0 has two distinct real roots if k 2 0 and the discriminant 0.

42 4 k 2

2

16 8k 2 20k 8 8k 2 20k 24

2k 1

8k 2 20k 24 0

0

k

2k 2 5k 6 0

The discriminant of 2k 2 5k 6,

5

1

2

4 2 6

25 48

23 0 and the

2

coefficient of k is positive, so 0 for all real values of k. Considering that k 2 , 2 the equation k 2 x 4 x 2k 1 0 has two distinct real roots when k 2 12. By the remainder theorem f 6

1

4

11 1

3

22

6 11 22 a 6

1

2

20

1 a

a

20 . Then: 1

6

20

1

13. By the remainder theorem, the polynomial p x

ax b

3 3

must satisfy the following 3

1 and p 2 27 . Then: a 1 b conditions: p 1 1 and a 2 b 27 which is equivalent to a b 1 and 2a b 3 . Subtracting the first equation from the 4 4 1 4 1 second we get 3a 4 and b a 1 . Therefore a . a 1 , b 3 3 3 3 3

14. By the remainder theorem f 2 23 3 2

2

8 12 2a

a 2

b

1 3 a

1

3

3a

3

f 1

18

1 . Then: 2

a

a

1

b

6

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x 4 ax 3 must satisfy f 1

15. By the remainder theorem, the polynomial f x 4

Then: 1

a 1

3 8

a

4.

x 3 ax 2 3x b . By the factor theorem, f 2

16. Let f x theorem, f

6 . We have: 23 a 22

1

3

8.

3 2

0 and by the remainder

b 0 and

2

1 a 1 3 1 b 6 . Simplifying, we get 4a b 2 and a b 4 . Subtracting the second equation from the first we get 3a 6 a 2 and b 4 a 4 2 6. 17. If x 2 4 x 3 is a factor, then x 1 is also a factor, thus f 1

1

a 4

3 4a

3 0

a 1

Alternatively, x 3 is a factor leading to the same answer. Another approach, but much longer is: x3

The polynomial f x

a 4 x2

3 4a x 3 can be written as

x 3 a 4 x 2 3 4a x 3 x 2 4 x 3 x m , where x m is the result of division of f x by x 2 4 x 3 . Now, after expanding and simplifying of the right-hand side:

x3

a 4 x2

3 4a x 3 x 3

m 4 x2

4m 3 x 3m .

a 4

m 4

The following conditions must be satisfied: 3 4a 4m 3 3 3m

a

m a m. 1 m

Therefore, a 1 . 18. By the factor theorem, f 2

3

2

2

2

5

2

2 k

0 . Then: 0

19. The equation kx 2 3x k 2 discriminant 0 . Then: 3 0

2

4k k 2

8 8 10 k

0

k

6

0 has two distinct real roots if k

0 and the

9 4k 2 8k

4k 2 8k 9 0

This is a parabola, concave downwards, and it is above the x-axis between its roots, as seen in the GDC screenshot:

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2

k

13 2

2

,

13

2.80, 0.803 (to 3 s. f.)

2

20. The equation 1 2k x 2 10 x k 2 0, k 1 and the discriminant 2

k

10

2

4 1 2k k 2

8k 2 12k 108 0

0

100 4 2k 2 3k 2

8k 2 12k 108

8k 2 12k 108 0

2k 2 3k 27 0

3

k

3,

0 +

0

k

1 or 2

0

Factorising, we get 2k 9 k 3

k 3 2k 9 k 3 2k 9

has real roots if 1 2k

+

0

0 9 2 + 0 0

+ + +

9 2

21. The inequality m x 1 x 2 can be written as x 2 mx m 0 . Since the coefficient of x 2 is positive, the inequality is true for all x , when the discriminant 0. 2 2 m 4 1 m m 4m m m 4

0

m m 4

4 0

m 4 m m m 4 m

0

+

+

0

0 + 0 0

+ + +

4, 0

22. Since both sides of the inequality 5 3x 2

x 1 are non-negative, we can write

2

5 3x x 1 . Now we have: 25 30 x 9 x 2 x 2 2 x 1 8x 2 32 x 24 0 x2 4 x 3 0

x 3 x 1

0

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1 0

x 1 x 3 x 1 x 3 x

+

3 + 0 0

+

0

+ + +

1, 3 3 x

23. The inequality x 2 4

3 x

x

0 can be written x 2

4

3 x

3 . By GDC: x

x 2 4 and g x

Let f x

x2 4

0, where x

2.30, 0

1, 1.30 (to 3 s. f.)

A more algebraic approach is also possible: x

2

3 4 x

x 1 x2

x3 4 x 3 0 x

0

x

x 3

0

This can be solved by setting up a table as in previous problems; the exact solution will be 1

x

13 2

,0

1,

1

13 2

24. Since both sides of the inequality x 2 2

2 x 1 are non-negative, we can write

2

x 2 2x 1 Now we have: x 2 4 x 4 4 x 2 4 x 1

3x2 8 x 3 0

3x 1 x 3

0

3

x 3 3x 1 3x 1 x 3 x

3,

0 +

0

+

1 3 + 0 0

+ + +

1 3

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x 4 x 2 x 4 x 2 or 0 . Now we have: x 1 x 4 x 1 x 4 x 4 x 4 x 2 x 1 x 2 16 x 2 x 2 0 0 x 1 x 4 x 1 x 4

25. f x

g x

x 14 x 1 x 4

x x x x x 1 x

0

1 4 14 14 x 4 , 1

1 0

+

4 + 0

X

+

X

+ +

14 + + 0

+ + +

0

+

4, 14

x 9 2, where x 9 , can be written as x 9 x 9 Since both sides of the inequality are nonnegative: 2 2 x 9 4 x 9

26. The inequality

2x 9.

x 2 18x 81 4 x 2 72x 324

3x 2 90 x 243 0

x 3 x 27 x 3 x 27 x

,3

27. The inequality For x 1: For x 1 :

x 2 30 x 81 0 3 0 +

+

0

x 3 x 27

27 + 0 0

0

+ + +

27, 2x x 1

1, where x 1 ,

2x 1 0 x 1 2x 1 0 x 1

3x 1 0 1 x x 1 0 x 1

x

1 . 3

1 x 1,

Therefore, the inequality is satisfied for x

no solution exists when x 1 .

1 3

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28.

2x 5 x x 2

2x 5 x 2 x 1

2

A

B

x 2

x 1 2x 5 x 1

Multiplying both sides by x 2 we get: Now, let x

2x 5 x 1

2 . Then A

2

2 5 2 1 2x 5 x 2

Multiplying both sides by x 1 we get: 21 5 1 2

Now, let x 1 . Then B 29.

2 x 48 x2 9

2x 5 x 3 x 3

A

B

x 3

x 3

2 x 48 x 3

3 . Then A

Therefore, A 9, B 30.

a b x a x b

2

2 x 48 x 3

2 3 48 3 3

42 6

B 3

1

x 2

x 1

B x 3 x 3

A

54 6

9

A x 3 x 3

B

7

7

A

B

x a

x b

, x

a and x

a . Then A

a b a b

B a a a b

Now, let x b . Then B a b x a x b

a b b a

A b b b a

1

1

x a

x b

b B x a x b

, x

a b x b

1

a b x a

Multiplying both sides by x b we get:

Therefore,

3

3 48 3 3

Multiplying both sides by x a we get: A Now, let x

9 3

A x 1 x 2

2 x 48 x 3

Multiplying both sides by x 3 we get: Now, let x 3 . Then B

B x 2 x 1

3 2x 5 1 Therefore, 2 3 x x 2

Multiplying both sides by x 3 we get: Now, let x

A

A x b x a

B

a b a b

1

a and x

b

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Exercise 3.1 1.

In parts (a) (d), substitute n (or k) = 1, 2, (a)

s(n) 2n 3

s(1)

(b)

g ( k ) 2k

g (1)

(c) (d) (e)

(f) 2.

3

f (n) 3 2

n

a1

5

an

an

1

3

bn

1

2n

1, g (2) 1, g (3) 5, g (4) 13, g (5) 29 3 , f (2) 2

a1 1, a2

a1

b1 3 bn

1, s(2) 1, s(3) 3, s(4) 5, s(5) 7

f (1)

( 1)n (2n ) 3

an

, 5 into the given formula.

5, a2

3 , f (3) 4

7, a3

3 3 , f (4) , f (5) 8 16

5, a4 19, a5

7, b3 13, b4

21, b5

In parts (a) (d), simply substitute n 1,1,3,5,7

(b)

2,6,18,54,162

(c)

2 2 6 4 10 , , , , 3 3 11 9 27

(d)

1,2,9,64,625

a50

29

8, a3 11, a4 14, a5 17

b1 3, b2

(a)

3 32

31

n = 50 into the formula.

97

b50

2 349 u50

a50

4.786 1023 100 2502

50 1251

5049 1.776 1083

In parts (e) (h), start with the first term and substitute it in the given formula to find the second term, and so on. To find the 50th term, we will use a GDC in Sequential mode. Be aware that some GDCs start with u n 1 rather than u n , as shown in the second set of screen shots. In this case, you start with n = 0 and end with n = 49. (e)

3,11,27,59,123

a50

4.50 1015

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0, 3,

(g)

2, 6, 18, 54, 162

(h)

3.

4.

3 3 21 39 , , 3 7 2 13 55 1 7

(f)

1, 1, 3, 5, 7

u50

1.00

b50

4.786 1023

a50

97

In this question, you need to observe and spot the pattern, perhaps using trial and error. (a)

un

1 un 1 , u1 4

(b)

un

4a 2 un 1 , u1 3

(c)

un

un

1 3

1 a 2

a k , u1

1

a 5k

In this question, you need to observe and spot the pattern, perhaps using trial and error. (a)

un

n2

(b)

un

3n 1

(c)

un

2n 1 n2

(d)

un

2n 1 n 3

3

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5.

For the Fibonacci sequence 1,1,2,3,5,8,13,21,34,55,89,... we have:

(a)

(b)

6.

a1

F2 F1

1 1, a2 1

a5

F6 F5

8 , a6 5

a9

F10 F9

55 , a10 34

an

Fn 1 Fn

Fn

1

5

1

For Fn (a)

F3 F2

2 1

F7 F6

1

Fn Fn

1

5

Fn 1 Fn

F9 F8

21 , a8 13

5 , 3

34 , 21

1 Fn Fn 1

1

1

1 an 1

n

we have:

2n

5

F5 F4

3 , a4 2

89 55

Fn

Fn n

F8 F7

13 , a7 8 F11 F10

F4 F3

2, a3

Substitute n You can use computer software for this as well. F1

F2

1 5

1

5

1

1

1 2 5 5

5

1 2 5 5

F4

1

4

5

F3

1

2

1 5

1 3 5 15 5 5

1 3 5 15 5 5

1

1 4 5 30 20 5

2

8 25

1 4 5 30 20 5 25

3

16

5

Similarly, F5

5, F6

8, F7

13, F8

21, F9

34, and F10

55

This sequence is equal to the Fibonacci sequence. (b)

By symbolic manipulation or by taking each term and multiplying it by a fraction whose numerator and denominators are the conjugates of the given number, or simplifying the right-hand side, you get: 3

(c)

Fn

5

1

1 2 5 5

6 2 5 2 Fn

1 5

1

2 1

5

2

2

n 1

2n

5

1 1

5

n 1

1 5

1

5

n

1

5

n

2n

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1

Factor out

and express both fractions with the same denominator, 5 then collect like-terms: Fn

1

1 21 5

Fn

1 5

5

n 1

1

n

5

21

n 1

5

2n

1

5

n 1

1

5

n

2n

2 1

5

1

n 1

5

2n

2 1

5

2n

Simplify and use result of (b) Fn

1

Fn

1 5

1

1 5

1

1 5

1

5

n 1

3

5

1

n 1

5

2n 5

2n

5

2n

n 1

1

2n

2

5

1

n 1

1

5

1

2n

5

n 1

1

2n

2

5

3

5

2

2

n 1

Fn

1

1

Exercise 3.2 1.

If there are four means between 3 and 7, there are six terms in this sequence such that a1

3, a6

7

Apply the nth term formula to find d and then simplify: a6

a1 5d

7 3 5d

The sequence is 3,

2.

d

4 5

0.8

19 23 27 31 , , , ,7 5 5 5 5

(a)

Arithmetic:

(b)

Arithmetic:

(c)

Arithmetic:

an

1

an

2(n 1) 3 a50

bn

1

bn

1, c2

a1 49d

b1 49d

1, c3

3

d

2

1 49 2 97

(n 1 2) (n 2) 1 b50

c1

(2n 3) 2

d 1

3 49 1 52 d

2

c50

c1

49d

1 49 2 97

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(d)

There is no constant common difference, so the sequence is not arithmetic. un

un

3un

1

2 3un

1

2

2

3 un

un

1

2

Also, u1 is not defined, so the sequence cannot be determined. (e)

e2

e1

5 2 3, e3

e2

7 5 2

There is no constant common difference, so the sequence is not arithmetic.

3.

f2

f1

f3

f2

f4

f3

(f)

Arithmetic:

(a)

(i)

a1

2, d

(ii)

an

2 (n 1) 4 4n 6

(iii)

an

an

(i)

a1

29, d

(ii)

an

29 (n 1) ( 4)

(iii)

an

an

(i)

a1

6, d

(ii)

an

6 (n 1) 9 9n 15

(iii)

an

(i)

a1 10.07, d

(ii)

an

10.07 (n 1) ( 0.12)

(iii)

an

an

(i)

a1 100, d

(ii)

an

100 (n 1) ( 3)

(iii)

an

an

(i)

a1

2, d

(ii)

an

2 (n 1) (

(iii)

an

an

(b)

(c)

(d)

(e)

(f)

f 50

an

1

1

1

1

1

1

f1 49d

4: a8

a1

4, a1

7

d

7

2 49 ( 7)

(8 1)d

341

2 7 4 26

2

4: a8

4, a1

a1

7d

29 7 ( 4) 1

4n 33

29

9 : a8

9, a1

a1

7d

6 7 9 57

6 0.12 : a8

a1

7d 10.07 7 ( 0.12) 9.23 0.12n 10.19

0.12, a1 10.07 3: a8

a1

7d 100 7 ( 3) 79 3n 103

3, a1 100

5 : a8 4

5 , a1 4

a1 7d 5 ) 4

2 7 (

5 ) 4

27 4

5 13 n 4 4

2

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4.

There are five means between 13 and 23, so there are seven terms in this sequence such that a1 13, a7

23

Apply the nth term formula to find d and then simplify: a7

a1 6d

23 13 6d

d

6

Thus, the sequence is 13,7,1, 5, 11, 17, 23 5.

Similar to question 4: a1

299, a5

300

a5

a1

4d

300 299 4d

d

1 4

0.25

The sequence is 299, 299.25, 299.5, 299.75, 300 6.

We need to find the first term and the common difference. a5 16, a14

a14

d

7.

42

4d 16 , and

a1

42, and solving the system

a1 13d

26 , a1 9

40 9

40 26 (n 1) 9 9

an

1 26n 66 9

Similar to question 6: a3

40

a1 2d

40

a9

18

a1 8d

18

142 3

an

d 8.

42

11 , a1 3

142 3

(n 1)

11 11 n 51 3 3

Use the nth term formula for each part. (a)

(b)

(c)

(d)

a1

3, d

6, an

525

an

a1 (n 1)d

a1

9, d

an

a1 (n 1)d

a1

1 3 ,d 8

an

a1

a1

1 ,d 3

an

a1 ( n 1) d

6, an

4

1 3

n 88

201 201 9 ( n 1) ( 6)

1 1 3 4 8

(n 1) d 1 2

525 3 (n 1) 6

9 3 , an 14 8 8 115 25 9 (n 1) 8 8 8

1 5 , an 2 6 6 17 1 1 (n 1) 6 3 6

n 36

n 11

n 16

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a1 1 k , d

(e)

an

a1 (n 1)d 20k

a1 15, a7

9.

a7

(1 k ) (1 k ) 2k , an 1 19 k 1 19k 1 k (n 1) 2k

( n 1) 2k

n 11

21

a1 6d

21 15 6 d

36 6

d

6

The sequence is 15,9,3, 3, 9, 15, 21

10.

a1

99, a5 100

a5

a1

4d

100 99 4d

1 4

d

The sequence is 99, 99.25, 99.5, 99.75, 100

11.

a3 11

a1 2d 11

a12

a1 11d

47

d

47

The sequence is defined by: a1

12.

a7

48

a1 6d

a13

10

a1 12d

3, an

48

d

10

The sequence is defined by: a1

a30 147, d

13.

14.

4, a1

3

3 (n 1) 4

19 , a1 3

86, an

4n 1 for n 1

86

86 (n 1)

19 19n 277 for n 1 3 3

4

a30

a1

29d

an

a1 (n 1)d

a1

7, d

3, an

an

a1 (n 1)d

147 a1

29 4

a1

31

31 (n 1) 4 4n 27

9803 9803

7 (n 1) 3

n 3271

Yes, 9803 is the 3271th term of the sequence.

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15.

a1

9689, a100

an

a1 (n 1)d

a110

a1 109d

an

1

8996 8996 9689 99d

d

7

9689 109 ( 7) 8926

9689 (n 1) ( 7) 1

n 1385

Yes, 1 is the 1385th term of the sequence.

16.

a1

2, a30 147

an

a1 (n 1)d

an

995

147 2 29d

2 (n 1) 5 995

d n

5

998 5

As a fractional result is not possible for n, we conclude that 995 is not a term of this sequence.

Exercise 3.3 1.

(a)

(b)

(c)

(i)

3, 3a 1 ,32 a 1 , 33a 1 ,...

(ii)

r

(iii)

u10

(i)

0,3,6,9...

(ii)

d

(iii)

a10

(i)

8,16,32,64,...

(ii)

r

(iii)

b10

3a 1 3

un un 1 u1r 9

an

bn bn 1

b1r 9

9

1

1

9d

16 8

39 a

(3n 3)

(i)

(e)

(i)

4,12,36,108,...

(ii)

r

(iii)

u10

3(n 1) 3

3

The sequence is geometric. ... 2

4096

1, 4, 10, 22,...

u1r 9

1

0 9 3 27

8 29

(d)

un un 1

3a

1

The sequence is arithmetic.

an

a1

33a 32 a

...

3 3a

The sequence is geometric.

Neither arithmetic nor geometric. The sequence is geometric.

12 ... 3 4

4 39

78732

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(f)

(g)

(h)

(i)

(j)

(k)

(l)

(m)

(i)

2,5,12.5,31.25,78.125...

(ii)

r

(iii)

u10

(i)

2, 5,12.5, 31.25,78.125...

(ii)

r

5 2

(iii)

u10

u1r 9

(i)

2,2.75,3.5,4.25,5...

(ii)

d

(iii)

u10

(i)

18, 12,8,

(ii)

r

(iii)

u10

(i)

52,55,58,61,...

(ii)

d

(iii)

u10

un un 1

u1r 9

12.5 5

... 2.5

2 2.59

7629.39

12.5 ... 5

The sequence is geometric.

2

2 ( 2.5)9

7629.39 The sequence is arithmetic.

2.75 2 3.5 2.75 4.25 3.5 5 4.25 0.75 u1

9d

2 9 0.75 8.75

16 32 , ... 3 9

12 18

The sequence is geometric.

8 ... 12

2 3 9

2 3

u1r 9 18

1024 2187

0.468

The sequence is arithmetic.

55 52 ... 61 58 3 u1 9d

52 9 3 79

1,3, 9,27, 81,...

(i)

3 1

(ii)

r

(iii)

u10

(i)

5 2

The sequence is geometric.

u1r 9

9 3

...

The sequence is geometric. 3

( 1) ( 3)9 19683

0.1,0.2,0.4,0.8,1.6,3.2,... 0.2 3.2 ... 0.1 1.6

(ii)

r

(iii)

u10

(i)

3,6,12,18,21,27,...

u1r 9

0.1 29

The sequence is geometric.

2

51.2

Neither arithmetic nor geometric.

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2.

(n)

(i)

6,14,20,28,34,...

Neither arithmetic nor geometric.

(o)

(i)

2.4,3.7,5,6.3,7.6,...

The sequence is arithmetic.

(ii)

d

(iii)

u10

(i)

Arithmetic: d

(ii)

an

3 (n 1) 5 5n 8

(iii)

a1

3, an

(i)

Arithmetic: d 15 19 ...

(ii)

an

(iii)

a1 19, an

(i)

Arithmetic: d

(ii)

an

8 (n 1) 11 11n 19

(iii)

a1

8, an

(i)

Arithmetic:

(a)

(b)

(c)

(d)

(e)

(f)

3.7 2.4 ... 7.6 6.3 1.3 u1

9d

2.4 9 1.3 14.1

2 ( 3) ... 5

an

a1 7d

a8

a1

a8

a1 7 d

3 7 5 32

5 for n 1

1

19 (n 1) ( 4) an

a8

4

7d 19 7 ( 4)

9

23 4n

4 for n 1

1

3 ( 8) ... 11

8 7 11 69

an 1 +11 for n 1

d

9.95 10.05 ...

0.1

a8

a1 7d 10.05 7 ( 0.1) 9.35

(ii)

an

10.05 (n 1) ( 0.1) 10.15 0.1n

(iii)

a1 10.05, an

an

(i)

Arithmetic: d

99 100 ...

(ii)

an

(iii)

a1 100, an

(i)

Arithmetic: d

(ii)

an

2 ( n 1)

(iii)

a1

2, an

0.1 for n 1

1

1

a8

a1

7d 100 7 ( 1) 93

100 (n 1) ( 1) 101 n

an

an

1

1 for n 1

1 2

2 ... 3 2

1

3 2

a8

a1 7d

2 7

3 2

17 2

7 3n 2

3 for n 1 2

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(g)

(h)

(i)

(j)

6 ... 2 3

(i)

Geometric: r

(ii)

an

3 2n

(iii)

a1

3, an

(i)

Geometric: r

(ii)

an

4 3n

(iii)

a1

4, an

(i)

Geometric: r

(ii)

an

5

(iii)

a1

5, an

(i)

Geometric: r

(ii)

an

3 ( 2) n

1

(iii)

a1

3, an

2an 1 for n 1

a8

a1 r 7

3 27

384

a8

a1 r 7

4 37

8748

1

2an 1 for n 1

12 ... 3 4

1

3an 1 for n 1

1

5 5

5 5

a8

a1 r 7

2

a8

1

7

5

an 1 for n 1

6 3

12 ... 6

a1 r 7

The sequence is neither arithmetic nor geometric.

(l)

(i)

Geometric: r

(ii)

an

2

3 2

(iii)

a1

2, an

(i)

1

n 1

(k)

(m)

5

3 ... 2 n 1

5 7

an

35

(iii)

a1

35, an

a8

a1 r 7

3 2

( 2)

7

384

2187 64

2

3 an 1 for n 1 2

25 ... 35

Geometric: r

(ii)

3n 1 ( 2)n

3 2

3 ( 2)7

n 1

5 7

a8

a1 r

7

35

5 7

7

3.32

5n 7n 2

5 an 1 for n 1 7

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(n)

(i)

(o)

(p)

(q)

3.

3 1 ... 6 2

Geometric: r 1 2

n 1

(ii)

an

6

(iii)

a1

6, an

(i)

Geometric: r

(ii)

an

9.5 2

(iii)

a1

9.5, an

(i)

Geometric: r

(ii)

an 100 0.95

(iii)

a1 100, an

(i)

Geometric: r

(ii)

an

2

(iii)

a1

2, an

3 8

1 2

a8

a1 r 7

( 6)

a8

a1 r 7

9.5 27

7

3 64

3 2

n 2

1 an 1 for n 1 2 19 ... 2 9.5

1216

n 1

2an 1 for n 1

95 ... 0.95 100

a8

a1 r 7 100 0.957

69.83

n 1

0.95an 1 for n 1 3 4

2

...

3 8

a8

a1 r 7

2

3 8

7

n 1

3 an 1 for n 1 8

The sequence will have six terms including 3 and 96: a1

3, a6 3r 5

a1r 5

96 r5

96

32

r

2

Thus, the sequence will be 3,6,12,24,48,96 4.

Similar to question 3: a1

7, a5 7r

4

a1r 4 4375

4375 r4

625

r4

54

r

5

Thus, the sequence will be 7,35,175,875,4375 or 7, 35,175, 875,4375

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5.

Similar to question 4: a1r 2

a1 16, a3

16r 2

81

81 16

r2

81

Thus, the sequence will be 16, 36, 81 or 16, 6.

9 4

r

36, 81

Similar to question 5: a1

a1r 5 1701

7, a6

7r 5 1701

r5

243

r 3

Thus, the sequence will be 7,21,63,189,567,1701 7.

Similar to question 6: a1

a1r 2

9, a3

9r 2

64

64 9

r2

64

8 3

r

Thus, the sequence will be 9, 24, 64 or 9, 24, 64 8.

There are 4 terms, thus:

a1

24, a4 a5

a1r 3

3

24

1 2

a1r 4

The nth term: an 9.

a1

a1r 2

24, a3

24r 3

3

24 16

3 2

4

a1r n

1

6

24r 2

1 2

24

3 24

r3

n 1

1 8

1 2

r

3 n 4

2

6 24

r2

6

1 4

1 2

r

There are two solutions: For r

1 : a4 2 1 : a4 2

For r

10.

a1r

3

24

a1r 3

1 2

3

1 2

24

24 3; an 8 3

24 8

1 24 2 3; an

n 1

3 n 4

2 24

1 2

n 1

We use the nth term formula for the 4th term. r

2 , a4 7 a3

a1r 3

a1r 2

14 3

2401 12

a1 2 7

2

a4 r3

14 2 / 3 7

3

2401 12

49 3

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11.

Let 118 098 be the nth term.

6,18,54,... an

a1r

a1

n 1

n 1

118098 6 3n 9

3

6, r 3

3

n 1 9

1

3n

1

19683

n 10

118 098 is the 10th term. 12.

Using the nth term formula for the 4th and 7th terms: a1r 3 18

a4

18

a7

729 8

(1)

729 8

a1r 6

(2)

We can divide (2) by (1) to obtain: r3

81 16

r

3

81 16

33 3 , and so a1 2 2

18 18 81 r 3 16

32 9

Now we can proceed as in the previous question: 59049 128 3 2

an

a1r

n 1

3 2

12

4n 4 3

59049 128

32 9

3 2

4n 4 12 3

4 3

n 1

3 2

531441 4096

n 10

Since n is a natural number, we can conclude that 13.

4n 4 3

59 049 is the 10th term of the sequence. 128

Using the nth term formula for the 3rd and 6th terms: a1r 2

a3

18

a6

243 4

18

(1)

243 4

a1r 5

(2)

We can divide (2) by (1) to obtain: r3

27 8

r

3 2

a1

18 r2

18 9 4

8

Now we can proceed as in the previous question: 19683 64 3 2

an n 1

a1r 3 2

n 1

19683 8 64

3 2

n 1

3 2

n 1

19683 512

9

n 1 9

n 10

Since n is a natural number, we can conclude that

19683 is the 10th term of the sequence. 64

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14.

The interest is paid n and the principal P r A P 1 n

nt

2 times per year. For t = 10 years, the annual interest rate r 0.04,

1500, and so we have:

Vitoria

A P 1

1500 1.0220

2228.92

er account.

P 500, r

15.

2 10

0.04 1500 1 2

0.04, n 4, t 16 r n

nt

0.04 4

500 1

4 16

945.23

Jane will have £945.23 on her 16th birthday.

A 4000, t

6, n 4, r nt

16.

17.

r A P 1 n 4000 P 0.05 1 4

0.05 P 1 4

4000 24

0.05 24

2968.79

This situation can be modelled by a geometric sequence whose first term is 7554 and whose common ratio is 1.005. Since we count the population of 2017 among the terms, the number of terms is 6. a6

7554 1.0055

7744.748

The population in 2022 would be 7745.

18.

19.

r

3 , a4 7

a1

7, r 3 3n

1

14 3

a2

a4 r2

14 3 2 3 7

an 137781 7 3n

19683

3n

1

39

686 27

1

n 10

137 781 is the 10th term.

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P 1000, n

20.

4, t 18, r nt

r n

A P 1

1000 1

0.06 0.06 4

4 18

2921.16

Erik will have £2921.16 on his 18th birthday.

Exercise 3.4 1.

First, we need to determine the number of terms.

a1 11, d an

6, an

a1 (n 1)d

365 365 11 ( n 1) 6

60 11 365 2

The sum of the sequence is S60 2.

n 60 11280

The series is geometric. First, we need to determine the number of terms. 3 , an 2

a1

2, r

an

a1r n 3 2

1

n 1

177147 1024

177147 1024 177147 2048

n 1

3 2

2 3 2

n 1

3 2

11

n 12

12

2

The sum of the series is S12

3 1 2 3 1 2

105469 1024

103

13

3.

(2 0.3k )

2 1.7 1.4 ... ( 1.9)

k 0

The series is arithmetic with 14 terms, a1 The sum of the sequence is S14

4.

2

4 5

2, and d

14 2 ( 1.9) 2

0.7

0.3 7 10

8 16 ... is an infinite geometric series with a1 25 125 2 10 The sum is S 2 7 1 5

2 and r

2 5

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5.

1 3

3 12

1 16

3 64

3 ... is an infinite geometric series with a1 256 1 4 4 3 16 4 3 3 39 3 3(4 3) 4 3 1 4

The sum is S

6.

1 and r 3

3 4

In each case we have an infinite geometric series: (a) 0.52 0.52525252... 0.52 0.0052 0.000052 ... 52 10

2

52 ,r 100

a1

4

52 10 1 100

6

52 10

0.52

S

... 52 100 1 1 100

52 99

(b) 0.453 0.453 53 5353... 0.4 0.053 0.00053 0.0000053 0.4 53 10 53 ,r 1000

a1

3

53 10

5

53 10

... 0.4 S

1 100 53 0.4 1000 1 1 100

0.453 0.4 S

7

4 10

53 990

449 990

(c) 3.0137 3.01373737... 3.01 0.0037 0.000037 0.00000037 ... 3.01 37 10 37 ,r 10000

a1

3.0137 3.01 S

7.

4

37 10

6

37 10

8

... 3.01 S

1 100

37 10000 3.01 1 1 100

301 37 100 9900

29836 9900

7459 2475

Maggie invests $150 (R) at the beginning of every month for six years, so we are calculating future value (FV) for annuity due. For an annual rate of r 0.06 and m 6 12 72 periods: 0.06 i 0.005 12 FV

R

1 i

m 1

i

1

1

150

1 0.005 0.005

73

1

1

13026.135

There will be $13,026.14 in her account after six years.

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8.

(a)

The series 9 13 17 ... 85 is arithmetic. We need to determine the number of terms in the series. a1 9, d 4, an 85

an

a1 (n 1)d

85 9 (n 1) 4

n 20

20 9 85 940 2 The series 8 14 20 ... 278 is arithmetic. We need to determine the number of terms in the series. a1 8, d 6, an 278

The partial sum of the series is S20

(b)

an

a1 (n 1)d

278 8 (n 1) 6

n 46

46 8 278 6578 2 The series 155 158 161 ... 527 is arithmetic. We need to determine the number of terms in the series. a1 155, d 3, an 527

The sum of the series is S46

(c)

an

a1 ( n 1)d

527 155 ( n 1) 3

The sum of the series is S125

9.

ak

2 3k

a1

Sn

n 5 2 3n 2

2 3 1 5, an

125 155 527 2

n 125 42625

2 3n

n(3n 7) 2

10.

For the arithmetic series 17 20 23... we have: n n n(3n 31) a1 17, d 3 Sn 2a1 (n 1) d 34 (n 1) 3 2 2 2 n(3n 31) Sn 678 678 3n2 31n 1356 0 2 The solutions of the quadratic equation 3n2 31n 1356 0 are 16.71 and 27.05, so the solutions of the inequality are n 16.71 or n 27.05 . Since n , we conclude that we need to add 17 terms to exceed 678.

11.

For the arithmetic series 18 11 4... we have: n n n(7 n 43) a1 18, d 7 Sn 2a1 (n 1)d 36 ( n 1) 7 2 2 2 n(7n 43) Sn 2335 2335 7n2 43n 4670 0 2 The solutions of the quadratic equation 7n2 43n 4670 0 are 29.08 and 22.94, so the solutions of the inequality are n 29.08 or n , we conclude that we 22.94 . Since n need to add 30 terms to exceed 2335.

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12.

First sequence: a, a 2d , a 4d ,..., a 98d 50 S50 (a a 98d ) 25(2a 98d ) T 2 Second sequence:

a d , a 3d , a 5d ,..., a 99d Combined sequence: a, a d , a 2d , a 3d ,..., a 99d 100 S100 (a a 99d ) 50(2a 99d ) S 2 Then: 2T 200 S 2 25(2a 98d ) 200 50(2a 99 d ) 100a 4900d 200 100a 4950d 50d 200 d

13.

(a)

For the arithmetic sequence 3,7,11,...,999 we have: a1 3, d 4, an 999

an S250

(b)

14.

4

a1 (n 1) d 250 3 999 2

999 3 ( n 1) 4

n 250

125250

The removed terms, 11,23,35,...,995 form an arithmetic sequence with 83 83 terms and b1 11 and d 12 S83 2 11 82 12 41749 2 The sum of the remaining terms is then 125250 41749 83501

We have the following system of simultaneous equations that can be solved by any method of your choice: a ( a d ) ( a 2 d ) ... ( a 9 d ) 235 (a 10d ) (a 11d ) ... (a 19d ) 735 10 a (a 9d ) 235 2 10 (a 10d ) (a 19d ) 2

15.

(a)

For

20

735

2a 9d 47 2a 29d 147

d

5, a 1

( k 2 1), using a GDC in Sequential mode:

k 1

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(b)

For

17

1 i

i 3

(c)

For

100

2

( 1)n

n 1

16.

an S60

6, an

367

a1 (n 1)d

367 13 (n 1) 6

60 13 367 2

a1

2, r

an

a1r n

11

a1

2 , an 3 1

1 r12 1 r

n 60

11400 4 3

For the geometric series 2

S12

18.

3 : n

For the arithmetic series 13 19 ... 367 we have:

a1 13, d

17.

:

3

8 9

16 4096 ..... we have: 27 177147

4096 177147 2 3

2 1 2 1

n 1

2 3 2 3

4096 177147

2 3

n 1

2 3

11

n 12

12

1.191

(3 0.2k ) 3 3.2 3.4 ... 5.2 is an arithmetic series with:

k 0

a1 3, d 0.2, n 12. So:

11

(3 0.2k ) k 0

S12

12 3 5.2 2

49.2

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19.

For the infinite geometric series 2 a1

20.

2 3

a1

S

2 1

2 3

For the infinite geometric series

1 2

a1

S

21.

2, r

(a)

1 , r 2

1 2

S4

(b)

we have:

2 2 3

1 2 3 3 3

2 ... we have: 9

2

3

3 2

2

3

3 3

2 2

3 3 25 125

93 125

2

3

6 2

3 we have a geometric series: 5n

3 3 3 18 3 ; S2 ; S3 5 5 25 25 5 3 3 3 3 468 5 25 125 625 625 1 5 3 1 1 5 1

Sn

16 27

6 5

3 2 3

1

For un S1

8 9

2 3

a1 1 r

1 r

4 3

n

15 1 1 n 4 5

For this series, we need to spot the pattern (or use partial fractions) and try to relate it 1 1 1 to the term number, n. Thus, vn 2 n 3n 2 n 1 n 2 S1 S3

1 1 1 1 3 ; S2 6 2 3 6 12 12 1 1 1 6 3 ; S4 6 12 20 20 10

1 2 4 8 1 1 6 12

1 20

1 30

10 30

1 3

4 12

n . Alternatively, you can list the 2n 4 terms and simplify as shown on the next page:

The pattern should be apparent by now: Sn

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n

2

1 3

1 3

1 4 1

1

n 1

n 2

1 3n 2

2

1 2

1

n 2 n we have:

n 1

2

1

2 1; S2

2

1

3

S3

2

1

3

2

4

3

S4

2

1

3

2

4

3

Sn

23.

1 12

For un S1

22.

1 2

1 3n 2 n

(c)

1 6

2

3 1

4 1 1 5

4

5 1

n 1 1

The heights that the ball reaches after each bounce form an infinite geometric sequence: 16 0.81,16 0.812 ,... (a)

After the10th bounce: 16 0.8110 1.945m

(b)

16 2 16 0.81 16 0.812 16 0.813 ....

16 2

12.96 152.42m 1 0.81

The shaded area in the first square is made up of two right-angled triangles that measure 8 cm on each side. When the two triangles are joined at their hypotenuse, they make a square with side length 8. Thus, the shaded area is 82 64 cm 2 . In each successive square, each of the shaded triangles is half the size of the previous ones, thus the new shaded area is one half of the shaded area in the previous square. 1 The shaded area in the second square is 64 32 cm 2 , in the third square it is 16 cm2, etc. 2 Total shaded area forms a geometric series with a1

(a)

S10

(b)

S

1 1 2 64 1 1 2

64 1 1 2

64, r

1 2

10

1023 127.875 8

128

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24.

(a)

The first shaded area is 4 2 2 1 6 The second shaded area is 1

(b)

1 1 1 1 4 8 8 16

Total shaded area is 6

3 3 8 128

(b)

6

3 128

819 128

32 5

1 16

Arithmetic series:

a1

7, d

5, an

an

a1 (n 1)d

S68

68 7 342 2

342 342 7 ( n 1) 5

n 68

11866

Arithmetic series:

a1 9486, d an S83

(c)

3 8

If the process were repeated indefinitely, the total shaded area forms an infinite 1 geometric sequence with a1 6, r : 16

1

(a)

1 1 2 4

The third shaded area is

S

25.

1 2

7, an

a1 (n 1)d

8912

8912 9486 (n 1) ( 7)

83 9486 8912 2

n 83

763517

Geometric series:

a1

2, r

an

a1r n

3, an 1 15

S15

a1

9565938

2 3n

1

9565938

3n

1

4782969

3n

1

314

n 15

15

r 1 3 2 14348907 r 1 3 1

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(d)

Geometric series: a1 120, r an

a1r

n 1

1 S9 120

1 , an 5

24 78125

120

1 5

1 5 1 1 5

n 1

24 78125

n 1

1 5

1 390 625

1 5

n 1

1 5

8

n 9

9

150

Exercise 3.5 Note: In all calculations, we will use the following equivalent notations interchangeably: n n! n Cx n Cx x x! n x ! 1.

(a)

Using the 5th x 2y

5

1x5

5 x 4 (2 y ) 10 x3 (2 y ) 2 10 x 2 (2 y )3 5 x(2 y) 4 1(2 y) 5

x5 10 x 4 y 40 x3 y 2 80 x 2 y 3 80 xy 4 (b)

32 y 5

Using the 4th row:

(a b)4 1a4

4a3 ( b) 6a 2 ( b)2

4a( b)3 1( b)4

a 4 4a3b 6a 2b2 4ab3 b4 (c)

Using the 6th row:

( x 3)6 1x6 6 x5 ( 3) 15 x 4 ( 3)2

20 x3 ( 3)3 15x 2 ( 3)4 6 x( 3)5 1( 3)6

x6 18 x5 135 x 4 540 x3 1215 x 2 1458 x 729 (d)

Using the 4th row:

(2 x3 )4 1 24

4 23 ( x3 ) 6 22 ( x3 )2 4 2( x3 )3 1 ( x3 ) 4

16 32 x3 24 x6 8 x9 (e)

x12

Using the 7th row: ( x 3b)7

1 x7

7 x 6 ( 3b) 21x5 ( 3b) 2

35 x3 ( 3b) 4 x7

21x 2 ( 3b)5

21x 6 b 189 x 5b 2 5103 x 2b 5

35 x 4 ( 3b)3 7 x( 3b) 6

945 x 4 b3

5103 xb 6

( 3b) 7

2835 x 3b 4

2187b 7

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(f)

Using the 6th row: 2n

1 n2

6

1 (2n)

6

1 6(2 n) 2 n 5

1 n2

15(2 n) 2

64n 6 192 n3

(g)

15(2 n)

4

2

1 n2

4

1 n2 160 60 240 n3 n6

5

6(2 n)

20(2n)

1 n2 12 1 n9 n12

3

1 n2

3

6

1

Using the 4th row: 4

3 2 x x

1

4

3 x

4

3 x

3

( 2 x) 6

3 x

2

( 2 x )2

1

3 4 ( 2 x )3 1 ( 2 x ) 4 x 81 x4 2.

216 x x3

216 96 x 16 x 2 x

(a)

8 3

(b)

18 5

(c)

7 4

(d)

5 0

5 1

5 2

5 3

5 4

5 5

(e)

6 0

6 1

6 2

6 3

6 4

6 5

8! 1 2 3 4 5 6 7 8 56 3!5! (1 2 3)(1 2 3 4 5)

18 13

7 3

18! 18! 0 5!13! 13!5!

7! 7! 4!3! 3!4!

7! 3!4!

2

1 2 3 4 5 6 7 (1 2 3)(1 2 3 4 )

2

352 1225

(1 1)5 32

6 6

(1 1)6

0

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3.

(a) 7 ni x ( 2 y )i i

7

( x 2 y)7

i 0

7 6 x ( 2 y )1 1

x7

7 5 x ( 2 y)2 2

7 4 x ( 2 y)3 3

7 2 x ( 2 y )5 5

7 1 x ( 2 y)6 6

( 2 y)7

x 7 14 x 6 y 84 x 5 y 2

280 x 4 y 3

560 x 3 y 4

672 x 2 y 5

7 3 x ( 2 y) 4 4

448xy 6 128 y 7

(b) 6 2a i

6

(2a b) 6

i 0

6 2a 0

6 i

( b) i

6 2a 1

6

6 2a 4

6 4

6 1

( b) 4

6 2a 2

( b)1 6 2a 5

6 5

6 2

6 3

( b) 3

6 ( b) 6 6

( b )5

64a 6 192a 5b 240a 4 b 2 160a 3b 3

6 2a 3

( b) 2

60a 2b 4 12ab 5

b6

(c) ( x 4)5

5

5 i 0

5 0

x 5 i ( 4) i

i

5 4 x5

5

x 5 i ( 4)i

1

x5 4 ( 4) 4

20 x 4 160 x 3

x 5 1 ( 4)1 5 5

5 2

x 5 2 ( 4) 2

5 3

x 5 3 ( 4) 3

x5 5 ( 4)5

640 x 2 1280 x 1024

(d) (2 x3 )6

6 i 0

6 6i 3 i 2 (x ) i

6 6 2 0

6 61 3 1 2 (x ) 1

6 64 3 4 2 (x ) 4 64 192 x3

6 62 3 2 2 (x ) 2

6 65 3 5 2 (x ) 5

240 x 6 160 x9

6 63 3 3 2 (x ) 3

6 3 6 (x ) 6

60 x12 12 x15

x18

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(e) 7 3x i

7

(3x b)7

i 0

7 3x 0

7 0

7 i

7 3x 1

( b) 0

7 3x 4 2187 x 7

( b) i

7 4

7 1

7 3x 5

( b) 4

7 3x 2

( b)1 7 5

5103x 6 b 5103x 5b 2

7 2

7 3x 6

( b)5

2835 x 4b3

7 3x 3

( b) 2 7 6

7 3

7 3x 7

( b) 6

945 x 3b 4 189 x 2 b5

( b)3 7 7

( b )7

21xb 6 b 7

(f) 2n

1 n2

6

6 i 0

6 (2n) 6 i

6 (2n)6 0

i

1 n2

i

6 (2n) 6 1

6 (2n)6 4

1 n2

4

64n 6 192n3

1 n2

1

4

6 (2n) 6 5

160 n3

240

1

6 (2n) 6 2 1 n2

5

60 12 n 6 n9

2

5

1 n2 6 6

2

6 (2n) 6 3 1 n2

6

1 n12

(g) 2 3 x x

4

4 i 0

4 0

4 i

2 x

96 x

( 3 x )i

4 0

2 x 4 3

16 x4

4 i

2 x

5 2

4 1

2 x

4 1

4 3

( 3 x )3 216 x

( 3 x )1

4 2

2 x

4 2

( 3 x )2

4 ( 3 x )4 4

216 x 81x 2

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3

1 n2

3

(h) 1

5

4

1

5

4

4

4

i

i 0

4

4 1

4 0

4 2

( 5) 2

2

5)i 4

( 5) 2

4 ( 5)1 1 4

( 5) 0

14 i (

( 5)1

4 ( 5) 0 0 2

i

i 0

( 5)0

0

4

4

14 i ( 5)i

( 5)3

3

4 ( 5) 2 2 4 4

4 4

( 5) 4

4 ( 5) 3 3

( 5) 4

4 ( 5) 4 4

2(1 30 25) 112

(i) 3 1

8

3 1

8 ( 3)8 i 1i i

8

8

i 0

i 0

8 ( 3) 7 1

( 3)8

8 ( 3)3 5

8 ( 3)8 i ( 1)i i

8

8 ( 3)6 2

8 ( 3) 2 6 8 ( 3)7 1

( 3)8

8 ( 3)3 5 2

8 ( 3) 7 1

8 ( 3)5 3

8 ( 3)1 ( 3) 0 7 8 ( 3)6 2

8 ( 3) 2 6

8 ( 3)5 3

8 ( 3) 5 3

8 ( 3) 4 4

8 ( 3)1 ( 3) 0 7

8 ( 3) 3 5

2 216 3 504 3 168 3 8 3 4.

8 ( 3) 4 4

8 ( 3)1 7 1792 3

(a) and (c) x

2 x

45

45 i 0

45 45 x i

45 45 x 0 ...

45 2 x 43

x 45 90 x 43

2 x

i

45 44 x 1 2 x

43

i

2 x

1

45 1 x 44

3960 x 41 ...

45 43 x 2

2 x

2

44

45 0 x 45

2 x

990 243 x 41

45 244 x 43

245 x 45

2 x

45

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(b)

For the constant term, i should be: x45

1 xi

i

x45

i

xi

45 i i

i

45 2

Since i is a natural number this is not possible, i.e. there is no constant term in the expansion. For the term containing x3 , i should be:

(d)

x 45

1 xi

i

x3

x 45

2i

x3

For i = 21, the 22nd term is:

5.

n k

2n

6.

(b)

(c)

n

(1 1) n

1n

0

n

1

(a)

45 21

1

1n

1

1 ...

2 x

21

45 21

221 x3

n k

n n

1n

n

...

1

n

x

45 21

i 21

n

n! n! k !(n k )! (n k )!(n (n k ))!

n 1

7.

45 2i 3

n n n

...

k! k k 1

2n 1

k 2

3 21 k k 1

k 2

3 21

k (k 1)!

(n k 1)! (n k 1) (n k ) 3 2 1 (n k 1) (n k ) 3 2 1 n

Cr

1

n

Cr

n! (r 1)!(n r 1)!

(n k 1)(n k )!

n! r! n r !

By multiplying the first fraction by

r n r 1 and the second by , r n r 1

both denominators will be equal: n

Cr

1

n

Cr

n! r n ! (n r 1) r (r 1)!( n r 1)! r ! n r ! (n r 1) n! r n ! (n r 1) r !(n r 1)! r !(n r 1)!

n ! r n ! (n r 1) r !(n r 1)!

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Now factor out n! in the numerator: n

8.

0

10.

11.

1

n

Cr

n! r n r 1

n! n 1

n 1!

r !(n r 1)!

r !(n r 1)!

r !(n 1 r )!

1 This appears to be the binomial expansion of 3

6

9.

Cr

1 3

6

2 5

8

1 7

n

8 0

n 0

x

1

8 1

n 1

6

1 x

2

6

1 3

5

2 5

7

1 7

n 1

6

6 i 0

i

x

2 3

6

3 5

8

2

1 3

4

2 5

6

2

n

6 7

1 7

2

1 x

2 6 i

2 3

2

3 5

2

n 2

6

...

6

8

...

6 7

2 3

8

2

Cr

6

, i.e.,

2 3

6

3 5

8

n

...

n 1

6 7

n

n

1 3

2 3

6

2 5

3 5

8

1 7

1

1

6 7

n

1n 1

i

To determine i, we have to solve the exponential equation: x2

i

1 x

6 i

x12

1

2i i

x0

12 3i 0

i

4

So, for i = 4, we have the term:

6 4

12.

2 3x x

8

4

1 x

6 4

x2

8 i 0

1 x4

15 x4

8 i

3x

2 x

8 i

15

i

To determine i, we have to solve the exponential equation: x

8 i

1 x

i

1

x8

i i

x0

8 2i 0

i

4

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So, for i = 4, we have the term:

8

3x

4 13.

8 4

8

3 x3

2x

2 x

4

8

8

i 0

16 x4

70 81x4

2x

i

3 x3

8 i

90720 i

To determine i, we have to solve the exponential equation: x

8 i

i

1 x3

x8

1

i 3i

x0

8 4i 0

i 2

So, for i = 2, we have the term:

8

2x

2

14.

1 x

10

3 x3

8 2

2

28 64 x6

10 x 1

1

10 2 x 2

(a)

1.0110

1 0.01

(b)

0.9910

1 0.01

10

9 x6

16128

1 10x 45x2 1 10 0.01 45 0.012 1.1045

10

1 10 0.01 45 0.012

0.9045

15. n r 1

2

n

n

r

r 1

n! 2n! (n r 1)!(r 1)! (n r )!r !

n! n r 1! r 1!

n! 1 (n r 1)!(r 1)! ( n r 1)(n r )

2 (n r )r

1 r (r 1)

n! r (r 1) 2(n r 1)( r 1) ( n r 1)(n r ) (n r 1)!(r 1)! (n r 1)( n r ) r (r 1) n !(n 2 3n 2) (n r 1)!(r 1)!

n !(n 1)(n 2) ( n r 1)!(r 1)!

(n 2)! (n 2) (r 1) !(r 1)!

n 2 r 1

The sum of the entry in row n column r 1 plus twice the entry in nth row and rth column, and the entry in the nth row and (r + 1)th column is equal to the entry two rows directly below the last entry.

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16.

In each case we have an infinite geometric series. 0.7 0.7777... 0.7 0.07 0.007 7 1 a1 ,r 10 10 7 7 10 0.7 S 1 9 1 10

(a)

0.345 0.3 0.0454545... 0.3 0.045 0.00045 45 1 a1 ,r 1000 100 45 3 45 1000 0.345 0.3 S 0.3 1 10 990 1 100

(b)

19 55

3.2129 3.21 0.00292929... 3.21 0.0029 0.000029 a1

(c)

29 1 ,r 10000 100

3.2129 3.21 S

9

(2 x 3)9

i 0

17.

9

i 3

3

29 10000 3.21 1 1 100

321 29 100 9900

7952 2475

9 (2 x)9 i ( 3)i i

(2 x )9 3 ( 3)3

84 64 x 6 ( 27)

145152 x 6

The coefficient of x 6 is 145152

(ax b)7

7 i 0

18. i

4

7 4

7 i

(ax) 7 i ( 3)i

( ax)7 4 b 4

35a 3 x3b 4

The coefficient of x3b4 is 35a3

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19.

2 z2

15

z

15

15

i

i 0

15 10

i 10

2 z2

15 i

( z )i

15 10

2 z2

( z )10

5 (3n) 5 ( 2 m)0 0

(3n 2 m)5

3003

9

9

9

i

i 0

21. i

720 n 2 m3

5 (3n) 0 ( 2 m)5 5 240nm 4

32m 5

i

49

i

9 4 4 3r 2 5

5

5 (3n)3 ( 2m) 2 2

5 (3n)1 ( 2m) 4 4

243n5 810n 4 m 1080n3 m 2

4 3r 2

96 096

5 (3n) 4 ( 2m)1 1

5 (3n) 2 ( 2 m) 3 3

20.

32 10 z z10

3r 2 5

126 256 243r 10

7838208 r 10

The coefficient of r10 is 7 838 208

22.

5

In the expansion of 2 kx , the coefficient of x 3 is 5 C3 22 Thus, 40k 3 1080

k3

27

k

k

3

40k 3

3

Exercise 3.6

1.

2.

5! 1 2 3 4 5 120 (5 5)! 1

(a)

5

(b)

5! 1 2 3 4 5 120

(c)

20 1

(d)

8 3

(a)

5

P5

P

P

C5

20! 1 2 3 ...19 20 (20 1)! 1 2 3 ...19 8! 1 2 3 4 5 6 7 8 (8 3)! 1 2 3 4 5

5 5

20

6 7 8 336

5! 1 (5 5)! 5!

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3.

C0

5 0

10

C3

10 3

10! (10 3)!3!

1 2 3 4 5 6 7 8 9 10 120 (1 2 3 4 5 6 7) 1 2 3

10

C7

10 7

10! (10 7)!7!

1 2 3 4 5 6 7 8 9 10 120 (1 2 3)1 2 3 4 5 6 7

(a)

7 3

7 4

(b)

8 4

(b)

5

(c)

(d)

5! 1 (5 0)! 0!

7! 7! 7! 1 2 3 4 5 6 7 2 2 (7 3)!3! (7 4)!4! 4!3! (1 2 3 4) (1 2 3)

70

8! 1 2 3 4 5 6 7 8 70 (8 4)!4! (1 2 3 4)2

10 6

10 7

10! 10! (10 6)!6! (10 7)!7!

(c) 1 2 3 4 5 6 7 8 9 10 (1 2 3 4)1 2 3 4 5 6

4.

(d)

11 7

11! (11 7)!7!

(a)

8 5

8 3

(b)

1 2 3 4 5 6 7 8 9 10 (1 2 3)1 2 3 4 5 6 7

210 120 330

1 2 3 4 5 6 7 8 9 10 11 330 (1 2 3 4)1 2 3 4 5 6 7

8! 8! 8! 8! 0 (8 5)!5! (8 3)!3! 3!5! 5!3!

11 10! 11 10 9...2 1 11! 39916800

(c)

10 3

10 7

(d)

10 1

10! 1 2 3 4 5 6 7 8 9 10 10 (10 1)!1! (1 2 3 4 5 6 7 8 9 ) 1

10! 10! 10! 10! 0 (10 3)!3! (10 7)!7! 7!3! 3!7!

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5.

(a)

(b)

(c)

10! 1 2 3 4 5 6 7 8 9 10 5! 1 2 3 4 5

(5!)2

10! 2! 5!

(1 2 3 4 5)2 14400

25! 1 2 3

101 8

30240; 2! 2

24 25 1.551121 1025

101! 101! 101 ; (101 8)!8! 93!8! 93

(5!)2

25!

101! 101! (101 93)!93! 8!93!

101 8

101 93

6.

Using the fundamental principle of counting, there are 3 2 4 24 different systems to choose from.

7.

Using the fundamental principle of counting, there are 3 4 2 3 72 different choices.

8.

Using the fundamental principle of counting, there are 8 3 13 312 different combinations of choices.

9.

Using the fundamental principle of counting, there are

4 4

4 412 16777216 ways to

12times

answer all the questions. 10.

Using the fundamental principle of counting, there are 2 2 6times

2 4 4

4 26 4 6

262144

6times

ways to answer all the questions. 11.

Using the fundamental principle of counting, each letter can be chosen in 26 ways and each digit in 10 ways, so there are 263 105 1757600000 different passwords.

12.

The first and last digit can be chosen in 9 ways (cannot be 0) and the three middle digits can be chosen in 10 ways, so there are 9 103 9 81000 such numbers.

13.

(a)

Eight people can be seated in a row in 8! 40320 different ways (permutations).

(b)

If every member of each couple likes to sit together, four couples can be seated in a row in 4! different ways, and for each member of a couple there are 2 possible places, so altogether there are 4! 24 384 different ways that they can be seated.

(a)

Eight children can be arranged in single file in 8! 40320 different ways

14.

(permutations). (b)

If the girls must go first, the five girls can be arranged in 5! ways and the three boys in 3! ways, so altogether there are 5!3! 720 different orders.

15.

In alphabetical order: AEJN, AENJ, AJEN, AJNE, ANEJ, ANJE, EAJN, EANJ, EJAN, EJNA, ENAJ, ENJA, JAEN, JANE, JEAN, JENA, JNAE, JNEA, NAEJ, NAJE, NEAJ, NEJA, NJAE, NJEA.

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16.

17.

ACG

AGC

CAG

CGA

GAC

GCA

ACI

AIC

CAI

CIA

IAC

ICA

ACM

AMC

CAM

CMA

MAC

MCA

AGI

AIG

GAI

GIA

IAG

IGA

AGM

AMG

GAM

GMA

MAG

MGA

AIM

AMI

IAM

IMA

MAI

MIA

CGI

CIG

GCI

GIC

ICG

IGC

CGM

CMG

GCM

GMC

MCG

MGC

CIM

CMI

ICM

IMC

MCI

MIC

GIM

GMI

IGM

IMG

MGI

MIG

(a)

Three letters can be chosen in 263 ways and four digits in 104 ways. Altogether there are 263 104 175760000 possible codes.

(b)

The letters can be chosen in 263 97 ways and the four digits in 104 ways. Altogether there are (263 97) 104 174790000 possible codes.

18.

(a)

The president can be chosen in 17 ways, the deputy in 16 ways, and the treasurer in 15 ways. Altogether there are 17 16 15 4080 ways.

(b)

If the president is male, he can be chosen in 7 ways, thus the deputy can be chosen in 16 ways and the treasurer in 15 ways; so, altogether, there are 7 16 15 1680 ways.

19.

(c)

The deputy (male) can be chosen in 7 ways, the treasurer (female) in 10 ways, and the president in 15 ways; so, altogether, there are 7 10 15 1050 ways.

(d)

If the president and deputy are both male, then they can be chosen in 7 6 ways; if they are both female in 10 9 ways. The treasurer can be chosen in 15 ways. Altogether there are (7 6 10 9) 15 1980 ways.

Since the order is not important, we have combinations. (a)

Three officers of the same specialisation can be chosen as 3 mathematicians out of 8 (in 8 C3 ways), or 3 computer scientists out of 12 (in 6 (in 6 C3 ways). Altogether there are 8 C3

12

C3

6

12

C3 ways), or 3 engineers out of

C3 56 220 20 296 ways.

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20.

21.

(b)

There are a total of 26 C3 ways to choose three officers from 26 people. Three officers that are not engineers can be chosen as 3 out of 20 mathematicians and computer scientists in 20 C3 ways. So, there are 26 C3 20 C3 2600 1140 1460 combinations with at least one engineer.

(c)

Two mathematicians can be chosen in 8 C2 ways, and the third member in 18 ways out of 12 computer scientists and 6 engineers; so, altogether, there are 28 18 504 ways. 8 C2 18

Three numbers, each in the range 1 to 50. (a)

There are 503 125000 different combinations.

(b)

There are

(c)

If the first and second number are matching, then the first number can be chosen in 50 ways, the second in 1 way (must be the same as the first), and the third in 49 ways; so, altogether there are 50 1 49 2450 combinations.

(d)

If two out of three numbers are matching, it may be the first and the second, or the first and the third, or the second and the third; so, altogether, there are 3 2450 7350 combinations.

50

P3

50 49 48 117600 combinations without duplicates.

Five couples can be permutated in 5! ways, but, as they are sitting around a circle, all circular 5! arrangements. For each couple, there are 2 different ways of sitting (male 5 5! 5 right or left of female), so, altogether, there are 2 768 different seating arrangements. 5

there are actually

22.

(a)

Two elements out of nine can be chosen in 9 C2 36 two-element subsets.

(b)

There are 9 C1 one-element subsets, 9 C3 three-element subsets, 9

36 ways, so there are

C5 five-element subsets, 9 C 7 seven-element subsets, and 9 C9 nine-element subsets.

Altogether, we have: 9

23.

C1

9

C3

9

C5

9

C7

9

C9

256 subsets with an odd number of elements.

(a)

Four out of 9 12 21 members can be chosen in

(b)

Two women out of 12 can be chosen in

chosen in

21 4

5985 ways.

12 ways, and two men out of 9 can be 2

9 12 ways; so, altogether, there are 2 2

9 2

66 36 2376 teams.

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(c)

(d)

More women than men can be realised if there are 3 or 4 women, so there are

12

9

12

3

1

4

220 9 495 2475 such teams.

(a)

Teams with only Tim or Gwen: 2

(b)

Two women (one is Gwen) can be chosen in 1

8

(not Tim) can be chosen in in

C3 2 1140 2280

11 1

8 2

11 8 1 2 1

ways, and two men

1 11 28 55 1 8 748 teams.

More women than men can be realised in the following cases:

3 women (not Gwen), 1 man (Tim)

4 women (one is Gwen)

Altogether: 1

11 2

8 1

3

11 11 1 1 3 3

(b)

All four defective disks could be in

96

11

8

2

1

1

3

A sample of 6 disks out of 100 can be chosen in

4560 1192052 400

11

1

11

1

(a)

(c)

1

ways; two women (not Gwen) can be chosen

2

3 women (one is Gwen), 1 man (not Tim)

24.

11

11 8 ways, and two men (one is Tim) can be chosen in 1 ways. 2 1

In total, there are 1

(c)

20

C2

4

C4

100

55 8 165 165 770 teams.

C6 1192052400 ways.

4560 samples, which gives us

0.00000383, i.e. 0.000383% of the total.

At least one defective disk could be in 96

C5

4

C1

96

C4

which gives us

4

C2

96

C3

265004096 1192052 400

4

C3

96

C2

4

C4

265004096 samples,

0.2223, i.e. 22.23% of the total.

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25.

26.

(a)

6 people out of 10 8 4 22 can be chosen in

(b)

Two members out of each party can be chosen in

22

C6

10

C2

74613 ways. 8

C2

4

C2

7560 ways.

If we place the boys in a row with spaces in between, we have the following setup: B

B

B

B

B

B

B

B

B

We have 10 spaces where a girl can stand. Thus, we have 10 P6 different ways of arranging the girls. In each case, the boys can be arranged in 9! different ways, and therefore there will be 54867 456000 ways of arranging the group with no two girls next to each other. 10 P6 9!

Chapter 3 practice questions 1.

a1 4, a4 19, an

A 3000, r

2.

3000

0.06, n

P 1

4 3d 19 4 (n 1)d 99

99

0.06 4

4, t

6; A

46

P

d 5 and n 20

r P 1 n

nt

3000 0.06 1 4

46

2098.63

You should invest $2,098.63 now.

3.

Nick studying hours form an arithmetic sequence with first term and common difference d = 2. Maxine s studying hours form a geometric sequence with first term b1 12 and common ratio r 1.1 (a)

a5

a1 4d 12 4 2 20

b5

b1r 4 12 1.14 17.57

In week 5, Nick studied for 20 hours and Maxine studied for 17.57 hours.

S arithmetic15

(b) S geometric15

15 2 12 (15 1)2 2 1.115 1 12 381.27 1.1 1

390

For the 15 weeks, Nick studied for a total of 390 hours and Maxine studied for a total of 381.27 hours.

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(c)

bn

40

n 1

12 1.1

40 12 1 log1.1

log 40

n

n 13.6

Maxine will exceed 40 hours of study per week in the 14th week. (d)

We need to determine n so that bn an . The easiest way of doing this is by entering n (as X), an (as Y1), and bn (as Y2) into TABLE in a GDC.

We see that in week 12 Nick will study for 34 hours while Maxine studies for 34.24 hours. So, Maxine will catch up with Nick in week 12. 4.

Plan A forms an arithmetic sequence with first term a1 1000 and common difference d = 80 Plan B forms a geometric sequence with first term b1 1000 and common ratio r 1.06 (a)

(b)

(c)

5.

(a)

b2 1000 1.06 1060 g b3 1000 1.062 1123.6 g a12

a1 11d 1000 11 80 1880 g

b12

b1 1.0611 1898.3 g

(i)

S A12

12 2a1 11d 2

(ii)

SB12

b1

6 2 1000 11 80

17 280 g

rn 1 1.0612 1 1000 16869.9 g r 1 1.06 1

The initial amount forms a geometric sequence with a1 500 and common ratio

r 1.06 (fixed rate 6% per annum). After 10 years, it will be worth a11 euro.

a1r10

500 1.0610

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(b)

The future value is a partial sum of a geometric sequence:

FV

6.

7.

r11 1 1 r 1

a1

500

1.0611 1 1 1.06 1

a1 6 and d 3.5

6, 9.5, 13, (a)

a40

a1 39d 6 39 3.5 142.5

(b)

S103

103 2a1 (103 1) d 2

For an (a)

3

103 2 6 102 3.5 2

19003.5

8 an3 1 :

a1 1, a2

3

8 13

3

3

7, a3

8 ( 3 7)3

{an} alternates between values of 1 and (b)

,986 to the nearest euro.

6985.82

a1

3

2, a2

8 23

3

0, a3

8 03

3

3

1, a4

8 13

3

7

7

2, a4

3

8 23

0

{an} alternates between values of 2 and 0 8.

The training program forms an arithmetic sequence with a1 (a)

an

20

2 (n 1) 0.5 20

2 and d 0.5

n 37

She first runs a distance of 20 km on the 37th day of her training. (b)

S37

37 2 2 36 0.5 2

407

The total distance run during 37 days of training would be 407 km. 9.

(a)

r

(b)

(i)

2400 1600

3600 2400

3 2

The number of new participants in 2022 is the 13th term in the sequence. a13

a1r12 1600

an

50000

(ii) n

3 2

1600

log 31.25 1 log1.5

12

207594

3 2

n 1

50000

3 2

n 1

31.25

n 9.489

The 10th term of the sequence will be greater than 50 000; therefore, the number of new participants will first exceed 50 000 in 2019.

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10.

11.

S13

(d)

This trend in growth would not continue due to market saturation.

a1

a1

25, a4 13

an

11995

(a)

MN

Area

(b)

(c)

(d)

12.

r13 1 1.513 1 1600 619582 r 1 1.5 1

(c)

a1 3d 13 a1

25 3d 13

n 1d 2

1 2

2

1 2

2 4

2

2 2

MNPQ

11995

25

d n 1

4 4

n 3006

2 2

1 2

1 2

2

2 2

(i)

RS

(ii)

1 1 1, , ,... 2 4

r

(i)

Area10 1

1 2

(ii)

S

1 2

1 2

9

a1

1

1 r

1 1 2

2

1 8

1 8

1 2

Area

RSTU

1 2

2

1 512 2

a1 200 and d a52

a1 51d

1 4

1 2

1 4 1 2

1

2 2

Aristede (a)

11995

20

200 51 20 1220

Aristede will swim 1220 metres in the final week. (b)

S52

52 2a1 51d 2

52 2 200 51 20 2

36920

Altogether, Aristede swims 36 920 metres.

13.

(a)

(b)

Area

Area

A

C

3 3

2

1 9

2

1; Area

B

1 3

2

1 9

1 81

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Shaded Area 2 1 8 (c)

Shaded Area 3

Shaded Area 2

Shaded Area

1

(d) Unshaded Area

14.

(a)

(b)

1 8 1 9 9

8 9

8 9

2

1

...

8 1 9 9 Shaded Area 0

2

8 9

9

(i) 4 3

8 9

16 ... is geometric with r 27

2 < 1; thus converging. 3

(ii)

The series 2

(iii)

The series 0.8 0.78 0.74 0.74 ... is arithmetic with d =

(iv)

The series 2

8 3

32 128 ... is geometric with r 9 27

2

For series (ii) we have S 1

15.

8 8 8 1 81 9

0.02.

4 1 ; thus diverging. 3

6

2 3

The Kell scheme forms an arithmetic sequence with a1 18000 and d

400

The YBO scheme forms a geometric sequence with b1 17000 and r 1.07 (a)

All answers are in euros. (i)

Kell: a2 18000 400 18 400, a3 YBO: b2 17000 1.07 18190, b3 Kell: S10

(ii) YBO: S10

(iii) (b)

(i)

a2

400 18800

b2 1.07 19 463.3

10 (2 18000 9 400) 198000 2 1.0710 1 17 000 234879.62 1.07 1

Kell: a10 18000 9 400 YBO: b10 17 000 1.07

9

21600 31253.81

From (a) (ii) we can see that b3

a3 , so Merijayne will start earning more

than Tim in the third year.

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(ii)

16.

17.

We can compare their total earnings with the help of a GDC. If X denotes the year, Y1 represents total earnings for Tim, and Y2 for Merijayne we have:

The number of seats in each row forms an arithmetic sequence with a1 16 and d (a)

a24

(b)

S24 16 18 ... 62

2

a1 23d 16 23 2 62 24 16 62 2

936

The values of the investment after each year form a geometric sequence with a1 7000, r 1.0525, and an 1 represents the value after n years. (a)

(b)

Value of investment = 7000 1.0525t

7000 1.0525t

10000

1.0525t

10 7

10 7 log(1.0525) log

t

6.97

The minimum number of years is 7. (c)

If the rate of 5% is compounded quarterly, the value of the investment over 7 years would be 7000 1

5 4 100

74

9911.95

For 5.25% compounded annually, the value of the investment would be 7000 1.05257 10015.04 Therefore, the investment at 5.25% annually is better. 18.

19.

(a)

S1 9 S2

20

a1

9

a1 a2

20

(b)

d

a2 a1 11 9 2

(c)

a4

a1 3d 9 3 2 15

a2

a d

7

S4

4 2a 3d 2

a d 12

2a 3d

9 a2

7 6

20

a2 11

a 15, d

8

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20.

We can expand the left-hand side to the second powers 5

1 x (1 ax)6

(1 5x 10x2

)(1 6ax 15a2 x2

)

Multiply, simplify and compare it to the right-hand side: 5

1 x (1 ax )6 1

1 5 x 6ax 10 x 2 15a 2 x 2 10 15a 2

5 6a x

30a x 2

30ax 2

1 bx 10 x 2

..... a 6 x11

By comparing the coefficients of the same powers of x we get:

5 6a b 10 15a 2

30a 10

a(a 2) 0

a

a5 a12

21.

6 13

a1 a3

a 4d a 11d

32

a1 a3 32

a1 5; a2

7

6 13

a 2d

a(a 2d ) 32

100 2 4 99 2 2

S100

22.

2, b

a1 d 13

2d 4d 32

d

2(all terms positive), a 4

10300

d 8

(a)

an

a1 (n 1)d 5 (n 1) 8 8n 3

(b)

an

400

8n 3 400

n 50.375

There are 50 terms that are less than 400.

10

(2 3 x)10

i 0

23. i

10

7

7

10 10 i 2 (3 x)i i

210 7 (3 x)7

120 8 2187 x 7

2099520 x 7

The coefficient of x 7 is 2099520.

24.

Sn

3n2 un

Sn

2n, Sn Sn

1

1

3 n 1

2

2 n 1

6n 5,

6n 5

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25.

Six people can be ordered in 6! ways, but, as they are seated around a circular table, 6! ways. As Mr Black and Mrs White should not 6 sit together, we must subtract all circular permutations in which this pair is regarded as one person, but multiplied by 2, because a male can be on the left or right of the female. 6! 5! Altogether, there are 2 120 24 2 72 ways. 6 5

are equivalent, so there are actually

26.

Firstly, we must determine which is the last positive term in the sequence.

a1 85, d an

7

85 (n 1)( 7) 0

For n 13

27.

x

7

7

i

i 0

n 13.14

13 2 85 12 ( 7) 2

S13

7

1 kx 2

92 7 n 0

x7

1 kx 2

i

559

i

The term in x corresponds to i = 2:

7

i 2

28.

S

S3

29.

x

2 a 1 r

a

7 2

27 2

1 r3 1 r

1 kx 2

2

21x k2

2 4

21 k2

7 3

k2

9

k

3, k

3

27(1 r ) 2

a

13

1 k x

21x5

27 (1 r ) 1 r 3 2 1 r

13

1 27

r3

r

1 ,a 9 3

Student A can get 1 book in 6 C1 ways. The rest of the books go to student B. Student A can get 2 books in 6 C 2 ways, 3 books in 6 C3 ways, 4 books in 6 C 4 ways, and 5 books in 6 C5 ways. Altogether, there are 6 C1

30.

C2

6

C3

6

C4

This is an infinite geometric series with: a1 S

31.

6

a1 1 r

12 2 1 3 n 1

For un

3 4

(a)

u1 48, r 4

,n

6

C5 12, r

62 ways. 2 3

36 5 :

(b)

Sn

48

4n 1 16(4n 1) 4 1

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32.

(a)

The series converges for r (b)

33.

2x 3

For this infinite geometric series a1 1, r

x 1.2

2 1.2 3

r

4 5

2x 3

1

1

S

x

3 2

5

4 5

1

3 2

1

A four-digit number cannot have a zero in the 10 000 position. There are 9 103 9000 four-digit numbers. Without digit 3, there are 8 93

5832 numbers.

So, with at least one digit 3, there are 9000 5832 3168 numbers. 34.

35.

(a)

For this series we: a1

(b)

Sn

(1

i

8

8 i 0

i

3n 2

8

1 x ;i 3 2

n 2 2 (n 1) 3 2

Sn

n(3n 1) 1365 2

1365

1 8 x) 2

2, d 3

1 x 2

3

n 2730 0 3

91 3

n 30,

1 3 x 8

56

n(3n 1) 2

7 x3

The coefficient of x 3 is 7.

36.

50

50

ln(2r )

r 1

37.

(a) (b)

r 1

an

(ii)

3an

(a)

a

2an

a ar 15 a 1 r

27 1

50 (2 49) 2

1275ln 2

4; a3 3a2 2a1 8; a4 3a3 2a2 16

27

27(1 r 2 ) 15

(b)

ln 2

2n

(i)

S

r r 1

a2 3a1 2a0

S2

38.

50

r ln(2) ln 2

2 3

1

3 2n

2 2n

1

a(1 r ) 15 a 27(1 r ) r2

4 9

r

3 2n

2n

2 2n

2n

1

an

1

27(1 r )(1 r ) 15 2 (all terms positive) 3

9

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39.

For the arithmetic sequence 2, a b,2a b 7, a 3b the difference must be constant. So, we have the following system of equations: (a b) 2 (2a b 7) ( a b)

a b 2 a 2b 7

(a 3b) (2 a b 7) (2a b 7) ( a b) 3b

9

a 3b

40.

a

7

2, b

a 4b 7

a 2b 7

3

There are 8 C4 ways of selecting teams of 4 members. Also, there are 6 C 2 ways of selecting teams that include both oldest members. Thus, the required number is 8 C4

41.

6

C2

55 ways.

As a, 1, b form an arithmetic progression, the difference is the same: 1 a b 1 As 1, a, b form a geometric progression, the ratio is the same:

a 1

b a

We solve the following system of equations: 1 a b 1 a b 1 a

a b a

2

2

a2

b

a 2 0

As a 1 gives b 1, the solution is a 42.

a 1 or a

2

2, b 4

We can see that: OB OA 1 OB1 OA cos

cos

OA1

OB2

OA2

OA1 cos

cos

OB3

OA3

OA2 cos

cos3

cos 2

cos

As the length of the arc is equal to

radius, the sum of the arc lengths is:

AB

cos

A1 B1

A2 B2

A3 B3 .....

cos 2

cos3

This is a geometric series with first term common ratio cos . Thus, the required sum is the u sum to infinity of this series: 1 1 r 1 cos 43.

For S n (a)

(b)

2n 2

n, n

we have:

u1

S1

2 12 1 1; u2

u3

S3

S2

un

Sn

Sn

2 32 1

2n2

S 2 u 1 2 22

2 1 5

3 6 9 n

2(n 1)2

(n 1)

4n 3

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44.

(a) 5 5i i 2 x i

5

(2 x )5

i 0

5 5 2 0

5 4 2 x 1

5 3 2 2 x 2

5 2 3 2 x 3

32 5 16 x 10 8 x 2 10 4 x 3 5 2 x 4

2.015

(b)

45.

5 1 4 2 x 4 x5

5 2 x5 5

32 80 x 80 x 2

40 x 3 10 x 4

(2 0.01)5 32 80 0.01 80 0.012 40 0.013 10 0.014 0.015 32 0.8 0.008 0.00004 0.0000001 0.0000000001 32.8080401001

Interest is paid yearly. For t years, r t

(a)

At

(b)

A 5 5000 1.063

(c)

5000 1.063

P1 r

n

0.063 rate, and principal P 5000 after t full years: t

5000 1.063 5

6786.35

10000

n

1.063

2

n

log 2 11.35 log1.063

The value will exceed $10,000 after 12 full years. 46.

4n 2

For an arithmetic sequence with Sn (4 22

u2

S2

S1

um

Sm

Sm

u32

S32

S31

1

2 2) (4 12

4m2

2 1) 12 2 10

4(m 1)2

2m

(4 3 22

2n we have:

2(m 1)

2 32) (4 312

8m 6

2 31) 250

As they are consecutive terms in a geometric sequence, the ratio is the same.

8m 6 10

47.

250 8m 6

(8m 6)2

2500

8m 6 50

m 7

u9 0 u1 8d and S16 12 8 2u1 15d This leads to a system of equations: 8 2u1 15d 0 u1 8d

x5

12

2u1 15d 2u1 16d

3 2 0

d

3 , u1 12 2

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n

1 x 48.

1

(a)

(b)

n

C1 x n

1 nx n (i)

n n 1 2

n

2 n

xn

C3 x n

2

n n 1 n 2 6

xn

3

n n 1

6

2

n3 9n2 14n 0

n 0,2,7 . Since there are 4 terms, n = 7.

(a)

We are selecting four students out of 11: 11C4

(b)

5

(c)

At least one junior is the complement of No junior .

C2

11

(a) (b)

6

C4

5

3 x 3.14

330

C2 150

4

C4 325

81 108x 54x2 12x3

3 0.1

4

x4

81 108 0.1 54 0.12 12 0.13 0.14

The number of ways the youngest will receives 3 books is 7 C3 2 students will share the 4 books left in 4 C2 all three in 35 6 210 ways.

52.

3

n n 1 n 2

n3 9n 2 14n 0

(ii)

51.

C2 x n

In an arithmetic sequence, the common difference is constant:

2

50.

n

1

n n 1

49.

1

(a)

92.3521

35 . Every time, the other

6 ways. Thus, the 7 books may be shared by

After 10, the first number divisible by 7 is 14 and the last number is 294. This is an arithmetic sequence whose first term is 14 and last term is 294. 294 14 7 n 1

41 14 294 2

n

41

(b)

S41

(c)

We can use the sum of the arithmetic sequence formula:

Sn

n 2u1 2

6314

n 1 d

n 2 1000 7 n 1 2 n 286.71 n 287

0

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53.

(a)

If the largest is a 5, there will be 3 cards to be chosen from the 4 numbers below it: 4

C3

4 . If the largest is a 6, there will be 3 cards to be chosen from the 5 numbers

below it: 5 C3 10 . If the largest is a 7, there will be 3 cards to be chosen from the 6 numbers below it: 6 C3 (b)

20. Thus, there will be 34 selections possible.

At least two even numbers is the complement of at most one even

.

We have 9 C4 126 selections altogether. There are 5 C4

5 selections with no even and 5 C3 4 40 with only one even.

Therefore, there are 126 5 40 81 selections with at least one even number. 54.

(a)

Consider the wives to be one unit. Then there will be 4 units to be arranged in 4! different ways. However, within the unit of wives, there will be 3! ways of arrangement. Thus, the number of ways this can happen is 4! 3! 144

(b)

There are 4 spaces around the husbands where the wives can sit, they can arrange themselves in 4 P3 24 different ways. However, for every arrangement the husbands can be arranged in 3! ways. Thus, the number of ways this can happen is 4! 3! 144

55.

(a)

(b)

2un 1 2un

(i)

vn 1 vn

(ii)

v1

2u1

2a

(iii)

vn

2u n

2a

2un 1

un

2d

n 1d

v n is a geometric sequence with first term v1 2a and common ratio 1 rn 1 r

2a

Sn

(ii)

The series converges if the common ratio is less than 1, i.e., 2d

(iii)

S

(iv)

S

1

d v1

1 r 2a 1 2d

2d

1 2nd 1 2d

(i)

v1

vn 1 vn

0 2a 1 2d 2a

1

1 1 2d

2

d

1

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Exercise 4.1 & 4.2 1. (a)

an , the function is equivalent to the horizontal line

(if ) (b) Domain: Range: (c) (i) When , the function is increasing for all values of x. y-intercept can be found at y-intercept at

(ii)

When , function is decreasing for all values of x. y-intercept can be found at y-intercept at

2. (a) Sketch:

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i. y-intercept can be found at y-intercept at x-intercept can be found at impossible no x-intercept Alternatively, asymptote of ii. Horizontal asymptote of iii. Domain: Range: (b) Sketch:

at at

, no x-intercept possible , as no upward translation

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i. y-intercept can be found at y-intercept at x-intercept can be found at

and x-intercept is ii. Upward translation of 8 units iii. Domain: Range: (c) Sketch:

horizontal asymptote of

at

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i. y-intercept can be found at y-intercept at x-intercept also at ii. Downward translation of 1 unit iii. Domain: Range: (d) Sketch:

horizontal asymptote of

at

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i. y-intercept can be found at undefined (this implies a vertical asymptote at no y-intercept x-intercept can be found at impossible as numerator can never equal zero (this implies a horizontal asymptote at no x-intercept ii. horizontal asymptote at

)

second horizontal asymptote at vertical asymptote at (see part i for further explanation on vertical asymptote) iii. Domain: Range: (e) Sketch:

i. y-intercept can be found at y-intercept at x-intercept also at ii. Downward translation of 3 units iii. Domain: Range:

horizontal asymptote of

at

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)

(f) Sketch:

i. y-intercept can be found at y-intercept at x-intercept can be found at impossible no x-intercept ii. Upward translation of 8 units iii. Domain: Function maximum occurs at Range: 3.

horizontal asymptote of

at

, has domain An upward translation of units, will shift the asymptote to If this will create a range of . Alternatively, if , the function will reflect about the x-axis, resulting in the range y-intercept occurs at y-intercept at

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.

4.

5. (a) (b) (c) 6. Given that however

for and both functions will be continually increasing, will be increasing at a faster rate. Therefore, will be steeper than .

7. Let growth factor per year be written as . Given that population triples after 25 years, , giving an annual growth factor of , where represents the number of years. (a) (b) (c) 8. Let growth factor per minute be written as . Given that population doubles after 3 minutes, , giving an annual growth factor of , gives the number of bacteria after minutes. (a) (b) (c) (d) © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

9. (a) Equation

where

is the initial investment amount

is the annual rate

and is years. Doubling in 10 years implies that (b) In equation

, where

would be the simple interest rate or

10. For investment problems, one can use the following expression:

, where

represents the present (original) value of the investment, represents the investment rate, represents the annual amount of compounding periods, represents the time in years, and represents the final value of the investment after years. Using this information we can calculate the sub-sections of this question: (a) As quarterly implies 4 compounding periods per year, the equations take the following form:

(rounded to 2 d.p.)

(b)

(rounded to 2 d.p.)

(c)

(rounded to 2 d.p.)

11. See question 10 for explaination of investment formula for compounding periods per year. (a) Monthly compounding periods implies 12 compounding periods per year, therefore: for years. (b)

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(c)

Intersection occurs at After 16 years, the investment will have a value greater than $20,000. 12. See question 10 for explanation of investment formula for (a) Annual compounding implies

compounding periods per year.

(rounded to 2 d.p.) (b) Monthly compounding implies (rounded to 2 d.p.) (c) Daily compounding implies (rounded to 2 d.p.) (d) Hourly compounding implies (rounded to 2 d.p.) © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

13. See question 10 for explanation of investment formula for In this question interest rate of 100% implies . (a) Annual compounding implies

compounding periods per year.

(b) Monthly compounding implies (c) Daily compounding implies (d) Hourly compounding implies (e) Compounding every minute implies

14. Increase of 3.2% implies growth rate of 1.032, meaning , where is the initial deer population and is the population after years. (a) , where is the population one year ago. (b)

, where

is the population eight years ago.

(rounded to the nearest unit) 15. The general formula for half-life decay is

, where

is the initial amount,

the half-life of the material, and is the final amount after years. Given the half-life of carbon is 5730, to find percentage remaining after 20 000 years, we compute of the original carbon will be left. 16. See question 15 for explanation of half-life formula. where is in days remaining in the blood stream 17. (a) Loss of 30% per week, (b)

20 weeks © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

is

18. Exponential growth is defined for functions as a value of would result in a constant function equivalent to . Alternatively, exponential decay can be defined for functions . Exponential functions cannot take the form as this would result in a discontinuous function. 19. Case I resembles an arithmetic sequence with and common difference Therefore, the sum of an arithmetic sequence can be used: Case II resembles a geometric sequence with common ratio, r Therefore, the sum of a geometric sequence can be used:

.

.

20. (a) From the graph, Also from the graph, By substitution: (Note: must be positive due to the nature of the function.) By substitution: Final function: (b) From the graph, Also from the graph, (Note:

must be positive due to the nature of the function.)

Final function: (c) From the graph, Also from the graph, By substitution: (Note:

must be positive due to the nature of the function.)

By substitution: Final function: (d) From the graph, Also from the graph, By substitution: By substitution: Final function: © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

Exercise 4.3 1. (a) Sketch:

i.

y-intercept can be found at y-intercept at

ii. iii.

x-intercept can be found at impossible no x-intercept Horizontal asymptote of at has been no upward translation Domain: Range:

as there is no x-intercept OR as there

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(b) Sketch:

i.

ii. iii.

y-intercept can be found at y-intercept at x-intercept can be found at is impossible no x-intercept Horizontal asymptote of at has been no upward translation Domain: Range:

as there is no x-intercept OR as there

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(c) Sketch:

i.

ii. iii.

y-intercept can be found at y-intercept at x-intercept can be found at is impossible no x-intercept Horizontal asymptote of at as there is no x-intercept OR as there has been no upward translation Domain: Range: (as there has been a reflection about the x-axis)

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(d) Sketch:

i.

y-intercept can be found at y-intercept at x-intercept can be found at

ii. iii.

and x-intercept are at and No asymptotes (just a minimum point) Domain: Range:

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(e) Sketch:

i.

y-intercept can be found at undefined no y-intercept (implies an asymptote at ) x-intercept can be found at impossible as numerator can never be zero no x-intercept (implies an asymptote at

)

ii. horizontal asymptote at

iii.

second horizontal asymptote at vertical asymptote at (see part i for further explanation on vertical asymptote) Domain: Range:

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(f) Sketch:

i.

y-intercept can be found at y-intercept at x-intercept can be found at

ii. iii.

and x-intercept is No asymptote (only a minimum) Domain: Range:

2. (a) (b)

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(c) (d) it is the reciprocal of 3.

As a result, the graph of

has an asymptote at

.

As a result, it will never reach there will never be an intersection between the two graphs. 4. (a) Bank A: interest (b) Bank A: interest 5. (a) Blue Star: Red Star: Blue Star account will result in the greatest total after 5 years (b) (see part a for explanation) (c) 6. (a) percentage remaining

%

percentage remaining

%

percentage remaining

%

(b) (c) (d) percentage remaining

%

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7. (a) (b)

kg percentage remaining

%

(c)

(d) Half-life occurs at half initial amount

kg

From observation, approximately 20 days © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

8. Option A:

increase of

Option B:

increase of

Option C: increase of Option A provides the best increase on the initial investment 9. (a) Given that (b)

is the growth rate per minute %

10. (a) Gradient

(b) Gradient

(c) Gradient

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(d) Gradient

11. (a) After 10 years: (b) years (c) years (d) The answers are the same, as the time to double depends only on the interest rate and the amount of compounding periods, and these are constant between (b) and (c).

Exercise 4.4 1. (a) (b) (c) (d) (e) (f) (g) (h) (i)

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2. (a) (b) (c) (d) (e) (f) (g) (h) (i) 3. (a)

(b)

(c)

(d)

(e)

(f)

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(g)

(h)

(i)

(j) (k)

(l) (m)

(n)

(o)

(p)

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(q)

(r) (s)

(t) 4. (a) (b) (c) (d) (e) 5. For the function (a)

, domain: is a translation of

by two units to the right OR

domain: (b) which is true for all real numbers not equal to zero domain: (c) is a translation of by two units downward (which has no effect on the domain) domain: (d)

(e) Due to square root: Due to logarithm:

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(f) Due to square root:

Due to logarithm:

6. (a) Due to denominator:

Due to logarithm: Function can never take the value of zero due to the numerator range: (b)

Function can never take a negative value due to absolute value range: (c) Due to denominator:

Due to logarithm: Function can never take the value of zero as a zero numerator is outside the domain range: 7. All functions in this question take the form (a) From the graph:

(b) From the graph:

(c) From the graph:

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(d) From the graph:

8. (a) (b) (c) (d) (e) (f) 9. (a) (b) (c) (d) (e) (f)

10. (a) (b) (c) (d) (e) (f) 11. (a) (b) (c) (d)

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12. (a)

(b)

13. 14. 15. For decibels 16. (a) (b) In the form

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Exercise 4.5 1. (a) (b) (c)

(d)

(e) (f) (g) (h)

(i)

(j)

(k)

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(l)

2. (a)

If

Resubstituting gives Either (impossible) OR (b)

If

Resubstituting Either

gives

OR

(c) If

Resubstituting Either

gives

OR

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(d) If

Resubstituting Either

gives

OR

3. (a) (b)

years It will take

years to double the investment.

4.

years 5. Doubling every hour implies a population

after hours

hours for the population to exceed 1 million bacteria 6. (a) years or

years

years or

years

(b) (c) years or

years

7. years Thus, in 6 years

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8. (a) (b)

years 9. (a) 37 dogs (b)

years 9 years 10. (a) (b)

litres

minutes and

minutes seconds

(c) minutes minutes 11. (a) (b)

kilograms

years 12. (a)

(b)

(c)

(d) ;

due to nature of the function

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(e)

(f)

(g)

(h)

due to nature of the function (i)

(j)

due to nature of the function (k) By observation

, OR

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13. (a)

OR (b)

(c)

OR

-value

(d)

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Chapter 4 practice questions 1. (a) Let point be as it is the x-intercept. x-intercept can be found at

(b) Let point be as it is the y-intercept. y-intercept can be found at

(c) Let point

point

be

as it is the intersection with the line

be

2. (a) grams (b)

years 3.

due to the nature of the function

4. Solving

Substituting into

, gives:

, gives:

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Rearranging

, gives:

Re-substituting into

, gives:

5. Let 1

2 2

-7 2 -5

7 -5 2

-2 2 0

6.

Either:

OR:

7. (a)

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Let

:

Resubstituting Either

OR

, (b)

due to the domain of log functions 8. (a)

(b)

9. years 10.

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11. (a)

(b)

is a reflection of

is a translation of

about the y-axis.

by one unit to the right.

12. (a)

(b) is remaining

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13.

14.

(to 3 s.f.)

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15.

The GDC table shows two asymptotes of OR

at

(to 3 s.f.)

16. (a)

Let

Either OR Resubstituting Either OR

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(b) i.

ii.

when

this equation has no solutions.

17.

Let

Either OR Resubstituting Either OR

gives:

impossible

This must be given in the form and 18.

Either: OR:

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19. (a) (b)

years months 20.

21. Either:

OR:

OR

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22.

(to 3 s.f.) 23.

is the growth factor When the growth is double.

24. (a)

(b)

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25. (a)

minimum of (b) As the range of

is

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Exercise 5.1 1.

(a) (b) (c) (d) (e)

Yes, since it is a sentence that asserts a true fact. No, since it may be true or false depending on circumstances. No, since it does not assert any fact. Yes, since it is a sentence that asserts a true fact. Yes, since it is a sentence that asserts a true fact.

2.

(a) (b) (c)

a = 7, since 3 < 4 is true, and both parts of this conjunction must be true. No value makes this conjunction true since one part is false. a = 7, since one part in this disjunction is false, the second one must be true.

(d)

Any value. This is a conditional of the form P

Q, and Q is true, therefore

any value of P will make it true.

3.

4.

(e)

a 7, since this is a conditional with a false consequent and only F

F can be true.

(f)

a 7, since this is a conditional with a false consequent and only F

F can be true.

Convention: a = antecedent; c = consequent. (a)

a, c

(b)

a, c

(c)

a, c

(e)

c, a

(f)

c, a

(g)

c, a

(a)

Converse: If triangles have four sides, then quadrilaterals have three sides. Contrapositive: If triangles do not have four sides, then quadrilaterals do not have three sides.

(b)

Converse: If

(c) (d) (e) (f) (g)

(d)

a, c

3 is a rational number, then the moon is made of butter.

Contrapositive: If 3 is an irrational number, then the moon is not made of butter. Converse: b divides 30 only if b divides 5. Contrapositive: b does not divide 30 only if b does not divide 5. Converse: f to be continuous is sufficient for the differentiability of f. Contrapositive: f to be discontinuous is sufficient for the non-differentiability of f. Converse: A sequence a is convergent whenever a is bounded. Contrapositive: A sequence a is not convergent whenever a is not bounded. Converse: A function f is integrable if f is bounded. Contrapositive: A function f is not integrable if f is not bounded. Converse: 3 + 3 = 6 is necessary for 3 + 2 = 5. Contrapositive: 3 + 3

6 is necessary for 3 + 2

5.

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5.

6.

We use the truth table for P

Q.

(a)

Q can be true or false, since F

(b)

Q must be true, since, in this case, only T

(c)

Q must be false.

We use the truth table for P

T and T

T are true. T will give true.

Q.

(a)

F, since T

F is false

(b)

T, since F

F is true

(c)

T, since F

T is true

(d)

T, since F

T or F

(e)

F, since T

F is false

F are true

Exercise 5.2 & 5.3 1.

(a) (b)

(c) (d)

(e)

This is an existence statement. It is enough to find such an integer, e.g. 20 = 4 + 6 + 10 Direct proof: Let an even integer be 2m where m is any integer. m can be written as the sum of any 3 integers, p, q and r. Thus 2m 2 p q r 2 p 2q 2r , which are even integers. Not true. Let (2n + 1) and (2k + 1) be any two odd integers. Then, (2n + 1) + (2k + 1) = 2(n + k + 1) which is even. Thus, the sum cannot be odd. Let m = 2n + 1 be any odd integer. 2n is even, thus it can be the sum of three even integers, so, m p q r p q r , which are three odd integers. Let n, n 1, and n 2 be three consecutive integers, then, n

(f) (g)

(h)

n 1

n 2

3n 3 3 n 1 , which is a multiple of 3.

Not true. A counter example is enough: n

n 1

n 2

n 3

4n 6 .

If n = 1, this sum is 10, which is not divisible by 4. True. Every three consecutive numbers should have at least one even number and every three consecutive numbers should have one multiple of 3 (multiples of 3 are periodic with period 3). Thus, the product should be a multiple of 6. Not true since a = b is another possibility.

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2.

(a) (b)

3.

If n

n n 4.

5.

We need to find the perfect square less than or equal to 1871. This is 1849 or 432. That is, he turned 43 in year 1849, so he was born in 1849 43 = 1806. Again, we need to find the perfect square less than or equal to 2018. The closest perfect square less than 2018 is 1936 = 442. This means that your friend turned 44 in year 1936, which means he/she will be 126 in year 2018. This is not likely given our life expectancy! (The greatest fully authenticated age to which any human has ever lived is 122 years 164 days by Frenchwoman Jeanne Louise Calment.) n! contains at least one even factor. Thus, for example we factor 2 out of both terms

n

2

2

n n

n

2Q , which is even.

(a)

Needs to be proven in both directions. If 5n + 3 even, then if you add any odd number to it, the resulting number will be odd. So, adding the odd integer 2n 5 to 5n + 3 will result in an odd integer. That is, 5n + 3 + 2n 5 = 7n 2 is odd. Similarly, if 7n 2 is odd, adding the odd integer 5 2n will result in an even number. That is, 7n 2 + 5 2n = 5n + 3 is even.

(b)

7n 2 will be even. If 5n + 3 is odd, then 5n must be even

m2 + n2 even

m2 and n2 are both even or both odd

n is even

7n 2 is even.

m and n have the same parity because if,

for example, m2 is even, m must be even as the product of two odd numbers cannot be even. Similarly, if m2 is odd, then m must be odd. 6.

By contradiction: Assume that

x

y

y . Square both sides and simplify, 0 = xy. Thus, one of x or y must

x

be zero, which contradicts the fact that both are positive. 7.

If x = 0, or y = 0, then xy = 0 is obvious. In the opposite direction: If xy = 0. Assume that x and y are both different from zero. But there are no nonzero real numbers that can have a product of zero, thus a contradiction.

8.

Let O be the set of odd numbers.

9.

(a)

x

O such that x = k2 where, k

(b)

x

O, x

k2 where, k

(c)

True. Since it is an existence statement, x = 81 = 92

(a)

Statement:

(b)

False, a counter example: x = 10, 13 10

x

+

, 13 | x. Negation:

x

+

, 13 x

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10.

By contradiction: Assume a2(b2 b) is odd, but at least a or b is even. 2 2 Say a is even, then, a = 2k and a (b b) = 4k2(b2 b) is even. Contradiction.

11.

Contrapositive: 5 | m and 5 | n

m = 5r and n = 5s

12.

Prove by cases: m + n is even

m and n must both be odd or both be even.

13.

1.

If both are odd: m2 is odd and n2 is odd

2.

If both are even: m2 is even and n2 is even m2

m2

mn = 25rs

25 | mn

n 2 is even. m2

n 2 is even.

n 2 = (m + n)2

2mn = 4k2 2mn = 2N

Or, direct: m + n is even

m + n = 2k

If n is even, then n = 2k

n2 + 2n + 9 = 4(k2 + k) + 9, which is odd.

If n2 + 2n + 9 is odd, then n2 + 2n must be even. But 2n is even, so n2 is even, and so, n must be even. (Or, by using the contrapositive method.) 14.

Let a = 2k + 1 and simplify. a2 + 3a + 5 = 2(2k2 + 5k + 4) + 1 = 2N + 1, which is odd.

Exercise 5.4 In the solutions to the following exercises, for convenience we have excluded some relatively obvious calculations. Conventions for this section: MI = mathematical induction. 1.

You can either recall arithmetic series, or by inspection the sum: S n Basis step: S1

n n 1

2 1 1 1 . So, the formula is true for n = 1.

Inductive step: assume the formula is true for n = k, i.e., Sk 2 4 2k k k 1 , and we prove the formula true for n = k + 1 Sk

1

2 4

2k

k k 1

2 k 1

2k

2

Sk

k 1 k

2k

2

2

This shows that S k 1 is true whenever S k is true, which completes the inductive step of the proof.

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2.

We will provide a proof using MI. Let P(n) be the statement that an 3n

1

Basis step: a1 1 31 1 . So, the formula is true for n = 1. Inductive step: assume the formula is true for n = k, i.e., ak 3k 1 , and we prove this to be true for n = k + 1. By definition of the sequence, ak 1 3ak and by assumption ak Therefore ak

1

3 3k

1

3k

1

3k

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for all positive integers.

3.

We will provide a proof using MI. Let P(n) be the statement that an 4n 3 Basis step: a2 1 4 5 4 2 3 . So, the formula is true for n = 2. Inductive step: assume the formula is true for n = k, i.e., ak 4k 3, and we prove this to be true for n = k + 1. By definition of the sequence, ak 1 ak 4, and by assumption ak Therefore ak

1

ak

4k 3

4 4k 3 4 4 k 1

3

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n 2.

4.

We will provide a proof using MI. Let P(n) be the statement that an 2 n 1 Basis step: a2

2 1 1 3 22 1 . So, the formula is true for n = 2.

Inductive step: assume the formula is true for n = k, i.e., ak 2 k 1, and we prove this to be true for n = k + 1. By definition of the sequence, ak 1 2ak 1, and by assumption ak Therefore ak

1

2 2k 1

1 2k

1

2k

1

1

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n

2.

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5.

We will provide a proof using MI. Let P(n) be the statement that an

n n 1

1 1 . So, the formula is true for n = 1. 2 1 1 Inductive step: assume the formula is true for n = k, i.e., k and we prove this to be true for n = k + 1. ak k 1 By definition of the sequence, 1 k ak 1 ak and by assumption ak k 1 k 2 k 1

Basis step: a1

Therefore ak

k 1

k k

1

k 1

k 1 k

2

2

2

1

k 1

k 1 k

2

k 1 k 2

k 1 k 2

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n 1.

6.

1 2

Either by inspection or realising that this is a geometric series: S (n) 1

n

We will provide a proof using MI. 1 1 Basis step: S (1) . So, the formula is true for n = 1. 1 2 2 Inductive step: assume the formula is true for n = k, i.e., S (k ) 1

1 2

k

and we prove this to be true for n = k + 1.

By definition of the sequence, the kth term is S ( k 1)

1 2

1 4

1 8

1 2k

1 , 2k

1 and by assumption S (k ) 2k 1

1 2

1 4

1 8

1 2k

1

Therefore S (k 1)

S (k ) 1

1 2

1 2 k

k 1

1 2

1

1 1 2 2

k

k

1

1 2k

1

1 2

k 1

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n 1.

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1 2

k

7.

We will provide a proof using MI. Let P(n) be the statement that S (n) Basis step: S (0) 1 20

1

2

+

n

n 1

1 . So, the formula is true for n = 1.

Inductive step: assume the formula is true for n = k, i.e., 2 k k 1 S (k ) + and we prove this to be true for n = k + 1. By definition of the sequence, ak 1 2k 1 and by assumption S (k ) therefore S (k 1)

2

+

k

k 1

2

S (k )

+

k

k 1

k 1

k 1

k 1

2 2k 1 1 2k 2 1 This shows that S (k 1) is true whenever S (k ) is true, thus, by the principle of MI,

S (n) is true for all non-negative integers.

8.

We will provide a proof using MI. Let P(n) be the statement that an a1r n Basis step: a1

a1r1 1

1

a1 , . So, the formula is true for n = 1.

Inductive step: assume the formula is true for n = k, i.e., ak a1r k 1 and we prove this to be true for n = k + 1. By definition of the geometric sequence, ak 1 rak and by assumption ak a1r k 1 , therefore ak

1

r ak

r a1r k

1

a1r k

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n 1.

9.

We will provide a proof using MI. Let P(n) be the statement that Sn

a ar n 1 r

a ar . So, the formula is true for n = 1. 1 r Inductive step: assume the formula is true for n = k, i.e., a ar k Sk and we prove this to be true for n = k + 1. 1 r By definition of the geometric sequence, a ar k ak 1 rak , and by assumption S k 1 r

Basis step: S1

a

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a ar k ar k 1 r a ar k a ar k 1 k 1 r ar 1 1 r 1 r 1 r This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI,

Therefore Sk

Sk

rak

P(n) is true for n 1.

10.

We will provide a proof using MI. Let P(n) be the statement that 2n n! Basis step: n 4,24 16 4! 24, . So, the formula is true for n = 4. Inductive step: assume the formula is true for n = k, i.e., P (k ) : 2k k ! and we prove this to be true for n = k + 1. We need to show that P (k 1) : 2k 2

k

k!

Thus, 2k

2 2 1

k

2 k!

1

(k 1)! is true and by assumption 2k

k 1 k ! since we know that k

3

k!

2 k 1

k 1!

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n

11.

4.

We will provide a proof using MI. Let P(n) be the statement that 2n n 2 Basis step: n 5, 25

32 52

25 . So, the formula is true for n = 5.

Inductive step: assume the formula is true for n = k, i.e., P(k ) : 2k k 2 and we prove this to be true for n = k + 1. We need to show that P (k 1) : 2 k 2k

k2

Thus, 2k

2k 1

1

2 2k

2k 2

k2

2k 2

k2

1

k2

k2

(k 1) 2 is true and by assumption 2k k 2 , but k

2k 1

k 1

4

k2

4k

2k 2k

k2

2k 1

2

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n

12.

5.

We will provide a proof using MI. Let P(n) be the statement that 1 1! Basis step: n 1, 1 1!

2 2!

3 3!

n n!

n 1! 1

1 1 ! 1 . So, the formula is true for n = 1.

Inductive step: assume the formula is true for n = k, i.e., P (k ) : 1 1! 2 2! 3 3! k k ! k 1 ! 1 and we prove this to be true for n = k + 1.

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We need to show that P (k 1) :1 1!

k k!

k 1

k 1!

k

2! 1

with the assumption that P(k ) is true. P (k 1) : 1 1!

k k!

k 1! 1

k 1 k 1

k 1 !1 k 1

k 1! k 1!

1

k

2! 1

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n 1.

13.

We will provide a proof using MI. 1 1 Let P(n) be the statement 1 2 2 3

1 3 4

1 n n 1

n n 1

1 1 . So, the formula is true for n = 1. 1 2 1 1 Inductive step: assume the formula is true for n = k, i.e., 1 1 1 1 k P(k ) : and we prove this to be true for n = k + 1. 1 2 2 3 3 4 k k 1 k 1

Basis step: n 1,

We need to show that P( k 1) :

1 1 2

1 k 1

k

1 k 1

k

2

k 1 k 2

with the assumption that P(k ) is true. P (k 1) :

1 1 2

k

k

1 k 1

1 k 1 k k

1

k 1

k 1

k

k

2

2 2

k 1

k

1 2

k 1 k 1

k

2

2

k 1 k 2

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n 1.

14.

We will provide a proof using MI. Let P(n) be the statement that n3

n 3q , i.e., n3 n is divisible by 3.

Basis step: n 1, 13 1 0, which is divisible by 3. So, the statement is true for n = 1. Inductive step: assume the statement is true for n = k, i.e., P k k3 k M and we prove this to be true for n = k + 1. We need to show that P(k 1) : k 1 3

P (k 1) : k k3

k

k3

k k2

k

3k 2 M

3

k

is divisible by 3 with the assumption P(k ) is true.

3k 1 k 1 N

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This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n 1.

15.

We will provide a proof using MI. Let P(n) be the statement that n5

n 5q , i.e., n5 n is divisible by 5.

Basis step: n 1, 15 1 0, which is divisible by 5. So, the statement is true for n = 1. Inductive step: assume the statement is true for n = k, i.e., P k k5 k M and we prove this to be true for n = k + 1. We need to show that P(k 1) : k 1 P (k 1) : k 1 k5

5

k

k5

k4

k

5 k4

2k 3

2k 2

5

is divisible by 5 with the assumption P(k ) is true.

k

k3 k

k2 5M

k

k 1

5N

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n 1.

16.

We can prove this by several ways. We choose proof by cases: n3 n n n n If n is even, then it is divisible by 2, and either n 1 or n + 1 must be divisible by 3. If n is odd, then n 1 and n + 1 are divisible by 2. Either n is divisible by 3 or it leaves a remainder of 2 when divided by 3. Thus n + 1 would be divisible by 3.

17.

We can prove this by several ways. We choose proof by cases: n2 + n = n(n + 1). If n is even, then it is divisible by 2. If n is odd, then n + 1 is divisible by 2.

18.

Using the binomial expansion: 5n

4 1

n

n

n

i

0

5n 1

n 1

n i

4i

n

1

1

4 4i

1

n

4 1

n i n i

4i 4i

1

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19.

We will provide a proof using MI. a 0 Let P(n) be the statement that 0 b a 0 0 b

Basis step: n 1,

1

n

an 0

0 bn

a1 0 . So, the statement is true for n = 1. 0 b1

Inductive step: assume the statement is true for n = k, i.e., a 0 P( k ) : 0 b

k

ak 0

0 and we prove this to be true for n = k + 1. bk

a 0 We need to show that P(k 1) : 0 b P(k 1) :

k 1

a 0 0 b

a 0 0 b

k

k 1

ak 1 0 with the assumption P(k ) is true. 0 bk 1 ak 0

a 0 0 b

0 bk

a 0 0 b

ak 1 0 0 bk 1

This shows that P(k 1) is true whenever P(k ) is true, thus, by the principle of MI, P(n) is true for n 1.

20.

In all three cases we will provide brief MI proofs. (a)

n 1:

1

2i 4

6 1 5

2i 4

k2

i 1

n

k:

k

5k

i 1

n k 1:

k 1

k

2i 4

i 1

k 1 (b)

n 1:

1

2i 4

2 k 1

4 k2

5k

2 k 1

4

i 1 2

2 31 1

31 1

2 3i

3k 1

5 k 1

1

n

k:

k

1

1

n k 1:

k 1

2 3i

k

1

1 k

3 3

2 3i

1

2 3k

3k 1 2 3k

1 k 1

1 3

1

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(c)

n 1:

1 2i 1 2i 1

1 1

n k:

k

1 2 1 2 1

1

n k 1:

k 1

k

1 2i 1 2i 1

1

1 2 1

k 2k 1

2i 1 2i 1

1

1 3

1

1

2i 1 2i 1

1

k 2k 1

2 k 1

1 2 k 1

1

1 2k 1 2k 3 2k 1 k 1

2k 2 3k 1 2k 1 2k 3

2k 1 2 k 3

Chapter 5 practice questions

1.

(a)

a2

3

a2

3

an

2.

3

7; a3

3

3

8

7

3

1; a4

3

8

1

3

3

7

8

2

3

0; a3

3

8

0

3

2; a4

3

8

2

3

0

2 16 is divisible by 4. k

n k: 5

3

2,0, 2,0,...

n 1, 51 91 k

1

1, 3 7,1, 3 7,...

an

(b)

8

9

k 1

2 4m

n k 1: 5

9k

5 5k

2 5 5k

9 9k

1

2 5 5k

9 5

k

9 9k

4 5k 9 9

9 4m 2

2 , now add and subtract 4 5k and simplify:

9 9k k

4 5 4 5k

k

4 5k

2

2 9 5k 2 36m 16

9 9k 4 5k

4 5k

2

4 9 m 4 5k

This last number is a multiple of 4.

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3.

n2

Sn un

Sn u2

n Sn

1

10, um

Sn

2

n

1

n

8n 6 8m 6, u32

250

Geometric sequence: u32 um

um u2

250 8m 6

8m 6

4.

50

8m 6 10

2

8m 6

502

11 2

m 7, m

We use MI: 12 1 1

n 1: 13

2

1 , statement is true for n = 1.

4

Assume true for n = k: 1 2

3

3

3

k

3

k2 k 1

2

4

Now, for n k 1 : 3

1 2

3

3

k

3

k 1

2

k2 k 1

3

k 1

4 k 1

2

k

2

3

k 1

2

k2

4k

4

4

2

4 5.

We use MI: For n = 0, 50 1

24 | 50 1

0

2k

2k

Assume true for n = k:

|

2 k For n = k + 1, 5

52 k 52 1, and by substituting 52 k

52 k

1

1

1

1

52 k 52 1

a 1 24a 1 and simplifying:

24a 1 52 1

24a 25 25 1 24 25a 1 24 | 52 k

1

1

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6.

We use MI: For n = 1: F12

F1 F1

12

1

Assume true for n = k: F12

1 1, which is a true statement. F22

F32

Fk2

Fk Fk

1

Now for n = k + 1: F12

7.

F22

F32

Fk2

Fk2 1

Fk Fk

1

Fk

Fk

1

Fk2 1 Fk

1

Fk 1 Fk

2

We use MI: For n = 1:

1 1 2!

1 1 1!

Assume true for n = k:

1 , which is true. 2

1 2 3 2! 3! 4!

k 1 k 1!

1 k 1!

Now for n = k + 1: 1 2 3 2! 3! 4!

k k 1!

k 1 1 k 2! 1

1 k 1! k

k 1 k 2!

2 k 1 1 k 2!

1 k

2!

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Exercise 6.1 1. 60 180 3 150 5 150 180 6 270 3 270 180 2 36 36 180 5 135 3 135 180 4 50 5 50 180 18 45 45 180 4 400 20 400 180 9

(a) 60 (b) (c) (d) (e) (f) (g) (h) 2. 3 4 7 (b) 2

(a)

(c) 2 2 (d) (e)

7 6 2.5

3 180 135 4 7 180 630 2 180 360 7 180 6 2.5

115

210 180

450

5 180 300 3 1 (g) 180 15 12 12 180 282.6 (h) 1.57 1.57

(f)

143

5 3

89.95 90.0

3. (a) e.g. 30 360 390 and 30 360 330 3 7 3 (b) e.g. and 2 2 2 2 2 2 (c) e.g. 175 360 535 and 175 360 185 11 13 2 2 (d) e.g. and 6 6 6 6 5 11 5 (e) e.g. and 2 2 3 3 3 3 (f) e.g. 3.25 2 9.53 and 3.25 2 3.03 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

4. When the angle is expressed in radians, 120 (a) s 6 12.6 cm r , so s 180 70 (b) s 12 14.7 cm r , so s 180 5.

When the angle is expressed in radians, s

r , so

6.

When the angle is expressed in radians, s

r , so r

When the angle is expressed in radians, 1 2 1 (a) A r 100 42 13.96 14.0 cm2 2 2 180 1 2 1 5 (b) A r 102 130.9 131 cm2 2 2 6 s 60 180 8. In radian measure, 3 rad, or 3 r 20 9. s r 2 16 32 cm 1 2 2A 2A r r2 r 10. In radian measure, A 2

s r s

12 1.5 rad or 8 15 45 7.16 2 2 3

1.5

180

85.9

7.

We therefore have r

2 24 60

180

2 24

172

6.77 cm

3

11. (a) Since one complete revolution corresponds to an angle of 2 rad, the angular velocity is 1.5 2 3 rad s 1. (b) Assuming no slipping, the contact point of the wheel with the ground is not moving, and the centre of the wheel (and therefore the bicycle) moves with respect to it with speed 0.70 v r 3 3.30 m s 1 or 3.30 3.6 11.9 km h 1 2 12. First of all we change units so everything is expressed using metres. 1000 70 cm 0.35 metres m s 1 6.94 m s 1, and r v 25 km h 1 25 60 60 2 v 6.94 r , we have Since v 19.8 rad s 1 r 0.35 13. The angle swept by the point in T seconds is given by T radians. The arc length and therefore the distance covered by the point in this second is given by s r Tr cm. The s rT linear speed is given by distance travelled divided by time taken, so we have v r. T T

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14. First of all we find the angle at the centre AOB .

Triangle MOB , where M is the midpoint of AB , is right-angled, so 26 MB 2 sin M OB 0.65 M OB sin 1 0.65 0.7076 rad. It follows that OB 20 AOB

2M OB 1.415 rad, and the length of arc AB is s

15. The total area swept by the pipe is given by A

R2

AOB OB 1.415 20 28.3 cm 4002

502655 m2. This area is swept in

A 20944 m2. 24 16. A 641 th portion of the polygon and its circumscribed circle is shown below (diagram not to scale).

24 hours, so in one hour the area covered is a

(a) The angle at the centre AOB is M OB

AOB 2

64

We have sin M OB

1 64

th of a round angle, so AOB

2 64

32

and

, where M is the midpoint of AB . MB OB

OB

MB sin M OB

3 2

30.6 cm

sin

64 (b) The circumference has length 2 r 192.0771 cm, while the polygon has perimeter 64 3 192 cm. The difference is d 0.0771 cm!

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17. From the values of the areas of the inscribed and circumscribed circles shown below, T

it follows that the radius OP of the circumscribed circle is double the radius OQ of the inscribed circle. In fact, from Area

r 2 we have OP

It follows from Pythagoras that PQ

200

OP2 OQ2

200

200 50

50

2 50 and OQ

50 .

150 and PR 2 150 . The

50 3 50 , so the area of height QT of the equilateral triangle is given by OP OQ 2 50 1 1 the triangle is A PR QT 2 150 3 50 3 50 3 50 150 3 cm2. 2 2 18. The area A of the segment is the difference between one quarter of the area C of the circle and the C 1 1 2 area T of the triangle, where C , so that r 2 and T r r r . It follows that T 2 2 2 C C C 4 A . Solving for C gives 4 A C A T 2 C 4 4 2 2

Exercise 6.2 1. Copying the triangle onto the unit circle, we have the following diagram for t (arc length on unit circle = angle in radians, and

6

6

30 )

Here the hypotenuse is the radius of the unit circle, so its length is 1. © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

It follows that the horizontal leg has length

1 3 and the vertical leg has length . Given the 2 2 3 1 , . 2 2

orientation of the triangle, these are also the coordinates of the terminal point, Repeating the process when the arc length is

from which we find the coordinates as

Equivalently, the coordinates when t 3 1 , . The coordinates when t 2 2

gives the following diagram

1 3 , 2 2

6 3

3

are given by x cos

are given by x cos

3

6

3 and y 2

sin

1 and y sin 3 2

6

1 , so 2

3 , so 2

1 3 , . 2 2

(a)

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The coordinates of the points are: 2 3

1 3 , , 2 2

4 3

1 , 2

5 6

3 , 2

3 1 7 , , 2 2 6

5 3

1 , 2

3 1 , 2 2

3 11 , 2 6

3 1 , 2 2

2. (a) (b) (c) (d) (e) (f) (g) (h)

cos50 0.64 sin80 0.98 cos1 0.54 sin0.5 0.48 sin 70 0.94 tan 70 2.76 cos70 0.34 cos1.5 0.071 sin 20 0.34 sin1 0.84 tan1 1.56 cos1 0.54

3. (a) First quadrant, cos

6

,sin

3 1 , 2 2

6

(b) Fourth quadrant, cos

5 5 ,sin 3 3

(c) Fourth quadrant, cos

7 7 ,sin 4 4

1 , 2

3 2

2 , 2

2 2

(d) Axis between third and fourth quadrant, cos (e) Second quadrant, cos 2,sin 2 (f) Fourth quadrant, cos (g) Fourth quadrant, cos (h) Second quadrant, cos

4

0, 1

0.416,0.909

,sin

1 ,sin

3 3 ,sin 2 2

cos

4

1

5 ,sin 4

(i) Third quadrant, cos3.52,sin 3.52

4

, sin

4

2 , 2

2 2

0.540, 0.841 5 4

cos

5 5 , sin 4 4

2 2 , 2 2

0.929, 0.369

4. (a) sin

3 5 (b) sin 6

(c) sin

3 1 ,cos , tan 3 2 3 2 3 1 5 3 5 3 ,cos , tan 2 6 2 6 3 3 2 3 2 ,cos , tan 4 2 4 2

3 4

1

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(d) sin

undefined 2 4 3 4 1 4 ,cos , tan (e) sin 3 2 3 2 3 (f) sin3 0,cos3 1, tan3 0 3 3 3 (g) sin undefined 1,cos 0, tan 2 2 2 7 1 7 3 7 ,cos , tan (h) sin 6 2 6 2 6 2

1,cos

(i) since 1.25

0, tan

2

5 5 , we have sin 4 4

3

3 3

2 5 , cos 2 4

2 5 , tan 2 4

1

5. 13 6

sin

13 6

2

13 6

cos

13 6

2

cos

10 3

sin

10 3

2

sin

10 3

cos

10 3

2

15 4

sin

15 4

2

sin

15 4

cos

15 4

2

cos

17 6

sin

17 6

2

sin

cos

17 6

cos

17 6

2

(a) cos

5 6

3 2

(a) sin

cos

(b) sin

cos

(c) sin

cos

(d) sin

sin

1 2

6 6

3 2

4 3

cos

3 2

4 3

1 2

7 4

2 2

7 4

5 6

cos

5 6

2 2

1 2 3 2

6.

(b) sin 315 3 2 5 (d) sec 3 (c) tan

(e) csc 240

2 2

undefined 5 cos 3 sin 240

1

1 2 1

1

2 3 2

1

2 3

2 3 3

7. (a) sin 2.5 0.598 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

(b) cot120

tan120

1

3

1

1 3

3 3

5 2 4 2 (d) sec6 1.04 (e) tan 0

(c) cos

8. (a) Since y-coordinate of point on unit circle is positive, either quadrant I or II (b) Since y-coordinate of point on unit circle is positive and x-coordinate is negative, quadrant II (c) y-coordinate of point on unit circle is negative. Tangent is positive, so x-coordinate must be negative as well (the ratio of two negative numbers is positive), so quadrant III (d) x-coordinate of point on unit circle is negative. Tangent is negative, so y-coordinate must be positive (the ratio of a positive and a negative number is negative), so quadrant II (e) Since x-coordinate of point on unit circle is positive, either quadrant I or IV 1 (f) Since sec , x-coordinate of point on unit circle is positive. Tangent is positive, cos so y-coordinate must be positive (the ratio of two positive numbers is positive), so quadrant I 1 (g) x-coordinate of point on unit circle is positive. Since csc , the y-coordinate is sin negative, so quadrant IV 1 (h) since cot , the x- and y-coordinate of point of unit circle must have different tan

sign (the ratio of two numbers with different sign is negative), so quadrant II or IV

Exercise 6.3 1. (a) (b) (c)

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(d) (e)

(g)

(f)

(h) (i)

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2. Detailed graphs in book. (a) 1 i. A ,T 2 2

ii. domain

, range

3.5, 2.5

(b) i.

A 3, T

ii. domain

2 3

, range

3.5, 2.5

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(c) i.

A 1.2, T

ii. domain

4

, range

4.3 1.2, 4.3 1.2

3.1,5.5

3. The meaning of the parameters A and B in the equation of the trigonometric function is A amplitude and B vertical translation (or y B is the equation of the principal axis). ymax ymin ymax ymin Therefore, we have A and B , where ymax and ymin are the 2 2 y-coordinates of the maximum and minimum points on the curve, respectively. So, we have: 10 4 10 4 (a) A 3 and B 7 2 2 8.6 3.2 8.6 3.2 (b) A 2.7 and B 5.9 2 2 4. The meaning of the parameters A and B in the equation of the trigonometric function is A amplitude and B vertical translation (or y B is the equation of the principal axis). ymax ymin ymax ymin and B , where ymax and ymin are the y-coordinates of the maximum A 2 2 6.2 2.4 and minimum points on the curve, respectively. So, we have A 1.9 and 2 6.2 2.4 B 4.3 2 5. (a) The parameter p is the amplitude of the trigonometric function, so from the graph we have p 8.

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(b) The parameter q relates to the period T of the trigonometric function as T graph we have T q

2 T

2

3

2 . From the q

since there are three full oscillations from x 0 to x

. This gives us

6.

3

6. Detailed graphs in book. (a)

(b) For the secant function, domain asymptotes at all odd multiples of , 1

1,

(in fact, sin k

For the cotangent function, domain at all multiples of

2

,x

2k 1

(in fact, cos

2

,k

since there are vertical

2k 1

2

0, k

) and range

.

For the cosecant function, domain at all multiples of

x

(in fact, sin k

x

,x 0, k x 0, k

k ,k

since there are vertical asymptotes , 1

) and range ,x

k ,k

) and range

1,

.

since there are vertical asymptotes .

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7. (a) The meaning of the parameters a and c in the equation of the trigonometric function is a amplitude and c vertical translation (or y c is the equation of the principal axis). ymax ymin ymax ymin Therefore, we have a and c , where ymax and ymin are the 2 2 y-coordinates of the maximum and minimum points on the curve, respectively. So, we have 1 3 1 3 a 2 and c 1 . The parameter b relates to the period T of the 2 2 2 2 trigonometric function as T . From the graph we have T . This gives us b 3 2 2 b 3. 2 T 3 1 (b) Setting up the equation y 0 , we have 2sin 3x 1 0 sin 3 x . This gives 2 1 1 . Finally, we have 3x sin 1 2k or 3x sin 1 2k , k 2 2 1 2k 1 5 2k x 2k 2k or x . 3 6 18 3 3 6 18 3 5 Point P lies between and , so its x-coordinate is . 18 6 3 8. The meaning of the parameters a and c in the equation of the trigonometric function is a amplitude and c vertical translation (or y c is the equation of the principal axis). ymax ymin ymax ymin Therefore, we have a and c , where ymax and ymin are the 2 2 y-coordinates of the maximum and minimum points on the curve, respectively. So, we have 2 4 2 4 a 3 and c 1 . The parameter b relates to a horizontal shift, and it is most 2 2 3 3 3 2 . This gives 3sin b 1 2 , or sin b 1, easily found by requesting that y 3 4 4 3 b 2k , k from which derive . 4 2 3 Solving for b we have b 2k 2k : there are multiple values for b since 2 4 4 horizontal shifts of a multiple of 2 do not affect the graph of the trigonometric function. 7 Possible values for b could come from k 0 b , k 1 b , etc. 4 4

Exercise 6.4 1. In this question we only need to consider solutions between 0 and 2 , therefore we will write down the general solution and then limit ourselves to the given domain. This is especially necessary for equations involving the inverse sine and the inverse tangent, since the range of these two functions includes negative angles. In all answers below, k . © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

5 1 1 or x 2k or x 2 cos 1 2k , giving x 3 2 2 3 1 1 1 2sin x 1 0 sin x x sin 1 2k or x sin 1 2k , giving 2 2 2 7 11 from x 2k or x 2k . Solutions in the given domain are therefore x 6 6 6 7 from k 0 k 1 and x 6 1 cot x 0 cot x 1 tan x 1 , giving x tan 1 1 k . Solutions in the given range are 5 or x x 4 4 3 3 3 , giving x sin 1 3 2sin x sin x 2k or x sin 1 2k . Solutions in 2 2 2 2 the given domain are x k 0 or x k 0 3 3 1 2 2 2 , giving x sin 1 2sin 2 x 1 sin 2 x sin x 2k , x sin 1 2k , 2 2 2 2 2 2 x sin 1 2k or x sin 1 2k . Solutions in the given domain are 2 2

(a) cos x (b)

(c)

(d)

(e)

1 2

x cos

1

7 5 3 k 1 or x k 1 k 0 ,x 4 4 4 4 (f) As an alternative method, we solve the next equation using the unit circle. 3 4cos 2 3 cos , giving the graph below where x cos . 2 x

k

0 , x

The four intersections between the unit circle and x cos given domain,

3 give the solutions in the 2

5 7 11 , , 6 6 6 6 ,

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(g) As a further method, we solve the next equation using the graph of the function f x tan x and its symmetries. tan 2 x 1 0

tan 2 x 1

tan x

1 , giving the graph below:

The four intersections between the graphs of y tan x and y 1 give the 3 5 7 , , , solutions in the given domain, x 4 4 4 4 1 1 2 2 (h) 4cos x 1 cos x , giving the graph below: cos x 4 2

The four intersections between the graphs of y cos x and y 2 4 5 , , 3 3 3 3 tan x 0 or tan x 1 0 tan x

the solutions in the given domain, x (i) tan x tan x 1

0

1 give 2

,

1 , giving the graph below:

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The four intersections between the graphs of y tan x and y 0, y 3 7 give the solutions in the given domain, x 0, , , 4 4

1

(j) sin cos 0 sin 0 or cos 0 . Using the unit circle with x cos and y sin we obtain the following diagram:

The two intersections between the unit circle and x cos

while the two intersections between the unit circle and y sin

(k) 5 sec x 3

sec x 2

cos x

3 , 2 2 0, 0 give

0 give

,

1 . Using any of the methods above, we obtain x 2

5 3 3 ,

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(l) csc 2

2

csc

2

1

sin

2

2 . Using the unit circle with y sin 2

we obtain the following diagram:

The four intersections between the unit circle and y sin give the solutions in the given domain,

2 2

3 5 7 , , , 4 4 4 4

2. (a) x1

sin 1 0.4 0.412

sin 1 0.4 2.73 since angles that add up to have equal sine values 1 (b) 3cos x 1 0 cos x 3 1 x1 cos 1 1.91 3 x2

x2

(c) x1

2

cos

1 3

1

4.37 since angles that add up to 2

have equal cosine values

tan 1 2 1.11

tan 1 2 4.25 since the period of the tangent function is 1 1 (d) sec2 x 3.46 3.46 cos 2 x cos 2 x 3.46 1 2 x cos 1 2k , k 2 x 1.28,7.56, where we wrote explicitly only solutions for 3.46 k 0,1 that when divided by two give values in the given domain 0 x 2 . So, x1,2 0.639,3.78 . x2

2x 2

cos

1

1 2k , k 3.46

2 x 5.01,11.29,

where we wrote explicitly only

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solutions for k 0,1 that when divided by two give values in the given domain 0 x 2 . So, x3,4 2.50,5.64 . (e) Graphing y cos x 1 and y 0.38 and looking for intersections in the given domain gives x 2.96,5.32 as shown below.

. (f) 3tan 2 x 1

tan x

1 3

3 . Graphing y 3

intersections with the help of symmetries gives x

(g) Graphing both hand sides of csc 2 x 3

tan x and y 5 7 11 , , 6 6 6 6 ,

3 and looking for 3

as shown below.

3 separately and looking for intersections gives 2

x 1.86,2.71,5.01,5.85 as shown below. © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

(h) Graphing y 3cot x and y 10 and looking for intersections gives x 0.291,3.43 as shown below.

3. (a) Here we look for integer values of k that make x 3 . Since 0

k

2

, we can start from k

3, 2, 1, 0, 1, 2 and correspondingly x

explicitly x

2

k

larger than 3

and smaller than

3 but we have to stop before k 3 . This gives

2

3 ,

2

2 ,

2

, , 2 2

,

2

2

5 3 3 5 , , , , , 2 2 2 2 2 2

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or

(b) As an alternative, we can solve the inequality 2 2

2k

2

2k

2

6

2

6

6

2k

2

for k . This gives

6 13 11 2k 6 6 13 1 11 1 k 6 2 6 2 13 11 , and the integers that satisfy this inequality are k 1, 0 . k 12 12 11 Correspondingly, we have x , 2 , so x 6 6 6 6 7 (c) Here, 0 , and since we are adding multiples of , the only acceptable values are 12 7 19 7 7 k 0, 1 , and correspondingly x . Finally, x , , 12 12 12 12 (d) As a last method, we can simply enter integer values of k, as long as the condition 0 x 4 is met. The first such value is k 1 , which gives x 0 , and from that value we keep adding 2 3 4 5 6 7 8 9 10 11 12 13 14 15 , obtaining x 0, , , , , , , , , , , , , , , 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 3 5 3 7 9 5 11 13 7 15 2 , , , ,3 , , , which simplifies to x 0, , , , , , , 4 2 4 4 2 4 4 2 4 4 2 4

4. 1 . When solving, we have to remember that angles that add up to 2 have 6 2 equal cosine values, so we have two families of solutions: 1 1 x cos 1 2k or x 2 cos 1 2k , so 6 2 6 2

(a) cos x

x x

6

3 3

6

2k or x 2k

6

or x 2

2

3 3

6

2k

2k

2k or x 2k 1 6 2 We now have to choose the values of k that satisfy 0 x 2 . We have 5 3 x (from k 1 ) or x (from k 0 ). 6 2 2 1 (b) tan x

tan 1 1 tan 1 1

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. We now have to choose the values of k

k

4

3 (from k 0 ) or 4

We have

4

that satisfy

.

(from k 1 ).

3 . When solving, we have to remember that angles that add up to 180 have equal 2 sine values, so we have two families of solutions: 3 3 2 x sin 1 k 360 or 2 x 180 sin 1 k 360 , so 2 2 2x 60 k 360 or 2x 180 60 k 360 that satisfy x 30 k 180 or x 60 k 180 . We now have to choose the values of k 0 x 360 . We have x 30 ,210 (from k 0,1 ) or x 60 ,240 (from k 0,1 ). 3 (d) sin 2 2 4

(c) sin 2 x

3 . When solving, we have to remember that angles that add up to 2 2 equal sine values, so we have two families of solutions:

have

sin

sin

2

2

satisfy

y

5

2

or

2k

or

2k

2

2

(e) 2cos2

3

3 2

1

2

1

3 2

2k , so

2k . We now have to choose the values of k

3

that

.

from k 0 , or 3 6 5cos 3 0 . Setting y 25 24 4

sin

2

5 7 4

3 or

from k 0 . 2 3 6 cos , this becomes 2 y 2 5 y 3 0 , giving

1 2

1 . The former has no solutions, since 3 is 2 2 4 outside the range of the cosine function, while the latter gives or . 3 3 (f) 3tan x 2cos x . First of all, we observe that in order for this equation to be satisfied, the angle x has to be either in the first quadrant, where both tangent and cosine are positive, or in the second quadrant, where both are negative. Recalling the definition of tangent, we have sin x 3 2cos x . Using Pythagoras, sin 2 x 1 cos 2 x , we have cos x This gives two equations, cos

3

1 cos 2 x cos x

3 and cos

2cos x

3 1 cos2 x 2cos2 x . Setting t cos2 x , we have 3 1 t Squaring both sides gives 9 1 t 4t 2 4t 2

9t 9 0 or t

9

81 144 8

9 15 8

3 or

2t .

3 3 , but only is an acceptable 4 4

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3 4

3 . The solutions to these 2 11 5 7 3 3 two equations are from cos x , and from cos x , but , , 2 2 6 6 6 6 recalling that the angle x must be either in the first or in the second quadrant we have 5 . x , 6 6

solution since t

(g) 2cos 3x 24

cos2 x 0 . This gives cos2 x

cos x

2

2 , so that 3x 24 45 k 360 or 3x 24 315 k 360 . 2 Solving for x gives 3x 21 k 360 or 3x 291 k 360 , and finally or x 97 k 120 . The solutions in the domain 0 x 360 are x 7 ,127 ,247 or x 97 ,217 ,337 cos 3x 24

(h) 9sec 2 sec2 sec cos cos x

12 12 4 9 3 2 3 11 3 , whose solutions are x , 2 6 6 3 . The solutions in the domain 0 2

from cos x are

5. The condition in the question amounts to the equation N t

3 and 2 5 . , 6 6

5 7 , from 6 6

90 , giving 74 42sin

12

t

90 .

42sin

t 16 12 16 8 . Since we are looking for the first time when N 90 , we consider the smallest sin t 12 42 21 8 12 8 t sin 1 sin 1 1.49 1.5 hours (assuming the argument of the solution giving t 12 21 21 sine function is in radians).

6. (a) This happens when H 12 , so when sin

2 D 80 365

when its argument is equal to integer multiples of

0 . A sine function equals zero

, so we have

2 D 80 365

0 , which

2 365 , which gives D 80 D 80 182.5 D 262.5 . 365 2 The days are therefore the 80th and, approximately, the 263rd of the year, so March 21st and September 20th.

gives D 80 , or

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(b) Setting up the equation H 15 and solving 12 7.26sin

2 D 80 365

15 gives:

D 105 and D 238 , so April 15th and August 26th.

(c) Requesting that H 17 and graphing the solutions to H 17 produces the following graph:

from which the number of days with more than 17 hours of daylight is 218 124 94 7. (a) 2cos2 x cos x 0 cos x 2cos x 1

0

cos x

two equations have solutions x

0 or 2cos x 1 0

3 or x 2 2 ,

cos x

1 . In the given domain, these 2

2 4 . , 3 3

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1 1 8 1 3 . 1 or 4 4 2 7 11 In the given domain, these two equations have solutions or . , 2 6 6 (c) tan 2 x tan x 2 tan 2 x tan x 2 0 . Solving for the variable tan x gives 1 1 8 1 3 tan x 2 or 1. In the given domain, these two equations have solutions 2 2 x tan 1 2 63.4 or x 45 (b) 2sin 2

sin

1 0 . Solving for the variable sin

1

gives sin

(d) 3cos2 x 6cos x 2 6 36 24 cos x 6

3cos2 x 6cos x 2 0 . Solving for the variable cos x gives 6 2 15 15 15 15 . Since 1 1 1 , the equation cos x 1 6 3 3 3 15 has no solutions. In the given domain, the remaining equation cos x 1 has solutions 3 15 15 x cos 1 1 1.87 or x cos 1 1 1.87 3 3

(e) 2sin tan

1

3 2

3 2

tan

3 (provided cos 2

0 ). Solving for

gives

56.3 . In the given domain this is the only solution.

(f) sin 2 x cos 2 x tan x

sin cos

3cos

sin 2 x 1 cos 2 x

tan 2 x 1

1 in the given domain gives x

tan x

1 (provided cos x

0 ). Solving

3 4 4 ,

(g) sec2 x 2sec x 4 0 . Solving for the variable sec x gives sec x

2

4 16 , therefore 2

there are no real solutions. (h) sin x tan x 3sin x . Here it is mandatory NOT to DIVIDE both hand sides by sin x , as this would lose us solutions. Instead, we FACTOR sin x and apply the null factor law, obtaining sin x tan x 3 0 sin x 0 or tan x 3 0 tan x 3 . Solving these two equations in the given domain gives x 0 ,180 or x tan 1 3 71.6 or x 180

tan 1 3 251.6

252 .

Exercise 6.5 1. In all of the following exercises, the key is expressing the given angle as a sum or difference of angles whose sine and cosine are known. 7 4 3 1 2 3 2 2 6 cos cos cos cos sin sin (a) cos 12 12 12 3 4 3 4 3 4 2 2 2 2 4 (b) sin165

sin 135

30

sin135 cos30

sin 30 cos135

2 3 2 2

1 2

2 2

6

2 4

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(c) tan

4 tan 12

12

3

3 1

3

5 12

(e) cos 255

tan

3 1 1 1

(d) sin

tan

3 12

sin

7 12

sin

cos

4

30 2

3

sin

4 2 2

cos 225 cos30 2

3 1

1 3 1 1 tan 3 4 2 3 4 2 3 2

2 cos 3 4

sin

3 1

4

3

1 tan 3

sin

2 3

4

(f) cot 75

4

3 1 3 1 3

12 12

cos 225 6

3

tan

3

2 3

4

1 2

3 2 2 2

2 4

6

2 3 2 2

sin 225 sin 30

2 1 2 2

6 4

1 tan 75

1 tan 30 45

1 tan 30 tan 45 tan 30 tan 45

3

3

3

3 3

3

3

3

3

3 3

3

9 6 3 3 9 3

3 1 3

1

3 1 3

12 6 3 6

2

3

2. (a) cos

12

cos

3

cos

4

2

cos

3

sin

4

3

sin

4

3 2 2 2

1 2 2 2

6

2 4

(b) cos 2 x 2cos x 1 , so letting 2x 2cos2

cos

2

1

cos

Applying this to cos cos

24

12 2

, we have: cos 1 1 2cos 2 cos 2 and finally cos 2 2 2 2

12 1

2

24

6

2 4

24

1 2

we have 6

1

2 4 2

2

chosen the positive sign since

cos

4 6

2

4

8

, where we have

is in the first quadrant where the cosine is positive.

3. tan

(a) tan

2

tan

2

1 tan

. Since tan 2

tan

consider what happens when

2

2

. tan

is undefined, we consider an angle

instead, and

tan tan

, where we

tan tan 1 tan tan

1 1 tan

have divided both numerator and denominator by tan . If we now let

tan

2

, we have

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.

, so

tan

1 0 1 cot 0 tan tan The same identity can be proved more directly using symmetry identities: tan

sin tan

(b) sin (c) csc

2

2

cos

sin

sin

2

cos

2

2

2

cos

cos

sin cos

cos

sin sin

2

sin cos

cos

2

1 2

sin

2

2

cos

cot

2

0 cos

1 sin

cos 0 0 sin

2

sin cos

1 cos

sec .

2

4. (a) cos x

1 sin 2 x

2

3 5

1

16 25

4 5

Since x is in the first quadrant we choose cos x 2

3 18 1 5 25 3 4 24 (c) sin 2 x 2sin x cos x 2 5 5 25

4 5

7 25

(b) cos 2 x 1 2sin 2 x 1 2

5. (a) sin x

1 cos 2 x

2 3

1

2

1

4 9

5 9

Since x is in the second quadrant, we choose sin x 2

5 3

(c) cos 2x 2cos2 x 1 2

2 3

(b) sin 2 x

2sin x cos x

2 3 2

1 2

5 3 5 3

4 5 9

4 1 9

1 9

6. 2 cos 1 sin 2 3 second quadrant. We then have:

5 . We then choose cos 3

(a) sin

sin 2

2sin cos

2

2 3

5 3

5 since 3

is in the

4 5 9

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cos 2

1 2sin 2

tan 2

sin 2 cos 2 4 5

(b) cos

4 5 9 1 9

2sin cos

cos 2

2cos2

1 2

sin 2 cos 2

24 25 7 25

2cos

2 2

1 9

4 5

3 since 5

sin 2

(c) tan

8 9

1

1 cos2

sin

We choose sin

tan 2

2

2 3

1 2

4 5 2

4 5

32 7 1 25 25

1

24 7

sin cos

2

sin

cos 2

1

cos 2

2sin cos 2

cos 2

2cos

tan 2

sin 2 cos 2

(d) sec

4

cos

2

2cos . Together with sin 2 1 5

1 2 4 5 3 5

4 3

5 5 2

2cos

5 since 5

4 5

10 1 25

1

1 , this gives

5 . We then chose cos 5 2 5 . We then have: 5

cos

20 25

cos2

15 25

3 5

1 . If follows that 4

2

1 cos

sin

15 since csc 4

2sin cos

2 5 5

5 5

sin

sin 2

3 . 5

24 25

is in the first quadrant, so that sin sin 2

9 25

is in the third quadrant. We then have: 3 5

2

2

4 5

1

1

2

1 4

2

1

0 implies sin 15 4

1 4

1 16

1 csc 2 15 16

15 . We then choose 4

0 . We then have: 15 8

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2

1 4

cos 2

2cos2

1 2

tan 2

sin 2 cos 2

15 8 7 8

1

1 1 8

7 8

15 7

7. (a) cos

x

(b) sin x

cos cos x sin sin x

sin x cos

2

2 tan x tan 1 tan x tan

(c) tan x (d) cos x

2

cos x cos

2

1 cos x 0 sin x

cos x

cos x sin x 0 1 cos x 2 tan x 0 tan x 1 tan x 0

cos x

sin

sin

2

sin x cos x 0 1 sin x

sin x

8. (a) sec

(b)

(c) (d)

sec tan

sec 1 cos 2

1 1 sin cos sin cos cos 1 1 csc 1 cos cos sin sin sin cos sin sin 2 sin cos 1 1 sin cos csc cos sin sin cos 2 2 2 2 1 1 sin 1 sin 2 cot 2 cos 2 cos 2 cos2 sin

1 sin 3

sin cos 2sin cos

sin cos sin 2

9. (a) cos

cos sin 2

cos

1 sin 2

cos cos2

cos3

1 cos 2 sin 2 1 2 sin sin 2 1 2sin 2 sin 2 1 sin 2 cos2 (c) cos 2 sin 2 sin 2 1 (d) tan 2 tan 2 2 tan 2 2 cos cot 2 sin sin cos sin cos sin cos (e) sin

(b)

1 cos 2 A 1 2cos 2 A 1 cos 2 A 2 2 cos cos cos (g) cos

sin cos

2sin cos

(f)

(h) 2cos2

cos 2

2cos2

2cos2

sin sin

1

cos cos

sin sin

2cos cos

1

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10. To prove the following identities, we manipulate both sides until they become equal. (a) cos 2 cos sin cos sin cos 2 sin 2 cos sin cos sin cos sin cos sin cos

cos sin

sin cos

cos

sin

sin

(b) 1 cos 1 cos 1 cos cos cos 2 cos sin 2 cos

1 sec sec

sin tan cos sec

1 cos cos cos 1 sin 2 cos cos 2 1 sin cos 2 sin cos

sin

sin cos

sin 2 cos

(c) 1 tan 2 x cos 2 x 1 tan 2 x sin 2 x 1 cos 2 x cos 2 x sin 2 x sin 2 x 1 cos 2 x cos 2 x sin 2 x cos 2 x cos 2 x sin 2 x 2 cos x sin 2 x cos 2 x cos 2 x sin 2 x cos 2 x sin 2 x 2 2 cos x sin x cos 2 x sin 2 x cos 2 x sin 2 x 1 cos 2 x sin 2 x cos 2 x sin 2 x

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(d) cos 4

sin 4

cos 2

sin 2

1 cos2 cos 2

cos 2 cos 2

sin 2 sin 2

sin 2

cos 2 cos 2

cos 2

sin 2

sin 2 sin 2

(e) cot

tan

2cot 2

1 1 tan 2 tan tan 2 2 1 tan 1 tan 2 2 tan 2 tan 1 tan 2 1 tan 2 tan tan

(f) cos cos

sin sin

cos cos

sin sin

cos 2

cos 2 sin 2 2sin cos

cos 2 sin 2 1 2sin cos

cos 2 1 sin 2 cos cos

sin sin

cos 2 sin 2 1 2sin cos

sin 2

cos 2 sin 2 1 2sin cos

cos 2 sin 2 1 2sin cos

(g) sec

1 1 sin

sec

tan

cos 1 sin 1 sin cos cos cos 1 sin 1 sin 1 sin 1 sin cos cos 1 sin 1 sin 2 1 sin cos cos 1 sin 1 sin 2 cos cos 1 sin 1 sin cos cos

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(h) tan A sec A sin A cos A

1 cos A 2

sin A 1 cos A

1 sin A 1 sin A

2

2

1 sin A 1 sin A 1 sin A 1 sin A 2

1 sin A 1 sin 2 A

sin A 1 cos 2 A

2

1 sin A cos 2 A

2

1 sin A cos 2 A

2

1 sin A cos 2 A

2

(i) tan 2 x tan x sin 2 x tan 2 x tan x 2 tan x tan x 1 tan 2 x 2sin x cos x 2 tan x tan x 1 tan 2 x 2 tan 2 x 1 tan 2 x 2sin x cos x 2 tan x tan x tan 3 x 1 tan 2 x 2 tan 2 x 2sin x cos x tan x tan 3 x 2 tan x 2sin x cos x 1 tan 2 x sin x 2 cos x 2sin x cos x sec2 x sin x 2 cos 2 x 2sin x cos x cos x 2sin x cos x 2sin x cos x

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(j) sin 2 cos 2 1 tan sin 2 cos 2 1 2sin cos 1 2sin 2 2sin cos

2cos

2sin cos 2sin cos 2sin cos 2cos sin

2sin 2 2cos 2 sin cos

sin cos

1

2

1

tan

1

tan tan

tan

(k) 1 cos sin 1 cos sin 1 cos sin

sin 1 cos 2 sin sin 1 cos 2 1 cos sin 2 sin 1 cos

2csc

1 cos 2

1 cos sin

2 2cos

1 cos sin

1 2cos cos 2 sin 1 cos

1 cos sin

1 cos sin 1 cos

1 cos sin

1 cos sin

1 cos sin

sin 1 cos

sin

1 cos

2

(l) 2csc

2

1 cos sin 2 sin 1 cos 1 2cos sin

cos 2 1 cos

2 sin sin 2

1 2cos 1 sin 1 cos

2 sin

2 1 cos sin 1 cos

2 sin

2 sin

2 sin

2 sin

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(m) cot x 1 csc x 1 tan x sec x cos x 1 cos x sin x sin x sin x 1 cos x cos x sin x cos x sin x cos x sin x sin x cos x cos x sin x cos x sin x cos x sin x cos x cos x sin x sin x

cos x sin x

(n) sin sin 2 sin 2 sin 2 sin 2 sin 2 sin 2

1 cos 2

2 2 2 2

1 cos 2 1 cos 2 2 1 cos 2 2 1 1 2sin 2 2 2sin 2 2 sin

2

11. Considering the two right-angled triangles in the diagram, we have 7 7 2 tan tan 1 tan 1 tan 1 . Taking the tangent of both sides we have tan x x x and the compound angle formula for the tangent gives us 7 2 7 2 5 tan tan 1 tan tan 1 5x x x x x x tan 2 2 7 2 x 14 x 14 7 2 1 1 tan tan 1 tan tan 1 x x x x x2

tan

1

2 , x

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12. (a) 2sin 2 x cos x 1

2 1 cos2 x

2

cos x 1 0 2

2 2cos x cos x 1 0 2cos cos x 1 0 1 1 8 1 3 1 or 1, which in the given domain gives: cos x 4 4 2 5 or x x , 3 3

(b) 1 cos 2 x 1 cos x 3 8

sec2 x 8cos x cos3 x

1 8

8cos x 1 2

In the given domain, this gives x

, 3 3

(c) 2cos x sin 2x 0

2cos x 2sin x cos x 0

2cos x 1 sin x 0 This gives 2cos x 0 or 1 sin x 0 given domain are x 90 ,90 or x

sin x 1. The solutions to these equations in the 90 , so finally x 90 ,90

(d) 2sin x cos 2 x 2sin x 1 2sin 2 x 2sin 2 x 2sin x 1 0 2 4 8 2 2 3 sin x 4 4 1 3 1 , while sin x since 2 1 3 sin 1 2.77 or x 2

1

3 1 3 . The equation sin x has no solutions 2 2 1 3 1 3 0.375 gives x sin 1 2 2

(e) cos 2 x sin 2 x cos 2 x sin 2 x sin x cos 2 x

2sin 2 x

sin 2 x cos 2 x

1 2 1 2

tan 2 x

tan x

1 2

Solving for x in the given domain gives x x

tan

1

1 2

2.53 , x

tan

1

tan

1

1 2

1 2

0.615 , x

tan

1

1 2

3.76 ,

5.67

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(f) 2sin x cos x 1 0 sin 2 x 2 x sin x

1 1

3 4

1

3 2

2k

2k

k

The solutions in the given domain are x

3 7 , 4 4

(g) 1 2

cos 2 x sin 2 x 1 2

cos 2 x 2 x cos

2 3

2x

x

3

1 2

1

2k

k

2k

or 2 x 2 3

or 2 x 2

or x

2 3

2

cos

2k

1

4 3

1 2

2k

2k , so

and finally

k

The solutions in the given domain are x

3

or x

2 3

(h) sec2 x tan x 1 0 1 sin x 1 0 2 cos x cos x 1 sin x cos x cos 2 x 0 cos 2 x Considering the numerator only, and rearranging it, we have: 1 cos 2 x sin x cos x

0

2

sin x sin x cos x 0 sin x sin x cos x

0

sin x 1 tan x 1 cos x The solutions to these two equations in the given domain are 5 x 0, or x , 4 4 sin x 0 or sin x cos x

0

sin x cos x

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(i) tan 2 x tan x 0 2 tan x tan x 0 1 tan 2 x 2 tan x 1 0 1 tan 2 tan x

2 1 tan 2 x 1 tan 2 x

0

2 1 tan 2 x 0 3 tan 2 x 0 tan x 1 tan 2 x The solutions to these three equations in the given domain are 4 2 5 x 0, or x , or x respectively. , , 3 3 3 3

which gives tan x 0 or

3

(j) 2sin 2 x cos3x cos3x 0 cos3x 2sin 2 x 1

0

This gives cos3x 0 or 2sin 2 x 1 0

sin 2 x

1 2

The solutions to the first equation are: 3x 90 k 360 or 3x 270 k 360 , giving x 30 k 120 or x 90 k 120 . The values in the given domain are x 30 ,90 ,150 . The solutions to the second equation are: 2x 210 k 360 or 2x 330 k 360 , giving x 105 k 180 or x 165 k 180 . The values in the given domain are x 105 or x 165 13. sin 3 x sin 2 x x

sin 2 x cos x sin x cos 2 x

2sin x 1 sin 2 x

sin x 2sin 3 x

2sin x cos 2 x sin x 1 2sin 2 x

2sin x 2sin 3 x sin x 2sin 3 x

3sin x 4sin 3 x

14. (a) Squaring the expression as suggested, we obtain

sin 2 x cos2 x

2

sin 4 x 2sin 2 x cos2 x cos4 x . On the other hand,

sin 2 x cos2 x

2

12 1 , so we have sin 4 x 2sin 2 x cos2 x cos4 x 1 .

Solving for sin 4 x cos 4 x , we have sin 4 x cos4 x 1 2sin 2 x cos2 x . Working on the right 1 1 2 1 2 hand side gives 1 2sin 2 x cos 2 x 1 2sin x cos x 1 sin 2 x 1 2sin 2 2 x . From 2 2 4 1 cos 2 and the double angle formulae, we have cos 2 1 2sin 2 , from which 2sin 2 2 2sin 2 x 1 cos4 x . Replacing this last identity, we have 1 1 1 1 1 2sin 2 x cos2 x 1 1 cos 4 x 1 cos 4 x cos 4 x 3 . So, we have proved that 4 4 4 4 1 sin 4 x cos 4 x 1 2sin 2 x cos 2 x cos 4 x 3 . 4 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

(b) The equation sin 4 x cos 4 x cos4x 3 2

cos4x

1 1 becomes cos 4 x 3 2 4

1 . This gives 4 x

The solutions in the given domain are x

2k

1 , so we have 2 x

4

k

2

.

3 5 7 , , 4 4 4 4 ,

Exercise 6.6 1. Recall that the range of the arcsine and arctangent functions is the first or fourth quadrant, while the range of the arccosine is the first or second quadrant. (a) The angle in the first or fourth quadrant whose sine is 1 is

2 1 (b) The angle in the first or second quadrant whose cosine is is 4 2

(c) The angle in the first or fourth quadrant whose tangent is

3 is

3

1 2 (d) The angle in the first or second quadrant whose cosine is is 2 3 (e) The angle in the first or fourth quadrant whose tangent is 0 is 0 3 (f) The angle in the first or fourth quadrant whose sine is is 2 3 2. (a) The angle in the first or fourth quadrant that has the same sine as

2 is 3

3

2 3

3 3 3 is in the first quadrant, so it is the range of the arccosine, so cos 1 cos 2 2 2 tan(arctan12) 12 2 2 The expression arccos is meaningless since 1 and the domain of the arccosine is 3 3 numbers between 1 and 1 inclusive. 3 3 The angle in the first or fourth quadrant that has the same tangent as is 4 4 4 The expression arcsin is meaningless since 1 and the domain of the arcsine is numbers between 1 and 1 inclusive.

(b) The angle (c) (d)

(e) (f)

3 3 y 3 and x cosarctan . It follows that , since y and x are the 4 4 x 4 3 sine and cosine of the arc whose tangent is . theorem, we also 4 have x 2 y 2 1 . Combining the two equations and solving for y we have: 4 16 2 x y y y2 1 3 9

(g) We define y sin arctan

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y2 1

16 9

1

9 25 3 , but we choose y 5

y2

3 3 since the angle arctan is in the first quadrant. 5 4 7 (h) Here we are looking for the cosine of the angle whose sine is . This is simply given by 25 7 7 7 , so x cosarcsin , y sin arcsin 25 25 25 2 7 49 576 2 x 1 x2 1 25 625 625 24 24 , but we choose x since the range of the arcsine function is the first or fourth x 25 25 quadrant where the cosine is positive.

y

(i) The domain of the arcsine function is 1 x 1 , but tan

3

3 1 , so the expression is

undefined. (j) tan

1

2sin

3

tan

1

2

3 2

tan

1

3

3

1 1 y 1 (k) We define x cosarctan and y sin arctan . It follows that , since y and x are the x 2 2 2 1 sine and cosine of the arc whose tangent is 2 have x 2 y 2 1 . Combining the two equations and solving for x we have: 1 1 2 y x x2 x 1 2 4 1 x2 1 1 4 4 x2 5 4 2 2 5 2 5 x , but we choose x since the range of the arctangent function 5 5 5 5 is the first or fourth quadrant where the cosine is positive.

(l) Here we are looking for the cosine of the angle whose sine is 0.6 . This is simply given by x cos sin 1 0.6 , y sin sin 1 0.6 0.6 , so x2

0.62 1

x2 1 0.36 0.64

4 since the range of the arcsine function is the first or 5 fourth quadrant where the cosine is positive. x

0.8 , but we choose x 0.8

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(m) We first use the compound angle formula 3 5 3 5 3 5 sin arccos arctan sin arccos cos arctan cos arccos sin arctan . 5 12 5 12 5 12 We then calculate the four pieces separately. 3 3 3 3 , we have sin arccos : setting y sin arccos and x cosarccos 5 5 5 5 2 4 3 16 4 2 2 . We choose y since the range of the x y 1 y2 1 y 5 5 25 5 5 arccosine function is the first or second quadrant where the sine is positive. cosarctan : 12 5 sin arctan y 5 5 5 12 tan arctan 5 setting x cosarctan and y sin arctan , we have 5 x cos arctan 12 12 12 12 12 2

12 5 169 12 . We choose x x 1 x2 1 x 13 12 144 13 since the range of the arctangent function is the first or fourth quadrant where the cosine is 5 5 12 5 positive. This also gives us y sin arctan 12 12 13 13 3 3 We finally have cosarccos . Replacing these four values in the first formula, we have 5 5 3 5 3 5 3 5 sin arccos arctan sin arccos cos arctan cos arccos sin arctan 5 12 5 12 5 12 4 12 3 5 63 5 13 5 13 65

and x 2

y 2 1 . This gives x 2

(n) We first use the compound angle formula 1 1 1 cos tan 1 3 sin 1 cos tan 1 3 cossin 1 sin tan 1 3 sin sin 1 . We then calculate the 3 3 3 four pieces separately. y cos tan 1 3 : setting x cos tan 1 3 and y sin tan 1 3 , we have 3 and x 2 y 2 1 . These x 1 1 1 2 give x 2 3x , so that x . We choose x since the 1 10 x 2 1 x 2 10 10 10 range of the arctangent function is the first or fourth quadrant where the cosine is positive. 3 This also gives us y sin tan 1 3 3x 10 1 1 1 1 , we have cossin 1 : setting x cossin 1 and y sin sin 1 3 3 3 3 2 8 2 2 2 2 1 8 2 2 2 which gives x . We choose x x y 1 x 1 x2 3 3 3 3 9 since the range of the arcsine function is the first or fourth quadrant where the cosine is 1 1 positive. Finally, sin sin 1 . 3 3 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

Replacing these values gives 1 1 1 cos tan 1 3 sin 1 cos tan 1 3 cos sin 1 sin tan 1 3 sin sin 1 3 3 3 1 2 2 10 3

3 1 10 3

2 2 3 3 10

3.

2 2 3

10

2 20 3 10 30

30

. (a) cos(arcsin x) : setting X X 2 Y2 1

X2

cos(arcsin x) and Y

x2 1

sin(arcsin x)

x we have

X 2 1 x2 , from which we have X

1 x2 .

cos(arcsin x)

We then choose cos(arcsin x) 1 x2 since the range of the arcsine function is the first or fourth quadrant where the cosine is positive. sin(arccos x) sin(arccos x ) (b) tan(arccos x) . To rewrite sin(arccos x) , we set cos(arccos x) x X cos(arccos x) x and Y sin(arccos x) so that we have X 2 Y2 1

x2 Y 2 1

Y 2 1 x2 , from which we have Y

sin(arccos x)

1 x2 since the range of the arccosine function is the first or

We then choose sin(arccos x)

1 x2 . x

second quadrant where the sine is positive. Finally, we have tan(arccos x) (c) cos(tan 1 x) : setting X

cos(tan 1 x) and Y

X 2 Y 2 1 . This gives X 2 1

We choose X

1 x2

xX

2

1

Y tan(tan 1 x) x and X 1 1 X . 2 1 x 1 x2

tan(tan 1 x) , we have

X 2 1 x2

1

X2

since the range of the arctangent function is the first or fourth

quadrant where the cosine is positive. We therefore have cos(tan 1 x) (d) sin 2cos 1 x

1 x2 .

1

. 1 x2 2sin(cos 1 x)cos(cos 1 x) , using the double angle formula. This gives,

according to (b),

sin 2cos 1 x

2sin(cos 1 x)cos(cos 1 x) 2 1 x 2 x 2 x 1 x 2 .

1 arccos x : here it is convenient to recall the half-angle formulae that can be derived 2 from the double angle formulae for the cosine (see Exercise 6.5, question 10). These state 1 cos 1 cos sin and cos , so that we can rewrite 2 2 2 2

(e) tan

1 x 1 1 x 2 tan arccos x . In the half-angle 2 1 x 1 x 2 1 formulae, the positive roots were chosen because the angle arccos x is necessarily in the first 2 quadrant where both cosine and sine are positive. 1 sin arccos x 2 1 cos arccos x 2

1 cos arccos x 2 1 cos arccos x 2

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(f) sin arcsin x 2arctan x

sin(arcsin x ) cos(2arctan x ) cos(arcsin x ) sin(2arctan x )

x 2cos2 (arctan x) 1

1 x 2 2sin(arctan x) cos(arctan x) , using first the compound

angle formula for the sine and then the double angle formulae for the cosine and the sine and 1 and sin(arcsin x) x , cos(arcsin x) 1 x2 . From (c) we recall cos(arctan x) 1 x2 x sin(arctan x) , so by replacing we have 1 x2 1 x sin arcsin x 2 arctan x x 2 1 1 x2 2 2 1 x 1 x2 x

4.

2 1 x2 1 x2

1 x2 2

x 1 x2

x x3

2x 1 x2 1 x2

Applying the cosine function to the left-hand side, we have 4 5 4 5 4 5 cos arcsin arcsin cos arcsin cos arcsin sin arcsin sin arcsin 5 13 5 13 5 13 1 x2 and sinarcsin x x , we have

Recalling that cosarcsin x sinarccos x 1

4 5

2

5 13

1

2

4 5 5 13

9 25

144 169

4 13

3 12 4 5 13 13

36 20 65

16 65

4 5 16 arcsin arccos 5 13 65

This proves that arcsin

5. Applying the tangent function to the left hand side and using the compound angle formula for the tangent, we have 1 1 1 1 5 tan arctan tan arctan 1 1 2 3 2 3 6 1 , where we have used tan arctan arctan 1 1 1 1 5 2 3 1 tan arctan tan arctan 1 2 3 2 3 6 1 1 arctan , this proves that tan arctan . The proof can tanarctan x x . Since arctan1 2 3 4 4 be equivalently demonstrated taking either the sine of the cosine of both hand sides. 6.

tan 1 x tan

1

1 x

tan tan 1 x tan

1

1 x

tan tan 1 x tan tan 1

1 tan tan x tan tan

x

1 x

1 1

4 3

1

tan tan

1 x 1 x

1

4 , taking the tangent of both sides 3

4 , using the compound angle formula for the tangent 3

4 , recalling that tan tan 1 x 3

1 x1 x 1 1 x x2

tan

x

4 3

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3 4 x2 3 x2

x 1 x 1

0

3 4x2 4 x 4 0 4x2 4 x 1 0 2x 1

2

0

2x 1 0

x

1 2

7. (a) 5cos 2 x

2 2 cos 2 x 5 2 2 2 x cos 1 2k or 2 x 2 cos 1 2k . This gives 5 5 1 2 1 2 x cos 1 k or x cos 1 k , which in the given domain gives 2 5 2 5 x 0.580 or x 2.56 (both from k 0 ).

(b) tan

x 2

2

x tan 1 2 k 2 x 2tan 1 2 2k , which in the given domain gives x 2.21 from k 0 .

(c) 2cos x sin x 0 sin x 0 cos x tan x 2 x tan 1 2 k , which in the given domain gives x 1.11,4.25

2

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(d) 3sec 2 x

2 tan x 4

3 1 tan 2 x

2 tan x 4 0 , using one of the Pythagorean identities

3tan 2 x 2 tan x 1 0 2 4 12 2 4 tan x 1 or 6 6

1 3

These give x tan 1 1 k

tan

x

4

k

or x

or x

1

1 3

k

, or

0.322 k , which in the given domain gives

5 or x 2.82,5.96 4 4

x

,

(e) 2tan 2 x 3tan x 1 0 1 3 9 8 3 1 tan x 1 or . These give 4 4 2 1 x tan 1 1 k or x tan 1 k , or 2

x x

4 4

k

or x 0.464 k , which in the given domain gives

or x 0.464

(f) tan x csc x 5 sin x 1 5 cos x sin x 1 5 cos x 1 cos x 5 1 x cos 1 2k or x 5 x 1.37 or x 4.91

2

cos1

1 2k , or 5

(g) tan 2 x 3tan x 0 2 tan x 1 tan 2 x

3tan x 0

tan x

2 1 tan 2 x

tan x

2 3 3tan 2 x 1 tan 2 x

3

0

0 , which gives either tan x 0 or

2 3 3tan 2 x 1 tan 2 x

0.

The first equation gives x tan 1 0 k , which in the given domain gives x The second equation gives 5 3tan 2 x 0

tan 2 x

5 , which gives either tan x 3

,2 . 5 , 3

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therefore x tan x

tan

1

5 3

k

5 , therefore x 3

0.912, 4.05 in the given domain, or tan

1

5 3

k

2.23,5.37 in the given domain.

(h) 2cos 2 x 3sin 2 x 2 2cos 2 x 3sin 2 x 2 cos 2 x sin 2 x 2cos 2 x 3sin 2 x 2cos 2 x 2sin 2 x 3sin 2 x

2sin 2 x

6sin x cos x 2sin 2 x 0 2sin x 3cos x sin x 0 , which gives either 2sin x 0 or 3cos x sin x 0 . The first equation gives x k , which in the given domain gives x 0, . The second equation gives sin x 3 0 cos x tan x 3 , which gives x tan 1 3 k . In the given domain, this gives x 1.89 8. From the given graph, we have 2 d tan , so solving for the angle we have

tan

1

2 . d

A complete graph of this function is shown below.

The function as it is now makes sense only for positive values of d, since d is defined as a distance and therefore it must be positive, so:

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If d is allowed to take negative values, the interpretation of which would be distance above P, then the meaning of the negative angles would be angles that open counterclockwise with respect to the line joining the lighthouse with P. A possible way to remove the ambiguity about the sign of 2 tan 1 , whose graph is shown below: would be to consider d

Chapter 6 practice questions 1. (a) The length after 2 seconds is L 2

110 25cos 2

2

110 25 135 cm.

(b) The minimum length is when the cosine function takes the value 1. This gives Lmin 110 25 85 cm. (c) This occurs the first time the cosine function takes the value 1, so we set up the equation cos 2 t 1 , which gives 2 t t 0.5 seconds. (d) The period T of the function cos bt is given by T

2 , so in our case T b

2 2

1 second.

2. 2sin 2 x cos x 1 0 2 1 cos 2 x

cos x 1 0

2cos 2 x cos x 1 0

Solving for cos x , we get cos x

3.

1

1 8 4

1 3 4

1 or 1 , which gives 2

5 and x . 3 3 Calling r the radius of the circle, the perimeter of the shaded section is given by p 2r 2 r . Solving for , we obtain

x

,

p

2r

2 r

r

Replacing values,

r

21

2r 1

p

36 6

2r 1

p r

21

p . r

2.28

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4. (a) i)

The function f is a cosine with a horizontal compression, so it is unaffected in its vertical properties and its minimum value is 1 . ii) The function g is a cosine with a horizontal stretch by a factor of two, so its period is twice the period of the cosine, 2 2 4 (b) Graphing both functions on a GDC, we obtain:

which shows that in the given domain there are four intersections, therefore four solutions to the equation f x g x . 5. (a) p plays the role of the principal axis of the oscillation, and in this case, it is the height of the dmax dmin 64 6 hub from the ground. From p we obtain p 35 cm. 2 2 (b) q plays the role of the amplitude of the oscillation, and in this case, it is the distance between the highest position of the reflector and the hub, q dmax p 64 35 29 cm. (c) m plays the role of the period of the oscillation, and it is equal to twice the time interval between a maximum and the next minimum, so m 2 0.75 0.5 0.5 seconds.

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6. To solve 1 sin 3 x cos 0.25 x , we graph both sides on a GDC for the given domain and look for intersections:

There are three intersections, therefore the solutions are x 0, x 1.06,2.05 7. (a) 2cos2 x 5cos x 2 0 5 25 16 5 3 1 cos x or 2 . The first value gives 4 4 2 1 1 1 cos x x cos 1 2k or x 2 cos 1 2k , so 2 2 2 2 4 2 4 . x 2k or x 2k . The values in the given domain are x , 3 3 3 3 The second value gives cos x 2 , which has no solutions since 2 is outside the range of the cosine function. (b) sin 2x cos x 0 2sin x cos x cos x 0 cos x 2sin x 1

0 which gives cos x 0 or 2sin x 1 0

In the given domain, the first equation gives x 8. Since

2 2x 2 negative.

x

sin x

1 2

3 ; the second equation gives x 2 2 ,

5 6 6 ,

, x is in the second quadrant where the sine is positive. We also have which means that 2x is either in the third of fourth quadrant, where the sine is

(a) sin 2 x cos2 x 1

1 cos2 x , but we choose the positive solution according to the

sin x

discussion above: sin x

1

8 9

1 9

1 3

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8 16 9 7 1 9 9 9 1 1 (c) sin 2 x 2sin x cos x 2 cos 2 x 2 3 3

(b) cos 2 x 2cos 2 x 1 2

8 3

solution according to the discussion above: sin 2 x

4 2 . We then choose the negative 9 4 2 9

9. (a) We look for a function of the form d t amplitude, p

dmax

dmin 2

a sin b t c

p , where a

d max

dmin 2

is the

is the principal axis, c is a horizontal shift, and b relates to the

2 5.8 2.6 5.8 2.6 . This gives a 1.6 and p 4.2 . The period of the T 2 2 oscillation is found considering that the time interval between the first maximum and the first 2 minimum, so 5.5 hours, is one half of the period, giving T 11 hours. This gives b . 11 2 t c 4.2 . In order to find the value of c, we set So far, the function is d t 1.6sin 11 period T as b

the condition that the tide is at a maximum 5 hours after midnight, so sin 2 5 c 11 2 2 9 1.6sin t 11 4

This gives

5 c

d t

4.2

(b) This is given by d 12

1.6sin

(c) Setting the inequality d t

11 4

c 5

2 9 12 11 4

11 4

2 5 c 11

1.

9 . The function is given by 4

4.2 3.15 metres.

3.5 and graphing both sides in the given domain gives:

The boat can dock safely from 12.5 hours after midnight, so around 12.30 pm, to 19.5 hours after midnight, so around 7.30 pm, and then the cycle repeats every eleven hours. © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

10. tan 2 x 2tan x 3 0 . Solving this quadratic equation for tan x gives 2 4 12 2 4 in the given tan x 1 or 3 . The first solution gives tan x 1 x 2 2 4 3 x tan 1 3 k . In the given domain, domain, while the second solution gives tan x this gives x

tan

1

3

1.89

11. (a) The length of arc ABC is given by s

r

3 10 15 cm 2

(b) The shaded region subtends an angle at the centre

3 . The area of the shaded region is 2

2

1 2 1 3 r 2 102 239 cm2 2 2 2 12. The solutions of the equation f x k are the intersections between the graph of f x and the given A

horizontal line y

k . Since the function f x oscillates around the x-axis with amplitude

there will be no solutions for k

5 or for k 2

13. The two points given enable us to state y 0

5 , 2

5 . 2 1 and y

3 2

3. The first condition gives

k a sin 0 1 k a 0 1 so that k 1 . The second condition gives 3 1 a sin 3 2 1 a 1 3 a

2

14. The equation 2tan 2 2

5sec 2

10 0 can be rewritten as 2 sec2

1

5sec

10 0 using

2

5sec 12 0 , which is a the identity 1 tan x sec x . So the equation becomes 2sec quadratic equation in the variable sec . This gives 3 5 25 96 5 11 sec 4 or . Since the angle is in the second quadrant where the 4 4 2 3 cosine and hence the secant are negative, the only acceptable solution is sec . 2

15. Using the compound angle formulae and the definition of sine, cosine and tangent in right-angled triangles, we have 15 8 8 6 60 24 84 (a) sin sin cos cos sin 17 10 17 10 85 85 8 8 15 6 32 45 13 cos cos sin sin (b) cos 17 10 17 10 85 85

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84 85 13 85

sin

(c) tan

cos tan tan 1 tan tan

tan

84 . Using the compound angle formula for the tangent 13

6 8

15 and tan 8

with tan

3 gives the same result. 4

16. First, we calculate the length of the hypotenuse as 12 22 5 . Then, using the double angle formulae, the compound angle formulae and the definition of sine and cosine in a right-angled triangle, we have 1 2 4 and sin 2 p 2sin p cos p 2 5 5 5

sin3 p sin 2 p

sin 2 p cos p cos2 p sin p 2sin p cos2 p

p

sin p 2cos 2 p 1 2sin 2 p

1 5

2

4 1 1 2 5 5

1 2sin 2 p sin p

11

11 5 25

5 5

17. y sin B 5 and tan B x cos B 12 cos 2 B sin 2 B 1 . Combining these two conditions into one equation for y ,

(a) Setting y sin B and x cos B , we have x2

y2

we have

12 y 5

2

y2 1

and its value is sin B

169 2 y 1 25

5 . Since B is obtuse, sin B is positive 13

y

5 13

(b) Same as above, with x cos B

12 y 5

(c) Using double angle formulae, sin 2 B

12 5 5 13 2sin B cos B

12 13 2

5 13

12 13

120 169

2

5 50 119 (d) Using double angle formulae, cos 2B 1 2sin B 1 2 1 13 169 169 2 tan 3 18. Using double angle formulae, we have tan 2 . This is a quadratic equation in 2 1 tan 4 tan , in fact rearranging it we obtain: 2 tan 3 0 2 1 tan 4 8 tan 3 1 tan 2 0 4 1 tan 2 2

8tan

3 1 tan 2

3tan 2

8tan 3 0 8 64 36 6

tan

0 8 10 6

1 or 3 3

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19. Given that sin x

, we can state that

k sin x

sin x cos

sin cos x

tan x cos

sin

tan x cos

1 k

k sin x cos

k tan x cos

sin

sin

tan x

1 k

2

20. We have tan 2 1 tan 2 The first solution gives

sin cos x . Dividing both hand sides by cos x we have

. Solving for tan x , we have:

sin

1 k

cos

1 k

tan

1

tan 1 1 k k 2 2 8 2 3 The solutions in the given domain are , 8 8 The second solution gives tan 1 1 tan 2 1 2 tan 1 1 k k 2 2 3 The solutions in the given domain are , 8 8 tan 2

1

2

1 k 1 k

tan 1 1 k

8

k

2

21. (a) Graphing both functions in the given domain, we obtain:

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(b) Looking for the intersection of the two curves, we obtain:

The solution to the equation f x

g x is x 0.412

(c) The range of g over the given domain is determined by the maximum and minimum values of

g . The maximum is given by g 0 g 1

g

1

cos 2 cos

1 , while the minima occur at the boundaries,

2 . The range is therefore cos 2

y 1

22. Here we can find two expressions for the length of AC, taking advantage of the fact that it belongs to two different right-angled triangles. By equating these two expressions, we will obtain an equation for D AC . 5 2 From triangle ABC, AC . From triangle ACD, AC . tan B AC tan D AC Using the information in the question, and calling x D AC , we have: 5 2 tan 2 x tan x 2tan 2x 5tan x 0 2 tan x 2 5tan x 0 1 tan 2 x 4 tan x 5 0 . Since tan x cannot be equal to zero in this problem, 1 tan 2 x the equation reduces to: 4 5 0 1 tan 2 x 4 5 5 tan 2 x 0 1 tan 2 x 1 5 5 tan 2 x 1 tan 2 x tan x . Since x is an angle in a triangle, we choose tan x 5 5 5 24.1 which gives x tan 1 5 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

5 5

23. The request that the water depth be 10 cm means that the water surface has equation y 16 10 6 . The width w of the water surface is the distance between the intersections of the channel boundary with the water surface, so the distance between the solutions to x 16sec 32 6 . Solving this equation gives: 36 x 16sec 26 36 x 26 13 sec 36 16 8 x 8 cos 36 13 x 8 36 arccos x arccos . The distance between these solutions is 36 13 13 36 36 72 w arccos arccos arccos 13 13 13

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Exercise 7.1 1. (a) Define A 3, 1, 5 , B Then: AB

4, 0, 2 , C 2, 2, 1

2

4 3

AC

2 3

2

BC

2

4

0 2 2

2

1 1 2 0

2 5

2

2

1 5

2

1 2

2

2

49 1 9

59

1 9 36

46

36 4 9

49

The triangle ABC is scalene. (b) Define A 2, 4, 3 , B 4, 3, Then: AB

4

2

AC

3

2

BC

3 4

2

2

2 , C 2

3 4 2

2 4 2

3

2

2

3,

2, 4

2

3

4

3

4

2

2

2

36 49 1

86

1 36 49

86

49 1 36

86

2

The triangle ABC is equilateral. (c) Define A 4, 5, 0 , B 2, 6, 2 , C 2, 3, 1 Then: AB

2 4

2

AC

2 4

2

3 5

2

1 0

BC

2 2

2

3 6

2

1 2

6 5

2

2

2 0

4 1 4

9

3

2

4 4 1

9

3

2

0 9 9

18

3 2

The triangle ABC is isosceles. (d) Define A a, b, c , B b, c, a , C c, a, b Then: AB

b a

AC

c a

BC

c b

2

c b

2

a c

2

a b

2

a c

2

b c

2

2

a b

2

b c

2

a b

2

a c

2

b c

2

2

a c

2

b a

2

a b

2

a c

2

b c

2

The triangle ABC is equilateral.

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2. A point on the y-axis has coordinates 0, y, 0 . The distance of this point to the point A 1, 2, 3 is

or 2 y

2

2

1 0

2

2

y

3 0

2

10 . It follows that 1

2 y

2

9 10

0 . Therefore, the point on the y-axis has coordinates 0, 2, 0 .

3. (a) Let A 1, 2, 3 , B 1, 4, 5 , C 5, 4, 0 . Then: 2

AB

1

1

AC

5

1

BC

5 1

2

2

4 2

4 4

Now, AB AC

2

4 2

2

2

2

5 3

2

0 3

0 5

4 4 4

2

36 4 9

16 0 25

2 3 7 10.5 and BC

AB BC

2 3

AC BC

7

12

2 3

49

7

41

41 6.40 (to 3 s. f.)

41 9.87 and AC 7 (to 3 s. f.)

AB AC

AB BC

41 13.4 and AB 2 3 3.46 (to 3 s. f.)

BC

AC

AC BC

AB

The triangle inequality is satisfied, so the points A, B and C are the vertices of a triangle. (b) Let P 2, 3, 3 , Q 1, 2, 4 , R 3, 8, 2 . Then: 2

PQ

1 2

2

PR

3 2

2

8

3

QR

3 1

2

8 2

2

2

3

2

4 3 2

2 3

2 4

1 25 1 2

2

27

1 25 1

4 100 4

3 3

27

3 3

108

6 3

Now, since PQ PR QR , the triangle inequality is not satisfied, so the points P, Q and R are not the vertices of a triangle. 4. Let A 0, 7, 10 , B 2

AB

1 0

AC

4 0

2

BC

4

1

1, 6, 6 , C

9 7 2

It is clearly seen that AB 2

2

2

6 7

4, 9, 6 . Then: 2

6 10

2

6 10

9 6

2

1 1 16

2

6 6

and AC 6 36 AB angled isosceles triangle.

2

BC

2

3 2

16 4 16

36

6

9 9 0

18

3 2

2

BC . Now: AB

18

2

BC

2

3 2

2

3 2

2

18 18 36

2

AC . Therefore, the triangle ABC is a right-

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5. (a) Let A 0, 1, 7 , B 2, 1, 9 , C 6, 5, 13 . Then: 2

AB

2 0

2

1

1

AC

6 0

2

5

1

BC

6 2

2

5 1

As AC

9

2

13

2

2

7

13

4 4 4 2

7 2

9

12

2 3

36 36 36 16 16 16

108 48

6 3

4 3

AB BC , the points A, B and C are collinear.

(b) Let A 2, 0, 4 , B 5, 1, 1 , C 4, 6, 3 . Then: 2

AB

5

2

AC

4

2

BC

4 5

2

2

1 0

2

2

6 0 6

9 1 9 2

3 4

2

1

2

1 4

4 36 1

2

3 1

AB

2

3 1

AC

2 1

2

BC

2

3

7 8

2

2

1 2

7 5

2

2

4 2

4 2

34

1 1 36

1

1, 2, 3 , C

2

AC AB ,

16 9 9

2

38

25 4 9

As AC AB BC , AB AC BC , and BC the points A, B and C are not collinear. (d) Let A 2, 3, 4 , B

30

3, 5, 1 , C 2, 7, 2 . Then:

5 8

2

41

1 25 4

As AC AB BC and AB AC BC and BC the points A, B and C are not collinear. (c) Let A 1, 8, 4 , B

19

AC AB ,

4, 1, 10 . Then:

2 3

2

3 4

2

1 3

2

10 4

AB

1 2

AC

4 2

2

BC

4

1

As AC

AB BC , the points A, B and C are collinear.

2

1 2

2

10

38

9 1 49 2

59

36 4 196 3

2

9 1 49

236

2 59

59

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6. (a) Let A 2, 6, 3 (i)

The origin of the coordinate system has coordinates O 0, 0, 0 , so the distance OA

2 0

2

2

6 0

3 0

2

4 36 9

49

7

(ii)

The distance from the point A 2, 6, 3 to the x-axis is equal to the length of the perpendicular segment joining the point A to the x-axis. This segment lies in the plane perpendicular to the xz-plane passing through the point A. The second endpoint of this segment lying on the x-axis has coordinates 2, 0, 0 . The distance, therefore, is: 2 2

(iii)

0 6

2

0

3

2

0 36 9

45

3 5

6.71 (to 3 s. f.)

Similarly, the distance from the point A to the y-axis is equal to the length of the perpendicular segment joining the point to the y-axis. Since the segment lies in the plane perpendicular to the yz-plane, its second endpoint has coordinates 0, 6, 0 . The distance from the point A to the y-axis is therefore 0 2

(iv)

2

2

6 6

2

0

3

2

4 0 9

13

3.61 (to 3 s. f.)

The distance from the point A to the z-axis is equal to the length of the perpendicular segment joining the point to the z-axis. Since the segment lies in the plane perpendicular to the xz-plane, its second endpoint has coordinates 0, 0, 3 . The distance from the point A to the y-axis is therefore 0 2

2

0 6

2

3

3

2

4 36 0

40

2 10

6.32 (to 3 s. f.)

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(b) Let A 2, (i)

3, 3

The origin of the coordinate system has coordinates O 0, 0, 0 , so the distance OA

2 0

2

3 0

2

3 0

2

4 3 9

16

4

(ii)

The distance from the point A 2, 3, 3 to the x-axis is equal to the length of the perpendicular segment joining the point A to the x-axis. This segment lies in the plane perpendicular to the xz-plane passing through the point A. The second endpoint of this segment lying on the x-axis has coordinates 2, 0, 0 . The distance, therefore, is: 2 2

(iii)

2

0

3

0 3

2

0 3 9

12

2 3 3.46 (to 3 s. f.)

Similarly, the distance from the point A to the y-axis is equal to the length of the perpendicular segment joining the point to the y-axis. Since the segment lies in the plane perpendicular to the yz-plane, its second endpoint has coordinates 0, 0 2

(iv)

2

2

3, 0 . The distance from the point A to the y-axis is 3

3

2

0 3

2

4 0 9

13 3.61 (to 3 s. f.)

The distance from the point A to the z-axis is equal to the length of the perpendicular segment joining the point to the z-axis. Since the segment lies in the plane perpendicular to the xz-plane, its second endpoint has coordinates 0, 0, 3 . The distance from the point A to the y-axis is therefore 0 2

2

0

3

2

3 3

2

4 3 0

7

2.65 (to 3 s. f.)

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7. The diagonals of the parallelogram PQRS bisect each other. Thus, the diagonals PR and QS must have the same midpoint M: M

2 6 2 2 4 4 , , 2 2 2

M 2, 0, 8

Let S x, y, z . Then M

2 x 4 y 8 z , , 2 2 2

Therefore x

2 x 2

M 2, 0, 4

4, z 16 and S 2,

2, y

2,

4 y 2

0,

8 z 2

4

4, 16

8.

(a) XY

2

3 2

2

7 2

5 3

2

1 25 4

30

XZ

1 2

2

4 2

2

2 3

2

1 4 25

30

YZ

1 3

2

4 7

2

2 5

2

4 9 49

62

Since XY = XZ, the triangle XYZ is isosceles. 1 3 4 7 2 5 11 3 , , A 2, , . 2 2 2 2 2 The segment XA is perpendicular to the segment YZ and its length is

(b) The midpoint of YZ, A

XA

2 2

2

11 2 2

2

3 3 2

2

0

49 4

9 4

58 4

58 2

Therefore, the area of the triangle XYZ:

AXYZ

1 YZ 2

XA

1 2

62

58 2

1 2

2 31

2 29 2

1 31 29 2

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899 2

9. The length of segment AB

6 2 1 AB 2

The radius of the sphere: r The surface area S the volume V

4 3

4 29

2

29 3

4 3

2

1

2

7

1 2 29 2

2

2

4

16 64 36

2 29

29

116 units2, 29 29

116 3

29 units3.

10. The length of the longest piece of straight wire that can be placed completely inside the box is less than or equal to the length of the diagonal of the box. Diagonal =

622 442 202

3844 1936 400

6180

78.6 (to 3 s. f.)

Therefore, the longest piece of straight wire that can be placed completely inside the box is equal to 78 cm. r 2 r 2 h12 , the surface area of the cylinder is

11. The surface area of the cone is S1

S2 2 r 2 h2 , and the surface area of the hemisphere is S3 r 3 cm, h1 5 cm, h2 16 cm 5 cm 3 cm 8 cm .

2 r 2 , where

Therefore, the surface area of the solid is: S

3

32 52

2

3 8

2

32

3

34 48

18

66

3

34

262 cm 2 (to 3 s. f.)

1 2 r h1 , the volume of the cylinder is V2 r 2h2 and the 3 2 3 volume of the hemisphere is V3 r . Therefore, the volume of the solid is 3

The volume of the cone is V1

V

1 3

32 5

32 8

2 3

33

15

72

18

105

330 cm 3 (to 3 s. f.)

12. Draw the triangle.

Let A xA , yA , z A , B xB , yB , zB and C xC , yC , zC

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When we join the midpoints of any two sides, the segment will be parallel to the third side. Thus, we have CMKL is a parallelogram. Then segment ML and segment CK 1 9 3 have the same midpoint. The midpoint of ML is , so, we can write , , 2 2 2 xC

2

1 2

2

yC

1 and

xC

3 2

9 2

yC

6 and

zC

4

3 2

2

zC

7

7 Similarly, the midpoint of KL is 1, ,1 , and it is the midpoint of BM 2 xB 1 1 2

yB

1 and

xB

5 2

7 2

yB

2 and

zB 1 1 2

zB

3

Similarly, A 3, 4,5 13. (a) Volume of the sphere: Vs Vc

r 2h r 2 2r 4 3 r Vs 3 2 3 Vc 2 r 3

2 r3

(b) Surface of the sphere: Ss

4 3 r , volume of the cylinder 3 The ratio:

4 r 2 , surface of the cylinder

2 r 2 2 rh 2 r 2 2 r 2r

Sc

6 r 2 . The ratio:

Ss Sc

4 r2 6 r2

2 3

14. Surface area of the building consists of surface area of the cone and surface area of the cylinder, excluding the bases of both solids. S

rl 2 rh

r l 2h

6 15 2 50

690

2170 m2

Volume of the building consists of volume of the cone and volume of the cylinder. V

1 2 r hc 3

hc

l2

V

r 2h

r2

1 2 2 2 r l r 3

r 2h

1 3

62

152 6 2

62 50

6170 m 3

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15. Volume of the spike consists of volume of the cube and volume of the pyramid: 1 V 183 182 h p 3 We need to calculate the height of the pyramid.

In the triangle BCD (the side wall of the pyramid): h1

152 92

225 81

144 12 cm

Therefore, the height of the pyramid:

Volume of the spike: V

hp

122 92

183

1 2 18 3 7 3

Surface area of the spike: S

5 182

144 81

4

63

3 7 cm

5832 324 7 m3

1 18 12 2

2052 m 2

Exercise 7.2 1. (a) sin

3 5

(i)

(ii) cos tan cot sec

AC

52 32

4

csc

(iii) AC AB BC AC AC BC AB AC AB BC

4 5 3 4 4 3 5 4 5 3

36.9 or

53.1

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5 8

(b) cos (i)

(ii) sin tan

cot sec

82 52

x

(c) tan

64 25

39

csc

(iii) BC AB BC AC AC BC AB AC AB BC

39 8 39 5 5

39

5 39 39

8 39

8 39 39

51.3 or

38.7

63.4 or

26.6

8 5

2

(i)

(ii)

sin cos cot

sec x

22 12

4 1

5

csc

(iii)

BC AB

2 5

2 5 5

AC AB AC BC AB AC AB BC

1 5

5 5

1 2

5 1 5 2

5

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1 3

(d) cot (i)

(ii)

sin cos tan

sec 32 12

x

(e) sec

10

csc

BC AB AC AB BC AC AB AC AB BC

3 3 10 10 10 1 10 10 10 3 3 1 10 10 1 10 3

71.6 or

18.4

44.8 or

45.2

11 61

(i)

(ii) sin cos

x

(iii)

112

61

2

60

2 15

(iii) BC AB AC AB

2 15 11 61 11

tan

BC AC

2 15 61

2 915 61

cot

AC BC

csc

AB BC

61 915 30 2 15 11 11 15 30 2 15

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(f) cos

4 65 65

4 65

(i)

(ii)

(iii)

BC AB BC AC AC BC AB AC AB BC

sin tan cot

sec

x

2. (a) cos

65

2

1 2

42

65 16

7

60 180

60 or

csc

2 2

45 or

45 180

4

(c) tan

3

60 or

60 180

3

(d) csc

2 3 3

(e) cot (f) cos

1

3

45 or

3 2

3 2

2 3

45 180

30 or 50 cos 60

30 180

7 4 4 7

60 180

60 or

29.7

3

4

6

x y

50 y

y

50 100 , sin 60 1 2

(b) sin 55

y 15

y 15sin 55

12.3 , cos 55

x 15

(c) sin 40

x 32

x

32sin 40

20.6 , cos 40

y 32

(d) tan 53

y 225

225 tan 53

60.3 or

65 4 65 7

3. (a) cos 60

y

7 65 65

3

(b) sin

sin

7 65

299 , cos 53

x

y sin 60

x 15cos 55 y 225 x

32 cos 40 x

225 cos 53

100 3 2 8.60 24.5 374

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50 3

(e) The triangle is a right-angled isosceles triangle, so x 18, y 18 2 (f) sin 30

4. (a) tan

100 x

x

100 sin 30

300 10

10 3 10

100 1 2

90

60 ,

3

y 100

200 , cot 30

y 100 cot 30

90

60

30

(b) sin

15 39

5 13

22.6 ,

90

90

22.6

67.4

(c) tan

44 121

4 11

20.0 ,

90

90

20.0

70.0

(d) sin

7 28

7 28

1 4

1 2

90

30 ,

90

100 3

30

60

5. Let the length of the tree be x. tan 70

x 41.5

x

41.5 tan 70

114 m

6.

tan

300 125

2.4

67.4 (to 3 s. f.)

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7. 1.62 2

tan

tan

x 5

x

5 tan

5 0.81

0.81

4.05 m

8. In the triangle ACD: cot 40

CD AD

CD

cot 52

CE BE

CE

AD cot 40 BE cot 52

10 000 cot 40 10 000 cot 52

In the triangle BCE: d

CD CE 10 000 cot 40

cot 52

4105 m

9. The triangle is isosceles, so the two angles at the base have the same measure. In the triangle BCD: 3 8

cos

and

1 2

68.0 (to 3 s. f.) 90

90

68.0

22.0

44.0

Therefore, the angles are 68.0 , 68.0 and 44.0

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10. In the triangle KMA:

KAM

KM AM

tan 86

KM 50

KM

50 tan 86

LM

50 tan 78

90

4

86

In the triangle LMA:

LAM

90

12

78

tan 78

LM AM

LM 50

x

LM

50 tan86

KM

tan 78

479.8 m

The boat moved 479.8 m in 5 minutes. Therefore, the speed of the boat was: v

479.8 m 5 min

0.4798 km 1 h 12

5.76 kmh

1

11.

In triangle ABC: tan 31 Now, 67 y tan31

x 67

y

x and y

. In triangle BCD: tan 55

x y

y

x . tan 55

x , therefore, tan 55

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67

x tan 55

tan 31

67 tan 31 tan 55

x

67 tan 31

x tan 55

tan 31 x tan 55

tan 31

x

x

67 tan 31 tan 55

x tan 31

tan 31 tan 55 67 tan 55 tan 31

x 69.5 (to 3 s. f.)

12.

In triangle ABC: tan 35 Now, 25 tan 35 y

y tan 35

y tan 65

y tan 65

tan 35

h y

h

y tan 65 .

25 tan 35

25 tan 35 tan 65 tan 35

In triangle ADC: cos 65 x

h . In triangle ADC: tan 65 25 y

y x

25 tan 35 cos 65 tan 65 tan 35

x

y cos 65

28.7 (to 3 s. f.)

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x tan 55

13.

30 x . In triangle ABD: tan 50 y

In triangle ABC: tan 55 Now, y tan 55 tan 55 x x tan 50

tan 55 tan 50

x y

y

x tan 50

30 x 30

1 x 30

tan 55 tan 50 x 30 tan 50

x

30 tan 50 tan 55 tan 50

151 m (to 3 s. f.)

14.

DAB In triangle ABD: ADB 180 2 and . It follows that and the triangle ABD is an isosceles triangle, where BD = AD = 60.

DBA

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In triangle BCD:

x2 102

15.

602

x2

3600 100 3500

In triangle ABC: AC

102 62

In triangle ABD: cos x

6 EB . Comparing, we get BD BD

In triangle CDE: sin 2x

4 . In triangle ABC: sin 2 x a

Therefore,

4 a

4 5

a

In triangle CDE: DE

16.

DE

2

BD

2

DEA cos

x 10 35 59.2 (to 3 s. f.)

EB

2

BD

DEA

6 BD 8 10

EB

EB . BD 6 and CE = 4.

4 . 5

5 2

a2

CE

2

25 16 9 and in triangle BED:

2

9 36 45

90

64 8 . In triangle BDE: cos x

BD

45 3 5

2x cos 90

From the picture: sin x Therefore cos

DEA

2x

sin 2 x 1 , cos x 10

2

1 10

2sin x cos x 3 10

3 10

6 10

0.6

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17.

The coordinates of point B and point C satisfy the equation of the line. We can write: an bq c

B

0

bq c ,q a

bq c and ap bm c a ap c and C p, b n

0

ap c . Therefore, b

m

Area of triangle ABC:

A

1 bq c p 2 a

ap c q b

1 ap bq c 2 a

ap bq c b

ap bq c 2ab

2

2

1 BC d 2

but also A

2

2

ap bq c bq c ap c BC p q a b a2 Therefore, the area of triangle ABC can be calculated as A

1 d 2

2

ap bq c a2

ap bq c 2ab

2

ap bq c ab

2

ap bq c ab

2

1 d 2 d

d

2

ap bq c b2

ap bq c a2 ap bq c

2

ap bq c b2

.

2

2

1 a2

a2 b2 ap bq c a 2 b2

. Comparing we get ap bq c b2

2

1 b2 2

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ap bq c ab

2

ap bq c

2

d

d ap bq c

a 2 b2 ab

d ap bq c a 2 b2

ap bq c

2

ap bq c

ap bq c a 2 b2

2

ap bq c

ap bq c a 2 b 2

a 2 b2

18.

x

In triangle ABC: tan Now, d tan x

x tan

x

x

y tan x tan

d tan

x tan

tan tan tan tan

1 1 tan

d x

1 tan

y

. In triangle BCD: tan

x y

y

x tan

.

tan d tan tan

x tan

tan

d tan tan

d . Multiplying numerator and denominator by

d

x

1 tan tan

:

d cot

cot

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19.

In triangle BOS, the angle OSB 90 and SOB . OB 6400 6400 Therefore, cos 0.970 (to 3 s. f) OS 6400 200 6600 and the angle of depression 14

Exercise 7.3 1. (a) Calculating the distance from the given point to the origin of the coordinate system we have: r

122 92

144 81

225 15 y 9 r 15 r 15 y 9

Now, by definition: sin x y

cot

12 9

4 , csc 3

3 x 12 4 , cos , tan 5 r 15 5 5 r 15 5 , and sec 3 x 12 4

y x

9 12

3 , 4

(b) Calculating the distance from the given point to the origin of the coordinate system we have: r

35

2

122

1225 144 y r

Now, by definition: sin x y

cot

35 , csc 12

r y

1369

37

12 , cos 37 37 , and sec 12

x r r x

35 , tan 37 37 35

y x

12 , 35

(c) Calculating the distance from the given point to the origin of the coordinate system we have: r

12

1

2

1 1

2

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y r

Now, by definition: sin tan

y x

1 1

r x

and sec

x y

1 , cot

2 1

1 2

2 , cos 2

1 1

x r r y

1 , csc

1 2 2 1

2 , 2 2,

2

(d) Calculating the distance from the given point to the origin of the coordinate system we have: r

75

2

5

2

75 25

y r

Now, by definition: sin

tan and sec

2. (a)

(b)

y x

5 75 r x

5

5 10

10 5 3

10

1 , cos 2

3 , cot 3

5 3

10 75

100

x r

x y

2 3

75 10

75 5

5 3 10

5 3 5

3 , csc

1 cos120

2 1

3 , 2

r y

10 5

2 3 3

3 2

sin120

sin 180

60

sin 60

cos120

cos 180

60

cos 60

csc120

1 sin120

2 3

2 3 , sec120 3

tan120

tan 180

60

tan 60

3

cot120

tan 180

60

cot 60

3 3

sin135

sin 180

45

cos135

cos 180

45

csc135

1 sin135

2 2

tan135

tan 180

45

tan 45

1

cot135

cot 180

45

cot 45

1

1 2

2 2

sin 45

2 2

cos 45

2 2 2

2

2 , sec135

1 cos135

2 2

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2

2,

(c)

(d)

sin 330

y r

y r

cos330

x r

csc330

1 sin 330

2

sec330

1 cos 330

2 3

sin 30

1 2

3 2

cos 30

2 3 3

tan 330

y x

y x

tan 30

3 3

cot 330

x y

x y

cot 30

3

sin 270

y y

1

cos 270

x y

0 y

0,

csc 270

1 sin 270

1

sec 270

1 cos 270

undefined

tan 270

y x

cot 270

0 y

y 0

undefined

0

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(e)

(f) sin

5 4

csc

5 4

tan

5 4

(g) sin

sec

6

sin

4

1 5 sin 4 tan

sin 30

1 6

(h) sin

7 6

csc

7 6

sin

cos

4

2 , sec

tan

4

4

1 , cos 2 6

5 4

y r

y r

sin 60

3 2

cos 240

x r

x r

cos 60

1 2

csc 240

1 sin 240

2 3 3

sec 240

1 cos 240

2

tan 240

y x

y x

tan 60

3

cot 240

x y

x y

cot 60

3 3

2 5 , cos 2 4

cos

1 5 cos 4 1 , cot

cos30

2 3 , tan 3 6

sin 240

5 4

4

2 , 2

4

2,

cot

3 , csc 6 2 3 , cot 6 3

tan 30

cos

cot

4

1 sin

4

1

2, 6

cot 30

3

6

sin

1 7 sin 6

sin

6

1 sin

6

7 1 , cos 2 6

2 , sec 6

7 6

cos

1 7 cos 6

cos

6 1 cos

6

3 , 2

2 3 , 3 6

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tan

(i)

(j)

7 6

tan

6

tan

7 3 , cot 6 3

6

cot

cot

6

sin

60

y r

y r

cos

60

x r

csc

60

1 sin 60

sec

60

tan

cot

cos 60

1

sin 60

6

3 3 2

1 2

2 3 3 2

cos

60

60

y x

y x

tan 60

3

60

x y

x y

cot 60

3 3

The distance of the point 0, y to the origin of the coordinate system is equal to r, therefore: sin

csc

3 2 3 2

sec

3 2

tan

3 2

y 3 0 1 , cos 0 y 2 y 1 1 3 sin 2 1 1 undefined 3 0 cos 2 y 3 0 undefined , cot 0 2 y

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0

(k)

sin

5 3

y r

y r

cos

5 3

x r

csc

5 3

sec

5 3

1 5 sin 3 1 5 cos 3

tan

5 3

y x

y x

tan

cot

5 3

x y

x y

cot

cos

3

sin

3

3 2

1 2

2 3 3 2

3

3

3

3 3

(l)

sin

210

sin150

cos

210

cos150

csc

210

1 sin 210

tan

210

tan150

cot

210

cot150

sin 180

cos 180

30

sin 30

1 2

30

cos30

210

1 cos 210

tan 180

30

tan 30

cot 180

30

cot 30

2 sec

3 2

2 3 3 3 3 3

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(m)

sin

4

y r

4

x r

cos

csc

sec

tan cot

(n) The distance of the point

y r

cos

sin

sin

4

2 2

2 2 2

2

4 1

4

4

2 2

1 4

2 2

cos

2 4

4

y x

y x

tan

4

x y

x y

cot

4 4

1 1

x, 0 to the origin of the coordinate

system is equal to x, therefore: sin

0 x

csc

1 sin

tan

0 x

x x

0 , cos

1,

undefined , sec x 0

0 , cot

1 cos

1,

undefined .

(o) The terminal side of the 4.25 angle forms an angle 4.25

2 2

4

radians with the x-axis

4

Therefore: sin 4.25

sin

4

2 , cos 4, 25 2

cos

4

2 , 2

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1

csc 4.25

sin tan 4.25

tan

2 2

cos

4 4

3.

1

2 , sec 4.25

1 , cot 4.25

cot

4

2 2

2,

4

1.

Since is an angle whose terminal side lies in the first quadrant of the coordinate system, there must be a point on the terminal side that is 17 units from the origin. Given that 8 we can assume the first coordinate of the point is cos 17 equal to 8. Then y 172 82 289 64 225 15 , so the coordinates of the point P are 8, 15 . Now: y 15 1 17 1 17 , sec , csc , sin r 17 cos 8 sin 15 x 8 y 15 , cot . tan y 15 x 8 Alternatively, we can use the Pythagorean identity: sin

1 cos2

6 and sin 0 , then the terminal side of the angle lies in the 4th quadrant 5 of the coordinate system. We can assume that the coordinates of P on the terminal side are y 6 6 61 2 and 52 6 25 36 61 , so sin 5, 6 . Then: r r 61 61

4. If tan

x r

cos 5. If sin

5 61

5 61 . 61

0 and cos

180 . Therefore, cos 1 undefined , sec 1 , tan cos180

1 sin180 1 tan180

csc cot

6. If sec

2 and

0 , then

cos180 tan180

1, 0,

undefined . 3 2

2 , then cos

1 . Therefore, 2

2

1 3 3 sin 1 cos 1 (the minus sign because the terminal 2 4 2 side of the angle is in the 4th quadrant of the coordinate system), 3 sin 1 2 2 3 1 1 3 2 3 , cot csc , tan . 1 cos sin 3 tan 3 3 3 2 2

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7. (a) The same sine ratio have angles in the 1st and 2nd quadrants. sin 30 , therefore (i) sin150 sin 180 30 30 , (ii) sin 95

sin 180

85

sin 85 , therefore

85 .

(b) The same cosine ratio have angles in the 1st and 4th quadrants. (i) cos 315

cos 360

(ii) cos 353

cos 360

45

cos 45 , therefore

7

45 ,

cos 7 , therefore

7 .

(c) The same tangent ratio have angles in the 1st and 3rd quadrants. tan , we have: Using the identity tan 180 (i) tan 240

tan 180

60

tan 60 , therefore

(ii) tan 200

tan 180

20

tan 20 , therefore

60 , 20 .

1 3 6 4 sin 60 12sin 60 12 6 3 2 2 1 (b) Area = 8 23 sin105 92sin105 88.9 (to 3 s.f.) 2

8. (a) Area =

(c) Area =

1 30 90 sin 45 2

9. Using the formula Area 43

1350sin 45

1 AB 2

1350

2 2

675 2

AC sin A we have

1 12 15 sin A 2

43 90sin A

sin A

43 90

A

28.5 (to 3 s.f.)

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10. 1 2 r , where is in 2 radians. The degree measure of 120 is equivalent 120 2 2 to . Therefore, 360 3 1 225 2 2 As 15 236 cm 2 (to 3 s. f.) 2 3 3

(a) Area of sector As

(b) Area of triangle AOB: 1 1 2 2 At OA sin120 15 sin 60 2 2 1 3 225 225 3 97.4 cm2 (to 3 s.f.) 2 2 4 11. (a) Area of the shaded region A A

1 2 r 2 3

A

1 10 2

1 2 r sin 2 3

2

3

3 2

Asector

1 2 r 2 3

Atriangle sin

3

9.06 cm2 (to 3 s.f.)

(b) First we need to convert measure of the central angle to radians: 135 is equivalent to 135 3 2 . Now, 360 4 1 2 3 1 2 3 A Asector Atriangle r r sin 2 4 2 4 1 2 3 3 r sin 2 4 4

1 12 2

2

3 4

2 2

119 cm2 (to 3 s. f.)

12. Let DE = h. Area of the parallelogram Ap ah . In the triangle AED: h sin h b sin . Therefore, b Ap ab sin .

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13.

The triangle ABC is equilateral, so ABC

CBD 180

ABC 180

60

60 and

120 .

In triangle BCD, angles BCD and BDC are 30 each. So, triangle ACD is right-angled and hence y

2 x cos30 3x

3 2

x 3

Alternatively, in triangle BCD by the cosine rule: y2

x2

x2 2 x x cos CBD 2x2 1 cos120

2 x 2 1 cos 60

Therefore, y

3x 2

2x2 1

1 2

2 x2

3 2

2x 2 1 cos 180

60

3x 2

x 3

14. Area of the triangle FGH is the sum of areas of triangles FGJ and JGH: AFGH AFGJ AJGH 1 1 1 1 1 AFGJ hx sin , AJGH xf sin and AFGH hx sin xf sin x h 2 2 2 2 2 1 1 hf sin 2 . Comparing: x h 2 2 hf sin 2 2hf sin cos 2hf cos h f sin h f sin h f

But also AFGH x

2x2 1 cos 60

f sin

1 hf sin 2 2

f sin

and

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15. In the triangle OAB by the cosine rule:

s2

r 2 r 2 2 r r cos

360 n

2r 2 1 cos

360 n

Using the formula for the cosine of a double angle: cos

360 n

s2

2r 2 1 1 2sin 2

Therefore, s

1 2sin 2

2r sin

180 n 180 n

, we get 4r 2 sin 2

180 n

180 n

1 6 8 sin x 24sin x 2 (b) Since the angle x is one of the internal angles in the triangle, the domain of the function A x is the interval 0 , 180 . The function describes area of the triangle,

16. (a) Area of the triangle A

so it must assume only positive values. Since its value depends on the value of sin x , the maximum of the function occurs when sin x 1 . Therefore, the range of the function is 0, 24 .

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(c) By GDC, the maximum value of the function occurs when x 90 . The coordinates of the maximum point are 90 , 24 . The triangle that corresponds to the maximum value is a right-angled triangle since sin x assumes its maximum value when x 90 .

17. (a) The length of the rod, L In the triangle AMB: sin In the triangle BCN: cos Therefore, L

3csc

AB BC BM AB BN BC 2sec

AB BC

3 sin 2 cos

3csc 2sec

(b)

(c)

In fact, L is equal to the length of the hypotenuse of the large right-angled triangle. This hypotenuse has to pass through the fixed point B. The minimum length of AC of 7.02 m (3 s.f.) 0.8527 happens when

3 , as gets smaller than 0.8527, AB the length of AB will get larger and tends to infinity as approaches 0. Similarly, when 2 approaches 90, cos will approach 0 making the length of BC infinite. Thus, the BC rod can have any length as long as it is not longer than 7.02 because then, it will not fit around that turn.

Since sin

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18. From the diagram it follows, that AC r AC

In the triangle ACD: r 383500 r r

AB r 383500 r

sin 0.2591 . We have:

sin 0.2591

.

383500 r sin 0.2591

r r sin 0.2591

383500sin 0.2591

Simplifying: r 1 sin 0.2591

383500sin 0.2591

r

sin 0.2591 383500 1 sin 0.2591

r 1740 km 19. (a) sin

x 1 cos

sec

(b) tan

sin 2 sin 2

1 sin 2

cos 1

, 1 x 1. 1 x2 sin y cos cos

y

sin 2 y2 1

1 y2

1

y2 1

20. In triangle OAP: cos In triangle OPB: tan In triangle OPC: OCP OP OC

1 OC

sin y

. Since sin 2

cos 2

1, then

1 . We have:

y

sin

sin

1 x2 . Therefore,

OC

In triangle OPB: cos

y2 1 2 sin y2

1

y2

sin 2

y2 1

and

y y2 1 . y2 1

OA OP PB OP

OA 1 PB 1

and cot 1 sin OP OB

OA PB CP OP

CP 1

OB

1 cos

CP . Also

csc 1 OB

sec

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21. (a) Let A 1, 4 and B 1, 2 . Then the gradient of the line passing through the points A y A yB 2 4 2 and B, m 1 . Therefore, the angle that the line makes with the xA xB 1 1 2 tan 1 1 45

positive direction of the x-axis,

(b) Let A 3, 1 and B 6, 5 . Then the gradient of the line passing through the points A yA yB 5 1 6 2 and B, m . Therefore, the angle that the line makes with xA xB 6 3 9 3 the positive direction of the x-axis, 2 tan 1 33.7 (or 180 33.7 146 ) (to 3 s.f.) . 3 (c) Let A 2,

1 and B 2

points A and B, m

4, 10 . Then the gradient of the line passing through the

yA xA

yB xB

1 2 4 2

21 2 6

10

21 12

tan

makes with the positive direction of the x-axis, 22. (a) If y

2 x , then the gradient m1

2

y x , then the gradient m2 1 angle between the lines: tan

1

m1

tan

1

m2

between the lines is 180

(b) If y if y

tan

1

tan

2

108.4

1

7 4

tan

2

1

tan

tan

1

1

60.3 (to 3 s.f.) tan 1

1

2 . Likewise, if

45 . Then the

1

108.4 . Therefore, the acute angle

71.6

3x 5 , then the gradient m1

2 x , then the gradient m2

7 . Therefore, the angle that the line 4

3

2

tan

tan

2

3

tan tan

1

1

3 . Likewise,

2 . Then the angle

between the lines: tan

1

m1

tan

1

m2

between the lines is 180

tan

135

1

3

tan

1

2

135 . Therefore, the acute angle

45

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Exercise 7.4 1. (a) Three angles are known infinite number of triangles. (b) Two sides and their included angle are known

(c) Since AB

AC sin 30

one unique triangle:

one right-angled triangle:

(d) One side and two angles one unique triangle:

(e) Since BC 7, AB sin 25 two triangles possible:

12sin 25

5.07 and AB sin 25

BC

AB , then there are

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(f) In the triangle ABC: BC AC that satisfies given conditions:

AB and AC BC

AB . There is one unique triangle

2. (a)

By the sine rule: ACB 180 AB sin115

BC sin 37

28

AC sin 28

AC sin 28

37

115

AB

14sin115 sin 28

BC

AC sin 37 sin 28

14sin 37 sin 28

17.9

27.0

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(b)

BAC 180 AB sin 47 BC sin 65

68

47

AC sin 68 AC sin 68

65 . By the sine rule: AC sin 47 sin 68 AC sin 65 sin 68

AB

BC

23sin 47 sin 68 23sin 65 sin 68

18.1

22.5

(c)

By the cosine rule: AB AB

68

2

59

By the sine rule: 59sin 71 74.1 Then BAC 180 sin B

2

2

BC

2

AC

2 68 59 cos 71

AC sin B

AB sin 71

0.7528 71

B

48.8

sin B

2

2 BC AC cos C

74.1 cm (to 3 s. f.) AC sin 71 , so AB

48.8 (or 131.2 which is impossible since C 60.2 (to 3 s. f.)

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71 ) .

(d)

By the cosine rule:

AB

2

cos C

BC

2

42

AC 2

2

2 BC AC cos C

2

37 26 2 42 37

BC

cos C

2

2

AC AB 2 BC AC

2

2

0.7905 . Therefore, ACB 37.8 (to 3 s.f.)

By the sine rule:

AB AC AC sin C sin B AB sin C sin B 60.6 37.8 60.6 and CAB 180

37sin 37.5 26

Then ABC

81.6 (to 3 s.f.)

0.8715 .

(e)

By the sine rule:

AC sin ABC

BC sin BAC

sin BAC

BC sin ABC AC

34sin 43 28

Then BAC 55.9 or BAC 124.1 Since 124.1 43 167.1 180 , there are two cases.

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0.8281

Case 1: Let BAC

55.9 . Then ACB 180

55.9

43

81.1 and by the sine

AB AC AC sin ACB 28sin 81.1 AB sin 43 sin ABC sin ACB sin ABC AB 40.6, BAC 55.9 , ACB 81.1 (to 3 s.f.)

rule:

Case 2: Let BAC 124.1 . Then ACB 180

124.1

40.6 . Therefore,

43

12.9 and by the sine

AB AC AC sin ACB 28sin12.9 AB sin 43 sin ABC sin ACB sin ABC AB 9.17, BAC 124.1 , ACB 12.9 (to 3 s.f.)

9.17 . Therefore,

rule:

(f)

By the sine rule:

BC sin BAC

AC sin ABC

sin ABC

AC sin BAC BC

0.55sin 62 0.51

0.9522

Then ABC 72.2 or ABC 107.8 Since 107.8 62 169.8 180 , there are two cases. Case 1: Let ABC rule:

AB sin ACB

72.2 . Then ACB 180

BC sin BAC

AB

62

BC sin ACB sin BAC

72.2

45.8 and by the sine

0.51sin 45.8 sin 62

0.414

Therefore, AB 0.414, ABC 72.2 , ACB 45.8 (to 3 s.f.) Case 2: Let ABC 107.8 . Then ACB 180 rule: AB

BC sin ACB sin BAC

0.51sin10.2 sin 62

107.8

62

10.2 and by the sine

0.102

Therefore, AB 0.102, ABC 107.8 , ACB 10.2 (to 3 s.f.)

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3.

Since DAB BCD 37 , then ABC ADC 143 In the triangle ABD, by the cosine rule: BD

2

AB

Now, BD

2

AD

18

2

2

14

2 AB AD cos DAB 2

2 18 14 cos37

10.8 cm (to 3 s.f.)

In the triangle ABC, by the cosine rule: AC

2

AC

AB

18

2

2

BC

14

2

2

2 AB BC cos DAB . Then

2 18 14 cos143

30.4 cm (to 3 s.f.)

4. By the cosine rule:

AB

2

AC

cos ACB

cos ACB

2

AC

8

2

BC 2

2

2 AC BC cos ACB 2

BC AB 2 AC BC 2

8 10 2 8 8

2

2

0.21875

ACB 77.4 Then CAB CBA

1 180 2

ACB

1 180 2

77.4

51.3

Therefore CAB 51.3 , CBA 51.3 , ACB 77.4 (to 3 s.f.)

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5. By the sine rule:

DF sin DEF sin DEF DEF

EF sin EDF

24sin 43 18

sin DEF

DF sin EDF EF

0.90933

65.4 or DEF 114.6

Therefore, DFE 180 or DFE 180

43

43

65.4

114.6

71.6 22.4

6. The smallest angle lies opposite the shortest side of a triangle. Then by the cosine rule:

42

62 92 2 6 9 cos 62 92 42 2 6 9

cos

0.9352

20.7 (to 3 s.f.)

1 RP RQ sin PRQ . We know, that PQR 180 78 40 2 RP RQ RQ sin PQR 15sin 62 By the sine rule: RP sin 40 sin RPQ sin PQR sin RPQ 1 15sin 62 15 sin 78 151 cm 2 (to 3 s.f.) Now, Area 2 sin 40

7. Area

62 .

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8. (a)

Length of the perpendicular from B to AC: x 5sin 36 (i) one triangle if BC 5 or BC

2.94

2.94

(ii) two triangles if 2.94 BC 5 (iii) no triangle if BC

2.94

(b) Length of the perpendicular from B to AC: x 10sin 60 (i) one triangle if BC 10 or BC (ii) two triangles if 5 3 (iii) no triangle if BC 9. In the triangle EAD: ADE

5 3

5 3

BC 10 5 3

8

Therefore, ADC 90 8 98 and by the cosine rule: x 2 352 502 2 35 50 cos 98

x

352 502 2 35 50 cos98

In the triangle BCD: CDB 90 By the cosine rule: y 2 y

64.9 m

8

82

352 502 2 35 50 cos 82

352 502 2 35 50 cos82

56.9 m

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10.

(a) x > 0 since it is a side of a triangle, hence x + 2 is the largest side. 2

By the cosine rule: x 2 x2

4x 4

4x

x2 4x x2 2x

2 x 2 10 x

x2

0

x2

x2

2

x 2

1 2

4x 4 2x x 2

x x 5

0

x

2x x 2 cos120 . Simplifying:

5

(b) Since x 5 , then AC 3 and BC 5 . Area of the triangle: 1 At AC BC sin ACB . We have 2

At

1 5 3 sin120 2

(c) sin C

sin120

sin B x

sin C x 2

sin A x 2

sin C x 2

15 sin 180 2

15 sin 60 2

60

15 2

3 2

15 3 4

7 3 14

15 3 14

3 . By the sine rule: 2

sin B

x sin C x 2

sin A

Now, sin A sin B sin C

5

3 2 5 2

x 2 sin C x 2 3 3 14

5 3 14

3

5 3 and 14 3 2 7 3 2

3 3 14 3 3 14

5 3 14

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11. By the cosine rule:

BC

2

cos A cos A

sin A

2

AB

AB

2

AC

2

2 AB AC cos A

2

2

AC BC 2 AB AC

82 62 72 2 8 6 2

1 cos A

17 32 2

17 32

1

7 15 32

Therefore, the area of the triangle At

12. (a) If c 2 (b) If c 2

1 AB AC sin A 2

1 8 6 2

7 15 32

21 15 4

a 2 b2 , then the triangle is obtuse.

a 2 b 2 , then the triangle is acute.

(c) By the cosine rule: c 2

a2 b2

2ab cos C

cos C

a2 b2 c2 . The denominator 2ab

is positive for all a, b 0 . Now, if c 2

a 2 b2 , then the numerator a 2 b2 c2

the angle C is obtuse. If c

2

a

2

2

0 and cos C

b then the numerator a

2

b

2

c

0 . It follows that 2

0 and cos C

It follows that the angle C is acute.

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0.

13. Let DF x . By the cosine rule: 2 2 15.1 x2 19.3 2x 19.3 cos 43.6 x2 2x 19.3 cos 43.6

x2 x

19.3

2

2

15.1

0

38.6cos 43.6 x 144.48 0 38.6 cos 43.6

38.6cos 43.6 2

2

4 1 144.48

x 6.84 or x 21.1

1 WZ YZ sin 112 2 8 WZ 112 WZ 14 cm

1 WZ 20 2

14. (a) Area

4 , then cos 5

(b) If sin WY

2

WY

2

WZ 142

2

YZ

1 2

4 5

2

9 25

x2

202 2 14 20

9 x2

3 5

260

1 2

260 10 x 2 3x 2

(d) As can be seen on the diagram: XYZ

WX sin XYW

WY sin WXY

XYW

1

sin

39 26

sin XYW

WY 2

2 65 cm

WX

260 10 x 2

2 x 3 x cos120

260 10 x 2 6 x 2

3 . In triangle WYZ, by the cosine rule: 5

2 WZ YZ cos

(c) In triangle WXY, by the cosine rule: WY We have: 260

4 5

2

XY

2

2 WX

6 x 2 cos 180

260 13x 2

x2

XY cos X . 260 10 x 2

60

20

x

20

2 5 sin120 2 65

39 26

13.9

In the triangle WYZ, by the sine rule:

WZ sin WYZ

WY sin WZY

sin WYZ

2 5 cm

XYW WYZ . In triangle WXY, by the sine rule: WX sin WXY WY

WZ sin WZY WY

4 5 2 65

14

28 5 65

28 65 325

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6 x 2 cos 60

WYZ

sin

1

28 65 325

Therefore XYZ

sin

39 26

1

sin

28 65 325

1

57.9

FG FH sin sin 2 FG FH Then sin 2sin cos FG FH sin 2sin cos

15. By the sine rule:

cos

FH 2 FG

cos

15 2 12 cos

1

5 8

5 8 51.3

PS PQ

16. (a) In the triangle PQS: cos P As PR (b) QR

PS RS , we can write RS

2

RS

2

QS 2

2

In the triangle PQS: PQ

2

p2

q r cos P

PQ cos P

PR PS

r cos P

q r cos P

2

QS

p2

PS

PS

2

QS

2

QS

2

r 2 r 2 cos2 P . Therefore,

q2 2qr cos P r 2 cos2 P r 2 r 2 cos2 P q2 r 2 2qr cos P

(c) Let PQR 60 . Then, by the cosine rule, we have: q2

p2

r 2 2 pr cos 60

q2

p2

r2

q2

p2 r 2

2 pr

1 2

pr

p2

pr r 2 q 2

0

We have a quadratic equation in variable p. Solving: r 2 4 1 r 2 q2

r 2 4r 2 4q2

4q2 3r 2

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p

4q 2 3r 2 2

r

1 r 2

4q 2 3r 2

17. This can be a Paper 3 type of question. 1 (a) We can write the expression A ab sin C as 2 A ab sin C and from 2 c 2 a 2 b2 2ab cos C , we have 2ab cos C a 2 b2 c 2 , or after squaring both sides: 2

2ab cos2 C

a 2 b2 c 2

2

Also, from 2 A ab sin C follows that 4 A2 4 ab

2

4 ab

2

1 sin 2 C

a 2 b2 c 2

2

4 ab sin 2 C

4a2b2 4 4 A2 16 A2

(b) s

4a 2b2

a b c 2

2

a2 b2 c2

a 2 b2 c 2 a2 b2 c2

a b c

2

ab sin2 C . Now,

2

2

2

2s

We can write: 16 A2

16 A2 16 A2 16A2

16A2 16A2

4a 2b2

a2 b2 c2

2

2ab a 2 b 2 c 2 2ab a 2 b 2 c 2 a 2 2ab b 2 c 2 a b

2

c2

a b

a 2 2ab b 2 c 2 2

c2

a b c a b c a b c a b c a b c b c a a c b a b c

Using the fact that b c 2s a , a c

2s b , a b 2s c we can write

16 A2

2 s 2s a a 2s b b 2s c c

16 A2

2s 2s 2a 2s 2b 2s 2c

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(c) Continuing previous calculations:

16 A2

2s 2s 2a 2s 2b 2s 2c

16 A2

2s 2 s a 2 s b 2 s c

16 A2

16s s a s b s c

Then A

A2

s s a s b s c

s s a s b s c

18. (a) In the triangle ACD: DAC 30 , ADC 90

ACD 60

We can write: ACD DCE ECB ACB 360 60

90

65

ACB 360

ACB 145

By the cosine rule: AB

AB

2

AC

2

BC

2

2 AC BC cos ACB

8502 5002 2 850 500 cos145

1291.8 km

(b) The bearing from A to B equals CAD CAB . In the triangle ABC by the sine rule:

BC sin CAB

AB sin ACB

sin CAB

BC sin ACB AB

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CAB sin

BC sin ACB AB

1

sin

1

500sin145 1291.8

CAB 12.8 Therefore, the bearing from A to B is equal to 30

12.8

42.8

Chapter 7 practice questions 1.

The triangle AOB is an isosceles triangle where OA OB 5 . The shortest distance from the chord AB to the centre of the circle is equal to the length of the altitude of the triangle. 2

2

OB OC 52 32 4 . Therefore, AB = 8. Area of In the triangle BOC: BC 1 1 the triangle AOB: A AB OC 8 3 12 . But also: 2 2 1 2 25 25 24 A 5 sin AOB sin AOB . It follows that sin AOB 12 sin AOB 2 2 2 25

2. If tan sin 2

cos

2

3 sin 3 3 , then sin cos . We know that for any angle 7 cos 7 7 cos 2 1 . Substituting for sin we get:

3 cos 7

2

1

cos 2

9 cos 2 49

Since

is an angle in the right-angled triangle, then cos

sin

3 7 58 7 58

,

1

58 cos 2 49

1

cos 2

49 58

49 58

7 58 and 58

3 58 58

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Now:

sin 2

2sin cos

cos 2

cos

2

2

sin

3 58 58

7 58 58

7 58 58

2

2

3 58 58

21 and 29 2

20 29

3.

The largest angle lies opposite the longest side. By the cosine rule:

AB

2

AC

2

BC

2

2 AC BC cos ACB

AC

cos ACB

2

2

BC AB 2 AC BC

2

Using the data given: cos ACB

42 52 72 2 4 5

1 5

ACB 101.5

5 1 sin 2 A 4. If sin A and the angle A is obtuse, then cos A 13 5 12 120 Therefore, sin 2 A 2 sin A cos A 2 13 13 169

5. (a) In triangle BQP: tan PBQ (b) In triangle ABQ: BQA 180

PQ BQ

PQ

BQ tan PBQ

QBA BAQ

180

40 tan 36

30

70

1

5 13

2

29.1 m (to 3 s.f.)

80 , and by the sine

rule: AB sin BQA

BQ sin QAB

AB

BQ sin BQA sin QAB

40sin 80 sin 70

12 13

41.9 m (to 3 s.f.)

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6. In triangle ABC using the cosine rule:

cos BAC Then BAC

2

2

2

AC BC 482 322 562 2 AB AC 2 48 32 1 cos 1 86.4 (to 3 s.f.) 16

AB

1 16

7. (a) The smallest angle A is opposite the shortest side. Using the cosine rule: 82 72 52 11 cos A A 38.2 (to 3 s.f.) 2 8 7 14 1 (b) Area of the triangle = 8 7 sin A 28sin 38.2 17.3 cm 2 (to 3 s.f.) 2 8. (a) In triangle 2, by the sine rule:

ACB sin

1

AB sin ABC AC

AB AC AB sin ABC sin ACB AC sin ACB sin ABC 20sin 50 sin 1 64.3 or ACB 116 17

From the picture it follows that ACB 116 (b) In triangle 1, ACB Area =

1 AB 2

64.3 and BAC 180

AC sin BAC

50

1 20 17 sin 65.7 2

64.3

65.7 . Therefore,

155 cm 2 (to 3 s.f.)

9.

After 2.5 hours boat A will be 20 kmh 1 2.5 h 50 km away from point P. Boat B will be 32 kmh 1 2.5 h 80 km away from the point P. The distance between the boats A and B: By the Cosine Rule in triangle APB: AB

PA

2

PB

502 802

2

2 PA PB cos APB

2 50 80 cos 70

78.5 km

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KL JL sin KJL sin JKL 25sin 51 sin 1 31 38

JKL sin

1

JL sin KJL KL

11. In triangle ABC: BAC 180 (a) By the sine rule:

ABC ACB

AB sin ACB

BC sin BAC

1 AB BC sin ABC 2

(b) Area A

180

AB

JL sin KJL . KL

sin JKL

10. By the sine rule in triangle JKL:

60

40

BC sin ACB sin BAC

1 3.26 5 sin 60 2

80

5sin 40 sin 80

3.26 cm

7.07 cm 2 (to 3 s.f.)

12. Let a denote the length of the side edge of the cube. The diagonal of the cube has length d

a2 a 2 a2

3a2

a 3

Applying the cosine rule in triangle ABC we

BC

have: cos ACB

a 3 2 2

2

a 3 2

a 3 2

Then ACB

cos

2

2

3 2 a 4

3 2 a a2 4 3 2 a 2

AC AB 2 BC AC

2

a

a 3 2 1

2

1 3

2

1 2 a 2 3 2 a 2

1 3

70.5 (to 3 s.f.)

Alternatively, look at the rectangle formed by AB and the opposite edge. Two sides are a each, while the other sides are a 2 each. Thus the area of the rectangle is a 2 2. 1 Area of triangle ABC is therefore a 2 2. But the area of the triangle is also 4 1a 3 a 3 sin ACB. Equating the two quantities will give us 2 2 2 2 sin ACB ACB 70.5 3

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13. (a)

BC

AB

2

2

AC

2 AB AC cos BAC

652 1042 2 65 104 cos 60 1 AC 2

(b) Area A

1 104 65 sin 60 2

AB sin BAC

1 AD AB sin BAD 2

(c) (i) A1 (ii) A2

1 AC 2

(iii) A

A1

1 x 65 sin 30 2

65 x 4

1690 3

1690 3 m 2

65 1 x 2 2

1 104 x sin 30 2

AD sin CAD

A2

91 m

52 x

1 2

169 x 1690 3 4

26 x

65 x 4 26 x x

(d) (i) ADC and ADB are supplementary angles. Therefore, ADC 180 sin ADC sin 180

ADB

40 3

ADB and

sin ADB

(ii) By the sine rule:

BD sin DAB

AB sin ADB

BD

AB sin DAB sin ADB

DC sin CAD

AC sin ADC

DC

AC sin CAD sin ADC

AB 2sin ADB AC 2sin ADC

BD DC

AB (since sin ADC AC

AB sin 30 sin ADB

AC sin 30 sin ADC

AB and 2sin ADB

AC . The ratio 2sin ADC

sin ADB ). Therefore,

BD DC

65 104

Note: This is a known result from geometry that the angle bisector divides the opposite side of an angle in the ratio of its adjacent sides. 14. (a) By the sine rule: x 2 sin 45 x 2 x

2

2 2 2 1

x

QR sin RPQ

PR sin PQR

x sin 30

x 2

x

2 1

2 2

2 1

2 1

2 1

2 2

x 2 sin 30

x sin 45

1 x 2

2 2 4 2 2

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5 8

(b) Angle PRQ 180

RPQ PQR

180

Since PRQ 180

30

45 , then

sin PRQ sin 30

45

1 2 2 2

3 2

30

2 2

105 .

45

2

6 4

Area of triangle PQR: A

1 x x 2 sin PQR 2 1 2 2 1 2 2 1 4 6 4 2 6 3 2

Alternatively, if x

1 4 2 2 2 2 2 2 1 6 2 2 2 2 2 6

2 4 2 2

6

2

1 4 3 2 2

6

2 6 2 2 3 3 3

4 2 2 , then the height of the triangle is half of that 2

The left part of the base is x cos30

2

6

4 2 2

3 2

2 2

2

2 3

2

6 and the right part is

2.

Thus, the area of the triangle is

1 2 2

6 2 3

3 2 2

2 6 3 3

15. Relative position of the two circles is shown on the following diagram:

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2

For better clarity, the diagram can be split in two parts:

The shaded area to be calculated is equal to the sum of areas of the two segments shown. Area of the larger segment: A1 Similarly, A2

AABD

AABC

A ABC , where AABC is the area of the sector ABC.

A ABD , where AABD is the area of the sector ABD.

In triangle ABC:

AC

cos ACB

A ABC

2

BC AB 2 AC BC

1 AC 2

Now, AABC

A1

2

2

ACB

2

1 7 2

1 2 AC sin ACB 2

7 2 7 2 92 2 7 7 2

1.396

17 98

ACB cos

1

17 98

1.396 rad

34.2 cm 2

1 2 7 sin 1.396 2

24.1 cm 2

34.2 cm2 24.1 cm2 10.1 cm2

In triangle ABD:

AD

cos ADB

A ABD

2

BD AB 2 AD BD

1 AD 2

Now, AABD

A2

2

2

ADB

2

1 5 2

1 2 AD sin ADB 2

52 52 92 2 5 5 2

2.240

31 50

ADB cos

1

31 50

28 cm 2

1 2 5 sin 2.240 2

9.80 cm 2

28 cm2 9.80 cm2 18.2 cm2

Therefore, the shaded area A A1

A2 10.1 cm2 18.2 cm2

28.3 cm2

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2.240 rad

16. (a) Let JLK . Then thus 180 60

180 . Both and 0 120

60 120

must be greater than 0,

(b) Area of triangle JKL: A

1 KL JL sin 180 2

1 KL 30 sin 2

60

60

15 KL sin

60

By the sine rule:

KL sin KJL

JL sin JKL

and A 15 20 3sin

KL

JL sin KJL sin JKL

sin

60

30sin sin 60

30sin 3 2

300 3 sin sin

20 3 sin

60

(c) By GDC:

The maximum area occurs for

17. (a) In triangle BMC: BM

2

Area of triangle ABC: A

BC

60

2

CM

1 AC BM 2

2

172 152 1 30 8 2

(b) Area of triangle ABC can be also written as 1 1 A ABC AB BC sin ABC 17 17 sin ABC 2 2 Comparing both formulas:

ABC

289 sin ABC 120 2

64

BM

8 cm

120 cm 2

289 sin ABC 2

sin ABC

240 . It follows that 289

56.1 or ABC 123.8 . Since the angle ABC is obtuse, then ABC 123.8

or, in radians, ABC 123.8

180

(c) The area of the shaded region R: 1 1 2 2 A AM AB ABC 2 2 (to 3 s.f.)

2.16 (to 3 s.f.)

A ABC

1 152 2

1 2 17 2.16 2

120

161 cm 2

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18. (a) By the cosine rule: L2 12 12 2 1 1 cos 2 2 cos It follows that L (b) Using the formula cos 2 1 2 sin 2

cos

1 2sin 2 ,

2sin 2

2

2

(c) As can be seen in the diagram, We know that L

L

2 2 cos

2 2 1 2sin 2

19. By the sine rule:

b sin 2

a sin

2

1 cos and sin

. Then

1 cos 2

2

is an acute angle. Therefore, sin

2

1 cos 2

.

a

cos

b 2a

. Using the result from (b):

2 2 4sin 2

2

.

2 2cos

2

b 2 sin cos

4sin 2 a sin

2sin

2

b 2 cos

2

20. In triangle ABC, we use the cosine rule.

AC

AC

2

AB

2

BC

2

2 AB BC cos ABC

552 112 2 55 11 cos97

In triangle ABC by the sine rule:

57.4 cm

BC sin CAB

AC sin ABC

sin CAB

BC sin ABC AC

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CAB sin

BC sin ABC AC

1

Therefore, DAC

sin

1

11sin 97 57.4

DAB CAB 76

11.0

In triangle ACD by the cosine rule: x2 x

532

57.4

2

In triangle ABD: tan ADB

65.0

AD

2 53 57.4 cos 65

21. In triangle ABC: tan ACB

11.0

AB BC AB BD

2

AC

2

2 AD AC cos DAC

59.5 cm

AB tan ACB AB BD tan ADB

12 tan 45 12 tan 60

BC

12 cm 12 3

4 3 cm

In triangle BDC applying the cosine rule: cos CBD sin CBD

BC

2

2

BD CD 2 BC BD

1 cos2 CBD

1

2

122

4 3

2

102

23 3 72

2 12 4 3

529 1728

0.8330

Now: Area of triangle ABC: A1

1 BC 2

AB

1 12 12 2

Area of triangle ABD: A2

1 BD 2

AB

1 4 3 12 2

Area of triangle BDC: A3

1 BD BC sin CBD 2

72 cm 2 24 3 cm 2

1 4 3 12 0.8330 2

34.6 cm 2

(to 3 s.f.) In triangle ACD: AC 12 2 cm, AD

12 sin 60

12 3 2

24 3

8 3 cm, CD 10 cm

Then, cos CAD

AC

2

2

AD CD 2 AC AD

Area of triangle ACD: A4

1 AC 2

2

12 2

2

8 3

2

2 12 2 8 3 AD sin CAD

102

0.8080

CAD 36.1

1 12 2 8 3 sin 36.1 2

69.3 cm 2

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22. Calculate lengths of the segments EF, DF and DE. These segments are diagonals of the respective rectangles, so: EF

62 32

45

DF

42 32

25 5 cm

DE

62 42

52

3 5 cm

2 13 cm

Using the cosine rule in the triangle DEF: DE

cos DEF

2

Therefore, DEF

2

EF DF 2 DE EF

cos

6 65 65

1

2

52 45 25 2 2 13 3 5

6 65 65

41.9 (to 3 s.f.)

23.

In triangle ACD: tan 58

x AC

AC

x tan 58

In triangle BCD: tan 27

x BC

BC

x tan 27

Triangle ABC is a right-angled triangle, so AC

2

x tan 58 x2 tan 2 58

AB

2

2

80 6400

BC 2

2

x tan 27

2

x2 tan 2 27

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x2 tan 2 27 x2

x

x2 tan 2 58

tan 2 58 tan 2 27

6400

tan 2 27 2 x tan 2 58

6400

6400 tan 2 27 tan 2 58 tan 2 58 tan 2 27

80 tan 27 tan 58 tan 2 58

tan 2 27

43.0 m

24.

The angle between two adjacent lateral faces is the angle between BN and DN (marked red in the diagram above). Method 1 We will not give details of some obvious calculations. Remember that ABCD is a square. 102 4 2 In triangle KLB: KB These are the edges of the pyramid.

2

132

KC

In triangle KBC and using the law of cosines: KB 2

BC 2

KC 2 2 BC KC cos BCK 132 64 132 2 8 2 cos BCK BCK 69.63 33

132 cos BCK

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In triangle BCN, BN

8sin BCN

8sin 69.63 7.5 DN

In triangle DBN and using the law of cosines: cos N

2

7.52 7.52 8 2 2 7.5 7.5

N

97.9

Method 2 1 AB KM 2

Area of a lateral face AL

In the right-angled triangle KLM: KM

2

KL

Therefore, AL But also, AL

LM

1 BN 2

KL

1 8 2

102

1 BN 2

KC

In triangle ABC: AC AL

2

KL

2

1 2 8 4 1 BN 2

8 2 and LC LC

2

2

1 BC 2

2

KL

2

1 BC 4

2

8 29 KL 1 AC 2

1 BN 2

102

2

LC

2

1 8 2 2

4 2

2

4 2 . Then

1 BN 2

132

BN

33

Comparing:

BN

33 8 29

BN

In triangle BDN: cos BND

Then BND

cos

1

4 29

8 29 33

BN

DN

2

2

DN BD 2 BN DN

2

8 29 2 33 2

2

8 29 33

8 2

2

2

97.9 (to 3 s.f.)

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4 29

25. (a) General equation of the straight line is L1 : y ax b . Since the line L1 passes through the origin, then b = 0 and L1 : y ax . The line L1 makes an angle of 30 with the 3 . Therefore, the equation of the line L1 : y 3 1 (b) The equation of the line L2 : x 2 y 6 can be written as L2 : y x 3. 2 Let the angle between the lines L1 and L2 be . Then:

positive x-axis, so a

tan

1

1 2

tan

3 x. 3

tan 30

1

3 3

26.6

30

56.6

56.6 (to 3 s.f.)

26.

In triangle PRQ: PR

PQ

2

QR

2

2 PQ QR cos PQR

150 2

2752

2 150 275 cos 75

277 km

PR QR . Then: sin PQR sin RPQ 275 275sin 75 sin 55 x sin 55 x 277

By the sine rule: 277 sin 75

It follows that sin 55

x

0.9590

55

x

sin

1

0.9590

x

sin

1

0.9590

55

73.5

55

The length of the flight from R to P is 277 km, the bearing is 18.5

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18.5

Exercise 8.1 1.

Since

4

(b) (c)

Since

7

6

6 0 i

(d)

49

(e)

81 9i

(a) (b)

3i 2 4i

(f)

1 i 2 3i

(h) (i) (j)

(k) (l) (m) (n) (o) (p) (q) (r)

2 2 3 3

5 9i 6 8i 15i 20 14 23i 6 21i 8i 28 34 13i 2 2i 3i 3 5 i

5i 5i 2i 2i

16 4 4 9

11i 16 11i 16 11 i 25 29 29 29 7i 4 7i 4 7 i 4 13 13 13

1 25 4 36 1 4 3i i 2 6 1 2 3i i 2 6

8 i i 3 7i 12 4 10i 13 5 12i 25 144 12i 3 4i 9 16 2 9i

7i

2 7i

2 7i 3 4i 2i 5i i 2i

7

5 5 i 0 i 4 4

3 4i 2 5i

3 2 2 3 1 4 9 2 3 1 3

7

0 9i

(e)

(g)

5 2i

7 0 i

1 i 3 4i 2 5i

(c) (d)

4

7i , 7

7

25 16

(f) 2.

2i , 5

(a)

1 18i 4 9

1 18i 13

1 18 i 13 13

7 3i

65 156i 169 48 36i 25

5 12i 13 48 25

5 12 i 13 13

36 i 25

68

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(39 52i)(24 10i) 416 1638i 8 63 i 242 102 676 13 26 1 7 4i 7 4i 7 4 i 7 4i 49 16 65 65 65 1 5 12i 5 12i 5 12 i 5 12i 25 144 169 169 169 3 2 3(3 4i) 2 6 8i 9 12i

(s) (t) (u)

12 16i 3 4i 6 8i 9 16 36 64 25 100 9 12i 3 4i 12 8i 12 8 i 25 25 25 25 25 54 19i 5 12i 54 19i 498 553i 498 553 i 5 12i 25 144 169 169 169 5 12i 3 4i 5 12i 33 56i 33 56 i 3 4i 9 16 25 25 25

(v)

(w) (x)

3.

7 i 17 19i 2 3i 13 Alternatively, we can find z by solving a system of equations: 2 3i a bi 7 i 2a 3b 3a 2b i 7 i 2 3i z 7 i

z

17 13 19 3a 2b 1 b 13 However, it is easier to find z using division, as we did.

2a 3b 7

4.

2x

y

a

xy 2 i 1 3i , using the fact that complex numbers are equal if their real parts as well

as imaginary parts are equal, we get a system of two equations: 2x

y 1

xy 2 3

From the first equation: y 2 x 1 ; hence: x 2 x 1 and y1

2 3

2x2

x 1 0

x1

1 , x2 1 2

2, y2 1

The solutions are: x1

1 , y1 2

2 or x2 1, y2 1

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5.

(a)

1

3i

3

13

3 12

3i 3 1

1 3 3i 3 3

3i

1

3 3

1 9 3 3i 3 3i

(b)

2

3

3i

i

8

First, write the number in the form: 1

6n

3i

Now, use the fact established in (a) that 1 1

(c)

3i

6n

1

n

3 2

3i

8

2 n

1 3

3i

8

3i

3 2

n

8 to carry out the calculations:

2n

Use the result in (b): 1

3i

48

1

68

3i

82 8

816

Caution: you may be tempted to use a GDC to evaluate high powers of complex numbers. Some GDCs do not have the capacity to perform such an operation. 6.

2 i 2

(a)

2

2

2

2

2 4i 2

(b)

2

1

2i

2i

4i

First, write the number in the form:

(c)

2 i 2

46

2 i 2

2 i 2

2 i 2 16

Let z

2 i 2

2

2

2

n

4i to carry out the requested

k

2

2

4i

2

k

16

k

Use the result in (b): 2 i 2

7.

4k

4k

2 i 2

Now, use the fact established in (a) that calculations:

2

x

yi . Then z 4i

Thus, z 4i 2

x2

y 4

3x 2

3 y 2 12

Therefore,

4 x2

x2

y 1

y 4 i and z i

y 4

2

x2

y2

4

y2

2

z

2 i 2

x2

4 11

2 i 2

2

1611 4i 4 23 i 246 i

4i

x

x2

2z i

11

44 2

2

2 x2

x

y 1

y 1i 2

y 2 8 y 16 4 x 2

4 y2 8y 4

2

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8.

First, simplify the fraction and then add 3: 2i

2 i 2

4i 2 2 4 2

2 i 2 2 i 2 z

9.

2i

3

2

9

2 3

3

x iy 4 7i

3 2i

2i

2 3

2i

9

2

2 i 3

3

x iy

3 2i 4 7i

Now simplify the RHS: 3 2i 4 7i 4 7i 4 7i

12 21i 8i 14 16 49 2 ,y 65

Therefore, x 10.

i z 1

3z 2

2 29i 65

29 65

3z iz

2 i

3 i z 2 i 3 i

Now, divide by 3 i and simplify: z

11.

12.

2 i z 1 2i

2 3i

Thus, z

3 2i

x iy

2

2 3i 1 2i

z 2

2 i

8 i 2 i

2 i 1 2

1 i 2 3 2i

5 12i x2

3 4i

y2

2xyi 3 4i

Two complex numbers are equal if their real parts are equal and imaginary parts are equal. We get a system of equations to be solved: x2

y2

2 xy

3 4

Solve for y in the second equation and substitute in the first: y

2 x

x

2

2 x

2

3

x2

4 x2

Therefore, we have two solutions: x, y

3

x 4 3x 2

4 0

2, 1 and x, y

x12

4, x22

1

2, 1

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13.

x2

(a)

y2

2 xyi

x2

8 6i

y2

8

2 xy 6

Solve for y in the second equation and substitute in the first: 2

3 8 x 4 8x 2 9 0 x12 1, x22 x So, either x 1 and y 3 , or x 1 and y 3 x2

(b)

9

Solving a quadratic equation with complex coefficients:

z

2

1 i

1 i

1 i

2 8 6i

4 2 2i

1 i

1 2i 1 8 8i 2

2 Since

1 i

z2

14.

Let z

1 3i , then z equals either z1

8 6i 1 3i

1 i

1 3i 2

2i , or

1 i

2

x iy.

z3

27i x3

x3

3x 2 yi 3xy 2

3 xy 2

i 3x 2 y

iy 3

27i

y3

27i

x x2

3 y2

0

3x 2 y

y3

Equate real and imaginary parts: x 3 3xy 2 0 3 x 2 y y 3 27

From the first equation, x = 0, or x 2

27

3y2

If we substitute these values into the second equation we get: x 0 y 3 27 y 3 z 3i x2

3 y2

Thus, x z

2

3 3 2

3 3 y2 y

3 3 2

y3

2

3 i or z 2

x 3 3 2

27

8 y3

27

y3

27 8

y

3 2

3 3 , and therefore, the solutions are: 2 3 i 2

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15.

Since the polynomial has real coefficients, then

1 2i 2

Thus x x f x

1 2i 2

x

4 x2

and x

1 2i 2

1 2i 2

x2

17 4

x

17 . So, the polynomial is of the form 4 4 x3

4 x 17 x c

By comparing the constant terms, we have 17c 16.

2i is also a zero.

are factors of the polynomial, and so is their product

x

4x2

x c

1 2

51

17c

c 3

Since a polynomial has real coefficients, 3 i 2 is also a zero. Hence, using the factor theorem, we have: a x

1 2

x 1 x 3 i 2

x 3 i 2

a x2

1 x 2

1 2

x2

6 x 11

Since we need integer coefficients, we let a = 2 (or any multiple of 2). After multiplication we have: 1 1 2 x2 x x 2 6x 11 2 x2 x 1 x 2 6 x 11 2 2 2 x 4 11x3 15 x 2 17 x 11

17.

Since a polynomial has real coefficients, 1 i 3 is also a zero. Hence, using the factor theorem, we have: a x 2

2

a x2

x 1 i 3 x 1 i 3

After multiplication we have: x2

4 x 4 x2

So, we let a 1 and the polynomial is: f ( x) 18.

4 x 4 x2

2x 4 x4

x4

2x 1 3

2 x3 8 x 16

2 x3 8 x 16

Since the polynomial has real coefficients, then 5 2i is also a zero. Hence, f x

x

5 2i

x

5 2i

x2 10 x 25 4 x c

x c

To determine c, we can check for the constant term: 87 29

c

c

3

Hence, the other zeros are 5 2i and 3 19.

Since the polynomial has real coefficients, then 1 i 3 is also a zero. Hence, f x

3 x

1 i 3

x

1 i 3

x c

3 x2

2x 1 3 x c

To determine c, we can check for the constant term: 8 3 4 Hence, the other zeros are 1 i 3 and

c

c

2 3

2 3

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20.

Let z

x iy. Then, a bi

x2 x2

a bi

2

y2 y2

x x

yi yi

x x

2

2 xy 2 x y2

yi x yi x x4

x2 x2

yi yi

2 x2 y 2

y4

x2

y2

y2 y2

2 xy i x y2

4 x2 y 2

2

x4

2

2 x2 y 2 x2

y2

y4 2

1

Note: Knowledge acquired in the next section will enable you to easily show this result without resorting to lengthy calculations: z Given that the number and its conjugate have the same modulus, then a bi 1 z* 21.

(a)

Using the binomial theorem, we have:

k i

4

k4

4 k 3i 6k 2

1

4k

Therefore, the number is real if 4k 3 (b)

4k

4k k 2 1

0

0

k

0, or k

1

Using the calculation from (a), we have: k4

6k 2 1 0

k2

6

36 4 2

3 2 2

Since both numbers are positive, k

22.

1 k 4 6k 2 1 i 4k 3 4k .

i

3 2 2

iz1 2 z2 3 i 2 z1 2 i z2 7 2i Multiplying the first equation by 2 , and the second by i, and adding the resulting equations: i2 z1 4 z2 2iz1

6 2i

2i 1 z2

7i 2

5 2i z2

8 9i

Substituting z 2 in the first equation, we have: iz1 Hence, the solutions are z1 1 i and z2 23.

iz1

1 i z2

3

2 i z1 iz2

4

z2 2 2 i

8 9i 5 2i 3 i

2 i iz1

1 i

z1 1 i

2 i

Multiplying the first equation by i, and the second by 1 i , and adding the resulting equations: z1 i 1 i z2 3i 2 i 1 i z1 i 1 i z2 4 1 i z1

1 3i z1

3i 4 4i

3iz1

4 7i

z1

7 4i 3

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Substituting z1 in the second equation, we have: 2 i

7 4i iz2 3

4

iz2

i 3

4 6

z2

7 4i and z2 3

Hence, the solutions are z1

1 2i 3 1 2i 3

Exercise 8.2 1.

(a)

r

22

4

(b)

r

22

. Hence, z 3

2

Hence, z (c)

r

22

(d)

r

6

12

2 2cis

2 , tan

2 cis

2

quadrant,

(f)

(g)

4 1

, and, since the number is in the first quadrant,

2 2

2 2 , tan

7 . Hence, z 4

2

3

6

6

2

2

2

2 2 cis

2 2 , tan 11 . Hence, z 6

fourth quadrant, (e)

2 1 , and, since the number is in the first quadrant, 2

2 2 , tan

1 , and, since the number is in the fourth 7 4

2 6

1 , and, since the number is in the 3

2 2 cis

11 6

2 3 3 , and, since the number is in the fourth 2 5 5 quadrant, . Hence, z 4cis 3 3 2 3 r 3 32 3 2 , tan 1 , and, since the number is in the second 3 3 3 quadrant, . Hence, z 3 2 cis 4 4 4 is not defined , and, since the number is on the positive r 42 02 4 , tan 0

r

22

2 3

2

4 , tan

y -axis,

. Hence, z 4 cis 2 2 Note: From the geometric interpretation, we can see that the distance from the origin is 4 and the angle is

2

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(h)

r

3 3

r

12 12

Hence, z (j)

r

3

2 cis 2

15

2

3

6 , tan

1

3 3

3

, and, since the number is in the third

7 7 . Hence, z 6cis 6 6 1 2 , tan 1 , and, since the number is in the first quadrant, 1

quadrant, (i)

2

4

4

02

0 15

15 , tan

0 , and, since the number is on the negative

x-axis, . Hence, z 15cis Note: From the geometric interpretation, we can see that the distance from the origin is 15 and the angle is . (k)

1

1

4 3i

1

4 3i

5cis arctan

r

2

4 25

quadrant, (l)

r

4 3i 2

2

3 4

1 cis 5.64 5

4 3i 25 3 25 4 25

1 , tan 5 3 4

1

3 4

arctan

3 , and, since the number is in the fourth 4

5.64 . Hence, z

1 cis5.64 5

3 3i

3

quadrant, (m)

tan

i 3 3i

r

3 25

1

1

Alternatively, 4 3i

1 cis 2 5

2

2

32

3 2 , tan

3 3

1 , and, since the number is in the second

3 3 . Hence, z 3 2 cis . 4 4 0 , tan 02 0 , and, since the number is on the positive x-axis,

0 . Hence, z cis0 . Note: From the geometric interpretation, we can see that the distance from the origin is and the angle is 0. (n)

r

02 e2

e , tan

not defined, and, since the number is on the positive y-axis,

. Hence, z ecis . 2 2 Note: From the geometric interpretation, we can see that the distance from the origin is e and the angle is

2

.

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2.

(a)

z1 z2

z1 z2

(b)

cis

z1 z2

z1 z2

(c)

z1 z2

(e)

cis

z1 z2

cis

z1 z2

3

cis

3

cis

6

2 3

cis

2 3

13 12

5 12

5 12

2 3 cis 3 4

reduced to z1 z2 3 3 cis 2 4 3

cos 3 2

4 3

25 12

2 2

4 3

2 3

9 cis 2

2 2 7 12

(f)

z1 z2

z1 z2

9 8

6

3 2 2cis

2

9 8

5 4

5 3

6

3 i 2

5 6

3 2

0 1i isin

isin

12

i

3 2

2 3

1 i 2

0 1 i

1 2

i

3 i 2

2 cos

12

2isin

12

and using trigonometric identities, this can be

4

i

9 17 cis 2 12 3 4

Using trigonometric identities for cos to

3

2cis

3

6

1 2

2

3 2

3 12

1 0i 1

i sin i sin

cos

2cis

12

6

isin

1 i 2

1 i 2

isin 2

2

cos

3 2

3 2

5 6

2 3

4

5 6

6

3

cos

2

cis

isin

cos

5 6

cis

i sin

cos 2

3

cis

5 6

6

cis 2

7 6

13 12

cos

cos

6

7 6

By considering that

z1 z2

5 6

5 6

6

cis

cis

3

5 6

cis cis

z1 z2

2

2

cis

z1 z2

(d)

cis

4 3

and sin

3 4

4 3

this can be reduced

2 i 35 12

6 2 cis

6 2 cis

11 12

This can be reduced as in previous questions to: z1 z2

z1 z2

6 3 6 2

3 2 5 cis 2 4

6 3 6 i 2

5 3

3 3 3

3 2 cis 2

5 12

3 3 3 i

3 2 19 cis 2 12

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This can be reduced as in previous questions to: z1 z2

(g)

z1 z2

z1 z2

(h)

cis 135

90

cis 135

z1 z2

z1 z2

(i)

3 3 3 3 3 3 i 4 4

90

5 8

z1 z2

2 2

cis 45

3 2cis 120

3 cis 120 2

2 2

cis 225

240

3 cis 225 2

2 i 2

6cis 360

3 cis 2

240

2 i 2

2 1 i 2

2 1 i 2

6 0 i

3 4

120

6

3 3 i 4

5 3 cis 195 16

30

5 6 15 2 64

Also, 195 = 240 45, so this can be reduced to z1 z2 5 8 cis 225 3 2

z1 z2

5 3 cis 12

30

z1 z2

and this can be simplified to (j)

z1

3 2 cis 315 , z2

z1 z2

and finally,

3.

(a).

For z1 : r

6

z1 z2

300

6 2 cis 615

2

. Hence, z1

For z 2 : r

22

fourth quadrant,

45

5 6 15 2 48

12

2 3

6 2

105

3 3 3 i 3 3 3

3 3 3 4

30 )

2 , tan

2cis

6 2 cis

3 2 cis(15 ) 2

300 )

3 2 cis(45 2 3

i

60

5 6 15 2 64

2cis300 , and hence:

3 2 2cis 315

3 2 cis(315 2

5 3 cis 12

5 6 15 2 48

which can be simplified to z1 z2 z1 z2

105

i

1 3

i 3 3 3 4

, and, since the number is in the first quadrant,

. 4 , tan

5 . Hence, z2 3

2 3 3 , and, since the number is in the 2 5 4cis or 4cis . 3 3

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1 z1

1 cis 2

6

1 z2

1 cis 4

5 3

z1 z2

z1 z2

(b)

2 cis 4 6

5 3

8

For z 2 : r

48

1 z1 1 z2

4 3cis

1 2 2 1 4 3

5 3

6

For z1 : r

z2

(c)

2 4cis

1 cis 4 3 8cis

1 cis 2

cis cis

z1 z2

2 cis 6 2 3

3 2

3

,

6

; hence, z1

2 2cis

6

.

5 ; hence, 3

3,

6 5 3

3 cis 12 3 5 3

6

8 6 cis

11 6

or 8 6cis

6 cis 6

3 2

6 cis 6 2

5 3 1 3

6

,

1,

6

; hence, z1

8cis

5 ; hence, z2 4

6

6

3 2cis

5 3 or 3 2cis 4 4

6

5 4

cis

z1 z2

24 2 cis

z1 z2

8 3 2

1

2 cis 4

For z 2 : r 3 2 , tan

1

1 cis 2 2

4 3 , tan

For z1 : r 8 , tan

1 z2

6

5 or 4 3cis 3 3

8 6 cis

1 cis 8

or 8cis

3 2

2 2 , tan

z1 z2

1 z1

11 6

cis

6

2 3 cis 6 4 5 4

6

5 4

24 2 cis

4 2 cis 3

17 12

13 12

or 24 2cis

7 12

4 2 11 cis 3 12

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(d)

For z1 : r

3 , tan

For z 2 : r

2 2 , tan

1 z1 1 z2

(e)

(f)

1 3

cis

1 2 2

is not defined,

2 4 3

cis

z1 z2

2 6 cis

z1 z2

3 cis 2 2 2

4 3

2

4 3

10 , tan

For z 2 : r

2 2 , tan

1 cis 10

1 z2

1 2 2

1 2 3

4

2

6 7 cis 4 6

1,

4

2

; hence, z1

is not defined,

4 5 cis

5 cis 2 3,

2 3 , tan

4 3

.

or 2 2cis

2 3

2

6

10cis

4

; hence, z2

2 2cis

2

3 4

4 3

0,

; hence, z1

0 ; hence, z2

2cis

3

2 3cis0

3

cis

z1 z2

4 3 cis

z1 z2

2 2 3

5 6

2

For z1 : r 2 , tan

1 z2

6 cis 4

2

10 cis 4 2 2

1 cis 2

or 2 6cis

2 cis 4

z1 z2

1 z1

11 6

4

4 5 cis

For z 2 : r

2 6 cis

4

z1 z2

2 2cis

2

2

10 cis 10

cis

3cis

2 2 cis 4 3

For z1 : r

1 z1

4 ; hence, z2 3

3,

3 cis 3

; hence, z1

2

cis

3 cis 0 6

0

3

3

0

0

4 3 cis

3

3 cis 3 3

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.

4.

(a)

Letting z x2

(b)

(i)

y 1

2

x2

y 2

2

y2

2 y 1 y2

4y 4

y

1 2

The points are on the unit circle and their y-coordinates are both

1 2

1 11 , and, since the number is in the fourth quadrant, 2 6 11 (or ). Hence, arg z1 . 6 6 1 7 For z 2 : sin , and, since the number is in the third quadrant, 2 6 5 7 (or ). Hence, arg z2 . 6 6 The diagram below shows the inequality. The book answers have another version of the drawing.

(ii)

5.

yi , we have:

x

For z1 : sin

y

y

z1+z2 z1+z2

z1 z1

z2

z2

x

x

In the above diagrams, z1 is represented by the red line segment(s), and z2 by the blue line segment(s). It is obvious that the line segment which represents z1

z2 (blue + pink) is shorter

same if z1 and z 2 are on the same line (on the same side of the origin), as shown on the second diagram.

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6.

z

1 3 i 2 2

3

3

(a)

(b)

(c)

7.

3

3

z

For z1 : z1

Since z3

2

3 3i 2

3

3 3 3i 2

3

1 3

4 4 2 2 , tan z1 z2 , z3

i sin

3 9

3 1

3i

3 1

3 3 3i 2

3 2

3i 2

2 3 3

3i 2

3 3

3 i

31

1 ; hence, arg z2 6

4

3 4i 1 3

3i

; hence, arg z1

8 2 , and arg z3

4 3

3 3i

3 3i 2

9 3 3i 2 3 3 3i 2

12 4 4 , tan

4 3

6

3 3i 3 3 3i 2

3 3 3i 2 3 3 3i 2

3

3 z2 3 z2

3 cos

3 3 3i 2

3

2z 3 z2

For z 2 : z2

8.

3 3i and z 2 2

3i

6 4

12

There are several methods available. Some will give you numerical approximate answers using a GDC and others may require knowledge from Chapter 9, or reference to the matrix chapter online. y

z3

z2

x

z1

Below are two such methods. Method I We can find the side lengths and the angles of the triangle: z1 z2

2 2 3

z1 z3

2 3 4

2

2

42

32 8 3

4 3 2

2

80

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z2 z3

4 3 2

4 3 6

2

136 32 3

between sides z1 z2 and z1 z3 is:

Angle

z1 z2

cos

2

2

z1 z3

2

z2 z3

2

76.6689...

2 z1 z2 z1 z3

1 z1 z2 z1 z3 sin 2

Hence, the area is: A

18.5

Method II There is a formula for the area of a triangle using the coordinates of the vertices. If the vertices are: A x1 , y1 , B x2 , y2 , C x3 , y3 , then the area is: x1 1 x2 2 x3

A

9.

y1 1 y2 1 y3 1

Let z

x iy

(a)

z

1 2

2

1

2 3

2

1 1

41

x2

3

2

y2

3

4

3 1

3

x2

y2

22 2 3 18.5

9

The set of points is the circle with centre (0, 0) and radius 3. (b)

z*

(c)

x

(d)

x yi ; then x yi yi

x

( x 3)2

yi 8

y2

2

x

x 0 . The set of points is the y-axis.

x 4 . The set of points is the line x 4 .

2x 8

x 3

yi

2

y2

4.

The set of points is the circle with centre (3, 0) and radius 2. (e)

We can look at this from a geometric perspective. In the diagram below, z 1 is shown in red, and z 3 in blue. The points representing 1, 3, and z, form a triangle. z 1 z 3 represent the sum of two sides of a triangle. Hence in all cases except one, z 1 z 3 2. The only exception is when z lies between 1 and 3, i.e., there is no triangle. In that case z 1

z 3

2

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It is much more involved to solve this algebraically:

( x 1)2

y2

( x 3)2

y2

2

( x 1)2

y2

2

( x 3)2

y2

Square both sides and simplify: 4 x 12

4 ( x 3)2

x 3

( x 3)2

y2

y2

Since the right side of the equation is negative, then x 3 0 Square both sides again: x 3

2

( x 3)2

y2

x 3

y 0

Substitute back y = 0 into the original equation:

( x 1) 2

( x 3) 2

2

x 1

x 3

2

1 x 3

x, y , 1 x 3, y 0 ; hence, the line segment between

The set of points is (1, 0) and (3, 0). 10.

Let z

x

(a)

yi

x2

y2

3

The set of points is the disk with centre (0, 0) and radius 3. (b)

x2

y 3

2

2

The solution is the set of points outside the disk with centre (0, 3) and radius 2.

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Exercise 8.3 1.

i

2 3

(a)

z

(b)

z 3e2

(c)

z 3e0.5

z

4e

i

3 cos 2 i

2

4

7 12

2

6

i

4

31 i 0

i sin

2.

z 13e 3

(f)

z 3e

(a)

r

(b)

r 2 , tan

(c)

r

(d)

r 4 , tan

(e)

r

(f)

r 4 , tan

(g)

r 6 , tan

(h)

z

1

8

8

3

13 cos i

3e e 3

3 i

2 2 , tan

1 3

i sin 3e cos

6 i

13

3 3

i sin

2

6

1 3 i 2 2 3

1 3

6

13 13 3 i 2 2

3e

1 , the first quadrant,

, the first quadrant,

2 2 , tan

3i

7 12

2

4

2 i2 3

3

3 0 i 1

i

(e)

1 3 i 2 2

4

isin 0.5

4 cos 6

2 3

isin

isin 2

3 cos 0.5 7 12

4cis

(d)

2 3

4 cos

4

1 3 i 2 2

; hence, z

; hence, z

, the fourth quadrant,

3 , the fourth quadrant,

3

3e 2

2e 6

6

3e 3 i 2

2 2e 4

i

i

; hence, z

; hence, z

4e

3

2 2e

6

i

i

3

3 2 , tan

18

is not defined, positive y-axis, 1 3

3 3i , r 3

z 3 2e 4

1 , the second quadrant,

, the third quadrant, 18

3 2 , tan

2

i 3 ; hence, z 3 2e 4 4

; hence, z

7 ; hence, z 6

4e 2 7

6e 6

1 , the second quadrant,

i

i

3 ; hence, 4

i

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3.

(i)

r

(j)

r

(a)

r

, tan e , tan

10

10

3

6

25 cis

4

6

9

9

6 cis 9

r

2 2 , tan

2 2i

12

12

i sin

r

6 , tan 3 i 3

(f)

r

18

8

8

7

4

32i

3 i

2cis

isin

6 64

6cis

3

; hence, 2 2i 2 2 cis

; hence,

2

3 4

cis 7

17 496 2

4

1 2

i sin i

1 2

isin 0

6 cis

4

1296

3 3 ; hence, z 3 2 cis 4 4

17 496 2 cis

5 4

17 496 2 cos

3 i 3

1296 cos 0

1 , the second quadrant, 7

i

3

4

1296cis

4

3 2

2

262144

1 , the fourth quadrant,

3 2 , tan 3 3i

4

isin

6 cis 8

1

e

10 077 696

262144cis

4

262144 cos (e)

i

10 077696cis 3

3

cis 12

2

; hence, 3 3i 3

1 , the fourth quadrant,

2 2

isin

2

64 cos

3

2 cis

; hence,

6

3 , the first quadrant,

10077 696 cos (d)

32 cos

64cis

6

e e2

; hence, 1 i

4 2

e2 i )

; hence, z

2

, the fourth quadrant,

2 cis 6

r 6 , tan 3 3i 3

cis 10

1

r 2 , tan 3 i

(c)

1 , the first quadrant, 2

e0 i (

0 , hence, z

is not defined, positive y-axis,

2 , tan

1 i

(b)

0 , positive x-axis,

5 4

5 4 17496

1 i

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(g)

r

6 , tan 3 i 3

(h)

1 , the fourth quadrant, 8

6 cis

1

4.

3 i

7

7

7

6

cis

1 3

7 6

1 cis 279936

3 1 i 2 2

1 559872

7

, the first quadrant,

(a)

n

4

7

2 2 cis 7

256cis

6

r cis

2k n

n

r 8 , tan

z

8 cis 3 2

z1

2 2 cos

z2

2 2 cos

or z

n

re

7 6

6 7 6

2 2 cis

isin

2 2

6

isin

7 6

3 i

3 2

256

i

2 cis

1 2

6

128 3 128i

r cis , we use the formula developed in the

3

k

6 3 2

2 2

3 i

2k i n n

3 , the first quadrant,

2k 2

6

; hence,

6

To find the nth roots of a complex number z chapter: z

6 cis

7 7 ; hence, 3 3 3i 6cis 6 6

, the third quadrant,

3

r 2 , tan 2

3 i 3

8

1 279936 (i)

; hence,

1 cis 2 4 1296 1 1 cos 0 i sin 0 1296 1296

8

r 6 , tan 3 3 3i

4

; hence, 4 4i 3 8cis

;k

0,1

1 2 3 2

3

6 i 2 1 2

6 i 2

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(b)

r 8 , tan

z

(c)

8 cis 3 3

3

z1

2cis

z2

2cis

z3

3 , the first quadrant,

9

2cis

4

2e

i

2cis

2k 3

9

;k

; hence, 4 4i 3 8cis

9

2cis

9

4 3

13 2cis 9

0,1,2

7 9

2e

i

7 9

2e

i

13 9

0 , on the negative x-axis,

1cis

4

2k 4

cis

cos

i sin

4

; hence, 1 1cis

k ;k 2

2 2

0,1, 2,3

2 2

z1

cis

z2

cis

3 4

cos

3 4

i sin

3 4

2 2

i

2 2

z3

cis

5 4

cos

5 4

isin

5 4

2 2

i

2 2

z4

cis

7 4

cos

7 7 + i sin 4 4

4

3

9

2 3

r 1 , tan z

2k 3

3

4

4

i

2 2

i

2 2

y

z2

z1

x

z3

1

z4

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(d)

r 1 , tan

2k 6

1cis 2 6

4

z

is not defined, on the positive y-axis,

z1

cis

z2

cis

2

6

12

4

cis

5 12

2

z3

cis

3 4

2 2

z4

cis

13 12

z5

cis

17 12

z6

cis

7 4

12 6

i

y

0,1,2,3, 4,5

6

2 4

2 2

i

2

6

6

i

4 6

2 4

6

i

4 i

2

2

i

4

2 2

; hence, i 1cis

4

6

2

k ;k 3

2

2 4

2 2

z2

z3

z1 x

z4 1

z6

z5

(e)

r

324 18 , tan

3 , the third quadrant,

9 9i 3 18cis

4 3

4 18 cis 15

2k 5

z

z1 z4

5

5

5

18e

i

18e

4 15

4 i 15

; z2 6 5

5

5

18e

18e 18e

i

i

4 3

i

5

18e

4 15

2 5

22 i 15

4 ; hence, 3

arctan 3

; z5

4 2k 15 5

5

;k

18e 5

i

18e

10 15

0,1, 2,3,4

; z3

4 8 i 15 5

5

5

18e

18e

i

4 15

4 5

5

18e

i

16 15

28 i 15

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5.

(a)

z5

z5

32 0

r 32 , tan z

(b)

5

32e

32 ; hence, we have to find the fifth roots of 32.

0 2k 5 5

i

0 ; hence, 32 32ei 0

0 , the positive x-axis, 2e

z1

2ei 0

2 , z2

z8

i 0

z8

i

2k 5

2e

i

;k

2 5

0,1, 2,3, 4

, z3

2e

i

4 5

, z4

2e

i

6 5

, z5

2e

i

8 5

i ; hence, we have to find the eighth roots of

i. 3

r 1 , tan i

z z1

(c)

8

3 2 8

2k 8

1e

e

e

3 i 16

, z2

19 i 16

i 3 ; hence, i e 2 or i e 2

is not defined, the negative y-axis,

e

, z6

7 i 16

e

i

, z3

23 i 16

z5

e

z3

4 3 4i 0

3 k 16 4

;k e

11 i 16

0,1, 2,3, 4,5,6,7 , z4

27 i 16

e

i

, z8

15 16

e

,

31 i 16

, z7

e

z3

4 3 4i ; hence, we must find the third roots of

4 3 4i .

r 8 , tan

z z1

(d)

3

3

5 6 3

i

2e

5 i 18

, z2

z1

i

16e

2ei0

2e z4

r 16 , tan 4

5 2k 18 3

i

8e

2e

, the second quadrant,

2k 3

z 4 16 0

z

5

1

17 i 18

, z3

; k 2e

0,1, 2

29 i 18

16 ; hence, we have to find the fourth roots of 16.

0 , the positive x-axis,

0 2k 4 4

2 , z2

i 5 ; hence, 4 3 4i 8e 6 6

2e

i

k 2

i

2e 2

;k

0 ; hence, 16 16ei 0

0,1, 2,3

2i , z3

2ei

2 , z4

2e

i

3 2

2i

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i

2

z 5 128 128i

(e)

z5

128 128i ; hence, we must find the fifth roots of

128 128i 15

2 2 , tan

r 128 2

5

z

(f)

3 4 5

i

15 2

2k 5

3

2 e i

22 e

3 20

, z2

z1

8e

z5

8e

z6

64i 0

i

7 4

8e

z

6.

(a) (b)

i

z5

2e

i

17 12

i

cis 6

cis 4

2e

5

n

1 3

cis 2n

cos

i

k 3

12

i

i

19 20

, z4

; k 8e

i

0,1, 2,3, 4 27 20

i

7 4

2e

cis

9 3

cis

2n n

i

3 4

1

2 1

i

5

cis 4

3

i

2

cos 2

i

2

2i , z4

2e

i

13 12

,

2i cos 4

cis 7 cos 3

cis 2

2 2

2

cis 3

1

2

2

4

Re e i e

2

; hence, 64i 64e

0,1, 2,3,4,5

1

2

cis 6

Re e

,

2 2i

;k

cis 9

cis 3

(d)

2

8e

5

i

, z6

cis

cis 9

1

2e 12 , z3

cis 9

(c)

7.

2e

2e 12 , z2

, z3

3 2k 20 5

i

is not defined, the positive y-axis,

64e

z1

8e

64i ; hence, we have to find the sixth roots of 64i .

2k i 2 6 6 6

11 20

2

z6

r 64 , tan

i

3 2k 20 5

i

1

2 2

3 ; hence, 128 128i 4

1 , the second quadrant,

cos 7

isin 4

isin 7

sin 3 i sin 2

i

Hence, we have to find e i e i : e ie

i

cos

cos cos

isin

cos

sin sin

The real part of e i e

i

i sin

i cos sin

is cos cos

sin cos sin sin

, so cos

cos cos

sin sin

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15

22 e

i

3 4

8.

(a)

cos 4

Re e 4

i

i 4

Hence, we have to find e e

i 4

cos

: (Using the binomial theorem)

4

isin

cos 4

i 4

Re e

4i cos3 sin

6i 2 cos 2 sin 2

4i 3 cos sin 3

i 4 sin 4

The real part of the number is: cos 4

6i 2 cos 2 sin 2

therefore: cos 4 (b)

cos 5

i 4sin 4

8cos

Re e 5

i

e

i 5

cos

cos 4

6cos 2

1 cos 2

cos 4

6cos 2

6cos 4

4

i 5

8cos 2

sin 4 1 cos 2 1 2cos 2

2

cos 4

1

i 5

: (Using the binomial theorem.)

5

i sin

cos5

6cos 2 sin 2

8cos 4 8cos 2 1

Re e

Hence, we have to find e

cos 4

5cos 4 sin( i) 10cos3 sin 2 ( i)2

10cos2 sin 3 ( i)3

5cos sin 4 ( i)4

sin 5 ( i)5

The real part of the number is: cos5 10 cos 3 sin 2 5cos sin 4 cos5

10 cos 3 (1 cos 2 ) 5cos (1 cos 2 ) 2

cos5

10 cos 3

16cos5

Therefore, cos 5 (c)

10cos 5

20 cos 3 16cos5

cos 4 cos 4

8cos 4 1 cos 4 8

2cos 2

1

10 cos 3

5cos 4

5cos 20cos3

8cos 2 a

4 cos 2

1

4cos 2

3

5cos

8cos 4

Using the formula from (a), cos 4 formula, cos

5cos

1

8cos 2

4 cos 2

cos 4

1 , and the double angle

1 , we have:

4cos 2

3 8cos 4

. Hence:

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9.

(a)

Since

1 z

cos

2

isin

, we have:

2

1 cos 2 isin 2 cos 2 z Using the even/odd property we have: cos z

sin

2

sin 2

1 z 1 z

z z

cos 2

(b)

zn

1 zn

zn

1 zn

1 zn

10.

isin 2n

1 3w 3w2

i sin

1 n z 2

2n

cos 2n

2cos 2n

1 zn

isin

2n

isin 2 n 1 n z 2i

sin 2n

2n

i sin 2n

cos

isin 2 n

2 2i sin 2

2n

cos 2n

2i sin 2n

1 3w 1 3w2

and

2cos 2

i sin 2

cos 2n

2cos 2n

cos 2n

So, z n

cos 2

isin

cos

i sin 2n

cos 2n

2

cos 2

i sin 2n

cos 2n 1 zn

2

isin 2

cos

isin 2

cos 2n

Hence, z n

cos 2

isin 2

cos 2

2

. Hence,

isin 2

cos 2

.

isin

2i sin 2n

1 . zn

9 w3 10 3w 1 w 1

1

w1 w 1 w 1 w

Since: w 1 w 1 3w 1 3w2

10 3 1

w w3 1 w

w w2

2 3

isin 1 2

2 3

3 i 2

1 , we have:

7

Note: We can establish the formula w w2 1, w cos

w 1 1 w

1 2

3 i, w2 2

1 2

3 i 2

cos

1 using the values of the cube roots of 1:

4 3

isin

4 3

1 2

3 i . Hence, 2

1

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11.

(a)

For the fourth roots of 1 1cis 0 : z

cis

2k 4

cis

k 2

,k

0,1, 2,3 2

z1

cis 0

z4

cis

1 , z2

cis

, z3

2

cis

2

(c) 12.

(a)

(b)

2

1

2

3

3

i 1 i

3

,

2

i,

1 i 1 1 i

2

. 3

1,

i , so:

1 i.

1 i

i

4 5

cis

1

For the fifth roots of 1 1cis 0 : 2k 5

z

cis

z1

cis 0

z4

6 cis 5

,k

0,1, 2,3, 4

1 , z2

2 5

cis

, z3 3

2 cis 5

cis 8 cis 5

z5

1

4

4

1 2 cos

4

2 cis 5

2

3

4

1

5

cos

5 1

2

3

1

.

5

isin

5 3

2

2

4

4

2

i sin

5

2 2cos

1

5

2

2 5

Therefore, we can denote the fifth roots as 1, , 2 , 3 , First multiply and then substitute the values from (a) 1

(c)

,

3

cis

We can solve the task using the values: 1

2

3

3 2

Therefore, we can denote the fourth roots as 1, , (b)

2

cis

5

5 2

4

1

1 1

5

1 1 1

0

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13.

1 i 3

2 cis

1 i 3

2cis

n

1 i 3

3

1 i 3

3

n 3

2n cis n

2n cis

n 3

Hence, 1 i 3

n

1 i 3

n

2n cos

n 3

i sin

n 3

cos

2n cos

n 3

isin

n 3

cos

2n 1 cos

n 3 219

524 288

n 3 n 3

n 3

i sin isin

n 3

Therefore, the number is real. For n 18 , the value is: 219 cos

14.

3

Since arg 2a 3i

18 3

135 , then arg 2a 3i

135 3

45 . Therefore, tan

1

3 2a

a

3 . 2

Chapter 8 practice questions

1.

Method I 1 i z 1 3i

z

1 3i 1 i

1 3i 1 i

1 3 i1 3

1 1

2

2 i . Hence, x = 2, y = 1

Method II Another method you are already familiar with is to equate the real and imaginary parts of two complex numbers. In this case, we have to solve system of equations: 1 i x x

y i y

yi

1 3i x

1 3i

Hence, x y 1 x y 3 x 2 y 1

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2.

(a)

w is a cubic root of 1 other than 1: If w cis

2 3

, then

1 w w2 1 cis

2 3

cis

1 cos

2 3

isin

1 3 i 2 2

1 If w cis

4 3

4 3 2 3

4 3

cos

1 3 i 2 2

isin

4 3

0

, then

4 8 4 2 cis 1 cis cis . 3 3 3 3 This is the same as in the previous case; hence, the value is 0. 1 w w2

1 cis

w

1

(b)

2

2

3

wx w y w x wy

But, since 1 w w2 2

w x

2

1

4

2

w xy w xy w3 y 2

0, then w w2

2

wx w y w x wy

3.

2

x

2

y

2

x2

y2

w w2 xy

1 , and thus,

w w2 xy

x2

y2

xy

1 2i i 2 1 2i 1 2i

(a)

1 i

(b)

Let P n be the statement: 1 i

4n

n

4 .

The basis step must be P 1 . 1 i

4

1 i

2

2i

2

22 i 2

Next, assume that for some k P k :1 i

Now, 1 i

4k

4

4( k 1)

4

1

4 ; hence, P 1 is true.

, P k is true.

k

1 i

4k 4

1 i

k

1 i

4

4

k

4

4

k 1

.

Therefore P k 1 is true whenever P k is true and by mathematical induction, P n must be true for all k

(c)

1 i

32

1 i

4 8

4

8

65 536

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4.

(a)

For z1 : z1

6 4

hence, z1

2 cos

For z 2 : z2

z2

(b)

isin

6

isin

1 3

, the fourth quadrant,

6

;

6 1 1

2 , tan

4

2 cos

1 , the fourth quadrant,

4

; hence,

4

isin

6

2 cos cos

isin

4 i sin

12

6

1 cos

6

isin

4

6

4

c)

4

12

In standard a bi form:

(c)

z1 z2

6 i 2 1 i 2 1 i 1 i

a cos b 3

6

2 i

1 1 2 4

6

2

4

, b sin

6 12

2 4

: 4

isin

4

a4 cos 4 b4 12 a4 cos b4 3

6i i 2 2

12

Using de M 4

2i 2

6

6

Hence, a cos

z1 z3

2 , tan

1 1

2 cos

2 2 6 2

In polar form: z1 z2

5.

2 4

isin i sin

3

3 12

a cos b 12

4

4

isin

12

4 a4 1 b4 2

i 3 2

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6.

Let z

x yi . Then:

x 16

2

y2

4

2

x 1 x 16

x2

y2

2

y 2 16 x 1

This implies that x 2 7.

8.

y2

5 i 2 i

5 i 2 i 2 i 2 i 11 3 Therefore, a ,b 5 5 a bi

arg x i

3x 2

2

3

2x

3

i z 2 z

10.

1 2z

1 2i 2 i

10 1 i 5 2

3

0

x

2x 3

x

2

16

4

3

1

3, or x

1 3

3

1 2i 5i 5 z3

i , i.e., a = 0, b = 1

(a)

z5 1

(b)

z 5 1 cis0 Hence, the zeros are the fifth roots of unity. 2k z cis , k 0,1, 2,3, 4 5

1,cis

z 1 z4

tan

3

y2

11 3i 5

4 1

x 3 1 x

1 2i 2 i 2 i 2 i

x2

16, and consequently z

arg x 2 1 2 xi

z 2 i

32 x 16 16 y 2

240

Since x is positive, the solution is x 9.

y2

y 2 16 x 2

32 x 256

15 x 2 15 y 2

2

z2

z 1

2 4 6 8 ,cis ,cis ,cis 5 5 5 5 cis

The solutions are: 1,cis

4 5

2 5

cis

,cis

.

2 5

4 5

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(c)

We use the factor theorem: z cis

2 5

2 5

z cis

z cos

2 5

i sin

2 5

z cos

z cos

2 5

i sin

2 5

z cos

2 5

z2

2cos

2

2 z cos 5 z2

2 5

1

z cis

4 5

2cos

2 5

sin 2

4 5

z cos

4 5

In the same way:

z2

Hence: z 4

11.

(a)

8i

(b)

(i)

z3

z2

z2

z 1

8, tan

2cos

2cos

2 5

2 5

i sin i sin 2 5

2 5 cos 2

sin 2

2 5

4 5

i sin 2

4 5

i sin

4 5

sin 2 4 5

z cos

z 1

z2

z 1

is not defined, positive y-axis,

2cos

2

4 5

z 1

,8i 8 cos

2

isin

2

Let z be a cube root of 8i. Then: z

3

8 cis 2 3

2k 3

2cis

6

2k 3

,k

0,1,2

For k = 0, the number is in the first quadrant: z 2cis (ii)

2 5

4 5

z cis

z cos

2 5

z

2 cos

6

isin

6

2

3 2

1 i 2

6

3 i

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4 5

12.

(a)

All the numbers are of modulus 1; hence, their product and quotient are of modulus 1 and thus z 1 . 3

cis z

8

cis

3

cis

cis

4

0

cos

4 3

4 cos 3

3

24

cis 0

cis

3

Hence, arg z

Since z

cis 8

24

cis 2

cis

z

cis 8

3

8

cis

(b)

cis 3

4

3 4 3

3

4 3

isin

, then

3

4 i sin 3

cos 4

isin 4

1

1

0

Hence, z is a cube root of 1. (c)

First expand the expression and then substitute the polar form to simplify calculations. 1 2z 2 z 2

z2

2 4z

2 z3

z2

4 4z

1

4 4 4cis 3 1 3 i 2 2

4 4

4 2 2 3i

13.

z z

2 1 4i 1 i 2 3i

2

1 5i 1 i 1 i 1 i

4 12i 9

2

4 cis 3

1 3 i 2 2

4 4cis

1 2

i 3 2

4 3

cis

8 3

3 2 3 3 i 2

2 3i

5 12i

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14.

(a)

Let P n be the statement: cos

n

isin

cos n

isin n

The basis step is P 1 and it is true, because both sides are cos Next, assume that P k is true for some k P k : cos

isin

cos

isin

k

cos k

k 1

.

isin k

cos

k

i sin

cos

i sin

cos

i sin

cos k +isin k

Now,

isin .

cos k cos cos(k

sin k sin

i cos k sin

) i sin(k

)

sin k cos

cos ( k 1)

i sin ( k 1)

Therefore, P k 1 is true and by mathematical induction P n is true for all k (b)

(i)

(ii)

Using de Moi 1 z 1 cos z z

n

z

zn zn

(c)

(i) (ii)

z

1 n

cos

cos z

z

cos

isin n

isin n

isin

n

1

isin

cos n

cos

isin n

cos

cos n

isin n

cos n

z5

5z 4 z

1

10 z 3 z

z5

5 z 3 10 z 10 z

n

isin

n

isin n

cos n

1 5

.

2

n

isin isin n

10 z 2 z 1

5z

n

3

3

5 zz z

4

2cos n

z

5

5

Using the result from (i), we have: z

z

1 5

5

2cos

32cos 5

and z5

z

Thus, 32cos5

5

5 z3

2 p 2iq

Hence:

q ip 2 1 i

2p 2 q 2q 2 p

3

10 z

z

2cos 5

5 2cos 3

2 cos 5

5cos 3

Therefore, cos5 15.

z

1 cos 5 16

2 p 2iq

2p q 2 p 2q 2

p

2 ,q 5

10 2 cos

10cos

5cos 3

2 q i 2

1

10cos

, and a 1, b 5, c 10

p

6 5

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16.

(a)

z15

25 cos 5

(b)

z12

4 cos

z14 16 cos

4 5

2 5

isin 5 isin

8 5

2 5

32 cos 2

4 5

; z13

8 cos

8 5

; z15

32 cos 2

isin

6 5

isin 2 isin

32

6 5

isin 2

32

(c)

17.

(d)

The transformation is a combination (in any order) of an enlargement of scale factor 2, 2 with the origin as the centre, and an anti-clockwise rotation of , again with the origin 5 as the centre.

(a)

Let z a bi . Then: a2

b2

a2

b 3

6b 9 0

(b)

2

b

a2

b2

a2

b2

6b 9

3 2

(i)

(ii)

Since arg z1

(iii)

arg z2

and sin arg z1

6

1.5 3 5 6

1 2

6

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(c)

arg

z1k z2 2i

Hence:

18.

arg z1k

k 6

2a b i 2 ab

arg z2

k

4

2a b

7

3 7 i

2 ab

k 6

arg i

5 6

2

k 6

3

1

Substituting b 7 2a (from the first equation) into the second equation: 1 2 a 7 2a 1 2a2 7 a 3 0 a1 , a2 3 2 Since a, b , the solution is a 3, b 1 19.

(a) (b)

See practice question 14 (b) Use binomial expansion: z

1 z

4

z4

4z3

z4 Since z

1 z

24 cos 4

(a) (b) (c)

(d)

1 z4

4 z2

4

4

2cos 2cos 4

Hence, cos 4

20.

1 1 6z2 2 z z

(ii)

1 z3

1 z2

6

, z4

1 z4

4 2 cos 2

1 cos 4 8

1 2i e 1 i z 2 i e e 2 1 , so it is less than 1. z 2 u1 Using the formula: S 1 r

(i)

4z

z4

4z2

2cos 4

6

4cos 2

1

1 z4

8cos 4

, z2

6 4

1 z2

1 z2

cos 4

1 z4

2cos 2

, we have:

4cos 2

3

3

ei 1 i e 2

cos isin 1 1 cos i sin 2 Change to polar form and use result in (i) S

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1 2i 1 3i e e 2 4 1 1 cos cos 2 cos3 2 4 cos isin 1 1 cos i sin 2 Taking the real parts, we have: S

ei

1 cos 2 2

i sin

1 sin 2 2

cos i sin 1 1 cos isin 2 Now, evaluate the RHS and simplify: 1 1 cos cos isin cos isin 2 1 1 1 1 1 cos i sin 1 cos i sin 1 cos 2 2 2 2 1 cos isin 2 5 cos 4 The real part of this number is 1 cos 2 1 4cos 5 5 4cos cos 4 cos

21.

1 cos3 4

...

1 sin 3 4

...

... Re

1 isin 2 1 isin 2

Method I If 3 2i is a root, then 3 2i is another root; therefore: P( z )

z 2 z 3 2i z 3 2i z 2

z 3

2

2i

2

z 2 z2

6 z 13

z3 8z 2

25 z 26

So, a 8, b 25, c 26 Method II P( 2) 0

P ( 3 2i) 0

8 4a 2b c 0

9 46i a 5 12i

9 5a 3b c

b

3 2i

c 0

46 12a 2b i 0

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Hence, we have to solve the system of equations: 4a 2b c 8 5a 3b c 12a 2b 22.

Let z

9 46

a 8, b

25, c

26

a bi :

a2 b2 25 a bi

a2 b2

2 5

20

25 a bi

15 1 8i a bi

15 a bi

1 8i

a bi a bi

10a 40bi 10a 40bi 1 8i 1 8i a 2 b2 20 10a 40bi 20 160i a 2, b 4 (by equating real and imaginary parts)

Hence, z 2 4i 23.

iz1 2 z2 3 , multiply the second equation by i and add the equations: z1 1 i z2 4 iz1 iz1

2 z2 3 1 i z2 4i

3 4i 7 1 i 1 i 2 2 Substituting in the second equation: z1 3 4i (2 1 i) z2

24.

(a)

z1,2

3 4i

4

z2

16 32 2

2 2 , tan

z1

4 4

z2

4 4 2 2 , tan

z2

2 2e

i

(c)

4 1 2 2

z z

2 2i

z1

z1 1 4i 2 2i, z2

2 2i

1 , in the first quadrant,

4

1 , in the fourth quadrant,

; hence, z1 4

2 2e

i

4

; hence,

4

4

(b)

4 4i 2

4

2 2 e 2

2 2 e i

4

i

2 2 e

z2 4

2 2 e

4 4

i

4 4

4

z14

i

4 4

2 4

8e

i

3 2

64ei 64e

8 cos

3 2

isin

3 2

8i

64 i

64

Thus, they are the same.

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(d)

(e)

z1 z2

z2 z1

2 2e 2 2e

2 2e

4

3n 2

2 2 e

2 e

(a)

z7

cos

2 7

4 i

e

i

4

4

e

i

4

4

e

i

e

2

i

i i 0

2

4

n 4

k

n 4k , k

7

2 7

isin

i

n 4

The number is real if

25.

i

2 2e

4

n i 4

n

z1n

i

i

cos

7 2 7

isin

7 2 7

cos 2

1,

isin 2 0

1

hence z 7 1 0 (b)

z 1 z6 z7

(c)

z6

z5

z4

z3

z2

z 1

z5

z4

z3

z2

z z6

z5

z4

z3

z2

z 1 z7 1

Using the result from (a), we have 0 z 7 1

z 1 z6

Since z 1 , then z 1 0 and hence z 6

z4

z5

z3

z5 z2

z4

z3

z2

z 1

z 1 0

Using the result from (b), we have: 0

z6

z5

cis

12 7

z4

z3 cis

z2

12 7

z 1 cis

10 7

2 7

cis

This implies that Re cis

12 7

cis

10 7

cis

2 7

cis

1

1

10 7

cis

2 7

1

Using the fact that real parts of two equal complex numbers must be equal:

Re cis

12 7

cis

But since cos we have: 12 cos 7

12 7

cos 2 cos

10 7 6 7

10 7

2 7

cos

2 10 ,cos 7 7

cos cos

cos

8 7 4 7

cos cos

12 7 cos

6 7

cos 4 7

cos

10 7

, and cos

4 7

cos

8 7

cos

2 7

cos

6 7

2 7

2 7

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Finally: 2 cos

6 7 cos

26.

(a)

27 z 3 8 0

z3

cos

4 7

cos

6 7

cos

4 7

8 27

8 cis 27

We need to find the cubic roots of z

3

8 27

2 ; arg z 3

3

2 7 cos

1 2 7

1 2

8 cis 27

2k ,k 3

0, 1, 2

2 2 2 5 cis ; z2 cis ; z3 cis 3 3 3 3 3 There several ways of finding the area. One is demonstrated in the diagram below. This is a sketch for the three roots in an Argand diagram. z1

(b)

The triangle is made up of three isosceles congruent triangles with a vertex angle of and sides of

2 3

2 . Thus the area of the triangle is 3 times the area of each. Using the law of 3

sines, we have: Area 3

1 2 2 2 3 3

3 2

3 3

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27.

2 3i must satisfy the equation, thus,

(a)

4

2 3i

3

4 2 3i

2

17 2 3i

16 2 3i

52

119 120i 184 36i 85 204i 32 48i 52 0

If 2 3i is a root, then 2 3i is also a root. This implies that one of factors is

(b)

z2

z 2 3i z 2 3i

The other factor, z 2

4 z 13

4, can be found either by inspection or long division.

Therefore, the other roots are 28.

(a)

2

a bi

a2 b2

i

2i. a 2 b2

2abi i

0 and 2ab 1 1

Solving the simultaneous equations gives a b

(b)

Using the quadratic equation, z

3 i

3 i

2 2

1 i

z

4 2 i

2

3 i

2

2i 2

Using the result from (a) z

29.

3 i

3 i

2i

1 i 2

2

2

3 i

2

Evaluating cos

isin

n

1 i

2 i or 1

2

usi

formula, and using the binomial theorem and

equating either real parts or imaginary parts will enable us to find expressions for cos n or sin n . cos

isin

4

cos4

isin 4

cos4 4icos3 sin 6cos2 sin 2 6cos 2 sin 2 sin 4

cos 4

cos 4

sin 4

4cos3 sin cos5

10cos3 sin 2

isin

30.

eln 2 1 i

6cos5 sin

eln 2eiln 2

20cos3 sin 3

eln 2 cos ln 2

5

5cos sin 4

Also, for sin6 we will need the imaginary part of cos Thus, sin6

sin 4

4cos sin 3

Similarly, for cos5 we will need the real part of cos Thus, cos5

4icos sin 3

isin ln 2

isin

6

6cos sin 5

2 cos ln 2

isin ln 2

real part = 2cos ln 2 and imaginary part = 2sin ln 2

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Exercise 9.1

1.

(a)

AB

OB OA

3

1

5

xB

xA

yB

yA

5

1

zB z A

2

2

2

2 2 0

1 1

(b)

AB

OB OA

xB

xA

yB

yA

zB

zA

1

AB

yB

OB OA

yA

(d)

2.

(a)

AB OB OA

yB

yA

zB

zA

Given PQ OQ OP

xQ

zQ 1 1

(b)

xA

3 2

1

3

1 2

1 2

1

zB z A xB

3

1 2

xB xA (c)

3 a a

2a

xP

yQ

yP

zQ

zP

3 2

a

xQ

xP

yQ

yP

zQ

zP

0

3 2

xQ

1 5 2 1

1 ; yQ 2

zQ 1 1 2 . So, Q

2 3

2a a a

a 2a

xQ

3

1 2 2

3 5

xQ 1

Given PQ OQ OP

2

5 2 1

1 2 1

yQ zQ

1 2

1

5 2

yQ

5 1 2 2

1 , 3, 2 2 3 2 1 2 1

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3

3 2 1 2 1

1 xP 5 yP 2 1 zP 5 2

(c)

1 2

yP

Given PQ

xQ

yQ

2a

5 2

yP

OQ OP

a, yQ

1 2

yQ

2;

xP

xP

a

yQ

yP

zQ

zP

2a a

2a

1

1 zP 1

xQ

2a , zQ

2a

3 2

1 xP

1, 2a, a

3 2

5 2

zP 1 1 0 . So, P

xQ a

4a ; zQ 2a a

2a 2a

a

5 , 2,0 2

xQ 0 ;

zQ a 2a 3a .

So, Q 0, 4a, 3a 3.

(a)

For points M, A and B to be collinear, it is sufficient to make AM parallel to AB . If the two vectors are parallel, then one of them is a scalar multiple of the other, for example, AM xM x A

AM

yM

yA

zM

zA

Therefore:

x y z 5

t AB .

x 0 y 0 ; z 5 1 t 1 5

AB

xB

xA

yB

yA

zB

zA

x y

t t 5t

z 5

1 0 1 0 0 5 x t, y t, z 5

5t

So, M t , t , 5 5t , where t Note: We can find M if BM xM x B x 1

BM

yM

yB

zM

zb

t AB . Then we have:

y 1 , AB z 0

1 1 5

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x 1

t

Therefore: y 1 z

t

x 1 t, y 1 t, z

5t

5t

So, M 1 t , 1 t , 5t , where t Both conditions describe the same set of points; for example, we can obtain point M (0, 0,5) by putting t 0 in

M (2, 2, 5) by putting t 2 in

(b)

1 in M 1 t,1 t, 5t ; or

M t , t,5 5t , or t

M t, t,5 5t , or t 1 in M 1 t,1 t, 5t .

For points M, A and B to be collinear, it is sufficient to make AM parallel to AB . AM

AM

AB

xM

xA

x

yM

yA

y 0

zM

zA

z 1

1

xB

xA

3 ( 1)

yB

yA

zB

zA

5 0 2 1

Therefore: So, M

t AB .

x 1 y z 1

4 t 5 3

x 1 y z 1

1 4t ,5t ,1 3t , where t

4t 5t 3t

x

1 4t , y 5t , z 1 3t

.

Note: If we start with the condition BM

t AB , we will have BM

x 3 y 5 ; z 2

therefore, from BM (c)

t AB , we will find x

3 4t , y

5 5t , z

2 3t.

For points M, A and B to be collinear, it is sufficient to make AM parallel to AB . AM

AM

xM

xA

yM

yA

zM

zA

t AB .

x 2 y 3 z 4

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AB

xB

xA

2 2

yB

yA

3 3 5 4

zB z A x 2 Therefore:

y 3

t

z 4

4

x 2

4t

6

y 3

6t

1

z 4

t

So, M 2 4t ,3 6t, 4 t , where t

x

2 4t , y 3 6t , z

4 t

. x 2

Note: If we start with the condition BM

therefore, from BM 4.

t AB , we will have BM

t AB , we will find x

2 4t , y

3 6t , z

If A is the midpoint of [BC], then a relationship between them can be BC Let C(x, y, z). (a)

BC 2 BA

x 1 y

4 2 4

8 8

z 1

1

2

1 ( 1) (b)

BA

BC

x so, C

1 2 1 5 3 3

x 1 1 y 2 1 z 3 1, y

1,

2 BA

5

1 2

y 3 ; z 5 5 t. 2 BA.

C 7, 8, 1

0

x 1

5 2 , BC 14 3

1 2 . Therefore: 1 z 3 y

0 5 28 3 11 ,z 2

1 3

28 3

29 ; 3

11 29 , . 2 3

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(c)

1 a 2 2a , BC 1 b

BA

x a y 2a . Therefore: z b

x a BC

2 2a

y 2a

2 BA

4 4a

z b x

2 a, y

;

2 2b 4 2 a, z

b 2;

so, C (2 a, 4 2a, b 2) 5.

Let G ( x, y, z ) : (a)

0

GA GB GC 0 0 0

So, G

(b)

0

1 3x 3 3y 1 3z

0 0 So, G 1,

1 x 2 y

1 x 2 y

1 z

1 z

3 z

1 , y 1, z 3

1 3

1 1 ,1, 3 3

GA GB GC 0

x

1 x 1 y

2 x 3 y 1 z

3 3x 5 3y 3 3z

x 1, y

1 x 2 y 5 z 5 ,z 3

0 x 0 y 1 z 1

5 , 1 3

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(c)

0

GA GB GC

6.

b x

c x

2a y

2b y

2c y

3a z

3b z

3c z

0

a b c 3x

0

2a 2b 2c 3 y

0

3a 3b 3c 3 z

x

So, G

a x

a b c ,y 3

2a 2b 2c ,z 3

a b c

a b c 2a 2b 2c , ,a b c 3 3

The relationship between points A, B, C , and D of parallelogram ABCD can be expressed using different vector relationships; for example, AB BA CD, . Here, we will use AD

BC ,

BC . Let D( x, y, z ).

C

D

B

A (a)

DC , AD

3 1 2 3 5 0

BC

Therefore:

3 1 1 5

x 3 y 2

, AD

z

3 1 x

3

x

1,

x 3 y 2 z 1

1 y

2

1

y

1, z 1

5

z

So, D( 1,1, 6) 2 2 3 2

(b)

BC

3

3

3 5

5

5 2 , AD

2 3 4 5

Therefore: 5 2

x

z

3 5. So, D

5

4 5

z

2

x

x

2

y

3

z

5

4 2, y

3 2 3

y 3 3,

4 2,3 3, 3 5

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6

7 1 2 2 1 2 3 3 1 5

BC

(c)

1 , AD 3 4

5 , y 1 2 3

x

2

7.

1 2 1 y 3 z

x

Therefore: 3 x 1 So, D

1 2 1 y 3 z 0

3

x

1 3

0, z

y

5 ,0, 4 . 2

Two vectors v and w have the same direction if, for t

m 2 m n

4

0,

v tw . Therefore:

2 t 4

2m n

m 2 m n

6

2t 4t

2m n

m 2t 2 m n 4t 6t

0

2m n 6t

0

Solve this system of equations by any method of your choice Therefore: m 5 , n 1. 8.

(a)

The length of the vector v

2i

2j

so the unit vector is 1 2i 2 j k

2 i 3

The length of the vector v

2k is

3

(b)

so the unit vector is (c)

1 2 14

(a)

6i

4j

2i

so the unit vector is 1 2i

j

The length of the vector v

2i

j

2k 2j

22 ( 1)2

2 j 3

3

6i 4 j 2 k

The length of the vector v 3

9.

22

k is

14 2k is

2 i 3 k is

3,

1 k. 3 62

4

2

i

14 22

1 j 3

9

22

1

j 1

2

14 2

56 2 14 ,

k.

( 2) 2

9 3,

2 k. 3

22 22 ( 1)2

9

3,

so the unit

vector in its direction is 1 2i 2 j k , and the vector of magnitude 2 is: 3

2 2i 3

2j

k .

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(b)

The length of the vector v

6i

unit vector in its direction is 4 2 14

(c)

2

6i 4 j 2 k

1

2i

(b) (c)

11.

i 3 j 2k

u v

3i 4 j 2k

u

i 3 j 2k

3u

3 i 3 j 2k

3u

3v

(d)

1 u u

1

(e)

1 u u

(a)

126

14

56 2 14 , so the

1

2

2

2

9 3 , so the unit

2k , and the vector of magnitude 2 is:

2k .

j

v

22

2k is

j

j

3

(a)

22

6i 4 j 2k , and the vector of magnitude 4 is:

2 14

vector in its direction is 1 2i

10.

2

4

6i 4 j 2k .

14

The length of the vector v

5 2i 3 u v

62

2k is

4j

1

u

1 14

14 3

i

14

2i

j

3i 4 j 2k

32 42

2

2i

2

29

12 32

j

2

3i 9 j 6k 45

22 12 02

3v

3 2i

j

14

5

6i 3 j

3 14 3 5

i 3 j 2k

j

2

2 14

k

1 14

3

i

1

14 2

14

2

j 3

14 2

14

k 2 14

2

1 9 4 14

1

Using B( x, y, z) for the terminal point and A( 1, 2, 3) for the initial point: x AB

1

y 2 z

3

x 1 Therefore: y 2 z 3

x 1 y 2 , AB z 3

4 2 2

w

x 3, y

4, z

5

So, the terminal point is (3, 4, 5)

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(b)

Using B( x, y, z) for the terminal point and A( 2,1, 4) for the initial point: x AB

2

x 2

y 1

y 1 , AB

z 4

z 4

Therefore:

x 2

2

y 1

3

z 4

1

v

x 0, y

2, z 5 .

So, the terminal point is (0, 2,5) 12.

(a)

A vector opposite in direction and a third the magnitude of u is 1 u 3

Therefore:

(b)

1

1 3 3 4

4 3 A vector in the same direction as w and whose magnitude equals 12 is 12 times a unit vector in the direction of w. Therefore, the vector is of form:

1 1 12 w 12 w 42 22 (c)

1 u. 3

-2

2

4 2 -2

4 12 2 24 -2

4 6 2 -2

If vectors are parallel, then one can be represented as t times the other. Therefore:

xi yj 2k t i 4 j 3k . From the z coordinate , we can find the value of t: 2 2 i 3

13.

3t

t

2 . So, x 3

2 1 3

2 and y 3

2 3

4

8 , and the vector is: 3

8 j 2k . 3

1 BC . 2 Let v be the vector from the vertex B to the midpoint of side AC;so, v BA 1 AC . 2 Let w be the vector from the vertex C to the midpoint of side AB; so, w CA 1 AB . 2

Let u be the vector from the vertex A to the midpoint of side BC; so, u

AB

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Adding the vectors: 1 1 1 1 1 u v w AB BC BA AC CA AB BC AC CA 2 2 2 2 2 1 1 1 1 1 BC CA AB BC CA AB 0 0 2 2 2 2 2 14.

The length of the vector v Hence, t 14 1

15.

The length of the vector v

The length of the vector v Hence, 0.25 3.25t 2 1

17.

(a)

a

f

1

0.5i

t2

2t

22

2t

2tj 3tk is

0.25 3.25t 2

1

t2

3 13

0 8 ,d

0 0 ,e

8 0

0

0

0

8

8

8

8

1 e 2

(b)

l

(c)

LM MN NL

f

8 4 ,m 8

m l

2

0.52

tj 1.5tk is

8 8 ,c

0 8

(3t )2

14t 2

t 14

(3t )2

4 13t 2

3 , so there is no solution.

13t 2

8 0 ,b

8 8 ,g

2

14 14

t

2i

4 13t 2

3tk is

2tj

1 14

t

Hence, 4 13t 2 1 16.

ti

1 AB 2

4 8 ,n 8

n m

t

2

(1.5t )2

0.25 3.25t 2

3 13

t

8 8 4

(l n) 0

Note: We can verify the statement using coordinates: 4 8 8 4 8 8 0

LM

MN NL

8 4

8 8

4 8

0

8 8

4 8

8 4

0

0

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8 18.

c

(a)

OE OA

0

0 ,d

OE OB

10

12 (b)

4

1 OA OB 2

f

AG

(c)

12 5 ,g 0 4 8

4

5 0

5

12 0

12

0 4

4

10 5 12 0

5 12

OG OA

FD

OD OF

4

1 OC OD 2

5 12

The vectors are the same because they connect a vertex and the midpoint of the parallel side in a parallelogram. 19.

i

1 j

2

1k

1

2

1

Hence, we have to solve the equation: 3 20.

We have to solve a vector equation: 4

3 1

3

3 2

1

3 2

2

3 2

2

2

2

2

3

2

2 4

4

1

1

3

2

1

3

0

1

1

2

1

a

2 3

6 3

4 1 1

A system of three equations which you can solve by a method of your choice 26 , 7

Therefore,

21.

11 , 7

3 7

We have to solve a vector equation:

1 1 5

3 2

3 2

1 1 5

1 0 1

3 2 0

0 1 1

1 1 5

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Therefore, 22.

2,

1,

3

2 1 1

We have to solve a vector equation: 2 1

3

4

3

4

1 1 0

3 0 1

4 1 1

2 1 1

1

An inconsistent system of equations. Hence, there are no such scalars 23.

(a)

, , .

u v and u+v are diagonals of a parallelogram. So, the parallelogram has diagonals of the same length; hence, it is a rectangle.

v1 u1 (b)

u v

v2 u2

u v

v1 u1

u v

v1 u1

2

v2 u2

2

v3 u3

2

v3 u3 v1 u1 u v

v2 u2

2

v2 u 2

2

v3 u3

2

v3 u3 Hence, v1 u1

2

v1 u1 v12

v2 u 2 2

2

v2 u2

2v1u1 u12 v2 2 v12

v3 u3 2

2

v3 u3

v1 u1 2

2v2u2 u2 2

2v1u1 u12

So: 0 4v1u1 4v2u2 4v3u3

v2 2

v1 u1 v32

2 2

v2 u2 v2 u2

2 2

v3 u3 v3 u3

2 2

2v3u3 u32

2v2u2 u22

v32

2v3u3 u32

v1u1 v2u2 v3u3 0

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24.

A summary of the information is shown below: D

C

T2

T1 30

A 45

B

If the traffic light is in equilibrium: AB AC AD We will express the vectors in component form:

AB

0 , AC is parallel with the unit vector 125

2 2 2 2

cos 45 AD is parallel with the unit vector sin 45

3 2 1 2

and its magnitude

and its magnitude is T2; hence,

2 2 2 2

0

Now, we have:

So, T1

cos 30 sin 30

3 2 1 2

is T1; hence, AC T1

AD T2

0

3 2

T2

125 2 2

0

T2

2 2 2 2

T1 3 T2 2

T1

T1

3 2 1 2

3 2 T2 2 2 1 2 125 T1 T2 2 2 T1

0

T2

0

2 3

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125 T1

1 2 T2 2 2

250 3

And: T1 T2

0

21 2 T2 2 32

125 T2

2T2

6T2

0

T2

2

125 3 2

6

3

2

250 3 2 2

125 3 2

6

2

6 125 2 3

3

3 1

2

2 3

Therefore, the cable tensions are T1 125 25.

0

3 1 N

and

125 T2

3 1 125 3 2 2

6

N

A summary of the information is shown below: C

D T2

T1 30

A 60

B

For vectors it holds: AB AC AD 0 We will express the vectors in component form:

AB

0 , AC is parallel with the unit vector 300

and its magnitude is T1; hence, AC T1

3 2 1 2

3 2 1 2

cos 60 AD is parallel with the unit vector sin 60

Hence AD T2

cos 30 sin 30

1 2 3 2

and its magnitude is T2.

1 2 3 2

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T1 3 T2

T2

T1 3

T1

300

3 1 T2 2 2

300 T1

T2

1 2 3 2

0

Now, we have:

So, T1

3 2 1 2

0

1 3 T2 2 2

0

300 2T1 0

0

1 3 T1 3 2 2

300 T1

3 1 T2 2 2 1 3 125 T1 T2 2 2 T1

0

0

T1 150

And: T2 T1 3 150 3 Therefore, the cable tensions are T1 150 N and T2 150 3 N .

Exercise 9.2 1.

(a)

u v 32

2

2

3

uv 2 cos

(c)

6

16

2

1

6 3 05

20

4

2

2

2

6

20 22

uv 35

6

2

02

1

2

3

2

52

2

16 29 41

20 40 35

117.65

122.31

1 2 13 13 2

3

u v 10 cos

2

2

1

cos (d)

4 16

cos (b)

1

1

2

5

2

2

13 10 29

2

3 5 0 2

15

15 2

1

3

2

0

2

40.24

0

2

5

2

2

2

15 10 29

151.74

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2.

3.

(a)

u v

u v cos

(b)

u v

u v cos

(a)

u v

2

3 4 cos

3 4

3 2 3 4 cos 3

3 4

1 2

6 1 2

6

6 3 4 5 0 . The dot product is zero; hence, the vectors are

1

orthogonal. (b)

u v 35

7 2 1 . The dot product is positive; hence, the angle is acute.

(c)

u v 10

3 6 6 3 0 . The dot product is zero; hence, the vectors are

orthogonal. 4.

(a)

v u w u

y x y x

0 . The dot product is zero; hence, v is orthogonal to u.

x y

x y 0 . The dot product is zero; hence, w is orthogonal to u.

Pay attention to the relationship between the coordinates of a two-dimensional vector and a vector that is perpendicular to it. (b)

The vectors perpendicular to u

2i

3 j are 3i

2 j and

Unit vectors in the direction of those vectors are: v1 v2

1 13

1 13

3i

2j.

3i 2 j and

3i 2 j . y

v1 x v2 u

5.

(a)

(i)

v

22

3

2

12

14

i v 12 0

3

01 2

cos

j v 0 2 1

3

01

k v 0 2 0

3

11 1

3

2 14 3

cos cos

14 1 14

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(ii)

cos 2

(iii)

cos 2

cos

1

cos

1

(i)

(ii)

(i)

(ii)

1

1

2

2

12

2

01 1

j v 01 1

2

01

k v 01 0

2

11 1

cos

cos

cos

1

cos

1

v

2

cos

1 6 1 6

32

2

74

2

6

6

2

1

2

2

6

65.9051...

66

,

2

1

6

66

12

6 1

cos

2

2

cos

65.9051...

2

1

cos

6

cos

1

1 4 1 6

2

1

144.7356...

6

145

14

i v 13 0

2

01 3

j v 03 1

2

01

k v 03 0

2

11 1

cos 2

cos 2

cos 2

143

6

i v 11 0

2

14

4 9 1 1 14

58

74.4986...

14

2

1

14

143.30077...

14

12

v

(iii)

(c)

3

2

3

14

57.6884...

14

cos

(b)

2

2

2

cos 2

cos 2

3 14 2

cos cos

3 14

2

14 1 14 2 14

2

1 14

2

9 4 1 1 14

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(iii)

(d)

(i)

cos

1

cos

1

cos

1

3 2 14

37

122.3115...

122

74.4986...

74

1 14

32 02

v

36.6992...

14

4

2

5

i v 1 3 0 0 0

j v

0 3 10 0

k v

(ii)

cos

6.

7.

cos

4 2 cos 3

3 cos cos

8.

(a)

4 4

cos 2

cos

1

3 5

3

0

cos

4

4

3 5

cos 2

53.1301...

3 5

cos

2

02

53 ,

0

cos 4 5

2

4 5 9 16 25

cos 1 0 90 ,

1

cos

1

4 5

143.1301...

1 2 2 2 1 2

3

cos

4

0 3 0 0 1

cos 2

(iii)

4

3

2 2 2 2 0

3 2 2 3 2 2 0

2

u v

3 m 2

5 m 3

0 0 8m 9

Vectors are perpendicular if their dot product is zero; therefore: 8m 9

0

m

9 8

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143

(b)

u v

2m

m 1

m 1 m

m 1

m 1

4 m 2 3m 1

Vectors are perpendicular if their dot product is zero; therefore: 4 m2

9.

u w

3m 1 0

u u mv

1 4

m 1, m

u u mu v

3

2

12

22

m

3 1 12 21

14 m

Vectors are orthogonal if their dot product is zero; therefore: 14 m 0 m 14 10.

(a)

cos

(b)

u v

27 45

1

1 5

2

6 4

2

2

3

45 45

4

18 45 36

2

1 5

63

3 2 0 4

18

45 6

45 6

63.4349...

5

02

27 45

127

63.4349...

1

1

27

2

4, 2, 4

45 4

v u v

cos

3

2 4 5 2 4 4

v u v

cos

4 0

4 2 62

6, 3,0

u u v 1

52

3

126.8698...

u u v

cos

11.

2

2

2,5, 4

cos

(c)

2 6 5

u v u v

cos

1 5

63

AB

3 1,5 2, 2

2,3,1

AC

m 1,1 2, 10m

(a)

The points A, B and C are collinear if AC is parallel to AB . Given this

3

m 1, 1, 10m 3

collinearity, AC t AB

2,3,1

t m 1, 1, 10m 3 . From the second

set of coordinates we can determine t: 3 2

1

3 m 1

3

10

1 3 3

t

t

3 . Hence:

1 . Checking with the third set of coordinates: 3

m

1

3

1 ; so, it fits and for m 1 the points are collinear. 3 3

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AC AB

(b)

2 m 1

3

1

1

10m 3

8m 2

Vectors are perpendicular if their dot product is zero; therefore: 1 AC AB 0 8m 2 0 m 4 12.

The vector equation of the line is an equation of the form: r

r0 tv , where r0 is the

position vector of any point on the line and the direction vector v is a vector parallel to the line. For the median through A , we can take the position vector of point A for r0 and the vector from A to the midpoint of BC for v. So: 3 3 5 1 1 2 1 4 m , , 3, 2, BC 2 2 2 2 r 2 , 0

1 3 4 2

v

1 0

2

1 2

3 2

1 4

mA : r

2 1

m

. Therefore:

1 0 3 2

For the median through B , we can take the position vector of point B for r0 and the vector from B to the midpoint of AC for v. So:

r0

3 5 , mAC 1

4 3 2 1 1 2 , , 2 2 2

v

7 3 2 1 5 2 1 1 2

1 2 9 2 3 2

7 1 1 , , 2 2 2

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3 Therefore, mB : r

5 1

1 2 9 n 2 3 2

For the median through C , we can take the position vector of point C for r0 and the vector from C to the midpoint of AB for v. So:

r0

5 1 1 4 3 2 , , 2 2 2

3,1, 2 , mAB 7 3 2 7 1 2 1 2

v

7 7 , , 1 2 2

1 2 9 2 3

3 1 2

Therefore, mC : r

k

1 2 9 2 3

The centroid is the point where all the medians meet. We will find the intersection of two lines, and then check that this point is also on the third line. If mA and mB intersect, then:

1

4 2 1

0 m 3 2

3 5 1

1 2 9 n . Therefore, we have: 2 3 2

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4 m 3

1 n 2

2 0m

5

1

9 n 2 3 1 n 2

3 m 2

Putting m

n

2 3

m

2 3

2 we can see that it fits the equation, so the point of intersection of 3

n

2 3

and mB is: 4, 2, 1

1, 0,

3 2

10 , 2,0 . Now, we have to check that 3

10 10 , 2,0 is on the third line as well: , 2,0 3 3 we can see that the centroid is

13.

(a)

AB

xA xB xC y A yB , 3 3 3 1 2 2 1 3

AD

3 BD

3

2 2 3 1

10 , 2,0 3

2 3

1 9 , , 3 . Hence, 2 2

4 3 3 2 5 1 1 1 2 , , . 3 3 3

yC z A zB zC , 3

4 0 ; AB 2

3 1 2 2 3 3

3,1, 2

10 , 2,0 . 3

Note: For the centroid, it holds that: The formula

mA

20; AC

2 0 ; AD 6 6 0 ; BD 4

holds in general.

1 1 4 2 3 3 1

40; BC

0 6 ; AC 0 3

4 2 3 1

3 1 52; CD

2

4 3 3

6

4 6 ; BC 2

2 6 ; CD 6

56

76

We will calculate the angles by finding the dot product and using the cosine angle formula:

AB AC 0 AB AD

angle 90 4

cos-1

4 20 40

98.1301

angle 180

98.1301

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82

AB BD

16

cos-1

16 20 52

119.7448

angle 180

119.7448

60

AB BC

20

cos-1

20 20 56

126.6992

angle 180

126.6992

53

AC AD 0 AC BC

angle 90

36

AC CD

36

cos-1

36 6 56

36.6992

36 6 76

angle 37

133.4915

angle 180

AD BD 36

cos-1

36 40 52

37.8749

angle 38

AD CD 40

cos-1

40 40 76

43.4915

angle 43

BC BD 16

cos-1

16 56 52

72.7525

angle 73

BC CD

40

cos-1

40 56 76

127.8168

36 55.0643 52 76 We will use the law of sines for this part. BD CD 36

(b)

cos-1

1 AB 2 1 AB 2 1 AC 2 1 BC 2

ABC ABD ACD CBD

cos-1

1 20 6 1 2 1 AD sin DAB 20 40 2 1 AD sin DAC 6 40 1 2 1 BD sin CBD 56 52 2

AC sin BAC

133.4915

angle 180

127.8168

angle 55

3 20

13.416

sin 82 14.005 18.974 sin 73

25.803

Thus, the surface area = 72.2 2 (c)

DC

6

DC

76; Thus, the angles it makes with the axes are:

2 76

103.26 , cos

6 cos

1

1

6 76

133.49 , cos

1

6 76

46.51

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47

52

(d)

Since DA DB AC

DA DB

BA and we already showed that AB AC 0.

0, thus

Alternatively, you can use coordinates of the

corresponding vectors. 2 DA DB

14.

cos

cos

k

2

2

16.

6

4

3

1

k 1

3 k

0 or k 3 1 1

k

1

1 1

k 1 2

1 2k k2 2

0

0 and AC 2

3 3k k k 2 10

k 2 10

1 1 1 k

Therefore: 1

2 x y

0

3 2k k 2 10

2

So, k

0

4

9 k2 1 1 9 k2

Therefore: 1

15.

6

1 k2

4 6

DA DB AC

2

3 4k

k k 1 cos k2 2 3

1 2

2

2 4k

k2

0

0

1 2

k2

6 4k

6

0

0

cos

0

4k

4

0

0

k

k

0, k

2

4

4.

6 x y

0;

2 x y

4 1 2

8 x 2y

0

Hence, we have to solve the system of equations: 6 x y 0

8 x 2y 0 Adding the equations: 14 y 0 17.

y

14 and x

20

Two vectors are parallel if one of them is a scalar multiple of the other, u tv . If one vector is a scalar multiple of another, then its components are also multiples of the components of the other vector. Put differently, the components of the two vectors are proportional. © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

1 x 2 x

2x 2 1 x

3 x ; 1 x

Take the first equality and solve for x: 1 x 2 x

2x 2 1 x

x2

6x 5

0

x

5 or x 1.

Substituting each value into the equalities, we get only x = 5 to satisfy both of them. 18.

(a)

ABC is the angle between vectors BA and BC. BA

2 3 1

OA OB 1 2 3

cos ABC

(b)

AC

3 5 4 1 4 0

1 4 9 1 16

AB BC

BA BC

1 2 3

1 2 3

1 8 0 14 17

1 4 0

ABC

cos

1

7 14 17

0 6 3

BAC is the angle between vectors AB and AC. 1 2 3

cos BAC 19.

(a)

0 6 3

1 4 9 36 9

0 12 9 14 45

BAC

cos

1

21 14 45

33

Vectors are orthogonal if their dot product is zero; therefore: b 1 1 3 b b 3b 2 4b 2 0 b 2 2 1 4

(b)

2 7 4b 2

b2 b 0 2b

4b2 2b 7 0 4b2 2b 0

0

b

0,

b 0,

1 2

1 2

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117

b2 0 , the vector b

For b

is a zero vector. A zero vector has no direction;

0

therefore, it is not orthogonal to any vector. Hence, the vectors are only orthogonal 1 . 2

for b (c)

Similar to the two questions above: 2b2 11b 15

(d) 20.

0

Similarly, 12 20 2b2

5 or b 3 2 0 b2 16

b

4

To determine the angle between two vectors, we are going to find their dot product:

p q p q

p2 q 2 .

Since, for any vector: v 2 p q

21.

b

p q

p2 q 2

v v cos 0 p

2

q

2

v

2

1

2

v we have:

0 ; therefore, the vectors are perpendicular.

We can find the z-component by transforming 300 m/min into km/h: 300 m/min

0.3 km/min

0.3 60 km/h

18 km/h

Since the heading is 45 northwest, the velocity vector in the xy-plane is parallel to the vector

1,1 . The unit vector in this direction is

1 2

1,1 . Since the airspeed is 200

km/h, and its vertical component is 18 km/h, then the speed of the xy-component is:

2002 182 39676 2

1,1

39676 199.188 199 km/h. So, the velocity vector in the xy-plane is: 19838

1,1

140.8,140.8 . Hence, the velocity vector is

140.8, 140.8, 18 . Note: If the 200 km h 1 is interpreted as the horizontal airspeed as seen by an observer on the ground, then the xy-component would be 200 hence the velocity vector would be 22.

1 2

1,1

141.4,141.4 , and

141.4, 141.4, 18 .

Vectors are perpendicular if their dot product is zero; therefore: 2t 4t 10 t 0 t 2 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

23.

Vectors are perpendicular if their dot product is zero; therefore: t 3t

24.

2

0

1 2

t

Vectors are perpendicular if their dot product is zero; therefore: 4t 2

2t

0

0

t

1 2

0, t

Note: For t 0 , the second vector is a zero-vector, but, according to the definition a zero vector is perpendicular to all vectors. 25.

Some of the diagonals are shown in the diagram below. Using the properties of symmetry, we can see that most of the angles are the same.

H

G

E

F D

C

A

B

We will use the component form of the diagonals (considering D to be at the origin): a 0 a a a

DF

AC

a , AF a EG

cos DF , AF cos DF , DE cos DF , DB cos DF , HA

DG

a , DE a

a a , EB 0

0 a

HC

3a 2 2a 2 2a 2 3a

2a

2

2a 2 3a 2 2a 2 0 3a

2

0 , DB a

HF

a , HA 0

GB

a

2a 2

2

CF

2a 2

2 6 2 6 2 6 0

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0 a

0

cos DF , AC

0

3a 2 2a 2 0

cos DF , EB

3a

2

0

2a 2

Hence, the angle is either 90° or cos 26.

2 . 6

1

To simplify the notation, denote the vector by v

a b b a . We will determine the

angles between v and a, and v and b. 2

2

Using the fact that a 2

a and b2

b , we have:

ab

a ba b a 2 v a

ba v

a b2

b a

cos

ba a v a

ab

cos

ba b v b

b ab

v b

b a v ab v

v

Since, the cosines of the angles are the same, and the angles are from 0° to 180°, then , and vector v is the angle bisector. 27.

The scalar product should be 0, so: 2 m n 0 If the magnitudes are equal, then: 1 m2 1 Hence, 2 n

28.

2

3 n2

4n

1

m 2 n.

4 1 n2 1 and m 4

n

m2 3 n2 2

1 4

7 4

For a set of three angles to be direction angles for a vector, the squares of their cosines must add up to 1.

2 cos 2 cos 2 cos 2 4 6 3

2 2

2

3 2

2

1 2

2

1 3 1 2 4 4

3 1; 2

hence, they cannot be the direction angles of one vector.

29.

cos

2

Hence,

cos

2

3

cos or

2

1 2

2

2 2

2

cos2

3 cos 2 4

1

cos

1 2

2 . 3

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30.

3cos2

If

1

Hence, the angles are cos

31.

1

3 3

u cos u cos

We can write u

1 3

cos

u

u u

cos cos

u cos u cos

u

cos

u cos

u cos

Hence, the direction vectors are:

32.

,

,

.

x y . Then, x 2 y 2 z 2 1 and z

Let the vector be: u

x 2y z 0 3x 4 y 2 z 0

Solution of the system two equation with two unknowns is: x 16 2 z 25

1 2 z 100

4 5 1 10 1

10 165

u

z2

1

z2

1 165

.

4 z, y 5

1 z ; hence: 10

100 . So, the vectors are: 165

8 1 10

Exercise 9.3 In calculating u v , you can use the formula given in the formula booklet, or the equivalent using determinants given in the book. We will use the determinant approach for convenience.

1.

(a)

i

i

j k

i j k 1 0 0 1 1 1

j k

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(b)

i i i

i j k 1 0 0 1 0 0

j i k

i j k 1 0 0 0 1 0

i j k 1 0 0 0 0 1

k

j

The results are the same.

i 2.

(a)

j

i

j k

j k

0 1 0

i k

1 1 1 i (b)

j i

j j

j k

j k

i

0 1 0 1 0 0

j k

0 1 0 0 1 0

i

j k

0 1 0 0 0 1

k i

The results are the same.

i 3.

(a)

(b)

k

i

j k

k i k

j k

0 0 1 1 1 1

i

j

i j k 0 0 1 1 0 0

j k k

i j k 0 0 1 0 1 0

i j k 0 0 1 0 0 1

j i

The results are the same.

4.

u

v w

i

j

k

u1

u2

u3

u1 (v3 w3 ) u3 (v1 w1 )

v3 w3

u1 (v2 w2 ) u2 (v1 w1 )

v1 w1 v2 w2

i u v u w

j

u2 (v3 w3 ) u3 (v2

w2 )

k

i

j

k

u2v3 u3v2

u2 w3 u3 w2

u1 u2 u3

u1

u2

u3

u1v3 u3v1

u1w3 u3 w1

v1

w1 w2

w3

u1v2 u2 v1

u1w2 u2 w1

v2

v3

u2 (v3 w3 ) u3 (v2 w2 ) u1 (v3 w3 ) u3 (v1 w1 ) u1 (v2 w2 ) u2 (v1 w1 ) Hence, u

v w

u

v w .

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u

v w

i

j

k

u2 (v3 w3 ) u3 (v2 w2 )

u1

u2

u3

u1 (v3 w3 ) u3 (v1 w1 )

v1 w1 v2 w2 v3 w3

i u v u w

j

u1 (v2 w2 ) u2 (v1 w1 )

k

i

j

k

u2v3 u3v2

u2 w3 u3 w2

u1 u2 u3

u1

u2

u3

u1v3 u3v1

u1w3 u3w1

v1

w1 w2

w3

u1v2 u2 v1

u1w2 u2 w1

v2

v3

u2 (v3 w3 ) u3 (v2 w2 ) u1 (v3 w3 ) u3 (v1 w1 ) u1 (v2 w2 ) u2 (v1 w1 ) Hence, u

5.

(a)

v w

u

i

j

2

3

2

3 2

3

v w .

k 13i 13k

For orthogonal vectors, the scalar product must be zero:

13i 13k 2i 3 j 2k

(b)

13i 13k

3i 2 j 3k

i 4

j 3

6 i 8 j 8k

0

2 2

k 0

6i 8 j 8k 4i 3 j

(c)

6i 8 j 8k

2 j 2k

i j 1 2 4 1

5 1 7

k 1 3

26 0 26 0 39 0 39 0

24 24 0 0 0 16 16 0

5 1 7

1 2 1

5 2 7 0

5 1 7

4 1 3

20 1 21 0

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i (d)

j k

5 1 2

i

j 3k

3 0 1

6.

j 3k 5i

i

j 3k 3i k

(c)

(a)

(b)

(c)

(d)

j 2k

5 1 6 0

3 0 3 0

2

i

j k

2

4

1

3

2 3

1

3 3m

m

m 2 1

m

6 2m

m 2

i j k 2 1 m

m 2

3 2m 6 3m

1

3 2

3

1

1

3 2

i j k m 2 1

3 2

2m 1 m2 2

3

2

3

m 4

(a)

(b)

7.

i

u

1 m

3 0 4

v w

u v

u v

v w

w

v w

u v

28 10 1

40 115 30

2 5 6

150 60 0

8 20 6

3m 2m 2 12 6m 1

2m 2 9m 11

3m 2m 2 12 6m 1

2m 2 9m 11

6m 3 2m 2 4 3m 12

8 20

28 10

80 160

6

1

640

28 10 1

8 20 6

2m 2 9m 11

80 160 640

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uw v

(e)

8.

u v

2

30

70

40

(6 0 24) 2

3 0 32 5

60

175

115

8

6

240

210

30

1

3

30

180

150

wv u (6 0 24) 2

2 10 48 0

60

0

60

8

4

240

240

0

uv w

wu v

(f)

1

6 4

3 1

1

5

19 33 , u v 18

1774 1774

There are two unit vectors: 9.

192

332 182

1774

19 33 18

We need a vector perpendicular to both AB and AC . 1 1 8

AB AC

1 3

3 1

4 4

2 Vector AB AC is parallel to 1 whose magnitude is 4 1 1 1 There are two unit vectors:

10.

(a)

(b)

u v

u v

6.

2 6 1 6 1

2 0 3

1 4 2

12 1 , Area 8

u v

122 12 82

209

3 4

0 3

u v

72

139

1

1

7 3 , Area 9

32

92

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11.

Denote the points as: A(2, 1,1), B(5,1, 4), C (0,1,1), D(3,3, 4).

5 2 1 1

Vectors AB

3 2 and DC 3

4 1

parallelogram. Since Area

3 0 3 1 4 1

3 2 are the same; hence, ABCD is a 3

AB AC , we have to find the vector product of those

vectors:

AB AC Area

12.

3

2

6

2

2

6

3

0

10

AB AC

36 36 100

172

2 43

The points are coplanar if the vectors PQ, PR, PS are coplanar. That means their scalar triple product must be zero.

1 1 , PR 1

PQ

1 1 2 3 4 5 13.

1 2 4

2

4

3 , PS 2

5 4

0

We will find the scalar triple product of the vectors: 2 m 4 m 3 m

AB

1

, AC

3 0

m 2 3 m 2 m 4 m

1 m 2 3 0 5 6

3 m ( 18) 8m2

5 6

0

2 m 6

30m 22

, AD

0

m 2 10 5m 12 3m m1 1, m2

0

11 4

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14.

(a)

The area of the triangle is half the area of the parallelogram formed with AB and AC

1 AB AC

1

5

1

2

A (b)

5 6

1 169 25 36 2

230 2

The area of the triangle is half the area of the parallelogram formed with AB and AC

1 4 8

3 0 10

1 AB AC 2

Area

40 34 12

1 402 342 122 2

2900 2

5 29

Check the scalar triple product on page 401 of the book. 3 2 2 (a)

(b)

(c)

16.

3

1 AB AC 2

AB AC

15.

13

(a)

5

2

2

1

2

6

2 1 3

1 4 2

128

3 3 2

3 1

2 1 3 1

5

1

21

1

2

Since volume

u v w , we have to find the scalar triple product:

3 1 3

78

5 5 2

3 1 3

So, volume 78

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(b)

Since volume

u v w , we have to find the scalar triple product:

4 2 3 5 6 2

63

2 3 5 So, volume 63 17.

(a)

Vectors are coplanar if their scalar triple product is zero: 2 1 2

4

1

6

3

1

30 0

1

So, they are not coplanar. (b)

Vectors are coplanar if their scalar triple product is zero: 4 2 1

9

6

6

6

1

0

1

So, they are coplanar. 18.

(a)

Vectors are coplanar if their scalar triple product is zero: 1 m 1

3

0

m

5

4

0

4m m

5m

Hence, 5m2 4m 12 0 (b)

12

m1

5m 2 4m 12

2, m2

6 5

Vectors are coplanar if their scalar triple product is zero: 2 3 2m

m

3

1

1

3

2

2 3

3 2m 1

2 m 3m 3

6m 2 3

Hence, 6m 2 3 0 has no solution, so they cannot be coplanar.

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19.

(a)

(b)

Since volume

u v w , we have to find the scalar triple product:

1 4 2 3 2 1 1 1 4

49

So, volume

49

u v

3

1

0

2

4

7

1

2

14

The area is: area

(c)

Since volume base h

1

(d)

0 72 142

u v

1 4

volume base

h

Vtetrahedron

cos 1 7 5 3

Hence, Vtetrahedron

(b)

7 5 5

7 14

49 7 10 30 18 7 5 21 10 So, the acute angle between the vector w and a vector perpendicular to the plane cos w , u v

plane is 90

(a)

49 7 5

7 5

0

1

determined by u and v is cos

20.

245

AB

3 1 , AC 3

1

7 10 . Hence, the angle between w and the 30

7 10 . 30 7 5 5

49 3

1 Vparallelepiped 3

1 u v w 3

2 2 , AD 1

4 4 3

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3 2

1 2

3 1

4

4

3 4 3

Hence, V 21.

From the definitions, we have: uv

u v cos 22.

u v sin

2

2

2

uv

2

2

u v sin 2

24.

cos

u v cos , u v

sin . Since 0

u v sin , so:

180 , it follows that

45 .

We will transform the right side of the formula:

u v

23.

4

2

u v

2

2

2

2

u v cos2

u v sin

2

u v 1 cos2

u v

We will transform the right side of the formula: AP AB

AP AB sin

AB

AB

(a)

d

In this case, the distance will be: d

BA BC

4 0 2

Hence, d (b)

AP sin

3 2 4

Hence, d

42

222

82

32

22

42

2

2

0

2 0

1 0

0 6

6 2

2

2

1

BC

4 22 8 564 29

In this case, the distance will be: d

BA BC

BA BC

6 5

564 29

BA BC BC

6 5 5

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(c)

BA BC

1

2

2

2

2

2

1

0

2

22

22

Hence, d 25.

2

2

22 2

BC

12 8

2

3 2

We will use the distributive property of the vector product and the fact that the vector product of parallel vectors is a zero-vector.

u v 26.

BA BC

In this case, the distance will be: d

v u

u v u u v v v u u v 0 0 u v 2(u v )

We will use the distributive property of the vector product and the fact that the vector product of parallel vectors is a zero-vector. 2u 3v 4v 5u 8u v 10u u 12v v 15v u 8u v 0 0 15u v

27.

We will use the distributive property of the vector product and the fact that the vector product of parallel vectors is a zero-vector. mu nv

28.

23(u v)

(a)

pv qu

AB AC

Hence, o (b)

mp u v

mq u u

np v v

mp u v

0 0 nq u v

a b

a 0

i j k a b 0

bc ac

0

c

a 0 c

ab

1 AB AC 2

1 2 2 bc 2

a 2c 2

nq v u

mp nq u v

a 2b 2 .

The faces are right-angled triangles, so we can find the area using the half product of the legs: A1

ab , A2 2

ac , A3 2

bc 2

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(c)

Hence,

A1

2

2

A2

A3

2

a 2b 2 4

a2c2 4

b2c 2 4

a 2b 2 a 2 c 2 b 2 c 2 4

a 2b 2 a 2 c 2 b 2 c 2 2 i 29.

k

2z 3 y

1 2 3

z 3x

x

y 2x

So,

i 30.

j y

z

2z 3y

1

3y 2z 1

z 3x y 2x

5 3

3x z 5 2x y 3

j

k

2z 3 y

1 2 3 x y z

z 3x y 2x

2z 3 y So, z 3x y 2x

1 5 0

3 y 2z 1 3x z 5 2x y 0

x

2

5 1 z, y 3 3

o2

1 2 z, z 3 3

z

1 z 0 3 2 y z 0 3 0 1 x

Trying to reduce the resulting system of equations, we notice that it is inconsistent. Hence, there is no such vector.

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Exercise 9.4 1 1.

(a)

In the equation r

r0 tu , vector r0 1 0

equation of the line is r

0 , and u 2

1 t 5

2 x y z

, so the vector

4

. For the parametric equations:

1 t

x

0 5t ; therefore, the parametric equations are y 2 4t z 1

3 1 and u 2

Substituting r0

the line: r

3 1

y 5

y 1 5

1 2 and u 6

1 2 6

t

3 5

.

2 4t

r0 tu , we get the vector equation of

x . The parametric equations are: y 1 z 2

of the line: r

5t

2 4

2 5 into r 1

The cartesian equations are x 3

Substituting r0

z

1 t

2 t 5

2

(c)

5

4

The cartesian equations are x 1

(b)

1

3 2t 1 5t . 2 t

z 2 . 1

into r

r0 tu, we get the vector equation

11

x 3 5 . The parametric equations are: y 11 z

The cartesian equations are x 1 3

y 2 5

1 3t 2 5t . 6 11t

z 6 . 11

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1 2.

(a)

In the equation r

r0 tu , vector r0 rA

4 , and u 2

1 4

so the vector equation of the line: r

AB

7 1

8

5 4

1 , 2

0 2

8 t 1

2

2

Note: For r0 we can use rB and for u we can use BA , or any vector parallel to this vector. Therefore, it is possible to find different, but correct, equations of the line.

(b)

Substituting r0

rA

4 2

and u

0 4 2 2

AB

3

4 4 into r 4

1 3 4

we get the vector equation of the line: r

2 3

Note: Since u is parallel to AB , we can use u

equation of the line would be r

4 2

(c)

Substituting r0

rA

4 t

4 4

1 AB 4

1 1 , and the vector 1

1 s 1

3

1 3 and u 3

r0 tu,

AB

we get the vector equation of the line: r

1

5 1 1 3 2 3

4 2 into r 5

1 3

4 t 2 .

3

5

r0 tu,

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3.

In the equation r

(a)

u

(b)

5 3 1 2

In the equation r

u

4.

AB

AB

5 0 0 2

3 and 2

a tb , we substitute a rA

2 , so the equation of the line is r 3

3 2

t

2 . 3

0 2

t

5 . 2

0 and 2

a tb , we substitute a rA

5 , so the equation of the line is r 2

Method 1 To determine the equation of the line in the required form, we need to find two points on the line r (2,1) t (3, 2) . One point is (2,1) ; another point we can find by letting, for example, t 1; therefore, the point is (5, 1) . The equation of the line through those two points is: y 1 x 2

1 1 5 2

3 y 1

2 x 2

2x 3 y

7

Method 2 We can write the equation of the line r x y

2 3t . From the first row: x 1 2t

row, we get: y 1 2t 1 2 x 2 3

5.

6.

In the equation r

2 1 2 3t

3( y 1)

3

t

in parametric form:

2 x

t

2 3

. Substituting t into the second

2( x 2)

3 j . So: r

4i

In the equation r

2 1 and u 4

r

t

7

r0 tu , the vector r0 rA 2i 3 j and u can be the same as the

direction vector of the given line; therefore, u

2 1 4

2x 3 y

r0 tu , the vector r0 rA

2i 3 j

4i 3 j

3 4 . So, we have: 7

3 4 7

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1 7.

(a)

1

The lines are not parallel since the direction vectors 3 and 4 are not a scalar 2 1 multiple of each other. For lines to intersect, there should be some point

2

x0 , y0 , z0 which satisfies the equations of both lines, r

2 3

2 3 4

r

1 t 3 and 1

1 s 4 , for some values of t and s. (Note: We have to change the 2

parameter in one of the equations so that they are not the same.) So:

x0 2 t 2 s y0 2 3t 3 4s z0 3 t 4 2 s From the first equation we see that t s. Substituting into the second equation: 2 3t 3 4t t 1 t s 1. Finally, substituting these values into the 2 2 . Hence, the lines intersect, and the point of third equation: 3 1 4 2

2,2,3

intersection is:

1 1,3,1

1, 1, 2 . 4

(b)

12 The lines are not parallel since the direction vectors 1 and 6 are not a 3 0 scalar multiple of each other. For lines to intersect, there should be some point 1 4

x0 , y0 , z0 which satisfies the equations of both lines, r 13 r

1 2

x0

1 4t

3

t 1

1

0

and

12 s 6 , for some values of t and s. So: 3

13 12s

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z0 1 0 t 2 3s 1 . Substituting into the second equation: 3

From the last equation, we see that s

1 3 equation: 1 16 3 t 1 6

of intersection is: r

t

4. Finally, substituting these values into the first 13 4

1,3,1

17

17 . Hence, the lines intersect, and the point

4 4,1,0

17, 1,1 . 7

(c)

The lines are not parallel since the direction vectors 1

1 and 0 2 3

are not a

scalar multiple of each other. For lines to intersect, there should be some point

x0 , y0 , z0 which satisfies the equations of both lines, r

1 3

7 t 1

5 r

4 6 7

and

3

1 s 0 2

, for some values of t and s. So:

x0 1 7t 4 s y0 3 t 6 0 s z0 5 3 t 7 2 s From the second equation, we can see that t 3 . Substituting into the first 18 . Finally, substituting these values into the last equation: 1 21 4 s s equation: 5 3 3 7 2 18

4

29 . Hence, the lines do not intersect;

they are skew. (d)

The lines have parallel direction vectors

2 1 and 1

4 2 , since 2

4 2 2

2 2 1 . 1

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To check whether the lines coincide, we examine the point 3, 4,6 , which is on

x the first line y z

3 4 6

x

5

4

y

2

s 2

z

7

2

3

5

4

So: 4 6

2

s 2

7

2

2 t 1 , and see whether it lies on the second line also. 1

3 5 4s 4

2 2s

s

1 2 s 3

6 7 2s

We can see that the point is not on the other line, so the lines do not coincide; therefore, the lines are parallel.

8.

(a)

A direction vector is:

2 1

r

t

3 2 2

1 x and 3 y

1

1 ; hence, the equations are: 3

2 t 1 3t

(b)

We know a point and a direction vector, so the equations are: 2 3 x 2 3t r t and 1 7 y 1 7t

(c)

For the direction vector, we can use any vector perpendicular to So we use vector

3 7

x y

7 3

3 . 7

7 as the direction vector of the line, since 3

21 21 0 . Therefore, the equations are: r

2 1

t

7 and 3

2 7t . 1 3t

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(d)

9.

(a)

We know a point and a direction vector, so the equations are: 0 2 x 2t and r t 2 4 y 2 4t

Substituting the point

0

x 11 9 0, , into the equation y 2 2 z

3

2

4

t 1 :

6

1

0 3 2t

11 2 9 2

3

2

4

t 1

6

11 4 t 2 9 6 t 2

1

3 . Check this value in the second 2 9 9 equation: 11 4 3 11 11 , and in the third equation: 9 6 3 . 2 2 2 2 2 2 2 2 So, the point is on the line when t 3 . 2

From the first equation, we can see that t

(b)

To check whether the point is on the line, we have to determine whether or not the system of equations has a solution: 1 3 2 1 3 2t

4 6

4 6

t 1 1

4 4 t 6 6 t

From the last two equations, we can see that t 0 , but this will not satisfy the first equation; hence, there is no solution to the system and the point does not lie on the line. (c)

We have to solve the system of equations: 1 2m 2 2m 3

3 4

2 t 1

6

1

1 2m 3 2t 2 2m 4 t 3

6 t

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From the last equation, we can see that t 3 . Substituting into the second equation: 2m 7 1 2

7 2

10.

(a)

(b)

(c)

11.

(a)

6 2

3 6

2

m

7 . Check using the first equation: 2

3. Therefore, the point will be on the line when m

(i)

The starting position is when t 0 , so the point is (3, 4)

(ii)

The velocity vector is v

(iii)

The speed is v

(i)

The starting position is when t 0 , so the point is ( 3,1)

(ii)

The velocity vector is v

(iii)

The speed is v

(i)

The starting position is when t 0 , so the point is (5, 2)

(ii)

The velocity vector is v

(iii)

The speed is v

72

25

5 12

12

2

13

24, 7 7

2

25

The direction of the velocity vector is given by the unit vector: 3 1 3 1

3

2

4

42

5 4

So, the velocity vector is: 160

(b)

242

52

242

7 24

1 3 5 4

32

3 4

96 128

The direction of the velocity vector is given by the unit vector: 12 1 1 12

122

5

2

5

13

5

1 12 So, the velocity vector is: 170 13 5

2040 13 850 13

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7 . 2

12.

(a)

The car is travelling from the point 3, 2 to 7,5 , so the direction vector of the velocity vector is given by the unit vector: Therefore, the unit vector is:

30 (b)

1 4 5 3

4 3

1 4

2

2

3

4 . 3

1 4 ; and the velocity vector is: 5 3

24 . 18 24 , so the 18

The starting point is 3, 2 and the direction vector of the line is

3 2

equation of the position of the car after t hours is r (c)

7 3 5 2

1 v , where v v

t

24 18

We have to determine the parameter of the point with position vector

7 5

3 2

t

24 18

4 3

t

24 18

7 5

1 6

t

Therefore, in 1 of an hour, i.e. in 10 minutes, the car will reach the traffic light. 6

13.

(a)

To be perpendicular to the vectors, both dot products must be zero.

1 a b

1 3 2

1 3a 2b

1 0 , and a b

So, we have to solve the system:

(b)

cos

v w v w

12

3

2

2 1

2 a b 0

1 3a 2b 1 a b 2

1 3 2

2 1 1

22

2

2

12

1

a

2

3, b

7 2 21

5

21 6

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(c)

2 Using the Pythagorean identity for sine, sin positive for angles from 0 180 we have:

1 cos 2

sin

21 36

1

14.

(a)

15 6

1 OP OQ sin POQ 2

Area of triangle OPQ is: A

1 15 14 6 2 6

So, Area

15 36

1 cos2 , and the fact that sine is

1 v w sin 2

v, w

35 2

First, we have to determine vectors AB and AC : AB

cos

1 1 3 2 5 3

0 1 , AC 2

0

1

1 2

3 2

02 12 22 12

Therefore,

cos

1

3

7 5 14

(b)

The area of the triangle is: A

(c)

(i)

0

2

1 1 2 1 3

2

2

1 3 2

0 3 4 5 14

7 5 14

147 1 AB AC sin 2

1 5 14 sin 2

2.29

Line L1 goes through the point 2, 1,0 and its direction vector is

AB

0 1 , so its equation is: r 2

Line L2 goes through the point

AC

1 3 , so its equation is: r 2

2 1 0

0 t 1 2

1,1,1 and its direction vector is 1 1 1

1 s 3 2

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(ii)

We have to solve the system of equations: 2 0 1 1 2 1 s

1

t 1

1

2

1

0

s

3

1 t 1 3s

2

2t 1 2s

From the first equation, we have s 3 , and from the second t Substituting these values into the third equation:

2 7 15.

(a)

1 23

14

5 . So, there is no point of intersection.

Let the direction vector be a vector parallel to AB : AB

6 1 7 3 8 17

5 10 25

1 5 2 5

Thus, we can use the vector

1 2 as the direction vector. 5

x Therefore, the parametric equations of the line are: y z (b)

7.

If point P is on the line, then vector OP

1 t 3 2t

1 t 3 2t 17 5t

. If OP is perpendicular to

17 5t the line, then OP and the direction vector of the line are perpendicular, and their scalar product is zero. 1 1 t 1

0 OP 2 5 So, OP

3 2t 17 5t

1 3 3 2 3 17 5 3

2 5

1 t

6 4t

85 25t

30t 90

t

4 3 and P 4, 3, 2 2

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3

16.

(a)

We will find a point and a vector on the line. Let y

0 , then x

p ; hence, point m

Let x

0 , then y

p ; hence, point n

p 0 Therefore, the vector m p 0 n

r

(b)

(i)

p m 0

n

t

m

0,

p m

n p n

p is on the line. n is parallel to the line. This vector is

p n

p m

mn parallel to the vector p

p ,0 is on the line. m

m

. So, a vector equation of the line is:

.

We already have one point on the line. To determine another point on the x0

line r

y0

t

a b

we can let, for example, t 1 ; therefore, the point is:

( x0 a, y0 b) . The equation of the line through those two points is: y y0 x x0

y0 b y0 x0 a x0

line is: bx ay (ii)

bx0 ay0 .

bx bx0

ay0

Parameterisation of a segment: r t

(a)

r t

b x x0 . Hence, an equation of the

We will write the equation of the line in slope-intercept form: ay

17.

a y y0

1 t

0 0 0

y

b x a

b x0 a

y0 . Hence, the slope of the line is

1 t OA tOB, 0 t 1

1 t 1 , 0 t 1 ; hence, r t 3

1 t 1 ,0 t 1 3

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b a

1 r t

(b)

1 t

1

0

t 1

1

2

1 t t t

, 0 t 1 ; hence,

1 t 2t

1 2t r t

t

,0 t 1

1 3t r t

(c)

1 t

1

0

0

t 3

1 18.

1 t 3t

0

1 t , 0 t 1 ; hence, r t

1 t

x y z

1 t

2 j 3k

t 2k . The parametric equations are:

0 2 3 2t

Note: We can write a vector equation in the form: r A direction vector of the parallel line is 2i 3 j whose equation we have to find is: r equations are:

2j

3 2t k

k ; hence a vector equation of the line

i 2j k

t 2i 3 j k . The parametric

x 1 2t y 2 3t z 1 t

Note: We can write a vector equation in the form: r 20.

,0 t 1

A direction vector of the parallel line is 2k ; hence, a vector equation of the line whose equation we have to find is: r

19.

3t

A direction vector of the line is x0 i

(1 2t )i

2 3t j

1 t k

y0 j z0 k ; hence, a vector equation of the line is

r 0 t x0i y0 j z0k , and the parametric equations are:

x y z

tx0 ty0 tz0

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21.

(a)

A direction vector of the line is a vector perpendicular to the xz-plane; hence, j . Therefore, a vector equation of the line is r

x equations are:

y

2 t 3

A direction vector of the line is a vector perpendicular to the yz-plane; hence, i . Therefore, a vector equation of the line is r

x equations are:

22.

tj , and the parametric

3

z (b)

3i 2 j 3k

3i 2 j 3k

ti , and the parametric

3 t y

2

z

3

A direction vector of the line is x0 i equations of the lines are:

x 0 x0

y0 j z0 k . Hence, the symmetric (Cartesian)

y 0 y0

z 0 z0

x x0

y y0

z z0

Note: We took the origin as a point on the line. We can take point A , then the equation will be:

x x0 x0

y y0 y0

z z0 z0 1

23.

(a)

0 1 and 1 are not a The lines are not parallel since the direction vectors 2 1 scalar multiple of each other. For the lines to intersect, there should be a point which satisfies the equations of both lines. x 3 t x 1 We will write the equations in parametric form:

y 1 t , y 4 z 5 2t z 2

and solve the system:

3 t 1 1 t 4 5 2t

2

From the first equation, we can see that t

2.

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From the second, 1 2 4

1; and, finally, substituting those values

into the third equation, we have: 5 2

2

intersect, and the point of intersection is:

2

1

1 1 . Hence, the lines

x 3 2 1 y 1 2 3 1,3,1 z 5 4 1

Note: If it is a Paper 2 question, we can solve the system using matrices, or any other GDC specific method available. First, transform the system of equations:

3 t 1 1 t 4 5 2t

t t 2 l

2 3

2t

3

and then use a GDC. Substitute back into one of the equations to find the intersection point. Notice that it is easier to solve the system without a GDC since, while preparing the system for the GDC, we would find the solutions.

1 (b)

3 and 6

The lines are parallel since for the direction vectors

2

2

2 it holds:

4

1

6

2 4

3 . To check whether the lines coincide, we examine the point 2

1, 2,1 , which is on the first line, and see whether it also lies on the second line x 2 2m y 1 6m z

4m 1 2 2m

So, 2

1 6m 1

4m

3 2m

m

3 6m

m

1

3 2 3 6

4m

We can see that the point is not on the other line, so the lines do not coincide; therefore, the lines are parallel.

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2 (c)

The lines are not parallel since the direction vectors 4

2 and 1 are not a 1 2

scalar multiple of each other. For the lines to intersect, there should be a point which satisfies the equations of both lines.

x 3 2t We will write the equations in parametric form: y

x 3 2

1 4t , z 2 t

y z

2

and

2 2

solve the system:

3 2t 3 2 1 4t

2

2 t

2 2

From the first equation, we can see that t . From the second, 1 4t 2 t t 1 ; and, finally, substituting those values into the third equation, we have: 2 1 2 2 1 0 . Hence, the lines do not intersect. They are skew.

1 (d)

1

1 are not a The lines are not parallel since the direction vectors 3 and 2 2 scalar multiple of each other. For the lines to intersect, there should be a point which satisfies the equations of both lines. x 1 t x 1 We will write the equations in parametric form:

y 1 3t , y 1 z 4 2t z 2

and solve the system:

1 t 1 1 3t 1 4 2t

2

From the first equation, we can see that t 1 3t

1 t

t

. From the second,

1 ; and, finally, substituting those values into the third 2

equation, we have: 4 2

1 2

2

1 2

5

1 . Hence, the lines do not

intersect. They are skew. © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

6 (e)

The lines are parallel since for the direction vectors 9

2

3

6

3 , it holds:

and

1

2

9

3 . To check if the lines coincide, we examine the point 1,2,0 ,

3 3

1

which is on the first line, and see whether it also lies on the second line.

x 2 2m y 3 3m z m 1 2 2m So, 2 3 3m 0 m

1 2m

m

1 2

0 m

We can see that the point is not on the other line, so the lines do not coincide; therefore, the lines are parallel. 5 3 (f)

The lines are not parallel since the direction vectors 1 and 3

3 are not a 4

scalar multiple of each other. For the lines to intersect, there should be a point which satisfies the equations of both lines. x 2 5t x 4 3 We will write the equations in parametric form:

y 1 t , z 2 3t

y

7 3

z 10 4

and solve the system:

2 5t 4 3 1 t 7 3 2 3t 10 4

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From the second equation, we can see that t

2 5 6 3 2, t

4 3

6 3 2

18

36

6 3 . From the first,

2 ; and, finally, substituting the values

0 into the third equation, we have: 2 3 0 10 4 2

2 2.

x Hence, the lines intersect, and the point of intersection is:

2 5 0 y 1 0 2,1, 2 z 2 3 0

1 (g)

2 2 and

The lines are not parallel since the direction vectors

1

9 are not a 6

scalar multiple of each other. For the lines to intersect, there should be a point which satisfies the equations of both lines. We have to change the name of the parameter in one of the equations and then solve the system:

1 t

2 2

2 2t 5 9 t 5 2 6 From the first equation, we can see that t 1 2 . From the third, 1 2 5 2 6 4 4 1; and, finally, substituting the values 1, t 1 2 1 3 into the second equation, we have:

2 23 5 91 intersection is:

24.

4 4 . Hence, the lines intersect, and the point of x 1 3 y 2 2 3 4, 4,8 z 3 5

The parametric equations of the line are:

x 2 3t y 3 t z 1 t

The distance from the origin to a point on the line is: d2

2 3t

2

3 t

2

1 t

2

14 4t 11t 2 . Since this is a parabola that

opens upwards, the distance is a minimum when

d 14 4t 11t 2 dt

0

4 22t

0

t

2 11

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2 is the closest to the origin. 11

Hence, the point on the line with t

x

2 11 2 y 3 11 2 z 1 11

16 11 35 11 13 11

2 3

A

16 35 13 , , 11 11 11

x 25.

The parametric equations of the line are:

t

y

4 3t

z

5 t

The distance from the origin to a point on the line is: d2

t2

4 3t

2

5 t

d 41 14t 11t 2

0

dt

2

41 14t 11t 2 . The distance is a minimum when

14 22t

Hence, the point on the line with t

7 11 7 23 y 4 3 11 11 7 62 z 5 11 11

0

t

7 11

7 is the closest to the origin. 11

x

26.

A

7 23 62 , , 11 11 11

x 5 t y 2 3t z 1 t

The parametric equations of the line are: The distance from the point d2

5 t 1

2

2 3t 4

1, 4,1 to a point on the line is: 2

1 t 1

2

40 24t 11t 2 .

The distance is a minimum when

d 40 24t 11t 2 dt

0

24 22t

Hence, the point on the line with t

0

t

12 11

12 is the closest to the origin. 11

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12 43 11 11 12 58 y 2 3 11 11 12 1 z 1 11 11 x 5

43 58 1 , , 11 11 11

A

Exercise 9.5 1.

For A : 3 3 2 ( 2) 3 ( 1) 8 11 ; hence, A does not lie in the plane. For B : 3 2 2 1 3 ( 1) 11 ; hence, B lies in the plane. For C : 3 1 2 4 3 0 11; hence, C lies in the plane.

2.

For A : i 3 j k 3i 2 j k

3 6 1 8

For B : i 3 j k 2i

2 3 2

j 2k

For C : i 3 j k i 4 j 0k 3.

(a)

1 12

6 ; hence, A does not lie in the plane. 3

11

6 ; hence, B does not lie in the plane. 6 ; hence, C does not lie in the plane.

A Cartesian equation for the plane is:

2 x 3

4 y 2

3 z 4

2x 4 y 3z

0

2x 4 y 3z 6 8 12

26

A vector equation for the plane is: 2 x 2 3 2 x

4 3 (b)

y z

4 3

2 4

4 3

y z

26

A Cartesian equation for the plane is:

2 x 3

0 y 2

3 z 1

0

2x 3z

6 3

2x 3z 3 A vector equation for the plane is: 2 x 2 3 2 x 0 y 0 2 0 y 3 3 z 3 1 3 z

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(c)

A Cartesian equation for the plane is:

0 x 0

0 y 3

3 z 1

0

3z 3

A vector equation for the plane is: 0 x 0 0 0 x

0 3 (d)

y z

0 3

3 1

0 3

y z

3

A vector perpendicular to the plane is the same as a vector perpendicular to the

5 1

parallel plane; hence, it is

2 So, a Cartesian equation for the plane is:

5 x 3

1 y 2

2 z 4

5x

y 2z

5

0

5x y 2z 15 2 8

A vector equation for the plane is: 5 x 5 3 5 x

1 2 (e)

y z

1 2

2 4

1 2

y z

5

A vector perpendicular to the plane is the same as a vector perpendicular to the 0

1 . 2

parallel plane; hence, it is

So, a Cartesian equation for the plane is:

0 x 3 1 y 0

2 z 1

0

y 2z

2

A vector equation for the plane is: 0 x 0 3 0 x

1 2

y z

1 2

0 1

1 2

y z

2

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2 (f)

The plane is parallel to a direction vector of the line 1 and the vector 2

3 2 4

1 2 5 3 2 4

r

2 0 . So, a parametric equation for the plane is: 1 2 1 2

2 0 1

From vectors parallel to the plane, we can find a vector perpendicular to the plane:

2 0 1

2 1 2

1 6 . Therefore, a vector equation for the plane is: 2

1

x

1

3

1

x

6 2

y z

6 2

2 4

6 2

y z

23

A Cartesian equation for the plane is: x 6 y 2 z 23 . Note: We can find other vectors parallel to the plane, or other points on the plane; hence, we can obtain different parametric and vector equations of the plane. 2 3 (g)

The plane is parallel to the direction vectors of the lines 1 and 2 . So, a 2 2 parametric equation for the plane is: 1 2 3

r

2 5

1 2

2 2

From vectors parallel to the plane, we can find a vector perpendicular to the plane: 2 3 2

1 2

2 2

2 . 1

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Therefore, a vector equation for the plane is: 2 x 2 1 2 x

2

y

2

2

2

y

1

z

1

5

1

z

1

A Cartesian equation for the plane is: 2 x 2 y z

1.

2 (h)

The plane is parallel to a direction vector of the line 1 and the vector 3

1 3

0 2

2

1 5 . So, a parametric equation for the plane is: 3

1 1 3 2

r

2 1 3

1 5 3

From vectors parallel to the plane, we can find a vector perpendicular to the plane:

2 1

1 5

3

3

18 3 . Therefore, a vector equation for the plane is: 11

18 3

x y

18 3

1 3

18 3

x y

11

z

11

2

11

z

5

A Cartesian equation for the plane is: 18x 3 y 11z 5 . (i)

p q ; hence, a Cartesian equation for the plane is: r

Vector OM

p x p

q y q

r z r

0

px qy rz

p2 q2 r 2

A vector equation for the plane is: p x p p p x

q r

y z

q r

q r

q r

y z

p2 q2 r 2

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(j)

The plane is parallel to vectors

3 1

2

7 3

4

1 2

3 and 2

0 1

1 . 2

0 2

2 0

So, a parametric equation for the plane is:

1 2 2

r

2 3 2

4 1 2

From vectors parallel to the plane, we can find a vector perpendicular to the plane: 2 4 8 4

3

1

4

2

2

14

2

2 . Therefore, a vector equation for the plane is: 7

4

x

4

1

4

x

2 7

y z

2 7

2 2

2 7

y z

14

A Cartesian equation for the plane is: 4 x 2 y 7 z 14 .

(k)

The plane is parallel to vectors

3 2 1 2

1 0 3 1 and 1 1 5 5 3

3 2

3 2 . 2

So, a parametric equation for the plane is: 2 1 3

r

2 2

1 5

2 2

From vectors parallel to the plane, we can find a vector perpendicular to the plane: 1 3 8 8

1 5

2 2

8 17 5

x y z

17 5 8 17 5

1 17 . Therefore, a vector equation for the plane is: 5 2 2 2

8 17 5

x y z

8

A Cartesian equation for the plane is: 8 x 17 y 5 z

8.

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1 (l)

The plane is parallel to a direction vector of the line 1 and the vector 3

2 1 5

1 2 3 1 2 3

r

1 1 . So, a parametric equation for the plane is: 2 1 1 3

1 1 2

From vectors parallel to the plane, we can find a vector perpendicular to the plane:

1 1 2

1 1 3

1 1 . Therefore, a vector equation for the plane is: 0

1

x

1

1

1

x

1 0

y z

1 0

2 3

1 0

y z

3

A Cartesian equation for the plane is: x y 3 . (m)

3 The plane is parallel to a direction vector of the line 2 and the vector 4

3 4

1 1

0

5 1 1 5

r

4 5 . So, a parametric equation for the plane is: 5 4 5 5

3 2 4

From vectors parallel to the plane, we can find a vector perpendicular to the plane: 4 3 30

5 5

2 4

1 . 23

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Therefore, a vector equation for the plane is: 30 x 30 1 30 x

1

y

1

23

z

23

1 5

1

y

23

z

86

A Cartesian equation for the plane is: 30x

y

23z

86 .

1 (n)

1

The plane is parallel to the direction vectors of the lines 0 and 1

1 . 1

So, a parametric equation for the plane is:

1 1 0

r

1 0 1

1 1 1

From vectors parallel to the plane, we can find a vector perpendicular to the plane:

1 0 1

1 1 1

1 0

x y

1

z

1 0 . Therefore, a vector equation for the plane is: 1 1 0

1 1

1 0

1 0 1

x y

1

z

A Cartesian equation for the plane is: x z 1 . 4.

(a).

The angle between the normals is given by: 3 1

cos

4 1 1

2 0

9 16 1 1 4

5 26 5

Since the angle between the planes is by definition the acute angle between the planes, the angle between the planes is: cos

1

5 26 5

63.98....

64.0

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(b)

The angle between the normals is given by: 4 3

7 1

2 2

0 16 49 1 9 4 4 66 17 Hence, the angle between the planes is: cos 1 0 cos

(c)

1

The angle between the normals is given by 1 1

cos

1

0

0

0

1

1 2

1 1 1

Hence, the angle between the planes is: cos

(d)

90

1

1 2

45

For the angle between the normal and the direction line, it holds: 1 6

cos

2

3

2

2

1

16 3 7

1 4 4 36 9 4 Hence, the angle between the plane and the line is: sin

(e)

1

16 21

49.6324...

49.6

For the angle between the normal and the direction line, it holds: 3 1

0 1

2 1

2 9 1 1 4 1 60 Hence, the angle between the plane and the line is: cos

1

sin

1

2 60

14.9632... 15.0

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(f)

The angle between the normals is given by: 1 0

cos

1 1 1

0 1

1 3

1 1 1 1

Hence, the angle between the planes is: 5.

(a)

cos

1

1 3

54.7356...

Parametric equations of the line are:

x 5 l y z

3l 2 4l

A Cartesian equation of the plane is: x 3 y 2 z For the intersection, we have:

5 l

3 3l

2 2 4l

35

18l

36

x 5 2 3 Hence, the point is: y 3 2 6 z 2 4 2 10 (b)

35

l

2

3, 6, 10

Parametric equations of the line are: x 2

y 4 3 z 3 For the intersection, we have:

4 2

2 4 3

33

30 0

15

30

2

2, 2, 6

x

Hence, the point is:

2 4 3 2

2

y z

3 2

6

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54.7

(c)

The direction vector of the line and the normal of the plane are perpendicular 1 2 (since 5 3

4 6

2 20 18 0 ); hence, the line and the plane are either parallel

or the line is in the plane. Since the point 3,4,6 is not on the plane, they are parallel and there is no intersection. Note: If we solve the system for intersection, we will have: Parametric equations of the line are: x 3 t

y z

4 5t 6 3t

A Cartesian equation of the plane is: 2 x 4 y 6 z For the intersection, we have: 2 3 t

4 4 5t

5

6 6 3t

5

26 5 ;

hence, there is no intersection. (d)

The direction vector of the line and the normal of the plane are perpendicular

(since

1 1 3 5 3

3 1 2

3

1 10 3 3

0 ); hence, the line and the plane are either parallel

or the line is in the plane. The point 0,4,5 is on the plane, so the line is in the plane. Note: If we solve the system for intersection, we will have: For the intersection, it holds: 3 t

1 t 3

4

2 5

5 t 3

6

4 10 6 ; hence,

all points from the line are on the plane.

6.

(a)

Solving the system: x 10

x y z 3

10 y z 3

y

7 z

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Hence, parametric equations of the line of intersection are:

x 10 y 7 t and the vector equation is r z t (b)

10 7 0

0 t 1 1

We have to solve the system: 2x y z 5

x y z

4

Adding the equations, we have: 3x 9 y z 1 y 1 z y z 1

x 3 ; therefore:

Hence, parametric equations of the line of intersection are:

x 3 y 1 t and the vector equation is r z t

3 1 0

0 t 1 1

(c)

A Cartesian equation of the first plane is: x y 2 z 1 , so the planes are parallel and there is no intersection. Note: If we solve the system for intersection, we will have: x y 2z 1 , and this is obviously inconsistent. x y 2z 5

(d)

A vector perpendicular to the first plane is:

1 2

3 2

16 8

0

8

8

2 x 1

y 0

2 8 1 ; hence, a Cartesian equation of the plane is: 1

z 2

0

2x y z 4 .

Now, we have to solve the system: 2x y z 4 . 3x y z 3

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Adding the equations, we have: 5 x 2 3

7 5

y z

7 5

y z

4

y z

3

6 5 6 5

y z

7

x

y

6 z 5

7 ; therefore: 5

Hence, parametric equations of the line of intersection are: 7 7 x 0 5 5 6 6 t 1 and the vector equation is r y t 5 5 1 z t 0 7.

A direction vector of the line is perpendicular to the normals of both planes:

2 1 1

1 2 1

3 3 3

1 3 1 . Hence, the normal to the required plane is 1

equation for the plane is: 1 x 1 2

1 1

y z

1 1

1 1 1

1 1

x y z

0

A Cartesian equation of the plane is: x 8.

1 1 , and a vector 1

y

z

0

A normal to the plane is perpendicular to the vector AB and the direction vector of the given plane: 3 1 4 2 4 2 1

2 2 1 3

1 2

0 2

1 2

12 2

2 6 . 1

Therefore, a vector equation for the plane is: 1 x 1 1 1 x

6 1

y z

6 1

2 3

6 1

y z

16

A Cartesian equation of the plane is: x 6 y z 16 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

9.

A Cartesian equation of the line is: x 1 3 1

y 2 1 2

z 5 1 5

x 1 2

y 2 1

z 5 ; hence, parametric equations of the line are: 4

x 1 2t y

2 t

z

5 4t

The distance from 2, 1,5 to a point on the line is: d2

1 2t 2

2

2 t 1

2

5 4t 5

The distance is a minimum when Hence, the point on the line with t

x 1 2

5 21

y

5 21

z 10.

2 5 4

A

2

21t 2 10t 10 .

d 21t 2 10t 10 dt

0

42t 10 0

t

5 21

5 is the closest to the origin. 21

31 37 85 , , 21 21 21

5 21

A normal to the plane is perpendicular to the direction vectors of both lines: 1 3 10 10

2 1

2 4

1 8

1 . Therefore, the plane contains the point 8

1, 2,3 and is

10 perpendicular to the vector 1 ; hence, its vector equation is: 8 10 1 8 10 x

x y z y 8z

10 1 8

1 2 3

10 1 8

x y z

32 , and Cartesian:

32

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11.

The plane contains points of the first line, so the point 1,1, 2 is on the plane. A normal to the plane is perpendicular to the direction vectors of both lines:

2 2

1

4 3 . Therefore, a vector equation for the plane is: 2

4

x

4

1

4

x

3

y

3

1

3

y

2

z

2

2

2

z

1

12.

1 2

The equation x

A

y B

z C

5 , and Cartesian: 4 x 3 y 2 z 5

1 can be written in the form BCx

ACy

ABz

ABC

BC and this is a Cartesian equation of the plane whose normal is vector

AC

and contains

AB the point A,0,0 . Note: The plane contains the points A,0,0 , 0, B,0 and 0,0,C . 13.

It is easier to write parametric equations of the plane, since we know one point and two non-parallel direction vectors: x 4 2 4

y

3

z

1

r

3

s 0

4

3

Note: If we need another form of the line, we can proceed as follows: The normals of both planes are parallel to our plane. Hence, their vector product is normal to our plane: 2 4 9

3 4

0 3

22 . Therefore, a vector equation for the plane is: 12

9

x

9

4

9

x

22 12

y z

22 12

3 1

22 12

y z

42 , and Cartesian: 9 x 22 y 12 z

42

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14.

It is easier to write parametric equations of the plane, since we know one point and two non-parallel direction vectors (a direction vector of the line and normal of the plane):

x

2

y

3

z

0

1 r

2

2

s

3

1

4

Note: If we need another form of the line, we can proceed as follows: The normal of the plane and the direction vector of the line are parallel to our plane. Hence, their vector product is normal to our plane:

2

1

5

3

2

4

1

2 . Therefore, a vector equation for the plane is: 1

5 2 1

x y z

5 2 1

5x

2y

z

2 3 0

5 2 1

x y z

16 , and Cartesian:

16

Chapter 9 practice questions 1.

2.

(a)

u 2v

(b)

The unit vector in the direction of u 2v is:

(a)

So, w

26

OA

62

OB OC

(b)

i 2j

6 52

2 3i 5 j

1 5i 12 j 13

02 11

AC OC OA

2 5i 12 j

10i

1 2

5

12

2

5i 12 j

1 5i 12 j 13

24 j

6 , so A lies on the circle.

02 2

i 2 j 6i 10 j 5i 12 j

2

6 , so B lies on the circle. 6 , so C lies on the circle.

5 11

6 0

1 11

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(c)

Method 1 Using a scalar product

cos OAC

1

6 0

AO AC AO AC

6

11 2

1

11

6 6 12

2

3 6

Method 2 Using a cosine rule in triangle OAC In triangle OAC , SSS is given: OA OC 6 , and AC

Hence, cos OAC (d)

62

12 2

2

62

12 2 12

12

1 11

12;

3 6

Method 1 Using the result from (c) Using the Pythagorean identity for sine, sin is positive for angles from 0 sin OAC

1

3 6

2

1

2

1 cos 2 , and the fact that sine

180 we have:

1 12

11 . Hence, 12

1 1 11 AB AC sin A 12 12 6 11 2 2 12 Method 2 Finding the area using side and height dimensions A

In triangle ABC , side AB 12 ; the height on this side is the second coordinate of point C , so: A 3.

u v

4i

(a)

6 11 .

11

3j

Then, a 4i 3 j

4.

1 12 2

8i

The speed of T: The speed of C:

b 2 j 18 24 36 16

4a 8

a 2

3a b 2 182 362

6 2 b

b 8

30 km/h

242 16

2

1552

39.4 km/h

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(b)

(i)

After half an hour, the vehicles have covered halve the distance: 9 1 18

2 24

12

1 36 2 16 (ii)

18 8

The vector joining their positions at 06:30 is

its position is

8

20

92

202

481

20

18 , so until k k

T must continue until its position vector is

9

12

9

the distance between the vehicles is: (c)

9 18

; hence, 21.9 km.

24 . At that point,

18 . To reach this position, it must travel for a total of one hour. 24

Hence, the crew starts work at 07:00. (d)

The southern (C) crew lays: 800 5 4000 m of cable. The northern (T) crew lays: 800 4.5 3600 m of cable. Their starting points were 24

8

32 km apart; hence, they are now

32 3.6 4 24.4 km apart. (e)

The position vector of T at 11:30 is The distance to base camp is:

18 24 3.6

18

18 20.4

182

20.4

20.42

740.16

27.2 km.

The time needed to cover this distance is: 27.2 60 54.4 54 minutes. 30

5.

(a)

(i)

Initially, Aircraft 1 is at position r origin is:

(ii)

16 12

162 12 2

The velocity vector is v 12 5

122

5

2

16 ; hence, its distance from the 12

20 km

12 ; hence, its speed is: 5 13 km/min.

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(b)

16 12

r

t

12 5

x y

16 12t 12 5t

From the first equation, we have t y

12 5

x 16 12

144 5 x 80 12

x 16 12t y 12 5t x 16 . Substituting into the second equation: 12

12 y

224 5 x

5 x 12 y

224

Note: If we multiply the vector equation of the line by the vector perpendicular to the direction vector, we can find the result quite quickly. x 16 12 5 x 5 16 5 12 . Now we have: t t y 12 5 12 y 12 12 12 5 5 x 12 y

(c)

(d)

5 16 12 12 t 0

5x 12 y

224

We have to determine the angle between the direction vectors: 12 2.5 0; hence, the angle between the paths of the aircrafts is 90° 5 6 (i)

23

r

5

t

2.5

x

23 2.5t

6

y

5 6t

Hence, x 23

x 23 2.5t y

5 6t

x 23 2.5 y 5 6

t t

y 5 6

2.5

Multiplying by 30: 30

x 23 2.5

30

y 5 6

12 x 23

5 y 5

12 x 5 y

301

Note: We could also have used the method from (b). (ii)

5 x 12 y 224 12 x 5 y 301

169 x 4732

x 28, y

12 28 301 7 5

Hence, the paths cross at the point 28, 7 . (e)

We will determine the time at which each of the planes is at 28, 7 . For Aircraft 1: 28 16 12t

7

12 5t

28 16 12t 7 12 5t

t 1

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For Aircraft 2: 28 23 2.5t

7

28 23 2.5t 7 5 6t

5 6t

t

2

So, the planes are not at the point where the two paths cross at the same time, i.e. the planes do not collide. 6.

Method 1

x 4 is the vector on the line, and it is y 1

If ( x, y ) is a point on the line, then the vector perpendicular to the vector

2 3

x 4 y 1

2 3

So, 0

2 . Hence, their dot product is zero: 3

0

x 4 y 1

2( x 4) 3( y 1) 2 x 8 3 y 3 2 x 3 y 5 and the equation of

the line is: 2 x 3 y

5

Method 2 If vector

2 is perpendicular to the line, then the vector 3 4

vector of the line. So, a vector equation of the line is r transform the equation into Cartesian form: x 4 3t x 4 3t 2 x 8 6t

y 7.

1 2t

y

(a)

At 13:00 t 1:

(b)

(i) (ii)

1 2t

x y

0 28

1

The velocity vector is:

3y

3 6t

6

6 20

8 x y

x t 1

y

t 0

3 ,or 2 1

t

3 , is a direction 2

3 2

. Now, we have to

2x 3 y 5

6

0

20

28

6 8

The speed is the magnitude of the velocity vector; therefore: 6 8

62

8

2

10 km/h

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(c)

x

0

y

28

6

t

x 6t y 28 8t (d)

6t 8

28 8t

4 3

4 x 24t 3 y 84 24t

4 x 3 y 84

The two ships will collide if the point (18, 4) is on the line. So:

18 4

0 28

t

6

18 6t 4 28 8t

8

t

3

Therefore, the ships will collide at t 12 3 15: 00 hours. (e)

x y

(f)

At t

18 4

t 1

5 12

3, Aristides is at

18 5t 5 4 12t 12

13 5t 8 12t

13 8

t

5 12

18 x and Boadicea is at 4 y

13 8

3

5 12

Therefore, their distance vector is:

102 242 8.

(a) (b)

9.

(a)

2x x 3 2 x2

676

28 28

18 4

0

2 x2 2 x 5 x 15 0

7 x 15

0

2x 3

(i)

OA

240 70

OA

So, the unit vector is:

(ii)

v 300

(iii)

t

250 300

0.96 0.28

10 ; hence, the ships are 24

26 km apart.

x 1 5

x

x 5

2402

702

1 240 250 70

28 . 28

2 x2 7 x 15 0 3 ,x 2

5

250

24 25 7 25

0.96 0.28

288 84

5 hour, or 50 minutes 6

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(b)

480 240 250 70

AB

cos

(c)

240 180

240

240

70 240

180 240

70

180

So,

cos 1 0.936

(i)

AX

240 180

4

(iii)

20.609

339 240 238 70

3

(ii)

70200 250 300

0.936

20.6

99 168

3 240 4 180 0; hence, n

AB.

The scalar projection of AX in the direction of 3 1 99 297 672 75; 5 168 4 5

n is

hence, the distance XY is 75 km. (d)

A to Y using the distances AX and XY . So, AX

1952 752

AY 10.

992 1682

38025 195; hence,

32400 180 km

Refer to the diagram below. y 15

U

10

T

V 5

-10

10

x

S -5

(a)

ST

7

2

9

7

2

9

, and, since STUV is a parallelogram, VU

ST

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9 9

VU

9 9

u v

5 15

5 15

9 9

Note: We can also use the direction vector

1 11

4

1 1

6

1 4 11 6

1 . So, r 1

(i)

EW

a 1 17 11

EW

a 1

So: a (ii)

2

For a

2

11.

cos

1

a2

a1 4

3 : EW 4

and

6

1 and initial point 5,15 . 1

5

36

6

2a 37

2 13

a2

3a 37

48 52

12 , 13

52

3, a2 5

, ET

7 1 7 11

6 4

6

6 4

4 6

6

4

12 13

24 24 16 36 16 36

157.38

157

The coordinates of the point of intersection should satisfy both equations. 5 3 2 4t 3 4t 7 3 8 4 7 1, t 1 So: 1 2 2 t t 2 1 Therefore, the position vector of point is OP

5 3 1 2

9 . 9 1 . 1

a 1 6

2a 15 0

So, cos

4

5 5

So, the point is on the line when (d)

v , and the

6

4,6 and the direction vector is parallel to

The line contains the point

So, for the direction vector, we can use the vector

(c)

4

v

4,6 .

coordinates of V are: (b)

9 9

v

2 3

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Note: We can transform the vector equations to Cartesian form ( 2 x 3 y 13 , and x 4y 10) , and then solve the system. 12.

(a)

(b) (c)

OR

10 1

PQ OQ OP

7 3

POPQ

cos OPQ

3 2

3 2

21 6 58 13

49 9 9 4

PO PQ

15 754

(i)

Since PQR OPQ 180 , cos PQR cos 180

(ii)

Using the Pythagorean identity for sine and the fact that the sine of angles in a triangle is always positive, we have: 1 cos 2 PQR

sin PQR

152 754

1

(iii)

13.

7 3

OPQ

cos OPQ.

1 cos 2 OPQ

529 754

23 754

Area of the parallelogram: Area

OR OP sin

1

8 9

7 3

2

13 53

23 13 53

23 square units

(a)

OB

(b)

To find D, we have to find the vector of the side of the parallelogram: 8 1 9 AD BC OC OB . Now, we can find the position vector of 9 7 2

7

, OC

3

D : OD OA AD (c)

BD

(d)

(i)

11 4 x y

1

9 2

11 . Hence, d 4

11 .

12 3

7 1 7

2 2

tBD

1 7

t

12 3

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4

Note: For the direction vector, we can use be (ii)

(e)

(f)

1 7

(a)

(b)

t

7

t

7

12 3

4 1

1 12t

8 9

1 12 4 3

BC OC OB

(ii)

OD OA AD OA BC

t

5i 5 j

2i

AC OC OA

5i 5 j

BD AC

7 8 12 2 3

7

i 3j

6i 2 j

4i 2 j

6i 2 j

2i

3i 3 j 4i 2 j

9i 7 j

6 18 130

6 2340

82.9

r i 3 j t 2i 7 j

(d)

We have to solve the vector equation: i 3 j t 2i 7 j 3 7t

12 3

BD .

(c)

Hence, 1 2t

0

2 3

t

i 3j

27 21 9 9 81 49

BD AC 82.87

t

1

1 4

(i)

cos

1

12 12 0; hence, CP

BD OD OB

. Then the equation would

.

5 7 3t

7 5

CP OP OC CP BD

14.

1

At point B, t 0 . We can see that

7 5

x y

1

4

s 2

4s

s

3 2t

3 7t

So, the position vector of the intersection is: r

4i 2 j s i 4 j .

2 12 8t

t

i 3 j 7 2i 7 j

7, s

11

15i 46 j.

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3 15.

(a)

1

The unit vector in the direction of 4 is: 0

32

42

02

3 4 0

3 1 4 , so the 5 0

3 18 4 . Therefore, the equation of the path of the velocity vector of the balloon is: 5 0 balloon is the same as the equation of the line through 0,0,5 with direction

3 18 4 :b vector 5 0 (b)

(i)

t

x

0

y z

0 5

3 18 t 4 . 5 0

49 32

0

0 (ii)

48 24 , so the speed is:

The velocity vector is

6 48 24

482

242 62

54 km/h

6 18 3 49 48t 5 18 4 t 32 24t 5 5 6t t

0

(c)

(i)

3 18 t 4 5 0

At R : 0 5

49 32 0

48 t

24 6

5 hour (50 minutes) satisfies all three equations. 6 Substituting t 5 into the expression for h (or b ): 6 t

(ii)

h

49 32 0

5 6

48 24 6

49 40 32 20 0 5

9 12 . Hence, R 9,12,5 . 5

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16.

(a)

(b)

200 400

600 200

800 600

(i)

AB OB OA

(ii)

AB

(i)

v

(ii)

At 13:00 t 1 , so:

(iii)

The distance from A to B is 1000 km, and, since the velocity of the aircraft

8002

250

6002

0.8 0.6

1000; hence, the unit vector is:

1 800 1000 600

0.8 0.6

200 150 600 200

1

is 250km/h, the time is 1000 250

200 150

400 50

4 hours; hence, the aircraft is flying over

town B at 16:00. (c)

Method 1 Evaluating the time needed Time taken to travel from A to B to C is 9 hours

81 hours . The warning light 9

will go on after 16 000 litres of fuel have been used. Time taken to use 16 000 litres = 16 000 1800

distance to town C is 1 250 9

80 1 hour remains and the . Hence, 9 9

27.8 km

Method 2 Evaluating the distances needed The distance from A to B to C is 2250 km. The distance covered with 16 000 litres of fuel is: 16000 250 1800

2222.22 km. So, the distance to town C is

2250 2222.22 27.8 km Method 3 Evaluating fuel usage Fuel used from A to B = 1800 4 7200 litres. Fuel remaining until the light goes on = 16 000

7200 = 8800 litres.

Number of hours before the warning light goes on: 8800

8 hours; therefore, 1800 9 the time remaining is 1 hour, and the distance to town C is: 1 250 27.8 km. 9 9 4

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17.

(a)

The vectors are perpendicular if their scalar product is zero. So, first, find the vectors: 1 4 3 1 3 2

QR

, PR

0 3

3

2c 5

2c 5 2

3

3

1

2c 5

2c 1

QR PR

0 1

1

2c 1

2c 1

6 3 (2c 5)(2c 1) 4c 2 8c 4

The vectors are perpendicular if: 4c 2 8c 4 0 (b)

3 1

PR

3 1 , PS

21

PS PR (c)

3

3 0 3

3 1 3

3

1 1 2 1

0 3

2

0

c 1

3 0 3

A vector equation of the line is: 3 3 3 3t

r (d)

1

1 4

4 c 1

3 5

t

1 3

3 t ,t 5 3t

We need one more direction vector (which is not parallel to the direction vector of the line) to determine a normal to the plane. We will take a point on the line and 3 1 2 point S : SQ

2 2 3

3 1 3

3 1 5 2 i 2 3

2 . Hence, the normal will be: 3

j k 2 3 1 3

9 15 4

Therefore, the equation will be:

9 x 1 15 y 1

4 z 2

0

9x 15 y 4z 2

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Note: We have a point and two vectors in the plane, so we can write parametric equations of the plane:

r

1

3

2

1

1

2 , , 3

2 (e)

3

Method 1

PQ n

Shortest distance is:

n

3 4 Since PQ

3 1 5 1 1 2 6

PQn n

1 2 , we have: 6 9 15 4

15 322

81 225 16

Method 2 Use the distance formula for a point x0 , y0 , z0 and a plane ax by cz d

d

0:

ax0 by0 cz0 d a 2 b2 c 2

Hence, for the point P 4,1, 1 and the plane 9 x 15 y 4 z 2 0 , the distance is: d

9 4

15 1 92 152

0 1

18.

(a)

(b)

AB

1

2

15 322

42

1

1 2 2 1

AB BC

4

3 , BC 1

1 0

1

0 1 2 2

1 0

1

1

i

j

k

3 1

1 0

1 1

3 1 1 0

1 1 2

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(c)

We can use the formula for the area of a triangle: A

Area

1 2

1 1

1 1 1 4 2

2

1 a b . Hence: 2

6 2 1

(d)

A normal to the plane is n the equation is: 1 x 1

(e)

AB BC

1 y 2

1 . Since point A is on the plane, 2

2 z 1

0

x y 2z 3

The normal n is parallel to the required line. Hence,

x 2 t y 1 t , where t z 6 2t (f)

The distance formula for a point x0 , y0 , z0 and a plane ax by cz d

d

ax0 by0 cz0 d a 2 b2 c2

Hence, for the point 2, 1 6 and the plane d

0 is:

1 2 1 1

2

1 1 4

6

3

18 6

x y 2 z 3 0 , the distance is:

3 6

1 6 , a unit vector in the direction of n is: 6

1 1 2

(g)

Since n

(h)

First, we will find the point of intersection of the plane and the line through D

1 1 4

perpendicular to the plane. Hence, we have to find the intersection of the plane P and the line from part (e).

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Since x

2 t, y

2 t

1 t

1 t, z

6 2t , we have:

2 6 2t

3

So, the point of intersection is

D and E . Hence:

19.

(a) (b)

u v

1

2

i

j

k

2 3

1 2

1 2

2 3 1 2

t 3

1, 2,0 . This point is the midpoint between points xE

1, 2, 0

6t 18

2 yE 1 z E 6 , , 2 2 2

2 3 i 1 2

1 3 j 2 2

E

1 k 2

4,5, 6

2 1

7 4 5

Method 1 2 w

2 3

2

The line of intersection of the planes is parallel to u v . So, 2 7 w u v 2 4 7 14 8 4 15 10 3 2 5

0 (for all

, ).

Hence, w is perpendicular to the line of intersection. Method 2 The line of intersection is perpendicular to the normals of both planes; hence, to vectors u and v . Therefore, it will be perpendicular to the plane containing those v w. two vectors, that is, to all vectors of the form u Method 3 The line of intersection is perpendicular to the normals of both planes; hence, to vectors u and v . Therefore, for a direction vector d of the line, it holds: d u 0 d u v d u d v 0 d v 0 and d is perpendicular to w .

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20.

(a)

OP

OQ

OR

OS

OA OB

OA OC

2 2 1 1

4 0

2 1

3

2 1

3

1 2 2 2

3 0

2 1

3

1 2 1 2

1 1

OB OC

P (4, 0, 3)

Q 3, 3, 0

Q 3,1,1

2 2 1 1 1 2

OA OB OC

5 2

2 1 2 2

(b)

OA OB

2

1 2

1 1

S 5, 2, 1

1

3

3

2 4

2 ; hence, the equation of the plane is: 4

3( x 2) 2( y 1) 4( z 2) 0

3x 2 y 4 z

0

Note: Parametric equations of the plane are: 1 4 3 r 2 0 3 , , 1 3 0 (c)

21.

(a)

V

AB

OA OB OC 1 1 3 2

0 1

5 3

2 0

cos

(b)

, AC

3 2

1 2

4

2

3 4 8

0 1 1 2

1 3

1 3

2

15

1

1

3

2

2

1 4 1 9 4

Method 1 1 Area AB AC sin 2

7 5 14

146.789

147

1 5 14 sin146.789 2

2.29 units2

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Method 2 1 Area AB AC sin sin 2 Using the Pythagorean identity for sine and cosine, we have:

sin

Area

1

2

7 5 14

1

1 AB AC sin 2

7 10

3 10

1 3 5 14 2 10

21 2

Method 3 Area

(c)

(i)

0 1 1 2 2

1 AB AC 2

For l1 : r

2 1

0 t 1

0

(ii)

1 1

s

1 2 x y

2

1

For l2 : r

1 3 2

4 2 1

1 16 4 1 2

21 2

2

1 t, t z 2t

1

x

3 2

y 1 3s , s

1 s

z 1 2s

The lines are not parallel, because the direction vectors are not parallel. Hence, we have to solve the system: 2 1 s s 3

1 t 1 3s 2t 1 2s From the first equation, s 3 , and substituting into the second equation:

t

2 33

7 and third equation: 2 7

1 2 3

14

5.

Therefore, the system has no solution and the lines do not intersect.

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(d)

(e d )(l1 l 2 )

The shortest distance is given by

(l1 l2 )

, where d and e are position

vectors of the points on the lines, and l1 and l2 are direction vectors of the lines. l1 l2

4 2

AB AC

1

Hence, 22.

(a)

1 1 1

(e d ) (l1 l2 ) (l1 l2 )

2 1 0

4 2 1

3 2 1

16 4 1

4 2 1 21

9 21

Method 1 Use matrices and their properties: 1 2 3

3 1 1

2 3 1

x y z

6 7 6

x y z

1 2 3

3 1 1

2 3 1

1

6 7 6

1 1 2

Method 2 Use row operations to solve a system of three equations (first and last steps shown). x 3y 2z 6 x 1

(b)

(c)

5y

7z

19

10 y

7z

24

1

2

i

3 2

1 3

1 3 2 1

v

u

vu

y z

1

2

m 3 2

n 1 3

11

m 2n

7 5

3m n 2m 3n

j

k

11

2 3

7 5

1 2

m 2n 3m n 2m 3n 11m 22n 21m 7 n 10m 15n

0

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(d)

3 1 . 1

The line is perpendicular to vector v and to the vector So, a direction vector of the line is: 11 3 i j k 12 7 1 11 7 5 26 5 1 3 1 1 10

6 2 13 , 5

and a vector equation of the line is: 1 6 r 1 13 2 5

23.

(a)

(i)

AB

1 1 2 3

0 1 , AC 3

4 1 AB AC

(ii)

Area

2 1 3 3

1 0

6 1

5

0

1

i

j

k

5

1 3

0 5

0 1

1 3 0 5

3 1

1 AB AC 2

5 3 1

1 2

1 25 9 1 2

35 2 5

(b)

(i)

The plane contains the point A and its normal is AB AC hence, for the Cartesian equation, it holds: 5( x 1) 3 y 3 1 z 1 0 5x 3 y z The equation is: 5 x 3 y z

3 ; 1

5 9 1

5 5

(ii)

The line contains the point D and its direction vector is AB AC

hence, the Cartesian equation of the line is:

x 5 5

y 2 3

z 1 1

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3 ; 1

(c)

We first write the equation of the line in parametric form, then solve the system: x 5 5t

y 2 3t z 1 t 5 5 5t 3 2 3t

1 t

5

25 25t 6 9t 1 t 5 x 5 51 0

The point is: y

2 31

z 1 (d)

1

35t

1

35

t 1

0,1, 2

2

The distance is the same as the distance between points D and P : d

5 0

2

2 1

2

1 2

2

25 9 1

35 1

24.

(a)

(b)

The line contains the point A and its direction vector is 1 ; hence, the Cartesian 1 equation of the line is: x 2 y 5 z 1 1 1 1 We first write the equation of the line in parametric form, then solve the system: x 2 t

y 5 t z 1 t 1 2 t 15 t

1 1 t

2 t 5 t 1 t 1 0 x

2

5 3

The point is: y

5

5 3

z

1

1 0 3t

5

t

5 3

1 10 8 , , 3 3 3

5 3

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(c)

Method 1 Denote the image point A . Then, the point of intersection

1 10 8 of the line , , 3 3 3

and the plane is the midpoint of AA . Hence, 1 10 8 2 x 5 y 1 z , , , , 3 3 3 2 2 2 1 2 x 2 4 x 2 3 2 3 3 10 5 y 20 5 y 5 3 2 3 3 8 1 z 16 z 1 3 2 3 4 5 13 A , , 3 3 3

13 3

Method 2 Parameters of the points of the line are: t 0 for A 5 10 t for the intersection; hence, t for the reflected point. 3 3 10 x' 2 3 10 3

Thus: y ' 5 z

1

4 5 13 , , 3 3 3

10 3

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(d)

We have: AB

2 2 0 5

0 5 , and a direction 7

6 1 1

vector of the line d 0 5 7

AB d d

1 . Hence: 1

d

1 1 1

12 7 5

1 1 1

3

218 3

654 3 3

25.

(a)

The plane contains the point P and its normal is equation of the plane is: 3 x 1 4 y 2 1 z 11

(b)

(i) (ii)

1

11

1 6 11

3x 4 y z 6 4 ; hence, P lies in

2

The intersection of the planes contains the point P and its direction vector is the vector product of the normal: 3 1 1 4 3 4 . Hence, a vector equation of the line is: 1 1 13 r

(c)

3 2

0

4 ; hence, the Cartesian 1

1

1

2 11

t 4 ,t 13

The angle between the normals is: 3 1 4 3 1 1 10 cos 1 9 16 1 1 9 1 26 11 Hence, the angle between the planes is: cos

10 26 11

53.7498

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53.7

26.

(a)

x 2 3

y 1

z 9 2

hence: M (b)

(i)

(ii)

x

(i)

x 4 3

y z 3 1 2 2 3 4

PM 3 1

PM

9

6

2

12

6

3 1

3 3

0 2

18

4

12

2

24 0

14

0

42

3 3:

The distance between the lines is equal to PM , where 3

6

3

12

3 ; hence, the distance is: 6

PM 2 d

(d)

3

0

2

(ii)

9 2 ;

,z

2 3 , ,9 2

9 2

(c)

2 3 ,y

9 9 36

54

3 6

3

3

12

A normal to the plane equals: 3 6

1 2

24 6

2 6

4 ; 1

hence, the Cartesian equation of the plane is: 2 x 4 4 y 0 1 z 3 0 2x 4 y z 5 (e)

The line is on

1

(from (d)).

Testing the line on

2

:

2 3

5

9 2

2 3

5

9 2

Therefore, the line is in both planes; hence, l1 is the line of intersection. Alternatively, solve the system: 2x 4 y z 5

x 5y z

11

So, the intersection is the line: 11.5 1.5 r 4.5 t 0.5 , t 0 1 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

11 .

A direction vector of the line is: 2

1.5 0.5

3 1

1

2

11.5

vector of a point on the line is:

4.5

9

0

27.

(a)

L1 : x

2 t, y

2 3t , z

, and, for t 9 , the position

1.5

2

0.5

0 , which is line l1 . 9

1

3 t and L2 : x

2 s, y 3 4 s, z

4 2s

Hence, at the point of intersection: x 2 t 2 s t s 0

y 2 3t 3 4 s z 3 t 4 2s

3t 4 s 1 t 2s 1

This is a system of three equations which can be solved using a method of your 1 and the point of intersection is 1, 1, 2 . choice. Hence, t s (b)

The normal to the plane is perpendicular to both direction vectors, hence: 1 1 2 3 4 1 1 2 1 Since, the plane contains the intersection point 1, 1, 2 , the Cartesian equation of the plane is:

2 x 1 1 y 1

(c)

1 z 2

0

The midpoint M of PQ is: M

2x y z 5 1 3 1 4 2 3 , , 2 2 2

3 5 2, , . 2 2 2

The vector MS is parallel to the normal to the plane

hence, S 2t 2, t

3 ,t 2

, so MS

t

1 1

5 . 2

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2t t ; t

Now, PS 2t 2 1

3 2

t

3 1 2

2

t

5 2 2

2

2t 1 6t 2

So, the possible solutions for S are: 1 3 1 5 S1 1 2, , 3,1,3 or S2 2 2 2 2 Note: We used the fact that PS

1 2,

1 2

2

15 2

5 2

t

t2

3

3 1 , 2 2

5 2

2

1 2

t 1 4

2

t

1 2

1, 2, 2

3 . The line L is the symmetry line of the

segment PQ ; hence, QS should be 3. That means that we will have the same equations if we use QS 28.

(a)

(i)

3.

Points on each line satisfy the following systems of equations:

x 2 2 x 2 s t L1 : y 1 3 ; L2 : y 0 2 s t z 1 8 9 z 1 s t Hence, at the points of intersection: 2 2 2 s t 1 3 2s t 1 8 9 1 s t Subtracting the third equation from the first, we have: 1 10 10 1 (ii)

If

for points on the plane L1, then those points are on the line:

x 2 2 y 1 3 z 1 8 9

2 1 2 1

Thus, a vector equation is: r

2

1

1 1

2 , 1

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(b)

The point 2,0, 1 from the line is on the plane. Hence, the Cartesian equation is:

3 x 2 (c)

2 y

Planes

1

and

z 1

2

0

3x 2 y z 5

intersect at the line r

intersection of this line and the plane

3

2 1

1 2 ,

1

1

, so we find the

. We write the equation of the line in

parametric form, and then substitute into the equation of

3 2

x 2 y 1 2 z 1 21 2

1

5

:

5 5

The equation is satisfied by any real value of

; hence, the plane

line, and intersection of the three planes is the line r

29.

3

3

2 1

1 2 ,

1

1

contains the

(a)

The angle between the two planes is the acute angle between their normals. 4 4 1 3 1 1 18 3 9 cos 13 18 26 3 2 2 13 13

(b)

(i)

The line of intersection between the planes is orthogonal to both of their normals. Thus, its direction vector is parallel to their vector product. 4 4 4 1 n1 n2 1 4 2 3 8

1 (ii)

(iii)

1

8

2

Substitute the coordinates of the point in each of the equations: 4(1) 0 4 8 and 4(1) 3(0) 4 0

r

1 0 4

1 t 2 3

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(c)

If B is on

AB

(e)

b 7 4a

a 1 7 4a , and if it is perpendicular L, then 3

OB OA

a 1 7 4a 3 (d)

then 4a b 1 8

1

1 9a 9 0

2 2

With values found in (c), AB

a 1 1 1 7 4 3

b 3 0 3

AB

9 9

3 2

3

With P as described, we have an isosceles right-angled triangle with right angle at A.

Thus, AB

AP

AP

3 2

Since P is on L, then its coordinates are 1 t , 2t , 4 2t , and AP t2

So, AP

1 30.

(a)

4t 2

4t 2

3 2

2, 2 2, 4 2 2 or 1

t

2 , and the possible positions are

2, 2 2, 4 2 2

For L to be perpendicular to , it has to be parallel to the normal vector. p 1 k 1 for some value k, which will give an inconsistent system with That is 2

1

3

p k no solution: 2 k . Therefore, L cannot be perpendicular to 1 3k (b)

t 2t . 2t

For L to lie in 2 q 3 9

then point 2, q,1 must be in the plane. q

Also, for L to lie in orthogonal: p 1 0 p 2 1

1

.

4

then its direction vector and the normal to the plane must be

5

3

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(c)

(i)

1 , then the angle between L and 11 the normal must be the complement of this angle. This means that the cosine of the angle between the normal and L is the same as 1 1 sin arcsin 11 11

If the angle between L and

Thus,

(ii)

p 2 1 p

2

1 1 3 5 11

1 11

is arcsin

p 5 p2 5

x 2 The equation of L is now: 2

1

y q 2

p

z 1

2

x y z

2 2t q 2t 1 t

1 t 2. Now substitute the coordinates of any point on L with the value of t just found into the equation of the plane: 2 4 q 4 3 9 q 10. If the intersection is at z

1, then 1 t

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Exercise 10.1 Answers may differ slightly from one person to another. This depends on GDC, software, or arrangement of data. General patterns will still look similar. 1.

(a)

The experimental unit would be a student. A sensible population would be all the students in a certain school (a large one), city, district, or country. The sample could be students from one class, or even a smaller group of students. The variable is qualitative as it describes a characteristic of a student (female or male) rather than a numerical quantity.

(b)

The experimental unit would be a final exam taken by a 10th-grade student. A sensible population would be all the exams taken by 10th-grade students in a certain school (a large one), city, district, or country. The sample could be the exams of students from one class, or even a smaller group of students. The variable is quantitative since we are counting the number of errors.

(c)

The experimental unit would be a new-born child. A sensible population would be all the new-born children in a certain city, district, or country. The sample could be new-born children born in the same hospital or born on the same day. The variable is quantitative since we are measuring height.

(d)

The experimental unit would be a child aged less than 14. A sensible population would be all the children aged less than 14 who live in a certain city, district, and country. The sample could be all the children aged less than 14 who live in the same building, block, or street. The variable is qualitative as it describes a characteristic of a child (blue eyes, brown eyes, green eyes, and so on) rather than a numerical quantity.

(e)

The experimental unit would be a working person. A sensible population would be all the working people in a certain city, district, or country. The sample could be people working in the same company, or people living in the same part of the city. The variable is quantitative since we measure the time it takes them to travel to work.

(f)

The experimental unit would be a country leader. A sensible population would be all the country leaders worldwide. The sample could be all the country leaders within a certain geographical region, or a continent, or even the leaders of the same country throughout history. The variable is qualitative as it describes a characteristic of a leader (excellent, good, fair, or poor) rather than a numerical quantity.

(g)

The experimental unit would be a student. A sensible population would be all the countries of origin of students at an international school, or a group of international schools within a certain country or a geographical region. The sample could be all the countries of origin of students from one grade of an international school. The variable is qualitative as it describes a characteristic of a student through their country of origin (Austria, Germany, Italy, Croatia, and so on) rather than a numerical quantity.

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2.

Note: Answers for this question are not unique. (a) (b)

3.

4.

Skewed to the right , and there are a few who are top scorers. It should be symmetric since the weights will be grouped around one particular weight. (Later on, we will find out that this weight is called the mean weight.)

(c)

Again, skewed to the right, since there are a few students who travel a lot and visit many countries.

(d)

In this case, we would again expect a distribution skewed to the right, because some students do receive a lot of emails (especially those who use social networking sites).

(a)

Quantitative, because we can measure the time taken to finish the essay.

(b)

Quantitative, because we can count the number of students in each section.

(c)

Qualitative, since the rating has descriptors rather than a numerical quantity.

(d)

Qualitative, since the country of origin is a name rather than a numerical quantity.

(a)

Discrete, since we can count the exact number of students from each country.

(b)

The weight of exam papers can be measured and therefore it is continuous.

(c)

Time can be measured and therefore it is continuous.

(d)

Discrete, since the number of customers must be counted.

(e)

Time can be measured and therefore it is continuous. We always measure time to a certain degree of accuracy.

(f)

The amount of sugar, as a mass, can be measured and therefore it is continuous. On the other hand, if we count the grains of sugar without breaking them apart, then we can say that it is discrete.

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5.

This analysis is done by grouping the grades into classes of length 0.2, since the range of the data is not very large. (Use a GDC or software.)

50

Cumulative frequency

40

30

20

10

0

G.P.A. 1.8

2

2.2

2.4

2.6

2.8

3

3.2

3.4

3.6

The data looks relatively symmetric, with no apparent outliers. 6.

We will group the data into classes of length 5 (not unique for example, you may start at 50 and end at 100) and count the frequencies for each interval. Interval 52.5 57.5 57.5 62.5 62.5 67.5 67.5 72.5 72.5 77.5 77.5 82.5 82.5 87.5 87.5 92.5 92.5 97.5

Midpoint 55 60 65 70 75 80 85 90 95

Frequency 3 4 6 4 5 1 8 5 4

We can say that this distribution is almost bimodal, where one group has a mode of 65 and the other group has a mode of 85. This may indicate that the class is split into two ability groups.

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7.

(a)

This set of data has a very large range and therefore we are going to group it into suitable intervals. The midpoint of each interval is shown in the table.

Midpoint 0 4 8 12 16 20 24 28 32 (b)

Frequency 10 14 12 3 3 4 2 1 1

The data is not symmetric but skewed to the right.

(c) Cumulative frequency 50

40

30

20

10 Months 4

8

12

16

20

24

28

32

36

From the diagram, we notice that there are 36 young drivers who will lose their licence; therefore, 72% may lose their licence.

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40

8.

(a)

We will use classes of length 0.6, having 0 as the first midpoint of the interval.

(b) Cumulative frequency 60 50 40 30 20 10 Minutes 0.6

1.2

1.8

2.4

3

3.6

4.2

4.8

5.4

6

6.6

Using the graph, we can say that there are approximately 49 customers who need to wait up to 2 minutes; therefore, 11 customers will need to wait longer than that. 9.

(a)

The data is skewed to the right, with the modal value as 6 8 days spent at hospital. A very few patients will stay longer than 20 days.

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(b)

Due to the poor scale on the frequency diagram, it is going to be difficult to estimate the frequencies. Our estimates are given in the table: Interval Frequency Cumulative frequency 0 2 750 750 2 4 450 1200 4 6 750 1950 6 8 950 2900 8 10 700 3600 10 12 425 4025 12 14 350 4375 14 16 225 4600 16 18 190 4790 18 20 180 4970 20 22 150 5120 22 24 100 5220 24 26 50 5270 26 28 40 5310 28 30 30 5340 30 32 30 5370 32 34 20 5390 34 36 10 5400

Cumulative frequency

6000 5000 4000 3000 2000 1000 0 0

5

10

15

20

25

30

35

40

Days

(c)

Using the table or graph, we can estimate the percentage of patients who stayed 1950 less than 6 days as: p 0.36111... 36% 5400

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(a) (b)

From the frequency graph, we can see that the right-most time recorded is 40 minutes. Therefore, the longest time is 40 minutes. In order to solve this problem, we need the cumulative frequencies. Interval Frequency Cumulative frequency 18 20 10 10 20 22 15 25 22 24 25 50 24 26 30 80 26 28 32 112 28 30 40 152 30 32 30 182 32 34 22 204 34 36 12 216 36 38 0 216 38 40 4 220 The percentage of time spent exercising more than 30 minutes is: p 1

152 220

0.30909... 31%

(c) 250 Cumulative frequency

10.

200 150 100 50 0 16

21

26

31

36

Minutes

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41

11.

(a)

The frequency table is given below. Speed

Frequency 20 70 110 150 40 10

(b) 180 Frequency 160 140 120 100 80 60 40 20 0 50

(c)

450

Speed 60

70

80

90

100

110

120

130

140

150

160

To draw the cumulative frequency graph, we take the endpoints of the intervals and calculate the corresponding cumulative frequencies: 20, 90, 200, 350, 390 and 400. Cumulative frequency

400 350 300 250 200 150 100 50 0 60

Speed 70

80

90

100

110

120

130

140

150

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160

12.

13.

(d)

To estimate the number of drivers exceeding the speed limit, we draw a vertical line from 130 km/h on the cumulative frequency diagram. Our estimation is 375; therefore, since there are 400 cars, 25 cars (or 6.2% of the cars) were exceeding the speed limit.

(a)

To draw the relative cumulative frequency graph, we take the endpoints of the intervals and calculate the corresponding cumulative relative frequencies, which are: 16, 116, 239, 343, 391 and 400. Dividing by 400 changes them to relative. The graph shows them as percent.

(b)

We need to draw two vertical lines, from 5.01 and 5.18, on the diagram above. Our estimate for the percentage of components of length up to 5.01 mm is 5% and up to 5.18 mm is 95%. Therefore, about 10% components will be scrapped.

(a) Frequency

100

50

Time 50

(b)

100

150

200

250

300

350

400

450

To draw the cumulative frequency graph, we take the endpoints of the intervals and calculate the corresponding cumulative frequencies: 12, 27, 69, 174, 240, 285 and 300.

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Cumulativerequency 300 250 200 150 100 50 Time 50

(c)

100

150

200

250

300

350

400

450

There are 300 customers, and 25% of 300 is 75. To find the waiting time that is exceeded by 75 customers, we draw a horizontal line at the cumulative frequency of 225 (see diagram above). Our estimate is 285 seconds (4 minutes 45 seconds).

Exercise 10.2 Note: In most of the cases, GDC/software can give the required answers. 1.

(a) (b)

x

5 4 7 8 6 6 5 7 8

48 8

6

In order to find the median, we need to list the observations in order of magnitude. Since there are eight observations, we need to take the two middle ones, which are the fourth and the fifth observations, and take their average. 4, 5, 5, 6, 6, 7, 7, 8 Since the middle observations are both 6, the median is 6.

(c)

The data is symmetric with respect to the mean value. A bar graph/histogram demonstrates the fact. 3 2 1 0 4

5

6

7

8

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2.

(a) (b)

x

5 7 8 6 12 7 8 11 4 10 10

78 10

7.8

4, 5, 6, 7, 7, 8, 8, 10, 11, 12 There are ten observations, so we need to identify the fifth and sixth observations. The fifth observation is 7 and the sixth is 8; therefore, the median is 7.5

(c)

This set of data is bimodal, with 7 and 8 as the modes, since these two observations have the highest frequency.

3. Number of cars (xi)

1

2

3

Number of households (fi) 12 24

8

6

xi × f i

0

0

50

24 16 18 58

We calculate the mean value by using the formula x

xi f i fi

58 1.16 50

There are 50 pieces of data, so to find the median value we have to identify the 25th and 26th observations, when listed in order of magnitude, and then take their average. From the table, we notice that all the observations from the 13th until the 36th are 1; therefore, the median is 1. The median is the measure that best describes this data since the data is skewed to the right and as such the mean is more influenced by it.

4.

We list the revenues in order of magnitude: 242137,

.

Since there are 10 observations, the median value is the average of the 5th and 6th observations: 265172 311870 x 288521 millions of dollars. 2 3070359 307 035.9 millions of 10 dollars. In this case, the median is more appropriate as there are extreme values.

The mean is the total sum of revenues divided by 10: x

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5.

For this question we will use a calculator since there are many observations. First, we input all the observations into a list and then, from the list menu, we can use the descriptive statistics features of the GDC.

As both measures are equal, and the data looks relatively symmetric, either measure looks good.

6.

7.

(a)

x

4460 90

(b)

y

4460 74 60 90 2

49.56 , correct to the nearest cent.

49.93, correct to the nearest cent.

We consider all the bags, measure their total weight, and divide it by the total number of bags.

2.15 144 309.6 x

2.15 144 1.80 56 144 56

1.80 56 100.80 , thus:

2.052

So, the mean weight of a bag of potatoes is 2.052 g

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8.

(a)

xi

x

25

749 25

29.96

(b) 1 89 2 0233445666777 3 34568 4 02266 Since there are 25 observations, we have to find the 13th

n 1 th observation. 2

Looking at the stem plot, the 13th observation is 27; therefore, the median is 27. (c)

To draw a histogram, we need to group the data into suitable intervals. We will use intervals of length 4, starting from 16. (Or use GDC/software.)

Grades Frequency 16 19 2 20 23 4 24 27 9 28 31 0 32 35 3 36 39 2 40 43 3 44 47 2

10 Frequency 9 8 7 6 5 4 3 2 1 0

Marks 16

20

24

28

32

36

40

44

48

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(d) 26

Cumulative frequency

24 22 20 18 16 14 12 10 8 6 4 2

Marks 0 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

An estimate for the median which corresponds to a cumulative frequency of 13 is approximately 27. 9.

(a) Year Fatal Serious Slight Total 1970 758 7860 13515 22133 1975 699 6912 13041 20652 1980 644 7218 13926 21788 1985 550 6507 13587 20644 1990 491 5237 14443 20171 1995 361 4071 12102 16534 2000 297 3007 11825 15129 2005 264 2250 10922 13436

Injuries 25000 20000 15000 10000 5000 0 1970

1975

1980

1985

1990

1995

2000

2005

From the bar graph, we notice that the number of injuries is decreasing year on year.

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(b)

Key: 1 denotes year 1970, 2 denotes year 1990, and 3 denotes year 2005

Fatal 1

2

3

17% 50% 33%

Serious 1

2

3

15% 51%

34%

Slight 1

2

28%

3 35%

37%

Alternatively, pie charts can be also produced for each year as given in the answers to the question in the book itself. 10.

(a)

Use of GDC/software is advisable 140

Number

130 120 110 100 90 80 70 60 50 40 30 20 10 0

Age 15

20

25

30

35

40

45

50

55

60

65

70

75

80

85

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(b)

To estimate the mean value, we need to use the midpoints of the intervals (17, 22, 27, ...,77) and the corresponding frequencies: x

xi fi

37.61

fi

(c) 900

Cumulative frequency

800 700 600 500 400 300 200 100 0 10

Ages 15

20

25

30

35

40

45

50

55

60

65

70

75

80

85

Since there were a total of 867 casualties, we need to draw a horizontal line at 433.5 to find an estimate of the median of the data. Our estimate is about 37 years of age.

To answer questions 11 15, we use our graphs from the previous exercise, together with GDC/software. 11.

Since there are 5400 patients, we need to draw a horizontal line at 2700 (on the cumulative frequency diagram) to estimate the median. Our estimate is 7.5 days. To estimate the mean, we use the midpoints of the intervals and the corresponding frequencies.

So, the estimate of the mean is 9.01 days (correct to 3 significant figures).

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12.

Since there are 220 recordings, we need to draw a horizontal line at 110 (on the cumulative frequency diagram) to estimate the median. Our estimate is 28 minutes. To estimate the mean, we use the midpoints of the intervals and the corresponding frequencies.

So, the estimate of the mean is 27.7 minutes (correct to three significant figures). The median according to calculations is 27, slightly different from the estimate using the graph. 13.

There are 400 cars and one of the cumulative frequencies is exactly 200; therefore, the median value is 105 km/h; (105, 200) is a point on the cumulative frequency diagram. To estimate the mean, we use the midpoints of the intervals and the corresponding frequencies.

So, the estimate of the mean is 103 km/h (correct to three significant figures). 14.

An estimate for the median (which corresponds to a cumulative frequency of 200) is 5.08. To estimate the mean, we use the midpoints of the intervals and the corresponding frequencies.

So, the estimate of the mean length is 5.09 (correct to three significant figures).

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15.

An estimate for the median (which corresponds to a cumulative frequency of 150) is 225 seconds. To estimate the mean, we use the midpoints of the intervals and the corresponding frequencies.

So, the estimate of the mean waiting time is 229 seconds (correct to three significant figures). 40

16.

xi

(a)

1664 40

1664

x

20

1664

x 20

1664 20

x

83.2 20 103.2

xi 12

4404

x 12

4404 60

x

73.4 12 61.4

i 1

41.6

20

xi

(b) i 1 60

17.

(a) i 1

(b)

Average score of the whole group of 100 students

61.4 60 67.4 40 100

63.8

Exercise 10.3 Note: In this part, answers may differ slightly from the answers in the book because of uses of different GDCs or software and expected variation in the accuracy of estimates from graphs. 1.

We use a GDC. Since there are only 15 patients, we will use a simple list with 15 elements. (a)

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Therefore, the mean pulse of the 15 patients is x

1072 15

71.5 (to 3 s.f.)

For the standard deviation, recall that for descriptive statistics, the syllabus uses

and not s.

Thus, standard deviation is 7.04 (to 3 s.f.) (b)

We use a GDC to plot the box-whisker diagram.

(c)

Since we can see that the data is skewed to the left, we need to check whether there are any outliers to the left. IQR = Q3 Q1 =79 68 11 . The outliers lie 1.5×IQR from the lower or upper quartile, so, in this problem, we calculate: Q1 1.5 IQR 68 1.5 11 51.5 56 xmin ; therefore, there are no outliers.

2.

(a)

For this question, we used a spreadsheet. We input the data into a column and used the functions to find the following: Mean value x

162.6; standard deviation sn

1

23.35

sn

23.12

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(b) 11

79

12

567

13

089

14

123679

15

033445689

16

02334568

17

1344789

18

0 2 2 5 5 7 79

19

8

20

9

21

08

Since there are 50 observations, we need to find the 25th and 26th observations and take their average. By counting the observations, we find the 25th and 26th observations are 162 163 162 and 163 respectively; therefore, the median is 162.5 2 (c)

To draw the cumulative frequency curve, we must group the data into suitable intervals.

Interval

Frequency

110 119

2

120 129

3

130 139

3

140 149

6

150 159

9

160 169

8

170 179

7

180 189

8

190 199

1

200 209

1

210 219

2

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50

Cumulative frequency

45 40 35 30 25 20 15 10 5

Passengers

0 105

115

125

135

145

155

165

175

185

195

205

215

225

The first and third quartiles correspond to the cumulative frequencies of 12.5 and 37.5 respectively; therefore, an estimate for Q1 is 148 and Q3 is 179. An estimate for the median (which corresponds to a cumulative frequency of 25) is 163. It is also possible to use the raw data that we entered into the spreadsheet to draw the box diagram. In that case, the measures are slightly more accurate. 6 Frequency

4

2

0 100

(d)

IQR

Passengers 110

120

130

140

150

160

170

180

190

200

210

220

230

240

Q3 Q1 179 148 31 . If there are any outliers, they would be outside of the

interval Q1 1.5 IQR,Q3 1.5 IQR

101.5,225.5 , which includes the whole range

of the number of passengers; therefore, there are no outliers. (e)

The empirical rule states that the whole range should lie within three standard deviations of the mean value 162.6 3 23.12 93.24, 231.96 ; therefore, there are no outliers.

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3.

(a)

Set up a cumulative frequency table as done before. Then use spreadsheet/GDC to graph. 100

Cumulative frequency

75

50

25

Marks 5

(b)

4.

10

15

20

25

30

35

40

45

50

55

60

We draw horizontal lines at 50, 25 and 75 to find estimates for the median and quartiles. The estimates are as follows: the median is 38, the lower quartile is 27, and the upper quartile is 47.

(a) 140 Cumulative frequency 130 120 110 100 90 80 70 60 50 40 30 20 10 0 25

(b)

Time 35

45

55

65

75

85

95

105

115

125

We draw horizontal lines at 65, 32.5 and 97.5 to find estimates for the median and quartiles. The estimates are as follows: the median is 67, the lower quartile is 55, and the upper quartile is 83. Therefore, IQR Q3 Q1 83 55 28

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(c) 160 Cumulative frequency 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0 25 35 45

Time 55

65

75

85

95

105

115

125

135

The horizontal line at 75 gives us an estimate of 71 minutes for the median finishing time for all 130 students.

26 22 84 32 110

5.

x

6.

(a)

326 11

29.6

To find the mean and standard deviation of the given data, we need to use the midpoints of the intervals.

The mean value is 72.1 and the standard deviation is 6.10 (correct to 3 s.f.) (b)

The mean will shift by 13 points and thus will become 85.1. Since we are adding a constant to each term, the standard deviation will not change.

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7.

(a) Number of empty seats x 10 x 20 x 30 x 40 x 50 x 60 x 70 x 80 x Days

15

65

165

335

595

815

905

950

90 980

x

100 1000

(b) Cumulative frequency 1000 900 800 700 600 500 400 300 200 100 Empty seats 10

(c)

20

30

40

50

60

70

80

90

100

(i)

To find an estimate of the median number of empty seats, we need to look at the cumulative frequency of 500 (see diagram above). The estimate is 47.

(ii)

For the first and third quartiles, we look at the cumulative frequencies of 250 and 750 respectively. Hence, an estimate for the first quartile is 35, while an estimate for the third quartile is 58. The IQR is the difference between the third and first quartiles, which is 23.

(iii)

From the previous estimates, we can see that the number of bumper days was about 250.

(iv)

The highest 15% corresponds to the cumulative frequency of 850. An estimate for the number of empty seats for that cumulative frequency is 65.

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8.

(a) 40 Employees

30

20

10

Time

0

30

35

40

45

50

55

60

65

70

(b) 220 Employees 200 180 160 140 120 100 80 60 40 20 0 28

Time 30

32

34

36

38

40

42

44

46

48

50

52

54

56

58

60

62

64

66

68

70

72

We draw horizontal lines at 106, 53 and 159 to find estimates for the median and quartiles. The estimates are as follows: the median is approximately 53, the lower quartile is 43, and the upper quartile is 59.5. Therefore, IQR 59.5 43 16.5 (c)

To estimate the mean and standard deviation, we will use the midpoints of the intervals.

The mean value is 49.7 and the standard deviation is 13.6 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

9.

(a)

We can solve this problem by using a calculator.

The minimum value is 152 and the maximum value is 193. The lower and upper quartiles are 165 and 178 respectively, whilst the median value is 168. (b)

(c)

The mean value is 170.5 and the standard deviation is 9.65, correct to 3 s.f. (d)

The heights are widely spread from very short to very tall players. Heights are slightly skewed to the right, bimodal at 165 and 170, with no apparent outliers. There is a very small range from the first quartile to the median. 25% of all the players have heights within those 3 cm, from 165 168 cm.

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(e) 130

Cumulative frequency

120 110 100 90 80 70 60 50 40 30 20 10 0 150

Height 155

160

165

170

175

180

185

190

195

The 90th percentile of 130 players corresponds to a cumulative frequency of 117. Using the graph, we estimate the 90th percentile as 184.5 cm.

10.

11.

22165 182 10 130 10

23985 171 , correct to the nearest cm. 140

(f)

x

(a)

The mean is not going to change since the added observation has the same value as the mean of the previous observations. Therefore, the new mean is also 12.

(b)

x

(c)

21

9 11 21 12 ; the new mean is 12. 10 9 11 x 10

210 99 x

x 111 ; therefore, the last observation is 111.

If the mean of all 10 data points is 30, then

x 30 10 300

(a)

If the value of 25 was incorrectly entered as 15, that means the total sum should increase 310 by 10; therefore, the correct mean value is: x 31 10

(b)

Since the added value is greater than the mean value, the mean is going to increase. 310 32 342 The new mean is: x 31.1 10 1 11

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12.

13.

You can take the size of the sample to be any number n: n n n 20 40 60 110 6 3 x 2 36.7 n 3

7 10 12 17 21 x 7 72 102 122 172 7

y

212

12 x2

67 x y2

122

y 84 172 7

x

y 17

15 x2 7

y2

172 7

x2

y 2 157

To solve the simultaneous equations, we will use the substitution method by expressing the variable y (from the first equation) in terms of x. x

y 17 x

y 17

x2

y2

x2

157

y 17 x x 6 or x 11

17 x

2

y 17 x 2x2

157

34 x 289 157

y 17 x x 2 17 x 66 0

y 11 or y 6 x 6 or x 11

Note that we have discarded one solution because of the condition that x 25

14.

xi

278

i 1 25

xi 2

3682

i 1

x

y.

278 11.12 25 sn 2

3682 11.122 25

23.6256

To answer questions 15 19, we use tables and graphs from the previous exercises, together with GDC/ software. In each case, enter the data (class midpoint to represent each class) into lists and use the menu of your GDC to calculate the standard deviation. For IQR, reading from the cumulative frequency graph to find the first quartile, third quartile and subtract them will give you the estimate, but the GDC output usually differs from that answer because the GDC is assuming all data in each class to have one value the class midpoint. So, the IQR from that reading may be less accurate than reading from the graph. 15.

Estimates of the upper and lower quartiles are 12.5 and 4.5 respectively; therefore, IQR 8 . The standard deviation is 6.63 (to 3 s.f.).

16.

Estimates of the upper and lower quartiles are 30.5 and 24.5 respectively; therefore, IQR 6 . The standard deviation is 4.46 (to 3 s.f.)

17.

Estimates: IQR

23 . The standard deviation is 16.7 (to 3 s.f.)

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18.

Estimates: IQR 0.1 . The standard deviation is 0.0569 (to 3 s.f.)

19.

Estimates of the upper and lower quartiles are 280 and 180 respectively; therefore, IQR 100 (or 60 from GDC). The standard deviation is 82.1 (to 3 s.f.)

Exercise 10.4 1.

The scatter plot below shows a weak positive linear relationship. The correlation coefficient is 0.26 which confirms the weakness of the relationship. The regression equation is: y = 6.56 + 0.29x. For every change of 1 unit in the x-values, the y-values will change, on average, by 0.29

2.

(a)

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3.

(b)

We chose speed as the explanatory variable because the car must first run to cause any fuel consumption. Hence the speed helps explain the fuel consumption. The relationship appears to be negatively sloped because the consumption is measured by the distance travelled per litre of fuel.

(c)

The relationship appears to be a relatively strong negative one without any apparent outliers. The correlation coefficient is 0.986 which is very close to 1, and is thus a very strong relationship.

(d)

The regression equation is: Fuel cons.km/L = 24.1 0.116 × Speed Km/h. For every increase of 1 km/h in speed, the average number of km per liter will decrease by 0.116 km/L. i.e. consumption will increase.

(a)

(b)

The relationship appears to be a positive one except for an outlier which can be traced to be Singapore. We chose the explanatory variable to be the Income because the income level dictates how willing are people to pay for goods.

(c)

The relationship is

(d)

The regression equation is: PPP = 24383 + 0.351 GNI/cap. For every increase of $1 in GNI/cap, the PPP will increase, on average by $0.351

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4.

5.

(a)

(b)

There is obviously a positive relationship between the number of visitors and consumption. As the number of visitors increases the consumption will also increase.

(c)

The relationship seems to be strong and there is an absence of outliers. The correlation coefficient is 0.978 which is very close to 1.

(d)

The regression equation is: Consumption = 40.0 + 0.777 × Visitors. For every increase of 1 visitor, we expect, on average, that consumption will increase by 0.777

The answers for this question are found at the end of each question from 1 to 4.

6.

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er = 20.2 + 1.03 × Before. test is expected to change by 1.03. The correlation coefficient is 0.97 indicating a very strong linear relationship. For a student with 6 7.

(a)

(b)

The regression equation is: Cost = 1066 + 47.1 × units.

(c)

For every increase of 1000 units in production, the cost, on average, will increase by 47100 euros. The correlation coefficient is 0.999, which is almost perfect association. This is a strong linear relationship.

(d)

Let number of 1000 units be x, then: Cost =1066 47.1x Cost 1066 47.1 x x

cost per unit

If the cost is 105, then 105

1066 x

47.1

x 18.411

Thus, the number of units will be 18 411

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8.

9.

(a)

r = 0.493. This is a relatively weak correlation between the two scores.

(b)

The regression equation is: Economics = 2.07 + 0.649 × Physics

(c)

Economics = 2.07 + 0.649 (4) = 4.7 (which can be rounded up to 5)

(a)

This appears to be a positively sloped trend. (b)

The regression equation is: Price ( ) = 2689 + 154 × points

(c)

The intercept is meaningless, as zero is not in the domain of the explanatory variable. On average, for every increase of 1 point, we expect the price to increase by 154 euros.

(d)

r = 0.93 indicating a strong association between points and price.

(e)

The average price of a 63-point diamond is predicted to be 2689 + 154 (63) = 7013 euros.

(f)

Residual = 9117 7013 = 2104

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10.

(a) Scatterplot of Blood volume vs Age 74 72 70 68 66 64 62 60 20

30

40

50 Age

60

70

(b)

r = 0.958. There is a strong negative correlation between the stroke volume and age of patients.

(c)

The regression equation is: Blood volume = 82.1 0.268 × Age. On average for an increase of 1 year, we expect blood volume to be decreasing by 0.268 ml per stroke. The interpretation of the intercept of 82.1 does not make sense in this situation.

(d)

On average, 45-year-olds may have 70 stroke volume. Using the model to predict the 90-year old volume is not advisable as it is an extrapolation of 17 years beyond the range of collected data.

80

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11.

(a) Scatterplot of Velocity vs Time 180 160 140 120 100 80 60 40 20 0 0

1

2

3

4

5

6

7

Time

Apparently, there is a strong association between time and speed as expected. However, it appears that there is a break point around 3 seconds.

(b)

The regression equation is: Velocity = 24.5 + 21.8 × Time Fitted Line Plot

Velocity = 24.46 + 21.83 Time 200

S R-Sq

11.3886 94.3%

150

100

50

0 0

1

2

3

4

5

6

7

Time

Apparently, the data do not follow a linear model through the whole range. There is a clear deviation from the line at both ends. © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

(c)

r = 0.97, which is a strong association indication. However, this number may not be of great validity since the data does not appear to be linear.

(d)

By splitting the data, we can clearly see that the new model fits the data better. The data clearly has two phases, one before 3 seconds and the other after 3 seconds. Scatterplot of Velocity vs Time 180

interval a b

160 140 120 100 80 60 40 20 0 0

1

2

3

4

5

6

7

Time

Chapter 10 practice questions 30

1.

(a)

yi

360 12 30

360

i 1

30

(b)

2

yi

925

sn

i 1

2.

xi f i fi

34

925 30

30.83 5.55

10 1 20 2 30 5 40 n 50 3 11 n

374 34n 350 40n

24 6n

34

350 40n 11 n

n 4

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3.

(a)

Use a GDC/software for the histogram. Time Frequency

(b)

(c)

(d)

1.6

2.1

2.6

3.1

3.6

4.1

4.6

5.1

5.6

2

5

5

5

14

6

6

2

3

There are 7 out of 50 measurements that are greater than or equal to 5.1 43 Therefore, the fraction of the measurements less than 5.1 is: 0.86 86% 50 There are 50 pieces of data, so to determine the median we need to find the 25th and 26th observations. We notice that these two observations are within the interval 3.6 4.1; therefore, the median is approximately 3.9 Using GDC with data entered in (a), we get:

The mean value is 3.68, whilst the standard deviation is 1.11, correct to 3 s.f.

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6.1 6.6 2

0

(e)

A cumulative frequency table is given below. A graph is given as we need it for part (f). Time Cu. Freq.

1.6

2.1

2.6

3.1

3.6

4.1

4.6

5.1

5.6

6.1 6.6

2

7

12

17

31

37

43

45

47

50

0

Cumulative frequency 50

40

30

20

10

0

4.

Time 2

3

4

5

6

(f)

Estimates for the minimum and maximum values are 1.6 and 6.6 respectively. The first and third quartiles correspond to the cumulative frequencies of 12.5 and 37.5 respectively; therefore, an estimate for the first quartile is 3.15 and the third quartile is 4.65. An estimate for the median (which corresponds to a cumulative frequency of 25) is 3.9

(a)

The median and the IQR would best represent the data, since the data is skewed to the right and there are a few outliers on the right.

(b)

First, we read the frequencies from the histogram and input them into the frequency distribution table on our GDC.

The mean value is 682.6 and the standard deviation is 536.2 (c)

Since we have grouped data, the endpoints of our intervals will be 150, 250, 350, ..., and so on. On a calculator, we can use the adding a number to the list feature. Alternatively, we can calculate the cumulative frequencies from the list menu.

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(d)

There are 460 cities, so to estimate the median, we draw a horizontal line from 230. To find the lower and upper quartiles, we need to draw horizontal lines from 115 and 345 respectively.

So, the median is about 500. The first quartile is about 330 and the third quartile is about 830. Therefore, the IQR is about 830 330 = 500 (e)

There are a few outliers to the right. The outliers are those points which are over Q3 + 1.5 × IQR, i.e. 830 1.5 500 1580, which gives us 50 cities from the histogram. We can also use a box-plot as shown:

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(f)

5.

The data is skewed to the right with quite a few outliers to the right (1600 and above). The data is also bimodal, with the modal values being 300 and 400. After the class with 900 places, there are a few gaps which makes the spread of the data between the third quartile and the maximum very large.

(a) cheapest (es

6.

7.

(b)

Red wines are generally more expensive in France as we can see that the median price is the highest; the minimum value in France is also the highest, but the upper 50% of wines are also within a very small range of appr This is also clear because the lower quartile in France is higher than the medians in both other countries. This indicates that at least 75% of the French wines are more expensive than at least 50% nes.

(c)

It appears that the wines are, on average, more expensive in France, where the prices are skewed towards the higher end. In Spain, you can find a higher percentage of cheaper wine than in the other two countries, but you also find the most expensive wines on the market; so, Spain has the widest range of prices. Italy seems to have the most symmetric distribution of wine prices.

Given the large size of the data, it is advisable to use software/GDC for calculations. (a)

The mean value of the data is 52.6 and the standard deviation is 7.60, both given correct to the three significant figures.

(b)

The median value is 51.3. The upper and lower quartiles are 49.9 and 52.6 respectively; therefore, the IQR is 2.65, all correct to three significant figures.

(c)

Apparently, the data is skewed to the right with a clear outlier of 112.72. This outlier pulled the value of the mean to the right and increased the spread of the data. The median and IQR are not influenced by the extreme value.

(a)

The distribution does not appear to be symmetric as the mean is less than the median, the lower whisker is longer than the upper one and the distance between Q1 and the median is larger than the distance between the median and Q3. The data are left skewed.

(b)

There are no outliers as Q1 1.5 × IQR = 37 < 42, and Q3 + 1.5 × IQR = 99 > 86

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(c)

8.

(d)

See (a)

(a)

Using the given cumulative graph, we draw a horizontal line at 50%. The cholesterol level corresponding to the point of intersection is approximately 225.

(b)

Again, by drawing horizontal lines at 25%, 75% for the quartiles, and 10% and 90% for the percentiles, we can make the following estimates: Q1

210, Q3

260. 10th percentile

185, and 90th percentile

300

(c)

IQR = Q3 Q1 = 260 210 = 50. The number of patients in the middle 50% is 50% of the 2000 subjects, which is 1000.

(d)

It appears that the minimum level is 100 and the maximum is 400; the quartiles are mentioned above and the median is 225. To decide whether we have outliers, we calculate the length of the whiskers. The left one goes as far as and the right whisker goes to 260 1.5 50 335 . Thus, on the left side we have a few outliers below 135 as the minimum is 100, and some outliers on the right side since the maximum of 400 is way beyond 335. So, an approximate boxplot of the data is shown below.

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(e)

9.

In addition to what was mentioned in part (d), we notice that the data is skewed to the right a bit, with more outliers on the right side, since the outliers lie outside of the interval 135 to 335. From the cumulative frequency graph, we see that there are almost 100 patients who have a cholesterol level greater than 335 mg/dl and only a few patients with a level less than 135 mg/dl.

(a)

(b)

Speed

Frequency

26-30

9

31-34

16

35-38

31

39-42

23

43-46

12

47-50

8

51-54

1

Histograms for this data are not unique. It depends on the choice of class width. Here is a sample GDC output:

Data is relatively symmetric with possible outlier at 55. The mode is approximately 37

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(c)

We will show the work by using the following table of calculations xi

fi

xi× fi

xi2× fi

28

9

252

7056

32.5 16

520

16900

36.5 31

1131.5 41299.75

40.5 23

931.5

37725.75

44.5 12

534

23763

48.5 8

388

18818

52.5 1

52.5

2756.25

100

3809.5 148318.8

= 38.1 s=

Here we used the formulas:

5.65

xi f xi n

and sn2

xi2 f xi n

x2

(d)

(e)

Speed

Cumulative frequency

30

9

34

25

38

56

42

79

46

91

50

99

54

100

In order to estimate the median and quartiles, we draw the cumulative frequency graph and draw horizontal lines at 50, 25 and 75. We use a GDC here.

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An estimate of the median is 37. Q1 is 34.5 and Q3 is 42; therefore, the IQR is 7.5 (f)

Since Q3 is 42 and 1.5 × IQR is 11.25, and the upper whisker goes up to 53.25 (which is smaller than the largest observation) we suspect that there may a possible outlier, given that our values are only estimates that may also differ from the real values. On the lower end, the whisker goes down to 23.25, which is less than the minimum observation, and hence, there are no outliers here. A box-plot may be used.

25

30

35

40 Speed

45

50

55

Using the raw data, we confirm that 54 is, in fact, an outlier.

10.

(a)

Using software/GDC the required values are: Mean = 1846.9, median = 1898.6, standard deviation = 233.8, Q1 = 1711.8, Q3 = 2031.3, IQR = 319.5

(b)

Q1 1.5 × IQR = 1232.55 > minimum, so there is an outlier on the left. Q3 + 1.5× IQR = 2510.55 > maximum, so there is no outlier on the right.

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(c)

Use a GDC/software:

sn

(d) (e)

1846.9 233.8

Germany will definitely be an outlier since Q3 + 1.5 × IQR = 2510.55 < 2758. Therefore, it will influence the mean and standard deviation. However, the median, the first and third quartiles, and the IQR will not change much.

90

11.

(a)

xi

4460 90

4460

i 1

(b) 12.

1613.1, 2080.7

New

49.6 minutes

4460 35 39 28 32 90 4

4594 94

48.9 minutes

670

1190

(a) Marks Number of candidates

(b)

30

130

330

1630

1810

1900

1960

2000

We used a spreadsheet to produce this graph: 2200

Cumulative frequency

2000 1800 1600 1400 1200 1000 800 600 400 200

Marks 10

(c)

(i)

20

30

40

50

60

70

80

90

100

By looking at the graph, we estimate that the median score (which corresponds to the cumulative frequency of 1000) is 47.

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n1

13.

(ii)

We draw a vertical line from 35 on the Marks axis and reach the cumulative frequency curve at the point at which the cumulative frequency is about 500. Therefore, 500 candidates had to retake the exam.

(iii)

The highest scoring 15% corresponds to the highest 300 results; therefore, we draw a horizontal line from 1700 on the cumulative frequency axis and reach the curve at the point at which the number of marks is about 61. Hence, a distinction will be awarded if 61 or more marks are scored on the test.

n2 n2

1

n1

2

72 179 28 162 72 28

17424 174 cm 100

25

xi

25

14.

(a)

xi

i 1

300

25

i 1

300 12 25 25

25

(b)

xi

2

xi 625

sn

i 1

15.

x

34

x

xi f xi n

350 40k 374 34k 16.

(a)

i 1

25

2

625 25

10 1 20 2 30 5 40 k 11 k

6k 24

5

50 3

34

k 4

Similar to what has been done earlier, to calculate an estimate for the mean, we will take the midpoints of the intervals (15, 45, 75, and so on) and the corresponding frequencies.

So, an estimate for the mean of the waiting times is 97.2 seconds.

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(b) Time Cum freq

30 5

60 20

90 53

120 74

150 85

180 92

210 97

240 100

(c)

17.

(d)

To find the three estimations asked for, we need to draw a horizontal line at 50 for the median, and at 25 and 75 for the quartiles. An estimation of the median value is 88, while the lower and upper quartiles are 66 and 124 respectively.

(a)

(i)

The bar between 50 and 60 has a height of 10, so the number of plants in that range is 10.

(ii)

There are two bars here that add up to 14 + 10 = 24, so the number of plants in that range is 24.

(b)

Reading frequencies as the heights of the bars, and representing every class with its midpoint, we create a frequency distribution table and use a GDC to do the calculations. (Alternatively, you can use the formulas too.)

xi 15 25 35 45 55 65 75 85 95 fi 1 5 7 9 10 16 14 10 8

So, an estimate for the mean is 63, and the standard deviation is 20.5

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18.

(c)

The data is skewed to the left; therefore, the mean is pulled to the left with the extreme small values. Thus, the median must be larger than the mean.

(d)

Since there are 80 plants, we know that the median is about 40. If we draw the first part of the cumulative frequency graph and draw a horizontal line from 40 to find the point of intersection with the graph, our estimate is approximately 65 cm.

(a)

Again, we will use the midpoints of the intervals (82.5, 87.5, 92.5, ..., and so on) and the corresponding frequencies. We put the two lists into a GDC and obtain an estimation of the standard deviation. An estimate for the standard deviation of the weights is 7.405

(b) Weight w w w w w w w (c)

(d)

Number of packets 5 15 30 56 69 76 80

By drawing horizontal lines at 40 for the media and at 20, and 60 for the quartiles, the estimates will be: (i)

median

(ii)

Q3

97

101

This is the total deviation from the mean; it has to be zero. Also, W1 W

W80 W

W1 W1

(e)

W80 W80

80 W 80

W1

W80 80

0

There are 71 packets that satisfy the condition 85 W 110 . There are 20 packets that satisfy the condition 100 W 110. Therefore, the probability is: P E

20 71

0.282 , correct to three significant figures.

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19.

(a)

Again, we will use the midpoints of the intervals (65, 75, 85, ..., and so on) and the corresponding frequencies. We put the two lists into a GDC and obtain an estimation of the mean. An estimate for the mean speed is 98.2 km h 1.

(b)

(i)

To find the value of m, we can either add 70 (the frequency of the speed interval 90 100) to the previous cumulative frequency, 95; or subtract 71 (the frequency of the speed interval 100 110) from the next cumulative frequency, 236. In both cases we get the same value: m 165 In a similar manner, we find the value of n: n 236 39 275

(c)

(ii)

Using a spreadsheet, we get the following graph:

(i)

We draw a vertical line from v 105 until we reach the cumulative frequency curve. This gives us an estimate for the cumulative frequency of 200. So, there 100 are 100 cars that will exceed the speed of 105 km h 1 and P 33.3% 300

(ii)

If 15% of the cars exceed this speed, then 85% do not exceed that speed. 85% of 300 is 255, so we draw a horizontal line from y 255 until we reach the cumulative frequency curve. This gives us an estimate of a speed of 115 km h 1.

20.

(a)

(i)

We take a horizontal line across from 100 on the vertical axis until it touches the graph. From that point, we take a vertical line down to the horizontal axis and read the value. An estimate for the median fare is $24.

(ii)

We take a vertical line up from 35 on the horizontal axis until we reach the graph. From that point, we take a horizontal line across to the vertical axis and read the value. An estimate for the number of cabs in which the fare taken is $35 or less is 158 cabs.

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(b)

40% of the cabs is 0.4 200 80. So, we take a horizontal line from 80 on the vertical axis until we reach the graph and then we estimate the x-coordinate, which is 22. Therefore, the fare is $22. To find the number of kilometres, we need to divide the fare by 0.55 (which is the fare per kilometre for distance travelled). Therefore, the distance travelled is 40 km.

(c)

If the distance travelled is 90 km, the driver will earn 90 0.55 49.5 dollars. We will use the graph to estimate the number of cabs that will earn less than 49.5 dollars there are about 185, and therefore there are 16 cabs that will earn more than that. So, the 15 percentage of the cabs that travel more than 90 km is: 0.075 7.5% 200

21.

Since the three numbers are given in order of magnitude a b c, we know that the middle one (the median) is 11. Given that the range is 10, we know that the difference between the minimum and maximum value is: c a 10 . Since the mean value is 9, we can establish another equation a 11 c in terms of a and c: 9 a c 16 . Solving these two equations (using the 3 elimination method), we get: 2a 6 a 3

22.

(a) 110 Cumulative frequency 105 100 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5 50

100

Price ( thousands of $) 150

200

250

300

350

400

450

500

(b)

By using horizontal lines at 25 and 75, we estimate the values of Q1 and Q3 as $130,000 and $250,000. Hence, the IQR is $120,000. (The answer here is different from the book because of the different accuracy of the graphs used.)

(c)

To find the frequencies m and n, we need to subtract two successive cumulative frequencies. fi

ci

ci

1

m 94 87 7, n 100 94 6

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(d)

We take the midpoints of the intervals in the first list and the corresponding frequencies in the second list, and then we use a GDC. So, an estimate of the mean selling price is $199,000.

(e)

23.

(a)

(i)

An estimate of the cumulative frequency for $350,000 is 90; therefore, there are about 10 houses that can be described as Luxury houses.

(ii)

Out of 10 Luxury houses, six were sold for $400,000; therefore, the probability that both selected houses have a selling price more than $400,000 is: 6 5 1 P E 0.33 10 9 3

(i)

To mark the median, we draw the horizontal line y 40 until it hits the graph, at which point we draw a vertical line down to the Diameter axis. An estimate for the median is 20 mm.

(ii)

To mark the upper quartile, we draw the horizontal line y 60 until it hits the graph, at which point we draw a vertical line down to the Diameter axis. An estimate for the upper quartile is 24 mm.

24.

(b)

The interquartile range is: IQR 24 14 10 mm

(a)

In this question, we can accept an error of 2 students. We need to read the cumulative frequencies at the endpoints of the intervals and then subtract the successive ones to obtain the frequencies. For 40, the cumulative frequency is 72, so the corresponding frequency is 72 22 50 For 60, the cumulative frequency is 142, so the corresponding frequency is 142 72 70 For 80, the cumulative frequency is 180, so the corresponding frequency is 180 142 38

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Mark Number of students

25.

[0,20[

[20,40[

[40,60[

[60,80[

[80,100[

22

50

70

38

20

(b)

40% of 200 students is 80, and then we draw a horizontal line and estimate the x-coordinate of the point. Our estimation is 43%

(a)

To find the median height, we draw a horizontal line y 60 until it hits the graph. We then estimate the x-coordinate of the point of intersection. We estimate 183 cm. For the lower and upper quartiles, we draw two horizontal lines y 30 and y 90.

(b)

Then we estimate the x-coordinates of the points of intersection. Therefore: Q1 175,Q3 189 IQR 189 175 14

26.

Since the modal value is 11, we know that c d 11 . Given that the range is 8, we can find the value of a: 11 a 8 a 3 . Finally, given that the mean value is 8, we can find the remaining 3 b 11 11 number b: 8 25 b 32 b 7 4

27.

(a)

We draw a vertical line x 40 until it hits the graph. We then estimate the y-coordinate of the point of intersection, i.e. 100. So, the number of students who scored 40 marks or less is 100.

(b)

There are 800 students, so the middle 50% is between 200 and 600 students. For a cumulative frequency of 200, the estimated mark is 55; whilst for 600, the estimated mark is 75. Hence, we say that the middle 50% of test results lie between 55 and 75 marks: a 55, b 75.

28.

Use the mean formula to find the first equation relating the unknowns: x y 90 13 104 x y 90 x y 14 8 Use the sum of the squares of the known observations and the variance formula to find the second equation relating the squares of the unknowns: 21

x2

y 2 1404 132 8

190

x2

y 2 1404 8

x2

y 2 116

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Now, use substitution to solve the simultaneous equations. y 14 x x y 14 y 14 x x2

y2

116

x2

y 14 x x 4 or x 10

Since x

29.

14 x

2

y 10 or y x

x 2 14 x 40 0

116

4

4 or x 10

y , we can discard the second solution. A GDC can also be used.

(a) Scatterplot of Yield vs Rainfall 90

80

70

60

50 10

15

20

25

30

Rainfall

The scatter plot shows an apparently linear association with two possible outliers: (7, 54) and (28, 78). Apart from these, the association seems to be relatively strong. (b)

r = 0.853, which is a relatively strong positive linear relationship.

(c)

Yield = 40.5 + 1.78 × Rainfall. On average, a change of 1 cm in rainfall corresponds to a change of 1.78 kg change in crop. The intercept is not useful in this case since 0 is not in the domain of the explanatory variable.

(d)

Since a rainfall of 19 cm lies within the domain of the explanatory variable, a prediction is appropriate: Yield = 40.5 + 1.78 (19) = 74.32 So, on average, the yield per tree in this region is expected to be about 74 kg.

(e)

Let the gradient of the regression of x on y be n, and that of the regression of y on x be m. r 2 0.8532 0.4088 We know that mn r 2 n m 1.78 The angle between 2 lines is given by tan 1 m tan 1 n tan 1 1.78 tan 1 0.4088 38

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Note: Some answers may differ from one student to another because of variations in graph accuracy.

Exercise 11.1 1.

(a)

S = left-handed, right-handed

(b)

S

h

: 50 h

(c)

S

t

:0 h

250 , where height (h) is in centimetres. 240 , if we decide that the night starts at 20:00

after midnight. 2.

S = (1, h), (2, h), (3, h), (4, h), (5, h), (6, h), (1, t), (2, t), (3, t), (4, t), (5, t), (6, t)

3.

(a)

S = 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , J , Q , K , A , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , J , Q , K , A , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , J , Q , K , A , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , J , Q , K , A

(b)

We need to list all the possible pairs from a deck of 52 cards, which is many more than the listing in part (a), so we are just going to initiate a possible listing. S = (2 (2 , 3 (3 , A (3 , 2

4.

, 3 ), (2 , 4 ), ...(2 , A ), (2 , 2 ), (2 , 3 ), ...(2 , A ), (2 , 2 ), ), ...(2 , A ), (2 , 2 ), (2 , 3 ), ...(2 , A ), (3 , 4 ), (3 , 5 ), ... ), (3 , 2 ), (3 , 3 ), ...(3 , A ), (3 , 2 ), (3 , 3 ), ...(3 , A ), ), (3 , 3 ), ...(3 , A ), ...(K , A

(c)

In the first experiment there are 52 outcomes, as there are 52 cards in the deck. In the 52 51 2652 second experiment there are 1326 outcomes, since the order of the pair 2 2

(a)

Since Tim tossed 20 coins 10 times, there are 200 possible outcomes. The sum of all the number of heads that appeared in the experiment is the number of favourable outcomes. 94 47 11 9 10 8 13 9 6 7 10 11 94 P H 0.47 200 100

(b)

Tim should expect any number between 0 and 20.

(c)

If he tossed 20 coins 1000 times, we would expect heads to be obtained exactly half the time; therefore, 10 000 heads.

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5.

6.

7.

(a)

S

1,1 , 1,2 , 1,3 , 1,4 , 2,1 , 2,2 ,..., 3,1 , 3,2 , 3,3 , 3,4 ,...,(4,4)}

(b)

We need to look at each pair from (a) and add 1 to the sum of the components. So, S 3,4,5,6,7,8,9 .

(a)

Since we are replacing the first ball drawn, there are three different colours possible for the first ball drawn and three different colours possible for the second ball drawn. S = (b, b), (b, g), (b, y), (g, b), (g, g), (g, y), (y, b), (y, g), (y, y)

(b)

A = (y, b), (y, g), (y, y)

(c)

B = (b, b), (g, g), (y, y)

(a)

Since we do not replace the first ball drawn, there are only two colours possible for the second ball drawn. S = (b, g), (b, y), (g, b), (g, y), (y, b), (y, g)

8.

9.

(b)

A = (y, b), (y, g)

(c)

B=Ø

(a)

S = (h, h, h), (h, h, t), (h, t, h), (t, h, h), (h, t, t), (t, h, t), (t, t, h), (t, t, t)

(b)

A = (h, h, h), (h, h, t), (h, t, h), (t, h, h)

Let H = Hungary, I = Italy, b = boat, d = drive, f = fly. Go on vacation: S = (I, f), (I, d), (I, t), (H, d), (H, b) Fly to destination: S = (I, f)

10.

(a)

S = (0, g), (0, f), (0, s), (0, c), (1, g), (1, f), (1, s), (1, c)

(b)

A = (0, s), (0, c)

(c)

B = (0, g), (0, f), (1, g), (1, f) Note: In this case we are not concerned as to whether or not the patient is insured.

(d)

C = (1, g), (1, f), (1, s), (1, c)

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11.

The study is investigating three different characteristics. There are 2 classifications for gender, 3 classifications for drinking habits, and 4 classifications for marital status; so, there are 2 3 4 24 different classifications for a person in this study. S (a)

G1 , K1 , M 1 ,..., G1 , K1 , M 4 , G1 , K 2 , M1 , G1 , K 2 , M 2 ,..., G1 , K 2 , M 4 , G1 , K 3 , M1 ,..., G1 , K3 , M 4 , G2 , K1 , M1 ,..., G2 , K1 , M 4 , G2 , K 2 , M1 ,..., G2 , K 2 , M 4 , G2 , K3 , M1 ,..., G2 , K3 , M 4

(b)

Set A containing G2 . A

G2 , K1 , M 1 ,..., G2 , K1 , M 4 , G2 , K 2 , M 1 , G2 , K 2 , M 2 ,..., G2 , K 2 , M 4 , G2 , K3 , M 1 ,..., G2 , K3 , M 4

Set B containing K2 or K3. B

G1 , K 2 , M 1 ,..., G1 , K 2 , M 4 , G1 , K 3 , M 1 ,..., G1 , K 3 , M 4 , G2 , K 2 , M1 ,..., G2 , K 2 , M 4 , G2 , K3 , M 1 ,..., G2 , K3 , M 4

Set B can also be described as the set that consists of all the triplets not containing K1. Set C

(c)

12.

(a)

triplets containing M2.

C

G1, K1, M 2 G1, K2 , M 2 , G1 , K3 , M 2 , G2 , K1 , M 2 , G2 , K2 , M 2 , G2 , K3 , M 2

(i) (ii)

Set A B can be described as the set that consists of male persons or persons who drink. Set A C can be described as the set that consists of single male persons.

(iii)

Set

(iv)

Set A B who drink.

(v)

The set A' B can be described as the set that consists of female persons who drink.

can be described as the set that consists of non-single persons.

C can be described as the set that consists of single male persons

Since we are taking four cars at a time, we will be recording quadruplets. For example, (L, L, L, L), (R, R, R, S), (L, S, R, S), and so on. Every car leaving the highway has three 4 options; therefore, there are 3

(b)

81 different quadruplets.

If all cars go in the same direction, we have (L, L, L, L), (R, R, R, R) and (S, S, S, S).

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(c)

If only two cars turn right, the remaining two cars will either turn left or go straight: (R, R, L, L), (R, R, S, S), (R, R, L, S), and now we have to find the remaining permutations. For example, let us take the first quadruplet and its permutations: (R, L, R, L), (R, L, L, R), (L, R, L, R), (L, R, R, L), (L, L, R, R). The same pattern works for the second quadruplet. For these two quadruplets, there are 12 different permutations altogether. The last quadruplet has more permutations since we have three possible ways of leaving the highway. There are 12 possibilities, so the remaining quadruplets are: (R, R, S, L), (R, L, R, S), (R, S, R, L), (R, L, S, R), (R, S, L, R), (L, R, R, S), (S, R, R, L), (L, R, S, R), (S, R, L, R), (L, S, R, R), (S, L, R, R). Therefore, there are a total of 24 outcomes where only two cars turn right.

(d)

Only two cars going in the same direction contains the previous part, and there are two more ways of the cars going in the same direction: L, L and S, S. So, altogether, there are 3 24 72 different outcomes.

13.

Since we have to look at three different components, we will be recording triplets. The first component, size of the vehicle, has three different classifications, whereas the remaining two components have just two different outcomes. Therefore, there are 3 2 2 12 different triplets. (a)

U

T,SY,O , T,SY,F , T,SN,O , T,SN,F , B,SY,O , B,SY,F ,

B,SN,O , B,SN,F , C,SY,O , C,SY,F , C,SN,O , C,SN,F

(b)

SY= T,SY,O , T,SY,F , B,SY,O , B,SY,F , C,SY,O , C,SY,F

(c)

C= C,SY,O , C,SY,F , C,SN,O , C,SN,F

(d)

C SY= C,SY,O , C,SY,F C'= T,SY,O , T,SY,F , T,SN,O , T,SN,F , B,SY,O , B,SY,F , B,SN,O , B,SN,F C

SY= T,SY,O , T,SY,F , B,SY,O , B,SY,F , C,SY,O , C,SY,F ,

C,SN,O , C,SN,F

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14.

15.

(a)

Since there are three components and each can work or not, there are 2 2 2 8 different outcomes.

U

0,0,0 , 0,0,1 , 0,1,0 , 1,0,0 , 0,1,1 , 1,0,1 , 1,1,0 , 1,1,1

(b)

X

0,1,1 , 1,0,1 , 1,1,0

(c)

Y

0,1,1 , 1,0,1 , 1,1,0 , 1,1,1

(d)

Z

1,0,1 , 1,1,0 , 1,1,1

(e)

Z'

(a)

0,0,0 , 0,0,1 , 0,1,0 , 1,0,0 , 0,1,1

X

Z

0,1,1 , 1,0,1 , 1,1,0 , 1,1,1

X

Z

1,0,1 , 1,1,0 , Y

Y

Z

1,0,1 , 1,1,0 , 1,1,1

U

Z

0,1,1 , 1,0,1 , 1,1,0 , 1,1,1

Y

Z

1,2,31,32, 41, 42,51,52,341,342, 431, 432,351,352,531,532,

451,452,541,542,3451,3452,3541,3542,4351,4352, 4531, 4532,5341,5342,5431,5432

There are 32 different outcomes in total.

(b)

A

31,32, 41,42,51,52 . There are six possible outcomes.

(c)

B

31,32, 41, 42,51,52,341,342, 431, 432,351,352,531,532, 451, 452,541,542, 3451,3452,3541,3542, 4351, 4352, 4531, 4532,5341,5342,5431,5432 .

There are 30 possible outcomes.

(d)

C

1,31, 41,51,341,431,351,531, 451,541,3451,3541, 4351, 4531,5341,5431 .

There are 16 possible outcomes.

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Exercise 11.2 1.

(a)

There are six multiples of 3 from 1 to 20. So, P( A)

6 3 20 10

Note: The number of multiples can be obtained by using the greatest integer function: 20 3 (b)

6.67

6

We will use the complementary event that the number is a multiple of 4. There are five multiples of 4 from 1 to 20. So, P( B) 1 P B

2.

3.

1

5 1 1 20 4

3 4

(a)

P A'

1 P A

(b)

P A

A'

(a)

(i)

There is one ace of hearts in a deck of cards, so: P( Ai )

(ii)

There is one ace of hearts and 13 spades in a deck, so: P( Aii )

(iii)

The ace of hearts is already included in the 13 hearts in a deck, so we only need 13 3 16 4 to add the three remaining aces. So, P( Aiii ) 52 52 13

(iv)

There are 12 face cards. We will use the probability of the complementary event: 12 3 10 P( Aiv ) 1 P( Aiv ) 1 1 52 13 13

(b)

(c)

1 0.37 0.63

P S

1 or P A

A'

P A

P A'

0.37 0.63 1

1 52 1 13 52

14 52

As the drawn card is not replaced, there are 51 cards remaining in the deck.

1 51

(i)

P( Bi )

(ii)

P( Bii ) 1 P( Ai1 ) 1

12 4 13 1 51 17 17

As the drawn card is replaced in the deck, there are no influences on the drawing of the next card. Therefore, the probability is the same as in (a). (i)

P(Ci )

1 52

(ii)

P(Cii ) 1 P(Cii ) 1

12 3 10 1 52 13 13

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7 26

4.

5.

The total number of students is 30. Looking at the table, we obtain: (a)

P A

4 12 8 30

24 30

4 5

(b)

P B

8 3 30

(c)

All of the students studied less than 6 hours; therefore, P C

11 30 1

There are 12 different possible outcomes: 6 possible outcomes for the dice and 2 possible outcomes for the coin, so by the counting principle we obtain 12. (a) whether a head or a tail is obtained we get: P A (b)

6.

7.

6 12

1 2 1 12

Obtaining a head and a 6 is just one possible outcome out of 12; therefore, P B

Let the probability of any other number than 1 appear be x, so the probability of 1 is 2x. 1 The sum of all the probabilities is 1; therefore, pi 1 7 x 1 x 7

1 7

(a)

P A

(b)

The odd numbers are 1, 3 and 5, so P B

(a)

(i)

2 1 1 7

4 7

There are 6 possible outcomes for the first dice and 6 possible outcomes for the second dice; therefore, there are a total of 36 possible outcomes. S

1,1 , 1,2 ,..., 1,6 , 2,1 ,..., 2,6 , 3,1 ,..., 3,6 , 4,1 ,..., 4,6 , 5,1 ,..., 5,6 , 6,1 ,..., 6,6

6 36

1 6

(ii)

There are six possible pairs with equal numbers; therefore, P A

(iii)

Looking at the sample space, we notice that there are eight such outcomes: B

1,3 , 2, 4 , 3,1 , 3,5 , 4,2 , 4,6 , 5,3 , 6, 4 P B

(iv)

8 36

2 9

This event is complementary to the event in part (ii), so P C

1

1 6

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5 6

(b)

The probability distribution for the sum of the numbers that appear is shown in the table. X (sum) P(X)

8.

(a)

2 1 36

3 2 36

4 3 36

5 4 36

6 5 36

7 6 36

8 5 36

9 4 36

10 3 36

(i)

There is no sum equal to 1 and therefore P D

(ii)

Looking at the table, we can read that P E

4 36

(iii)

Looking at the table, we can read that P F

5 36

(iv)

The largest sum is 12 and therefore P G

11 2 36

12 1 36

0

1 9

0

Since the sum of all the probabilities is equal to 1,

P AB USA

1

0.43 0.41 0.12

1 0.96 0.04

(b)

Since the events are mutually exclusive, the probability of their union is the sum of their probabilities. Therefore, P O B USA 0.43 0.12 0.55

(c)

Since we have to independently select two people, their probabilities should multiply; therefore, P China

(d)

P O|USA

P O China

0.43 0.36 0.1548 0.155 (3 s.f.)

Since we have to independently select three people, their probabilities should multiply:

P O

P O USA

P O China

P O Russia

0.43 0.36 0.39 0.060372 0.0604,

correct to three significant figures. (e)

First, we need to find the probability of type B in Russia: P B Russia 1 0.39 0.34 0.09 1 0.82 0.18 Similarly, as in (d), we need to calculate the probability of only one blood type:

P A

P A USA

P A China

P A Russia

P B

P B USA

P B China

P B Russia

P AB P S

P AB USA P O

P A

P AB China P B

P AB

0.41 0.27 0.34 0.037638 0.0376 0.12 0.26 0.18 0.05616 0.00562

P AB Russia

0.04 0.11 0.09 0.000396

0.104022 0.104 (3 s.f.)

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9.

10.

11.

(a)

Yes, since the sum of all the probabilities is 1.

(b)

No, since four mutually exclusive events are given, and their sum exceeds 1.

(c)

No. There is the same number of cards in each suit, but by looking at the probability distribution we notice that one heart and one diamond are missing, and there are two extra spades.

(a)

Since the sum of all the probabilities is 1, P Other 1 0.58 0.24 0.12 1 0.94 0.06

(b)

We need to use the complementary event, so P not German

(c)

P GG

(d)

The two Swiss that we select could have German, French, or Italian as their mother tongue. Therefore: P GG +P FF +P II +P OO 0.582 0.242 0.122 0.062 0.412

(a)

We use the probability of the complementary event, so: P A 1 0.165 0.142 0.075 0.081 0.209 0.145

(b)

1 0.58 0.42

0.58 0.58 0.3364 0.336 (3 s.f.)

1 0.817 0.183

Again, the complementary event will be used: P B 1 0.165 0.145 1 0.31 0.69 n!

12.

f x

n Cx 1

x 1 ! n x 1!

n Cx

n!

n x 1 x 1

n x

x 1

x

n 1 2

x! n x !

13.

n n 1

(a)

190 n 2 n 380 n 20 n 19 0 2 Since n must be a positive integer, the only possible solution is n 20

(b)

We know the symmetric property of binomial coefficients, n Cr

n C2

190

n Cn r

4 8 12 . Alternatively:

n C4

n C8

n

n C4

n C8

n! 4! n 4 !

n! 8! n 8 !

4! n 4 n 5 n 6 n 7 n 4 n 5 n 6 n 7

4! n 4 ! 8! n 8 ! n 8 ! 8 7 6 5 4! n 8 ! 8 7 6 5

n 12

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14.

There are 36 different outcomes that we will present as ordered pairs. The first component will represent the outcome of the white dice, and the second component will represent the red dice. U

1,1 , 1, 2 , 4,1 ,

, 1, 6 , 2,1 ,

, 4,6 , 5,1 ,

, 2,6 , 3,1 ,

, 5,6 , 6,1 ,

, 3,6 ,

, 6,6

(a)

By inspection, we can see that the sum is greater than 8 on ten different outcomes; 10 5 therefore, P A 36 18

(b)

A number greater than 4 means that 5 or 6 will appear on the first dice. There are 12 such possible outcomes; therefore, P B

15.

12 36

1 3

(c)

At most a total of 5 means that the sum could be 2, 3, 4 or 5. By inspection, we can see 10 5 that again there are 10 possible pairs; therefore, P C . We see that it is the 36 18 same as the probability of obtaining a sum greater than 8, which is true due to the symmetric property of the outcomes.

(a)

There are 9 books on the shelf altogether. We select 3 books, and one of the books must be selected, we have to select 2 out of 8 books. 28 1 8 C2 1 P A 84 3 9 C3

16.

(b)

We need to select 3 books from the 9 books on the shelf. There are 5 C 2 ways to select two novels and 3 ways to select one science book (from three). 10 3 5 5 C2 3 C1 P B 84 14 9 C3

(a)

We have to select 5 cards (from 52) and we need 3 kings (from 4). That means the remaining 2 drawn cards can be any of the 48 non-king cards. P A

(b)

4 C3 48 C2 52 C5

94 54145

0.00174

Again, we need to select 5 cards altogether. We have to select 4 hearts (from 13) and 1 diamond (from 13). P B

13 C4 13 C1 52 C5

143 39984

0.00358

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17.

(a)

We have to select 6 students from a class of 22. We would like to have 1 out of the 12 boys and 5 out of the10 girls. P A

(b)

12 C4 10 C2

12 C5 10 C1

12 C6 10 C0

22 C6

22 C6

22 C6

495 45 792 10 924 74613

943 0.417 2261

We need to select both married couples, that is, all 4 from the group of 4, and then we look at the remaining 11 people from which we need to select a further 2 people. P A

(b)

4 C4 11 C2 15 C6

1 55 5005

1 0.0110 91

If we select the three youngest members in the group, we can select any of the remaining members of the group for the final three places. P B

3 C3 12 C3 15 C6 10 C3 15 C3

1 220 5005

4 0.0440 91

120 455 177100

78 253

(a)

P A

(b)

At least 3 means 3, 4, 5 or 6. We will use the complementary event since it has fewer calculations. The complementary event is at most 2, which is 0, 1 or 2 colour laser printers. C C C C C C P B 1 P B 1 10 0 15 6 10 1 15 5 10 2 15 4 25 C6 25 C6 25 C6

25 C6

1

20.

0.0405

In questions where conditions are given, we need to fulfil the conditions first and then we have to see what the possibilities of the remaining elements are. We select 6 people from a group of 15. (a)

19.

22 C6

144 3553

It could be 4, 5 or 6 boys and 2, 1 or no girls respectively in the team of 6 students. P B

18.

12 C1 10 C5

(a)

0.308

1 5005 10 3003 45 1365 689 1 177100 1265

0.445

Since there are 30 buses and we need to select 6 for inspection, there are 30 C6

593775 ways to select six buses.

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(b)

Half means that 3 buses have cracks on the instrument panel. P B

(c)

10 C3 20 C3 30 C6

1 P C' 1

1

10 C0 20 C6

10 C1 20 C5

10 C2 20 C4

30 C6

30 C6

30 C6

38760 155040 218025 593775

2426 7917

0.306

At most half means 0, 1, 2 or 3. Again, we will calculate the probability by using the complementary event, which is 4, 5 or 6 buses have cracks. P D

1 P D' 1

21.

0.230

At least half means 3, 4, 5 or 6. Again, we will calculate the probability by using the complementary event, which is 0, 1 or 2 buses have cracks. P C

(d)

608 2639

1

10 C4 20 C2

10 C5 20 C1

10 C6 20 C0

30 C6

30 C6

30 C6

39900 5040 210 593775

1045 1131

0.924

There are 67 workers in the factory altogether and we have to select 9. 30 C9

10005 29900 492

(a)

P A

(b)

The same shift means either from the day, evening or morning shift. P B

(c)

30 C9

22 C9

15 C9

67 C9

67 C9

67 C9

0.000335

269265 777 412792

0.000346

We will calculate the probability that at least two of the shifts are represented by considering the complementary event, which is that only one of the shifts is represented, as in part (b). P C

(d)

67 C9

1 P B

1

269 265 777 412792

0.9997

The probability that at least one of the shifts is unrepresented means that one shift is not selected, either from the day, evening or morning shift. That is, when a shift is not represented, all 9 are selected from the two other shifts. P D

morning

evening

day

52 C9

45 C9

37 C9

67 C9

67 C9

67 C9

468942155 42757703560

0.1097

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22.

(a)

(b)

(c)

Since we have to select 2 out of these 8 chips, there are 8 P2 56 different outcomes. A sum of 7 is obtained in the following pairs: (1, 6), (6, 1), (2, 5), (5, 2), (3, 4) and (4, 3); 6 3 0.107 therefore, the probability is P A 56 28 Since we have to select 2 out of these 20 chips, there are 20 P2 380 different outcomes. If we take the smaller number first, we notice that there are 17 possible outcomes, 3. If we start with the 34 larger numbers, there are also 17 ways. So, the probability is P B 0.0895 380 We have 380 different outcomes and we are going to use the complementary event to find the required probability. Again, we take the smaller number first and need to find all of those pairs in which the numbers differ by 1, 2, or 3. As in the previous part, we will count the number of pairs that differ by 1, which is 19 2, as before. ( such a number in the box). The number of pairs in which the numbers differ by 2 is ). The number of pairs in which 18 2 . (19 the numbers differ by 3 is 17 2 . (18, )

38 36 34 272 0.716 380 380 Note: It is impossible to draw two equal numbers. P B

23.

1

To solve this problem, we are going to use geometric probability. Let x and y 20:00, with time in minutes on both axes. Now, we can conclude that the range of both variables will be 0 x, y 60 . In order to have dinner together, they have to arrive at the restaurant within 30 minutes of each other; therefore, they have to satisfy the inequality x

y

30 . The shaded

area represents the favourable outcomes. 60

30

0

30

60

We notice that all the possible outcomes are represented by a square with an area of 60 2 , while the unshaded part consists of 2 triangles, which, when put together, make a square of side 30.

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The area of the shaded region is the area of the large square minus the unshaded area: 602 302 3 P D 0.75 4 602 24.

Let x represent the time of the tram and y the time of the bus at the station. Since both stay for 3 minutes in the station, we are going to use minutes as units on both axes. Now, we can conclude that the range of both variables will be 0 x, y 20 . In order to be at the station at the same time, they have to arrive at the station within 3 minutes of each other; thus, they have to satisfy the inequality x y 3 . The shaded area represents the favourable outcomes. 20

3 0

20

3

We notice that all the possible outcomes are represented by a square with an area of 202 , whereas the unshaded part consists of 2 triangles, which, when put together, make a square of side 17. The area of the shaded region is the area of the large square minus the unshaded area. 202 17 2 202

Thus P D

25.

111 400

0.278

We have to select 6 from 30 slips. There are 8 C2 ways of selecting a film, 10

C2 and

12

C2 of the others:

(a)

P A

(b)

P B

(c)

8 C2

10 C2

12 C2

30 C6 8 C6

10 C6 30 C6

12 C6

264 1885 166 84825

0.140

0.00196

Selecting only films and songs means that we have 8 12 20 slips for the favourable 2584 20 C6 0.0653 event. Hence, P C 39585 30 C6

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26.

There are 1000 cubes altogether. Eight cubes are painted green at three faces (vertices of the original cube). There are eight cubes painted green at two sides per edge. There are 12 edges, so there are 96 such cubes. There are 64 cubes on each of six faces that have only a green face; therefore, there are 384 such cubes.

96 1000

12 125

(a)

P A

(b)

P B

(c)

From the above explanation, we notice that there are 8 96 384 488 cubes with at least one face painted. So, to find the probability that no face is coloured, we use the complement which is the cube with at least one face is green: 488 64 P C 1 P C P C 1 100 125

8 1000

1 125

Exercise 11.3 1.

Using the addition rule: P A

2.

B

P A

P B

P B

4 5

B

3 4

7 20

3 10

B

B

P A

P B

P A

B

P A

9 3 7 10 10 10

(b)

First, we need to write the event in a different form: B ' A

P B' A

P A

P A

P B

B

B

P B

(a)

A

P A

P A

1 2 A\ A

B , and since

A we can now calculate the probability: P A\ A

B

A

P A

P A

B

7 3 10 10

4 10

2 5

B

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(d)

By using a Venn diagram, we can spot that the intersection of the complements of two sets can be written as the complement of the union of the sets. Hence: 1 B ' A' A B ' P B ' A' P A B ' 1 P A B 10 1 P B A' 2 5 Using the conditional probability formula: P B A ' 3 P A' 3 10

A'

P B\ A

B

P B

P A

Given the addition rule, we can calculate the intersection. 1 2 P A B P A P B P A B P A B 3 9

B

4 9

1 9

It is obvious that the events are not mutually exclusive since P A and also, they are not independent since P A

4.

1 5

Similarly, P B

(e) 3.

5 3 10 10

(c)

P B

1 2 3 9

2 27

B 1 9

1 9

0,

P A

B

Since the events are independent, we can use the multiplication rule of independent 3 7 events to find the probability of event B. P A B P A P B P B 103 10 7 From the addition rule, we obtain:

P A

5.

P A

P B

P A

B

P A

B

3 7 3 7 10 10

58 70

29 35

In order to pass the test without having to wait 6 months, the new driver has to pass either the first time or fail the first attempt and pass the second time.

P A

6.

B

(a)

pass 1st attempt

fail 1st attempt

pass 2nd attempt

0.6

0.4

0.75

Using the complementary event, we get P O '

(b)

0.9

(i)

1 P O P O O

P O '

1 0.08 0.92 (92%)

0.08 0.08 0.0064 0.64%

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(ii)

The complementary event of at least one of them has O blood is none of them has O blood. P O 'O '

(iii)

(a)

P O 'O

2 0.08 0.92 0.1472 14.72%

1 0.928

1 P 8O '

0.486781 48.7% (3 s.f.)

10 different digits can be used to make each four-digit number; as such, there are 10 4

(b)

1 0.8464 0.1536

Using the complementary event that none has O blood: P C

7.

P B

If only one of them has O , that means that the other is not O , so: P O O '

(c)

0.92 0.92 0.8464

10 000 different possible PIN numbers.

Since the first digit cannot be zero, there are 9 digits that can be in the tenposition, while each of the other digits can be any of the 10 numerals. Therefore, there are 9 103 numbers without a zero as a starting number. So, the probability is: 9000 9 P B 10000 10

(c) P C

8.

1

94 104

10000 6561 3439 10000 10000 P D

C

(d)

Using the conditional probability formula: P D C

(a)

We need to use the complementary event that no red ball is drawn: P A

1 P BB

1

2 8

2 8

64 4 64

60 64

Using the conditional probability formula: P B A

(c)

P C A

P A

C

6 8

2 8 15 16

1000 3439

3439 10000

P C

15 16

(b)

P A

1000 10000

P A P B

B

2 8

6 8

6 8

6 8

15 16

1 5

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48 64 15 16

4 5

9.

(a)

U

1,1 , 1,2 , , 1,6 , 2,1 , , 2,6 ,

5,1 , , 5,6 , 6,1 , , 6,6

(b)

(c)

x

2

3

4

5

6

7

8

9

10

11

12

P(x)

1 36

1 18

1 12

1 9

5 36

1 6

5 36

1 9

1 12

1 18

1 36

There are 11 pairs with at least one dice showing 6, so P C

(ii)

For a sum of at most 10, we will use a complementary event that is 11 or 12. 1 1 1 11 1 Looking at the table, we obtain: P D 1 18 36 12 12

(iii)

Using the addition formula: P X Y P X P Y P X P X

(iv)

11 36

Y

3 36

Y 2 36

12 36

1 3

Using the conditional probability formula: 2 P X Y 36 2 P Y X P Y X 3 P X 3 36 The reason is that X

10.

11 36

(i)

Y

6,4 , 4,6

and X

6,4 , 4,6 , 5,5 700 1500

7 15

(a)

There are 1500 students altogether, 700 of which are female, so: P A

(b)

P B

(c)

Now the sample space contains only female students, so: P C

(d)

In this problem, we need to use the addition formula. Let T be the event of selecting a

220 1500

11 75 180 700

9 35

student from grade 12, while F is an event of selecting a female student. P T

F

P T

P F

P T

F

P T

F

400 700 180 1500

46 75

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(e)

There are 400 grade 12 students altogether, 220 of which are male, so: P E

(f)

220 400

11 20

In order for these two events to be independent, we need to see if the probability of the intersection of the two events is equal to the product of the probabilities of G) and female (F). P G

400 1500

4 ,P F 15

7 15

P G

P F

By looking at the table we notice that P G

4 7 28 15 15 225 180 9 27 F 1500 75 225

28 225

Therefore, the events are not independent. 11.

(a)

Using the table, we can read off the results: P Need 0.41 0.15 0.56 (i) (ii)

(b)

(c)

P Need

No Use

0.15

In this case, we use the conditional probability formula: 0.15 15 P No Use Need 0.56 56 A way of showing independence is by considering whether P Use Need

P Use .

If we use the probabilities from the table, we obtain: P Use

0.41 0.04 .045, P Need

P Use Need

P Use

Need

P Need

0.41 0.56

0.56 41 56

45 100

P Use

Hence, the events are not independent.

12.

If the events are mutually exclusive, their intersection is an empty set and its probability is zero: P A B P 0 If they are independent, then the probability of their intersection is the product of their probabilities: P A B P A P B , and P A B P A To find the probability of the union, we need to use the addition formula: P A B P A P B P A B The conditional probability uses the formula P A B

P A

B

P B

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P(A) P(B) Conditions for events A and B P(A B) P(A B) P(A B) 0.4 Mutually exclusive 0 0.7 0 0.3 0.4 Independent 0.12 0.58 0.3 0.3 0.5 Mutually exclusive 0 0.6 0 0.1 0.5 Independent 0.1 0.6 0.2 0.2 Note: Numbers in bold were given.

13.

(a)

Since the condition is that the chosen student is doing Economics SL, we have to use the conditional probability formula:

n Physics

P Physics Economics (b)

14.

3 10

We use the addition formula: P M

(b)

We use the complementary event: P M

(c)

Exactly one acceptable card means that the person cannot have both cards.

V \ M

V

V

Physics

(a)

0.21 0.57 0.13 0.65 65%

V '

1 0.65 0.35 35%

0.65 0.13 0.52 52%

Let S be the set of patients taking a swimming class and A be the set of patients taking an aerobics class. The following probabilities are given: P S

16.

n Economics

12 40

If doing Physics and Economics are independent, then by definition 3 30 P Physics Economics P Physics 10 100 Alternatively, we can use the multiplication rule: 40 30 3 12 P Economics P Physics P Economics 100 100 25 100 So the events are independent.

P M

15.

Economics

A

132 300

11 ,P S 25

78 300

13 ,P A 50 1

11 25

P S

A

84 300

7 25

14 25

(a)

P S

(b)

P S

(a)

In each attempt, the probability of rolling a two is 1 out of 6.

A ' A

1 P S P S

P A

A

P S

A

13 50

7 25

11 25

1 10

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P A

1 6

1 6

1 6

1 216

(b) P B

17.

s are 5 5 5 125 91 1 1 6 6 6 216 216

1 P B'

(c)

Exactly one two can be rolled in three different ways. A two can be rolled at the first, 1 5 5 25 second or third rolling; therefore, the probability is P C 3 6 6 6 72

(a)

She needs to miss the centre with her first shot and then hit the centre with her second shot; therefore, the probability is P A 0.7 0.3 0.21 21%

(b)

She can hit the centre with her first, second or third shot; therefore, the probability is: number of ways

P B (c)

3

twice no hit

hit

0.3 0.7 2

0.441 44.1%

We will calculate th no hit in 3 attempts

P C 18.

1 P C'

1

0.73

1 0.343 0.657 65.7%

Since each dice has 12 different outcomes, rolling two dice has 12 12 144 outcomes. (a) the P A

1 P A'

12 12 shows on either dic 11 11 121 23 1 1 12 12 144 144

(b)

A sum of 12 can be achieved as follows: (1, 11), (2, 10), (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), 11 (8, 4), (9, 3), (10, 2), (11, 1). Therefore, the probability is P B 144

(c)

A total score of at least 20 can be achieved as follows: (8, 12), (9, 11), (9, 12), (10, 10), (10, 11), (10, 12), (11, 9), (11, 10), (11, 11), (11, 12), (12, 8), (12, 9), (12, 10), (12, 11), 15 5 (12, 12). Therefore, the probability is P C 144 48

(d)

If 12 shows on a dice, there are 23 different outcomes. From these 23 outcomes, we look

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for those that have a sum of at least 20. Looking at the previous part, we notice nine such 9 pairs, so P D 23

19.

(a) follows: (10, 1), (10, 2), (10, 3), (10, 4), (10, 5), together with these pairs in reverse order. 10 5 P A B 144 72 (b) obtained in 23 different ways, while the sum of at most 15 can be obtained in 99 ways (144 45, where the 45 outcomes represent a sum of greater than 15). By using the addition rule, and the result from the previous part, we get: P A

(c)

B

P B

P A

23 99 10 144

B

112 144

7 9

This is the complementary event of the event in part (a). We need to list all the remaining pairs that are not listed in part (a). The probability is: P A

(d)

P A

B '

1 P A

B

1

5 72

67 72

This is the complementary event of the event in part (b). We need to list all the remaining pairs that are not listed in part (b). The probability is: P

A

B '

1 P A

B

1

7 9

2 9

(e)

By using D

(c).

(f)

By using D

(d).

(g)

By using the symmetrical difference property, we obtain A' B

A

B'

A

B

A

B , and, since A

B

A

B,

we calculate the probability in the following way. P A' B

A

B'

P A

B

P A

B

7 9

5 72

51 72

17 24

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20.

(a)

We can use the addition rule: P A

B

P A

P B

P A

Since we know that P A (b)

B

B

P A

B

P A

P B

1 , we can conclude that P A

P A

B

B

P A

P B

1

We can use the addition rule twice: P A

B

C

P A

B

P C

P A

P B

P A

P A

B

B

C

P C

P A

B

C

The last probability can be handled separately by using the distribution properties of set operations and the addition rule once again.

P A

B

C

P A C P A C

B

C

P B

C

P A C

B

C

Now we can finish the proof. P A

B

Note: A 21.

C

C

P A

B

P A

P B

P A

B

P C

P A

B

C

P A

P B

P A

B

P C

P A

C

P B

P A

P B

P C

P A

B

C

P C

A

B

P A

B

C

B

P A

C

P B

C C

P A P A

B B

C

When we roll three fair 6-sided dice, there are 216 possible outcomes. 6 216

1 36

(a)

Triples can be rolled in 6 different ways; therefore, P A

(b)

A sum of 8 or less can be rolled in 56 different ways and only 2 are triples. The possible combinations containing a triple are those with 1 or 2 since a triple with 3 adds to more than 8. Then there are 10 possible combinations with two equal numbers and each combination will appear 3 times, for example, (1, 1, 2), (1, 1, 3)...(2, 2, 3)... (2, 2, 4), (2, 3, 3). At the end there are 4 combinations, each with different numbers, for example, (1, 2, 3), (1, 2, 4), (1, 2, 5) and (1, 3, 4); and each of these combinations appear 6 times. So, we have a total of 56 different combinations. P B

2 56

1 28

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C C

(c) P C

1 P C'

1

5 6

5 6

5 125 1 6 216

91 216

Note: The result is the same as that in question 16 (b) since there is no difference in the appearance of a 2 or 6 on a dice.

(d)

If all three dice have different numbers, there are 6 C3 20 different combinations and each will appear 3! 6 times; therefore, we have 120 different ways. Again we will use

P D

22.

1 P D'

1

5 C3

120

3!

1

60 120

1 2

Let the coins be denoted as A with two heads, B with two tails, and C and D are normal. Let H be the event of getting heads, and T be the event of getting tails. A tree diagram can help. 0.5

A 0.25 0.25 0.25

B

0.5 0.5 0.5 0.5

C

0.25

D

0.5 0.5 0.5

H H T T H T H T

To choose a coin at random means that the probability of choosing any coin is 0.25 The event T of the experiment where the outcome is tails contains the following mutually exclusive events: B T , C T , and D T . Thus, P T

P B

T

P C

T

P D

T

0.25 (0.5 0.5) 0.25 0.5 0.25 0.5 0.5

The subset H with P H

T

T containing heads in the opposite face contains the events C

T and D

0.25 0.5 0.25 0.5 0.25

Now, the question can be stated as P H T

P H P T

T

0.25 0.5

0.5

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T

23.

There are five different ways of rolling a sum of 6: (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1). 5 So, the probability that either of the players will roll that sum is 36 (a)

K : no sum 6

G: no sum 6

K : sum 6

31 36

31 36

5 36

K : no sum 6

G: no sum 6

K : no sum 6

G: sum 6

31 36

31 36

31 36

5 36

P A

(b)

P B

(c)

Kassanthra wins if: she wins on her first roll:

4805 46656

0.103

148955 1679616

0.0887

5 36 K: no sum 6

G: no sum 6

K: sum 6

31 36

31 36

5 36

or, she wins on her second roll:

5 36

31 36

2

or, she wins on her third roll: K: no sum 6

G: no sum 6

K: no sum 6

G: no sum 6

K: sum 6

31 36

31 36

31 36

31 36

5 36

5 36

31 36

4

and so on. Thus, to calculate the probability that Kassanthra wins, we need to find the sum of an 5 and a common ratio r 36

infinite geometric sequence with a first term of a

Therefore, P(Kassanthra wins) =

24.

(a)

1 r

5 36 31 1 36

2

36 67

0.537

The day has no effect on the observation, so we just need to find the probability that more than two requests will be made using complements: P A

(b)

a

1

0.1 0.3 0.5

0.1 10%

Since the days are independent, we need to multiply the probabilities: P B

0.15

0.00001

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31 36

2

25.

The class has 11 students altogether and we need to select 4 students at random. (a) P A

11 C4 6 C3

1

15 330

5 C1

21 22

6 C4

100 15 330

23 66

(b)

There must be either 3 or 4 girls: P B

(c)

Let C be the event that the boys are in the majority. The conditional probability formula

P C

gives us: P C A

P C

26.

6 C4

1

A

5 C3

A

P A 6 C1 11 C4

5 C4

11 C4

. First, we will find the probability of the numerator.

13 66

P C A

13 66 3 21 22 1

13 63

A Venn diagram may help identify the calculations involved.

B2 (0.25) B1(0.22)

0.11 0.07

0.05 B3 (0.28)

0.01

(a)

P B1

(b)

Use De

B2

P B1

P B2

P B1

P B1

B2

B2

P B1

0.22 0.25 0.11 0.36

B2

1 P B1

B2

1 0.36 0.64

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(c)

Notice that the part of B3 that does not intersect B1 and B2 is a subset of B1 and therefore we only need to add B1 count the event B1 P( B1

B2 )

B3

B3 and B2

B2

B3 , making sure not to double

B3 shown in colour. Thus:

B2

P B1

B2

P B1

B3

P B2

B3

P B1

B2

B3

0.64 0.05 .07 0.01 0.75

There are also alternative approaches to this. (d)

Notice that this event consists of B3 which does not contain B1 and B2 . Thus: P B1

B2

B3

P B3

P B1

B3

P B2

B3

P B1

B2

B3

0.28 0.05 .07 0.01 0.17

(e)

P B2

B3 B1

(f)

P B2

B3 B1

P B1

B2

B3

0.01 0.22

P B1 P B2

B3

1 22

0.0455

B1

P B1

First, we need to find the probability of the numerator: P B2

B3

B1

P B1

B2

P B1

B3

P B1

B2

B3

0.11 0.05 0.01 0.15 Therefore, P B2 27.

B3 B1

0.15 0.22

15 22

Let N and D be the sets of all the joints found to be faulty by Nick and David respectively. Given the information in the question, we can find the probabilities: 1448 1502 2390 P N ,P D and P N D 20000 20000 20000 (a)

P

N

D

1 P N

D

1

(b)

P N

D

P N

P D

P N

P D

N

P D

P D

N

239 2000

D

1761 2000

0.8805

1448 1502 2390 20000

1502 560 20000

942 20000

471 10000

560 20000 0.0471

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Exercise 11.4 1.

2.

(a)

(b)

P A

(c)

P EA

2 3 1 1 3 50 3 25 P E P A

1 25 4 75

3 4

Let C be the event that a person is diagnosed with cancer, and E1 and E2 the events that the person (A tree diagram can help!) (a)

This is what we call total probability a person is diagnosed with cancer if he/she has it or not. P C

(b)

26 99974 0.78 0.06 0.0601872 0.0602 100000 100000

This is a case of Bayes theorem.

P E1 C

3.

A

4 75

0.00026 0.78 0.0601872

0.0033694872 0.00337

Let S be the event that a driver is spotted speeding, and E and W the events that the driver used the east or west entrance. (a)

P S

(b)

P W S

0.4 0.4 0.6 0.6 0.52

0.6 0.6 0.52

0.692 (correct to 3 s.f.)

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4.

Let G be the event that a drawn ball is green, and E1, E2 and E3 the events that the ball is drawn from box 1, 2 and 3 respectively. (a)

(b)

5.

6.

1 4 3 20

P G

1 1 3 2 1 3

P E2 G

(a)

P H

1 2 5 3 3 10

1 3

1 2

0.5 0.6 0.5 0.5 0.55

(b)

P E2 H

0.5 0.5 0.45

5 9

0.556

Let C be the event that the question was correctly answered, and E1 and E2 the events that the student was well-prepared and unprepared respectively. 0.7 0.6 0.6 0.3

0.7 well prepared

8.

1 6 3 20

Let H be the event that the selected coin lands on heads, and E1 and E2 the events that the selected coin is biased and unbiased respectively.

P E1 C

7.

1 8 3 16

correct answer

not prepared

0.2

0.42 0.48

7 8

0.875

guessing answer

Let T be the event that Nigel gets to his morning class on time, and E1 and E2 the events that the alarm clock was set and was not set respectively. (a)

P T

0.85 0.9 0.15 0.6 0.855 85.5%

(b)

P E2 T

0.15 0.6 0.855

2 19

0.105 10.5%

We are going to use a probability tree. Let F represent a successful first serve, S a successful second serve, and W and L winning and losing the point respectively. A tree diagram will help visualise the situation:

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Second serve

0.75 0.6

0.25 0.95

0.4

L S

0.05

9.

10.

0.5

W

0.5

L

F

(a)

Cheung wins either from the first serve or from the second as shown. 16 P W 0.6 0.75 0.4 0.95 0.5 0.64 25

(b)

P First W

P First

W

P W

0.6 0.75 0.64

45 0.703125 0.703 64

Let F be the event that the first of February is a fine day, and S the event that the second of February is a fine day. (a)

P S

(b)

P F S

0.75 0.8 0.25 0.4 0.7

P F P S

S

0.75 0.2 0.3

1 2

0.5

Let A be the event that the person has arthritis, and P that the test is positive.

P AP

11.

W

P A

P

P P

0.33 0.87 0.33 0.87 0.67 0.04

0.915

Let H represent students in the HL class, and L and A students studying locally and abroad respectively. (a)

P H

L

0.05 0.72 0.036

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(b)

From the table, we know that: P L

0.67

Therefore, P H ' L

L

0.67 P H

On the other hand: P H ' L

12.

(c)

P H L

(d)

P H A

L

P L P H

0.036 0.67

P A H P A

L

P H' L

0.67

0.67 0.036 0.634

P L H'

P L H'

0.634 0.95

0.667

0.0537 0.05 0.28 0.33

0.014 0.33

0.0424

Let U represent athletes who are users, and P and N a positive and negative test respectively.

PU

PU P

13.

P H

P H'

P H

P

0.1 0.5 0.1 0.5 0.9 0.09

P P

0.382

Let A, R and S represent the estimates made by Antonio, Richard and Sarah respectively, and E an error in the estimation. We would like to find the largest from P A | E , P R | E and P S | E . We notice that all three expressions have the same denominator, and therefore the largest one will be the one with the largest numerator. So, we calculate the following: P A E 0.3 0.03 0.009, P R E 0.2 0.02 0.004 and P S

E

0.5 0.01 0.005

Thus, Antonio is probably responsible for most of the serious errors.

14.

15.

Let A represent the event that an aircraft is present, and S that there is a signal. (a)

P S

(b)

P A S

P A

P S|A

P A

P S P S

P A A

P S|A

0.05 0.99 0.95 0.1 0.1445

0.05 0.01 0.000584 1 0.1445

Let N, E and M represent a game against a novice, an experienced player and a master player respectively, and W the probability of winning the game.

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(a)

PW

P N

PWN

P E

PWE

P M

PWM

0.5 0.5 0.25 0.4 0.25 0.3 0.425

16.

P W M

0.25 0.3 0.425

P M W

(a)

Let A be the event of passing the test:

P A

17.

P M

(b)

P W

pass 1st

fail 1st

pass 2nd

fail 1st

fail 2nd

pass 3rd

0.8

0.2

0.5

0.2

0.5

0.3

P 2nd A

(a)

P F

56 26 18 250

P F

A'

(b)

0.93

0.2 0.1 0.108 0.93

(b)

P F' A

0.176

100 250

2 5

84 52 56 18 26 250 P F'

14 250 14 26 250

A

P A

T

56 250

118 125

0.944

0.4, P F 236 250

14 40

T is independent of F since P F

7 20

28 125

0.224

0.35

56 140

0.4 P F T

The event M is mutually exclusive to F since the probability of their intersection is zero.

(c)

Let C be the event of owning a car. (i)

P C P C

(ii)

P TC

PT

P C T

P A

14 0 25 0

9 10

P T

P CT P C

40 25 0

8 10 126 250 179 250

P C A 70 25 0 126 179

P S 3 10

P C S

126 32 21 250

179 250

0.716

0.704

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18.

Let H, M and L represent high risk, medium risk and low risk drivers respectively, and A those drivers who will have an accident. (a)

P H

A

P H

(b)

P A

P H

P AH

P A| H

0.2 0.06 0.012

P M

P A| M

P L

P A| L

0.2 0.06 0.5 0.03 0.3 0.01 0.012 0.015 0.003 0.03

(c)

P H

P H A

P AH

0.012 0.03

P A

0.4

Chapter 11 practice questions 1.

(a)

Since the events are independent: P A k

B

P A

0.6 k

0.3

P B 0

0.18 k k

k 0.3

0.6 or k

k 2 0.3k 0.18 0

0.3

Algebraically we get two solutions, but only 0.3 can be a probability value, since probability cannot be negative. (b)

Using the addition formula: P A

(c)

B

(a)

P B

P A

B

0.3 0.6 0.18 0.72

Since the events are independent, the complementary events are independent too.

P A' B '

2.

P A

P A'

P A' B '

1 0.3 0.7

Since the tests are taken independently, we multiply the probabilities: P A

0.02 0.02 0.0004

(b)

Using the complementary event: P B

(c)

P C

1 P A

1 0.0004 0.9996

0.02 0.02 0.0004

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3.

4.

Since they work independently of each other, we need to multiply the probabilities and then use the complementary event. P A

1 P A

(a)

(i)

1 0.002 0.01 0.99998

Using the addition formula:

P S (ii)

5.

P F

P S

F

120 200

60 200

10 200

17 20

F

P S

F

170 200

10 200

160 200

4 5

Does not take French or Spanish is the complementary event of the union, so: P S

(b)

P S

Either but not both means that we exclude the intersection from the union, so:

P S (iii)

F

F

1 P S

F

1

17 20

3 20

Using the conditional probability formula: P F S

P F P S

S

10 200 120 200

1 12

It would be a good idea to find the total sums first. There are 126, 84 and 160 disks produced after one run on machines I, II and III respectively. There are a total of 20 defective and 350 non-defective disks. That means there are 370 disks produced in total. Let D denote defective and ND non-defective. (a)

(i)

P I

126 370

(ii)

P D

II

(iii)

We need to use the addition formula:

P ND (iv)

63 185 4 370

I

2 185

126 350 120 370

356 370

178 185

Since this is a conditional probability, our sample space is defective disks and the favourable outcomes are the defective items produced by machine I. 6 3 P ID 20 10

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(b)

If the quality is independent of the machine, then, for example P I D =P I . However, by comparing our answers to (i) and (iv), this is not true. So, the quality and machine used are not independent. Alternatively, you can use the multiplication rule where you 2 20 84 84 P D ×P II notice for example that P D II 185 370 370 6845

6.

7.

8.

126 200

63 100

(a)

There are 126 envelopes which satisfy your wish, so: P A

(b)

There are 68 red envelopes without a prize, so: P B

(a)

P AB

(b)

The events are independent since P B

(c)

Given that A and B are independent, then

and B are independent too, so:

P B

0.42

(a)

P A

B

0.3

P B

A

P B

P A

0.18 P B

0.6

P B

0.18 0.3

68 70

34 35

0.6

0.6 P B A

1 0.3

Since we know that there are 74 students who took the test, the number of boys who failed is 74 32 16 12 14. There are 6 girls who are too young to take the test and, since there are 10 students altogether that are too young to take the test, the number of boys who are too young is 10 6 = 4. Since the total numbers of boys and girls are 70 and 50 respectively, we calculate all of those who were training but did not take the test as: 70 32 14 4 20, and 50 16 12 6 16. Boys Girls 32 16 Passed the ski test 14 12 Failed the ski test 20 16 Training, but did not take the test yet 4 6 Too young to take the test

(b)

74 120

37 60

(i)

P Test

(ii)

P Test Girl

(iii)

These two events are independent, so we multiply the probabilities:

16 12 50

P passed Boy

28 50

14 25

passed Girl

32 16 70 50

128 875

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9.

10.

11.

12.

1 3 4 8

3 32

(a)

P A

B

P AB

(b)

P A

B

P A

(c)

P

(a)

Probability that she will miss both serves is P A

(b)

To win a point, she will make the first serve and win the point, or she will miss the first serve, make the second serve and win the point, so: P B 0.6 0.75 0.4 0.95 0.5 0.64 64%

(a)

P X

Y

P X

(b)

P X

Y

P

A

B

P B B

P A

1 P A

B

P Y

X

Y

P B 1

P X

27 32

Y

1 P X

2 P A

9 16

B

3 3 2 8 32

3 4

5 32

0.4 0.05 0.02 2%

0.6 0.8 1 0.4

Y

1 0.4 0.6

Using the given information, here is a complete table.

News Sport Total

38 100

(a)

P news

(b)

P news man

Men Women Total 13 25 38 33 29 62 46 54 100

19 50

13 46

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13.

(a)

(b)

(i)

(ii) (c)

Since n X

Y

n X

Y

n X

P X

Y

2 36

nU n Y

n

X

Y

n X

Y

(a)

11 6 15 2.

1 18

Events X and Y are not mutually exclusive since there are two elements in the intersection of the two sets: X

14.

36 21 15, then

Y

.

There are 90 females, so there are 110 males. If 60 were unemployed, then 140 were employed. If 20 males were unemployed, then 40 females were unemployed, and so on.

Males Females Totals 20 40 60 Unemployed 90 50 140 Employed 110 90 200 Totals

(b)

15.

(i)

PU

F

(ii)

P M E

40 200 90 140

1 5 9 14

There are three possible combinations of different colours: WR, WG, or RG. In total, there are 26 chips in the bag. The WR chips can be chosen in 10 10 100 ways, the WG in 60 ways and the RG in 60 ways. In total there 220 ways of choosing 2 chips with different colours. There are 220 44 325 ways of choosing any 2 chips, and thus, P A 26 C2 325 65

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16.

(a)

The student can take Chemistry and Biology or not take Chemistry but take Biology. So, P B

17.

0.4 0.6 0.6 0.5 0.54

P C

B

0.4 0.6 0.54

4 9

(b)

This is a case of Bay

(a)

The probability distribution table for the sum of the 2 numbers on a pair of dice is:

P B

s

2

3

4

5

6

7

8

9

10

11

12

P(S)

1 36

2 36

3 36

4 36

5 36

6 36

5 36

4 36

3 36

2 36

1 36

1 2

P S 7 (b)

P CB

6 36

21 7 36 12

There are 11 possible pairs when at least one dice shows a 3: six pairs with 3 showing on the first dice and six pairs with 3 showing on the second dice, but the pair (3, 3) should 11 only be counted once. So, P B 36

(c)

Now the event from (a) becomes a sample space and we need to find the favourable pairs (out of 21 pairs) found in (b). The pairs that satisfy the condition are (1, 3), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4) and (4, 3).

P B S

7

7 21

1 3

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18.

19.

3 4 6 1 11 11 11 11

(a)

P A

B

P A

P B

P A

(b)

P A

B

P A

P B

3 4 12 11 11 121

(a)

Since the probability of A

B

B occurred, P A B

P A.

Therefore, the events are independent.

20.

(b)

If P A

(c)

Given that P A

B

0

A B

B

; therefore, the events are mutually exclusive.

P A

A

B

A

A

B , neither.

(a)

(b)

P L

(c)

P W L

7 1 8 4

1 3 8 5

P W P L

L

7 32 7 32 47 160

3 40

35 12 160

47 160

35 47

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21.

(a)

(b)

22.

(a)

(i)

P R

G

0.4 0.9 0.36

radish

germinates

beans

germinates

0.4

0.9

0.6

0.8

0.36 0.84

3 7

(ii)

P G

(iii)

P RG

(i)

P A

(ii)

P A

(iii)

To confirm whether the events are independent we first find P B

P R

G

P G

80 210 B

0.36 0.48 0.84 0.429

8 21 35 210

1 6 100 210

10 . 21

Now we look at the product of the probabilities: 8 10 80 1 P A P B P A B . Hence, they are not independent. 21 21 441 6 Alternatively, we can look at the conditional probability.

50 85

10 17

(b)

P Y1 H

(c)

We can select a student from Year 1 in 110 ways and a student from Year 2 in 100 ways. We can select 2 students from the college in Therefore:

P C

100 110 210 C2

11000 21945

210

C2 ways.

200 399

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23.

Let O be the event that an odd number is chosen and E an even number is chosen. If an odd and an even number have to be selected in any order, we can say that the chosen slips can be odd and even or even and odd. 3 4 4 3 1 P OE P EO P O P E P E P O 9 8 9 8 3

24.

P X

Y

P X

Y

P X

P X

can use the multiplication law. P X P X

25.

Y

P X

Y

P X

P Y

P Y

P X P X

Y

0.3 0.5 0.6

0.6 0.5 0.3 0.8

P R

1 4 2 3

P LR 1 4

P L

2 3 3 1 4 5

10 19

Let F be the event that the umbrella is left at the first at the second . P SF

27.

P X

Y

Let R be the event that it is raining, and L that Sophia is late.

P RL

26.

P Y

0.3 0.3 0.6 . Since the events are independent, we

(a)

S

(i)

2 3 1 3

1 3 2 3

P L1

1 3

2 9 5 9

, and S that the umbrella is left

2 5

Cath no win

Lucy win

5 6

1 6

5 36

Cath no win

Lucy no win

Cath win

1 6

(ii)

P C2

5 6

5 6

P Cn

5 5 6 6

n 1

(iii)

1 6

25 216

1 25 6 36

n 1

1 5 6 6

2n 2

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(b)

We have an infinite geometric series: p

1 6

25 1 36 6

25 36

2

1 6

25 36

3

1 ... 6

1 6

25 36

p

1 6

(c)

25 1 36 6

25 1 36 6

2

25 36

3

1 ... 6

1 6

25 36

p

First, we will calculate the probability that Catherine wins the game:

p

1 6

25 p 36

25 p 36

p

1 6

11 p 36

1 6

p

1 36 6 11

p

6 11

The probability that Lucy wins the game is complementary to the above, so: 6 5 . Alternatively, we can look at the process of Lucy winning: P Lucy wins 1 11 11 Cath no win

Lucy win

Cath no win

Lucy no win

Cath no win

Lucy win

5 6

1 6

5 6

5 6

5 6

1 6

Which is an infinite geometric series with first term

(d)

5 1 5 and common ratio 6 6 6

2

If Catherine wins more games than Lucy, that means that she has to win 4, 5 or 6 times. We also need to find the number of sequences with that number of wins. For example, if Catherine wins four out of six games played, she can do that in 6 C 4 different ways (we are using the binomial coefficients). Cath wins Lucy wins 4 times twice

P D

28.

(a)

6 C4

6 11

4

5 11

Cath wins Lucy wins 5 times once

2 6 C5

6 11

5

5 11

Cath wins 6 times

6 11

6

0.432

If the first two selected balls are white, then 1 white and 22 red balls remain in the box. Therefore, the probability that the next ball will be red is

22 23

0.957

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(b)

There are three different selections that will give the result of exactly two red balls: RRW, WGR or WRR. So, the probability is calculated as follows:

P B

22 21 3 25 24 23

3

693 2300

0.3013

Alternatively P B

22

C2 25

29.

30.

3

C1

231 3 2300

C3

0.3013

Let F be the event that the chosen day is Monday, and B that Roberto catches the 08:00 bus. (a)

P B

(b)

P FB

(a)

P E1

P F

P BF

P F

P F

P BF

0.132 0.732

P B

4 6

P BF

0.2 0.66 0.8 0.75 0.732

11 0.180 61

2 3

Eric no win

2 6

4 6

P H1

(c)

We will have an infinite geometric series: Eric win

Eric no win

Harriet no win

Eric win

2 3

1 3

1 3

2 3

2

1 9

P E

2 1 1 3 9

31.

2 9

(b)

(a)

P RR A

2 1 5 4

1 9

1 1 3 3 3

...

2

2 3

2 1 3 1 1 9

1 1 3 3

3

2 9 3 8

2 ... 3 3 4

1 10

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(b)

P B

4

3

n 4

n 3

12 n 4 n 3

n 4 n 3

2 15

90

This leads to a quadratic equation n2 13 or n 6 . Discard n

n

Two red

Bag B

Two red

1 3

1 10

2 3

2 15

3 8 90

11 90

(c)

P RR

(d)

Recall that 1 or 6 means that we draw from bag A. Let RR be the event that two red balls are drawn, and A that the balls are drawn from bag A. P A

(a)

E X

(b)

(i)

1 1 3 10 11 90

P RR A P RR

xP x

1

3 11

1 2 3 2 3 6 6 6

7 3

To get a sum of 5, the numbers would be either A = {1, 2, 2} which can be arranged in 3 different ways, or B = {1, 1, 3} which can also be arranged in 3 different ways. P A

(ii)

13 because the number of balls cannot be negative.

Bag A

P A RR

32.

7n 78 0 .

B

3

1 2 2 6 6 6

3

1 1 3 6 6 6

7 72

To get a median of 1, the numbers would be either C = {1, 1, 2} that can be arranged in 3 different ways, D = {1, 1, 3} which could also be arranged in 3 different ways, or E = {1, 1, 1}. P C

(c)

D

E

3

1 1 2 6 6 6

The probability of selecting a 2 is

1 3

3

1 1 3 6 6 6

1 1 1 6 6 6

2 27

and the probability of not selecting a 2 is

each number of 2s selected, say 3, there will be probability in question is

10

2 3

. Also for

C3 ways of doing that. Thus, the

# of 2 s # of no 2 s choices 10

(d)

C0

1 3

0

2 3

10

1 10 C1 3

1

2 3

9 10

C0

1 3

2

2 3

8 10

C0

1 3

3

2 3

7

0.559

Let the number of balls be n. The complement of having at least one 2 is having no 2.

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Thus, if the probability of taking at least one 2 is greater than 0.95, then the probability of having no 2 must be smaller than 0.05. 1 3

n C0

0

2 3

n

Now, since ln n

(e)

ln 0.05 2 ln 3

2 3

0.05 2 3

n

0.05

n ln

2 3

ln 0.05

0, dividing by it will change the sense of the inequality, i.e.,

7.39 . Since n is a whole number, the least possible number will be 8.

Again, to get 1 is a binomial. If the expected value is 4.8, then np1

4.8

p1

0.6.

If the variance of the number of 2s is 1.5, then Variance = np2 1 p2

1.5

8 p2 1 p2

1.5

This is a quadratic equation with roots 0.75 or 0.25. We cannot use 0.75 since the probabilities for 1, 2, and 3 must not add up to more than 1. Thus, 2s constitute 25% of the balls in the bag. This leaves 15% for the 3s. We need the number of each kind to be a whole number. Putting the probabilities in fraction form and using the same denominator gives us an idea:

0.15

3 ;0.25 20

5 ;0.60 20

12 20

The smallest number of balls in the bag that will work (trial and error) is 20. This means the smallest possible number of balls with 3 is 3.

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Exercise 12.1 1.

(a)

lim n

1 4n n

lim( n

1 n

4n ) n

lim n

1 lim 4 n n

0 4

4

1 4n n 4.1 4.02 4.01 4.001 4.0001 4.000001

n 10 50 100 1000 10000 1000000

1 4x is the line y 4 , the behavior x of the graph towards the positive end confirms that the limit value is 4.

The dotted line included in the graph of y

(b)

The variable in the limit is h, so x is treated as a constant:

lim(3x2 2hx h2 ) lim3x 2 lim 2hx lim h2 h

(c)

0

h

h

0

h

0

The variable in the limit is d, so x is treated as a constant: ( x d )2 0 d

x2

lim d

d

lim x

3

0

x2 9 x 3

d

lim d

lim 2x lim d

(d)

0

3x 2 0 0 3 x 2

0

2 xd d 2 d

x2

2 xd d 2 0 d

lim d

2 xd 0 d

lim d

d2 0 d

lim d

2x 0 2 x

( x 3)( x 3) 3 x 3

lim x

0

x2

lim( x 3) 3 3 6 x

3

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The graph has a hole at (3, 6) , as seen in the graph. 2.

3x 2 . The required values are presented in the table. x2 3 The values of f decrease towards 0 as x x f(x) 10 0.329897 becomes infinitely large, it follows that: 50 0.060873 3x 2 lim f ( x) lim 2 0 100 0.030209 x x x 3 1000 0.003002 10000 0.000300 1000000 0.000003

(a)

Let f ( x )

(b)

Let f ( x )

(c)

Let f ( x )

5x 6 . The required values are presented in the table. 2x 5 The values of f increase towards 2.5 as x x f(x) 10 1.760000 becomes infinitely large, it follows that: 50 2.323810 5x 6 5 lim f ( x ) lim x x 100 2.409756 2x 5 2 1000 2.490773 10000 2.499075 1000000 2.499991

3x 2 2 . The required values are x 3 presented in the table. The values of f increase without bound as x

becomes infinitely large, it follows that: lim f ( x) x

3x 2 2 lim x x 3

x

f(x)

10 43 50 160 100 309 1000 3009 10000 30009 1000000 3000009

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3.

In many cases where substituting x a into the function results in the indeterminate case 0 of , this means both numerator and denominator have a common factor x a. The task 0 then is to extract this common factor and cancel it for values of x a, and then reevaluate the remaining expression for x a. x 4 0 (a) lim 2 , which means that the common factor is x 4, which has to be x 4 x 16 0 cancelled before we can evaluate the function. lim x

4

x 4 x 2 16

x2

1x 4 4 (x 4 )( x 4)

lim x

x 2 x 1

lim

lim x

( x 1)( x 2) ( x 1)( x 1)

4

1

1

x 4

4 4

x 2 1 2 x 1 1 1

3 2

(b)

lim

(c)

When substituting x 0 into the expression of the limit, the undefined case

2

x 1

x 1

lim

1 8

x 1

0 is 0 obtained. This means that the expression of the limit needs to be manipulated further in order for the limit to be evaluated. The solution is to multiply both the numerator and the denominator by the conjugate of the numerator:

2 x x

lim x

0

lim x

0

2

( 2 x 2)( 2 x 0 x( 2 x 2)

2)

lim x

2 x 2 x( 2 x 2)

lim x

0

1 2 x

x

1 2 0

2

( 2 x ) 2 ( 2)2 0 x( 2 x 2)

lim

1 2

(d)

(e)

To find this limit, another limit has to be used, namely lim

x3 (1

lim x

3 4 x2

1 0 4 0 0

1 x3 x

(You are allowed to use this result.) sin x tan x sin x 1 lim lim cos x lim x 0 x 0 x 0 x x x cos x (f)

1

1 x3

2 2

1 ) 3 x 1 x lim 3 lim x 4 x 3x 1 x x3 (4 3 1 ) x 2 x3 3

x

0

sin x x

sin x 1 lim 0 x 0 cos x x

lim

2 4

1 4

1

1

1 1 cos 0

For solving this limit, the same limit as in part (e) has to be used: sin 3 x sin 3 x 3 x sin 3 x lim lim lim lim 3 1 3 3 x 0 x 0 x 3x x x 0 3x x 0

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x

4.

1 Let f ( x) 1 . The graph of y f ( x) and the corresponding table of values, x obtained from a GDC, are shown below:

By inspection, the graph has a horizontal asymptote, as confirmed by the table of values. The value entered in Y2 and displayed as 2.7183, is e 2.7182818... (The GDC can only display six characters in the table, so all values are rounded such that they fit.) It follows that the horizontal asymptote is y e . 1 x

This means that lim 1 x

value such that lim g ( x) x

5.

x

e , or, more general, lim 1 x

x0

1 g ( x)

g ( x)

e , where x0 is a

.

x0

If function f has a horizontal asymptote with equation y

3 , then the values of f

approach 3 when x becomes infinitely large either positively or negatively. It follows that: (a) (b) 6.

lim f ( x ) 3

x

lim f ( x) 3

x

If function g has a vertical asymptote with equation x a , then it has a denominator containing a factor of ( x a ) . When the values of x approach value a, the values of ( x a ) decrease towards 0 (as g ( x) 0) . This causes the denominator of the function to

have a factor which is almost 0, which in turn causes the values of g to increase without bound. It follows that lim g ( x) . x

a

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7.

(a)

x(3

3x 1 lim x 1 x

lim

1 ) x

3

1 x

lim 3 x 1 1 x( 1) 1 x x The value of the limit is finite, so the function has a horizontal asymptote, y x

3.

When investigating the existence of vertical asymptotes, the denominator of the expression must be considered:

x 1 0

x

1

The limit when x 1 has to be evaluated when the values of x increase and decrease towards 1: 3x 1 lim x 1 1 x x 1 (The numerator is negative, while the denominator takes negative values increasing toward 0). 3x 1 lim x 1 1 x x 1 (Because the numerator is negative, while the denominator takes positive values decreasing toward 0). Consequently, due to the fact that both limits are infinite, there is a vertical asymptote, x 1. A graph confirming these findings is shown below, where the asymptotes are drawn in red:

(b)

lim

x

1 ( x 2) 2

0

The function has a horizontal asymptote, y

0. x 2 0

x 2

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To investigate the existence of a vertical asymptote, the limit of the function when x 2 has to be evaluated: 1 2 (x 2)2 2

lim x x

Consequently, there is a vertical asymptote, x

2

The graph is shown below:

(c)

lim (

x

1 x a

b)

lim

x

1 x a

lim b

x

0 b

b

the function has a horizontal asymptote, y

x a 0

x a

The limit of the function when x increase and decrease towards a: lim x a x a

lim x a x a

1 x a 1 x a

b

b

lim

b

lim

x a x a

x a x a

1 x a 1 x a

a has to be evaluated when the values of x

lim b

b

x a x a

lim b x a x a

b

Consequently, due to the fact that both limits are infinite, there is a vertical asymptote, x a .

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(d)

3 ) x2 lim x 9 x 2 (1 2 ) x x 2 (2

2 x2 3 lim 2 x x 9

3 x2 lim x 9 1 2 x 2

the function has a horizontal asymptote, y x2 9

0

x

2

2

3

The limits when x 3 have to be evaluated when the values of x increase and decrease towards 3 and 3, respectively: 2x2 3 lim 2 x 3 x 9 x 3 2x2 3 3 x2 9 3

lim

2x2 3 3 x2 9 3

lim

lim x x

lim

x x

2x2 3 3 x2 9 3

lim

x x

2 x2 3 lim x 3 ( x 3)( x 3) x 3

x x

x x

2 x2 3 3 ( x 3)( x 3) 3 2x2 3 3 ( x 3)( x 3) 3 2x2 3 3 ( x 3)( x 3) 3

lim

x x

2 x2 3 1 lim lim x 3 x x 3 3 x3 x 3 x 3

15 6

2 x2 3 1 lim 3 x x 3 3 x3 x 3 3

15 6

lim x x

2x2 3 1 lim 3 x 3 x 3 x 3 3 x 3

15 6

2x2 3 1 lim 3 x 3 x 3 x 3 3 x 3

15 6

lim

x x

lim

x x

Consequently, there are two vertical asymptotes, x 3 and x

3

The graph is shown below:

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(e)

lim

x

5 3x x 2 5x

5 3 x lim x 5 x2 1 x

5 3 x lim x 5 x 1 x

x

the function has a horizontal asymptote, y x2 5x

0

x ( x 5)

0

x

0, x

5 3 1 x lim lim x 5 x x 1 x

0 ( 3) 0

0

5

The limits when x 0 and when x 5 have to be evaluated when the values of x increase and decrease towards 0 and 5, respectively: lim

5 3x x 2 5x

lim

5 3x x( x 5)

lim

lim

5 3x x 2 5x

lim

5 3x x( x 5)

lim

lim

5 3x x 2 5x

lim

5 3x x( x 5)

lim

lim

5 3x x 2 5x

lim

5 3x x( x 5)

lim

x 0 x 0

x 0 x 0

x 5 x 5

x 5 x 5

x 0 x 0

x 0 x 0

x 5 x 5

x 5 x 5

1 5 3x lim x 0 x x0 x 5

( 1)

1 5 3x lim 0 x x 0 x 5 0 x 0

( 1)

x 0 x 0

x x

x 5 x 5

x 5 x 5

1 x 5 1 x 5

5 3x 5 x 5

( 2)

5 3x 5 x 5

( 2)

lim x x

lim x x

Consequently, there are two vertical asymptotes, x 0, x 5 The graph is shown below:

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2

(f)

lim

x

x 4 x 4

4 x2 lim x 4 x 1 x x2 1

4 x2 4 1 x

x 1 lim

x

4 x2 lim x lim x x 4 1 x 1

1

the value of the limit is infinite, so the function does not have a horizontal asymptote. x 4 0 x 4 The limit when x 4 has to be evaluated when the values of x increase and decrease towards 4: lim

x2 4 x 4

lim( x 2 4) lim

lim

x2 4 x 4

lim( x 2 4) lim

x 4 x 4

x 4 x 4

x 4 x 4

x 4 x 4

x 4 x 4

x 4 x 4

1 x 4 1 x 4

12 (

)

12 (

)

Consequently, due to the fact that both limits are infinite, there is a vertical asymptote, x 4 . The graph is shown below:

8.

(a)

(i)

The values displayed by the GDC when given the function

f ( x)

x2 5 3 x2 5 3 lim suggest that x 2 x2 2 x x2 2 x

0.3333 , or

x2 5 3 1 , when taking into account the limitations x 2 x2 2 x 3 of the calculator regarding rounding. lim

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(ii)

x2 5 3 x2 2 x

lim x

2

lim x

2

lim x

(b)

(i)

2

lim x

2

x2 5 3 x2 2 x

x2 4 x( x 2) x 2 5 3 x 2 x( x

2

lim x

2

x2 5 3 x2 5 3

5 3)

2( 2

x

2

x2 5 9 x( x 2)( x 2 5 3)

( x 2)( x 2) x( x 2)( x 2 5 3)

2 2 2

lim

5 3)

4 2 6

1 3

Based on the values generated by a GDC for the function f ( x)

4x 1 x2 2

the value of the limit can be estimated to be 4.

(ii)

lim

x

4x 1 x2

2

1 x 2 x2 1 2 x x 4

lim

x

1 x lim x 2 x 1 2 x x 4

1 x lim x 2 1 2 x 4

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4

,

9.

The variable in the limit is h, so x is treated as a constant. Multiply both the numerator and the denominator of the expression by the conjugate of the numerator, to remove the 0 undefined case : 0 lim h

0

x h h

x

lim h

0

x h h

x

x h

x

x h

x

x h x 0 h( x h x)

lim h

10.

This is an indeterminate case of the form 1 lim x h h 0 h

1 x

h

h 0 h( x h

lim h

lim 0

( x h )2 ( x ) 2 h( x h x)

lim h

0

x) 1 x h

1 x

0 . 0

x ( x h) x x h h x( x h) x( x h) x ( x h) lim lim lim h 0 h 0 h 0 h h h h 1 1 1 lim lim h 0 hx ( x h) h 0 x ( x h) x( x 0) x2

Exercise 12.2 1.

(a)

From first principles: f ( x h) f ( x ) 1 ( x h)2 (1 x 2 ) f '( x) lim lim h 0 h 0 h h 2 2 2 1 x 2 xh h 1 x 2 xh h 2 lim lim h 0 h 0 h h h( 2 x h) lim lim( 2 x h) 2x 0 2x h 0 h 0 h

(b)

f '( x) lim h

0

f ( x h) h

f ( x)

( x h)3 2 ( x3 2) 0 h

lim h

x3 3x 2 h 3xh 2 h3 2 x3 2 3x 2 h 3xh 2 h3 lim lim h 0 h 0 h h 2 2 h(3x 3xh h ) lim lim(3x 2 3xh h2 ) 3x 2 0 0 3x 2 h 0 h 0 h © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

2 x

(c)

f '( x) lim h

0

f ( x h) h

f ( x)

lim h

0

x h h

x

lim h

0

x h h

( x h )2 ( x )2 x h x lim h 0 h( x h 0 h x) h( x h x) 1 1 1 lim h 0 x h x x 0 x 2 x lim

(d)

f '( x) lim h

x2 lim h

0

lim h

2.

(a)

0

0

f ( x h) h

f ( x)

x 2 2 xh h 2 x 2 ( x h) 2 h

2x h x ( x h)2 2

lim h

2x 0 x ( x 0)2 2

1 ( x h) 2 lim h 0 h

0

2 xh h 2 1 x 2 ( x h) 2 h 2x x4

lim h

0

x

x h x h

h h( x h

x)

x 2 ( x h) 2 x 2 ( x h )2 lim h 0 h

1 x2

h ( 2 x h) 1 0 x2 (x h) 2 h

lim h

2 x3

The slope of the curve at the point where x 1 is f '(1)

2(1)

2.

The graph of the function and the required tangent are shown below:

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x x

(b)

The slope of the curve at the point where x 1 is f '(1) 3 12

3.

The graph is shown below:

(c)

The slope of the curve at the point where x 1 is f '(1)

1 2 1

1 . 2

The graph is shown below:

(d)

The slope of the curve at the point where x 1 is f '(1)

2 13

2.

The graph is shown below:

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3.

(a)

(b)

(c)

(i)

y ' 3 2x 4 1

y ' 6x 4

(ii)

y '(0)

y '(0)

4

(i)

y ' 0 6 1 2x

y'

6 2x

(ii)

y '( 3)

(i)

y

(ii)

y '( 1)

2x

6(0) 4

6 2( 3)

3

y '( 3)

y ' 2( 3) x

6 ( 1) 4

4

y '( 1)

0

y'

6 x4

6

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(d)

(e)

(f)

(i)

y ' 5 x 4 3x 2 1

(ii)

y '(1) 5 14 3 12 1

(i)

y

(ii)

y '(2)

(i)

y

(ii)

y '(1)

x2

4 x 12

y ' 2x 4

2 2 4

2x x

2

1

1 12

y '(1) 1

y' 0

3x 9 14

3

y ' 2 1 ( 1) x

2

3( 3) x

4

y'

2

1 x2

y '(1) 10

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9 x4

(g)

4.

x3 1 x2

(i)

y

(ii)

y '( 1) 1

y'

x x

2 ( 1) 3

2

3

y ' 1 ( 2) x

y' 1

2 x3

y' 3

The point with coordinates (2, 4) is on the curve representing the given function, this means its coordinates satisfy the equation y 4

2

2

a(2) b

2a b

x2

ax b :

8

Next, the first derivative must be found: y x 2 ax b y ' 2x a The slope of the curve at x y '(2) 1 2 2 a 1

2 is 1: a 5 To find b, substitute the value of a into the first equation: 2( 5) b 8 b 2 5.

In the following questions, first find the expression of the first derivative, then equate it to the given value to obtain an equation, and solve for x. Finally, substitute the x-value(s) into the original function to find the y-coordinate of the required point(s). (a)

y

x 2 3x

3 2x 3

y ' 2x 3 2x

0

0

y

02 3 0

2

y

( 2)3

x

y

0

The required point is (0, 0) (b)

y ' 3x2 12 3 x 2

x2

4

x

y

8

The required points are (2,8) and ( 2, 8)

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(c)

y ' 2x 5

5 0 2x 5 2x 5 x y 2 5 21 , The required point is 2 4

(d)

5 2

2

5

5 2

1

y

21 4

y ' 2x 3

1 2x 3

2x

2

x 1

y 12 3 1

y

2

The required point is (1, 2) 6.

(a)

Between A and B is where the curve is the steepest, which means the slope of the curve is the greatest.

(b)

(i)

When the instantaneous rate of change is positive, the values of the function are increasing, so the points are A, B and F.

(ii)

When the instantaneous rate of change is negative, the values of the function are decreasing, so the points are D and E.

(iii)

The rate of change is 0 at a turning point, so the required point is C.

(c)

7.

B and D, and E and F, as the line segments joining the points in each pair have approximately the same slope.

The slope of the curve y y ' 2x 4

x2

4 x 6 is given by the first derivative:

At the point where x 3 the slope is: y '(3) 2 3 4 y '(3) 2 The same applies for the second curve: y 8x 3x2 y ' 8 6x y '( a ) 8 6a The two slopes are equal: 8 6a 2 6a 6 a 1 To find the value of b, substitute a 1 into the equation of the second curve (the required point lies on this curve): b 8(1) 3 12 b 5

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8.

Find the first derivative of the given function: y ax 3 2 x 2 x 7 y ' 3ax 2 4 x 1 Substitute x 2 into this equation to find the slope at the indicated point: y '(2) 3a 2 2 4 2 1 y '(2) 12 a 9 It is given that the slope is 3, so equate the expression for the slope to 3 and solve for a: 3 12a 9 12a 12 a 1

9.

It is known that two parallel lines have the same gradient, this means that, at the point ( a, b) on the graph of the function, the gradient of the curve (which is the same as the gradient of the tangent line), must be the same as the gradient of the given line, namely 5. The gradient of the curve is given by the first derivative: y x2 x y ' 2x 1 The gradient of the curve at x 5 2a 1 2a 6 a 3

a is 5:

The y-coordinate of the required point is calculated by substituting a

1 into the 3

expression of the original function: b 32 3 b 6 The required point is (3, 6) 10.

(a) (b)

f ( x)

x 3 1, h

The expression point where x f '( x)

3x 2

0.1

f (2 0.1) 0.1

f (2)

(a)

f ( x)

f

x

10.261 9 12.61 0.1

f (2 h) f (2) gives the value of the first derivative of f at the h 2, f '(2) . f '(2) 3 2 2

f '(2) 12

The given expression approaches 12 when x 11.

2.13 1 (23 1) 0.1

2 and h 0.1 .

f ( x h) f ( x ) h 0 h 2 2 a( x h) b( x h) c (ax bx c ) lim h 0 h 2 2 ax 2axh ah bx bh c ax 2 bx c lim h 0 h 2 2axh ah bh lim lim(2ax ah b) 2ax b h 0 h 0 h

ax 2 bx c

f '( x)

lim

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(b)

When using the expression found in (a), the respective derivatives are: a 1, b 0, c 0 f '( x) 2 1x 0 f '( x) 2 x a 3, b 4, c 2 f '( x) 2 3x 4 f '( x) 6 x 4 When calculating the derivatives directly, the following expressions are obtained: a 1, b 0, c 0 f ( x) x 2 f '( x) 2 x a

3, b

4, c

2

f ( x)

3x 2 4 x 2

f '( x)

6x 4

In both cases, the expressions found using direct differentiation are the same as the ones found by using first principles. 12.

(a)

When t 1, C 2 13 17 19 C and when t 4, C 2 43 17 33 C . The required average rate of change is the gradient of the line going through points (1,19) and (4, 33) . 33 19 14 4.6 degrees Celsius per hour. 4 1 3 The first derivative of C has to be found:

average rate of change

(b)

3

C

(c) 13.

(a)

3 t

2t 2 17

C'

14 3

t

3 12 t 2

2

14 9

t

C' 3 t 196 81

t

2.419...

t

2.42 hours

h ( x h) h ( x ) h The derivative of h( x) can be found from first principles, remembering

The derivative of h( x) is h ( x) lim h

0

that the denominator is the difference between the two x-values: h '( x) lim h

h

x h

h(( x h)) h( x) h 0 h h h '( x) , which shows that is odd.

0

This means h '( x) (b)

h( x )

lim

h x

Use a similar method: p '( x) lim h

0

p

x h

p( x) h

lim h

0

p(( x h)) h

p( x)

p(( x h)) p( x) h 0 h This means p '( x) p '( x) , which shows that p ' is even. lim

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14.

cos x 1 x cos x 1 0 The expression is undefined when x 0 , as it results in x 0 One of the trigonometric identities for the cosine of the double angle has to be used, namely: x x cos x 1 2 sin 2 cos x 1 2 sin 2 2 2 x x 2sin 2 sin 2 2 2 L lim lim x 0 x 0 x x 2 sin x The given limit lim 1 has to be used now: x 0 x

Let L

lim x

0

sin 2 L

lim x

0

x 2

x 2

x 2 sin x x 2 2

sin lim x

0

sin lim x

x 2

0

x 2

lim sin x

0

x 2

1 0 0

cos x 1 0 0 x

lim x

15.

(a)

Applying first principles: 1 f ( x h) h x x h x( x h) lim h 0 h

f '( x ) lim h

0

f ( x)

1 x

x ( x h) x ( x h) lim h 0 h

lim x h h 0 h h h x ( x h) lim lim h 0 h 0 hx ( x h h)

lim h

0

1 x ( x h)

1 x2

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(b)

(c)

2 ( x h) 2 x f ( x h) f ( x ) 3 ( x h) 3 x f '( x ) lim lim h 0 h 0 h h (2 ( x h))(3 x) (3 ( x h))(2 x) (3 x )(3 ( x h)) lim h 0 h 6 3( x h) 2 x x( x h) 6 2( x h) 3x x ( x h) lim h 0 (3 x )(3 ( x h)) 3( x h) 2 x 2( x h) 3x 1 5h lim lim h 0 (3 x )(3 ( x h)) h h 0 (3 x)(3 ( x h)) 5 5 5 lim h 0 (3 x )(3 ( x h)) (3 x )(3 ( x 0)) (3 x) 2 f '( x) lim h

0

f ( x h) h

0

0

( x 2) 2

lim h

x 2 x h 2

0

lim h

0

lim h

2

0

lim h

0

1 x 2

x 2 x h 2 1 x 2 x h 2 h

x 2 x 2

x h 2 1 x h 2 h

x h 2

2

x 2 x h 2 x 2 x h 2 x 2 x h 2

0

lim h

h

x 2 x h 2 x 2 x h 2

lim h

lim

x 2 x h 2 x 2 x h 2 h

lim h

f ( x)

1 x h 2 h

0

x 2

x h 2

h x 2 x h 2

x 2 1

x h 2

x 2 x h 2 x 2 1 1

x h 2

x 2

3

1 h 1 h

1 h 1 h 1 h

2 ( x 2)3

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16.

c , where c is a constant, c

Let f ( x) f '( x )

lim h

0

f ( x h) h

f ( x)

lim h

0

c c h

lim h

0

0 h

lim 0 h

0

0

Exercise 12.3 1.

(a)

The vertex is a stationary point; it follows that its x-coordinate can be found by solving the equation y ' 0 : y

x2

2x 6

y ' 2x 2

0 2x 2 x 1 The corresponding y-coordinate is: y 12 2 1 6 y 7 the vertex is (1, 7) (b)

y

4 x 2 12 x 17

3 2 3 3 4( )2 12( ) 17 y 8 2 2 x2 6 x 7 y' 2x 6

0 8 x 12 y

(c)

y

0 y 2.

(a)

(b)

y ' 8 x 12

x

2x 6 x 3 32 6 3 7 y

2

the vertex is (

3 ,8) 2

the vertex is (3, 2)

(i)

f ( x)

x 2 5x 6

(ii)

f '( x )

0

2x 5 0

x

(iii)

f '( x)

0

2x 5 0

x

(i)

f ( x)

7 4 x 3x 2

f '( x)

(ii)

f '( x)

0

4 6x

0

x

(iii)

f '( x)

0

4 6x

0

x

f '( x )

2x 5 5 2 5 2

f is increasing for x f is decreasing for x

5 2 5 2

4 6x 2 3 2 3

f is increasing for x f is decreasing for x

2 3 2 3

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(c)

f ( x)

(ii)

f '( x) 0 x

(d)

1 3 x 3

(i)

x

x2 1

f '( x )

x2 1 0

x

1

x2 1 0

x

1

x

1, x 1

f is increasing for

1, x 1

(iii)

f '( x) 0

(i)

1 x 1 f ( x) x 4 4 x 3

(ii)

f '( x )

(iii)

f '( x ) 0

0

f is decreasing for

4 x 3 12 x 2

f '( x)

4 x 2 ( x 3)

1 x 1

0

x 3 0

x

3

f is increasing for x

3

4 x 2 ( x 3) 0

x 3 0

x 3

f is decreasing for x

3

but it has a stationary point at x = 0. 3.

(a)

(i)

To find the stationary points, the equation y ' 0 must be solved. y

2 x3 3 x 2 72 x 5

y ' 6 x 2 6 x 72

0 6 x 2 6 x 72 6( x 2 6( x 4)( x 3) 0 x

x 12) 4, x 3

To find the y-coordinates, substitute the two x-values into the expression of the original function: x 4 y 2( 4)3 3( 4) 2 72( 4) 5 y 213 x

3

y

2 33 3 32 72 3 5

y

130

The stationary points are ( 4, 213) and (3, 130) (ii)

To determine the nature of the two stationary points, the second derivative is used: y ' 6 x 2 6 x 72 y '' 12 x 6 Substitute the x-coordinates of the stationary points into the expression of y '' , to determine the concavity of the function: y ''( 4) 12( 4) 6

42 0

the graph of the function is concave

down. So, there is a maximum at ( 4, 213) . y ''(3) 12 3 6

42 0

the graph of the function is concave up, so

there is a minimum at (3, 130) .

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(iii)

(b)

(i)

(ii)

The graph is shown below:

1 3 1 2 1 2 x 5 y' x 0 x x 0 6 2 2 To find the y-coordinates, substitute this x-value into the expression of the original function: 1 3 x 0 y 0 5 y 5 6 The stationary point is (0, 5) . y

As the first derivative is always positive ( 2 x2

0 for all x

,x

0 ),

the point where x 0 is neither a maximum nor a minimum is a point where the tangent to the graph is horizontal. (iii)

The graph is shown below:

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(c)

(i)

y

x( x 3) 2

x3 6x 2 9 x

y

0 3x 2 12 x 9 x 1

y 1(1 3) 2

x

y

3

y ' 3 x 2 12 x 9

3( x 2 4 x 3) y

3(3 3) 2

0

3( x 1)( x 3)

0

x 1, x

4

y

0

The stationary points are (1, 4) and (3, 0) (ii)

y ' 3 x 2 12 x 9 y ''(1) 6 1 12

y '' 6 x 12

the graph of the function is concave down,

6 0

so there is a maximum at (1, 4) . y ''(3)

6 3 12 6

the graph of the function is concave up,

0

so there is a minimum at (3, 0) .

(d)

(iii)

The graph is shown below:

(i)

y

x4 2 x3 5 x2 6 0

4 x3 6 x 2 10 x

x

0, x

x

0

y

x

5 2

y

x

1

y

y ' 4 x 3 6 x 2 10 x 2 x(2 x 2 3 x 5)

5 ,x 1 2 0 4 2 03 5 0 2 6

5 2

4

2

5 2

3

5

y

5 2

2 x(2 x 5)( x 1)

6

2

6

( 1) 4 2 ( 1)3 5 ( 1) 2 6

The stationary points are (0, 6) ,

0

y y

279 16 4

5 279 , and ( 1, 4) 2 16

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0

3

(ii)

y ' 4 x 3 6 x 2 10 x

y '' 12 x 2 12 x 10

y ''(0) 12 0 2 12 0 10

the graph of the function is concave

10 0

down, so there is a maximum at (0, 6) . y ''

5 2

12

5 2

2

12

5 2

the graph of the function is

10 35 0

5 279 , . 2 16

concave up, so there is a minimum at y ''( 1) 12( 1) 2 12( 1) 10 14 0

the graph of the function is

concave up, so there is a minimum at ( 1, 4) . (iii)

(e)

(i)

The graph is shown below:

1 2

1 12 y x x y' 1 x 2 1 1 0 1 1 2 x 2 x

y

1 4

1 4

x

1 2

1 2 x x

1 4

1 4 1 1 , The stationary point is 4 4 x

1 4

y' 1

y

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(ii)

y' 1 y ''

1 x 2

1 4

1 2

1 4

1 1 ( )x 2 2

y ''

1 4

3

2

0

3 2

y ''

1 4 x3

the graph of the function is concave up, so

1 1 , . 4 4 The graph is shown below:

there is a minimum at

(iii)

4.

(a)

v(t )

s '(t ), a(t )

s (t ) t 3 4t 2 t v(t )

(b)

s ''(t )

s '(t ) 3t 2 8t 1

3t 2 8t 1 , a(t )

s ''(t )

6t 8

6t 8

The displacement time graph is:

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The velocity time graph is:

The acceleration time graph is:

(c)

Displacement is a maximum when s '(t ) 3t

2

8t 1 0

t

0.1314..., t

0

2.535...

t

0.131, t

2.54

From graph: the maximum occurs when t 0.131, the displacement at this time is: s (0.1314...) 0.1314...3 4 0.1314...2 0.1314... s(0.1314...) 0.06460... smax

(d)

0.0646

Velocity is a minimum when v '(t )

0:

4 t 1.3 3 The minimum velocity is: 6t 8 0

4 v 3

4 3 3

t

2

8

4 3

1

v

4 3

13 3

vmin

4.3

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(e)

The object moves in the positive direction with decreasing velocity until it stops at t 0.131, after which it accelerates in the opposite (negative) direction until it reaches its maximum velocity at t 1.3 . The object continues to move in the same direction with decreasing velocity until it rests at t 2.54 , after which it turns and accelerates in the positive direction.

5.

(a)

(i)

To find the stationary points, the equation y ' 0 must be solved. x 3 12 x

f ( x)

f '( x ) 3 x 2 12

0 3x 2 12 x2 4 x 2 To determine the nature of the two stationary points, the second derivative is used:

f '( x ) 3 x 2 12 x

2

f ''(2)

f ''( x )

6x

the curve is concave up, hence the point

6 2 12 0

is a minimum point. x 2 f ''( 2) 6( 2)

12 0

the curve is concave down, hence

the point is a maximum point. To find the x-coordinate of the point of inflection, solve the equation y '' 0 : f ''( x) 6 x 0 6 x x 0 (ii)

The corresponding y-coordinates are: x 2 y 23 12 2 y 16 x x

2 0

y y

( 2)3 12 ( 2) 03 12 0

y

y 16

0

The graph of the function has a maximum at ( 2,16) , a minimum at (2, 16) and a point of inflection at (0, 0) .

The graph is shown below:

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(b)

(i)

Solve y ' 0 : f ( x) 0

x3

1 4 x 2x2 f '( x ) x 3 4 x 4 4x x( x 2 4) 0 x 0, x

2

Use the second derivative to determine the nature of the three stationary points: f '( x) x 3 4 x f ''( x ) 3x 2 4 x

f ''(2) 3 22

2

the curve is concave up, hence the

4 8 0

point is a minimum point. x 2 f ''( 2) 3( 2)2 4 8 0 the point is a minimum point. x 0 f ''(0) 3 0 2 4 4 0

the curve is concave up, hence the curve is concave down, hence the

point is a maximum point. To find the x-coordinate of the point(s) of inflection, solve the equation y '' 0 : f ''( x) 3x 2 4

0 3x 2 4

4 3

x2

2 3

x

x

2 3 3

are two points of inflection. (ii)

The corresponding y-coordinates are: 1 4 x 2 y 2 2 22 y 4 4 1 x 2 y ( 2) 4 2( 2) 2 y 4 4 1 4 x 0 y 0 2 02 y 0 4 x x

2 3 3 2 3 3

1 2 3 4 3

y y

1 4

4

2 3 3

2 3 2 3 4

2

2

y 2 3 3

20 9

2

y

20 9

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there

The graph of the function has a maximum at (0, 0) , two minimum points, at (2, 4) and ( 2, 4) , and two points of inflection, at

2 3 20 and , 3 9

2 3 20 . , 3 9

The graph is shown below:

(c)

(i)

f ( x)

x 4x

1

f '( x ) 1 4( 1) x

2

f '( x ) 1

4 x2

Solve y ' 0 : 4 4 1 x2 4 x 2 2 2 x x Use the second derivative to determine the nature of the stationary points: 8 f '( x ) 1 4 x 2 f ''( x ) 4( 2) x 3 f ''( x ) x3 8 the curve is concave up, hence the point x 2 f ''(2) 1 0 23 is a minimum point. 8 x 2 f ''( 2) 1 0 the curve is concave down, hence the ( 2)3 0 1

point is a maximum point. Solve y '' 0 to find the x-coordinate of the point(s) of inflection: 8 8 0 3 x x3 has no solutions. f ''( x)

there are no points of inflection, as this equation

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(ii)

The corresponding y-coordinates are: 4 x 2 y 2 y 4 2 4 x 2 y 2 y 4 2 The function has a minimum at (2, 4) , a maximum at ( 2, 4) and no points of inflection. The graph is shown below:

(d)

(i)

f ( x)

3x5 5 x3

f '( x)

3 5x 4 5 3x 2

15 x 4 15 x 2

f '( x)

Solve y ' 0 : 15 x 4 15 x 2

0

15 x 2 ( x 2 1)

0

x

0, x 2

1

x

0, x

1

Use the second derivative to determine the nature of the three stationary points: f '( x ) 15 x 4 15 x 2 f ''( x) 15 4 x 3 15 2 x f ''( x ) 60 x 3 30 x x

0

f ''(0)

minimum. x 1 f ''( 1)

60 03 30 0

0

x 0 is neither a maximum or a

60( 1)3 30( 1) 30 0

hence the point is a minimum point. x 1 f ''(1) 60 13 30 1 30 0

the curve is concave up,

the curve is concave down,

hence the point is a maximum point.

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Solve y '' 0 to find the x-coordinate of the point(s) of inflection: 60 x 3 30 x

f ''( x )

1 x 2 points of inflection. x 0, x 2

(ii)

0

60 x3 30 x

1

0, x

30 x ( 2 x 2 1)

x 0, x

2

2 2

0

there are three

The corresponding y-coordinates are: x 1 y 3 15 5 13 y 2 x x

x

1

3( 1)5 5( 1)3

y

0

y

2 2

3 0 5 5 03

y

3(

y

2 5 ) 5( 2

y

2

0

2 3 ) 2

y

7 2 8

2 2 2 7 2 y 3( )5 5( )3 y 2 2 2 8 The function has a maximum at (1, 2) , a minimum at ( 1, 2) and inflection x

2 7 2 2 7 2 , ) and ( , ) 2 8 2 8 The graph is shown below:

points at (

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(e)

(i)

3 x 4 4 x 3 12 x 2 5

f ( x)

f '( x) 12 x 3 12 x 2

f '( x ) 3 4 x 3 4 3 x 2 12 2 x

24 x

Solve y ' 0 : 0 12 x( x 2

x 2)

12 x( x 1)( x 2)

0

x

0, x

1, x

2

Use the second derivative to determine the nature of the stationary points: f '( x) 12 x 3 12 x 2 24 x f ''( x) 12 3 x 2 12 2 x 24 36 x 2

f ''( x) x

24 x 24

f ''(0) 36 0 2 24 0 24

0

24 0

the curve is concave down,

hence the point is a maximum point. x 1 f ''( 1) 36( 1) 2 24( 1) 24 36 0 hence the point is a minimum point. x 2 f ''(2) 36 2 2 24 2 24 72 0

the curve is concave up,

the curve is concave up,

hence the point is a minimum point. Solve y '' 0 to find the x-coordinate of the point(s) of inflection: 36 x 2 24 x 24

f ''( x )

0 12(3x 2 2 x 2)

( 2)2 4 3 ( 2) 2 3 two points of inflection. ( 2)

x

(ii)

2

x

x x

x

1

y

2

y

1

7 3

1

7 3

y

3 2

y

4

4( 1)

4 2

3

3

1

12( 1)

12 2

7

2

4

4

3

2

y

1

7

7 3

there are

5 5

5

1

y

0

27 3

12

3

1

7

2

5

3

140 80 7 27

y

x

3( 1)

3

x

6

The corresponding y-coordinates are: x 0 y 3 0 4 4 03 12 0 2 5 y 4

28

y

3

1

7 3

4

4

1

7 3

3

12

1

7 3

2

5

80 7 149 27

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The function has a maximum at (0, 5) , a minimum at ( 1, 0) and (2, 27) , and 1

inflection points at

7 3

140 80 7 27

,

(or (1.22, 13.4) ) and

1

7 80 7 149 (or ( 0.549, 2.32) ) , 3 27 The graph is shown below:

6.

(a)

v(t )

s '(t ), a(t )

s (t )

t (8t 2 33t 27) 8t 3 33t 2

s ''(t ) v(t )

s ''(t )

27t

s '(t )

24t 2 66t 27

48t 66 24t 2 66t 27 , a(t )

48t 66

Substitute t 0 into expressions of the velocity and acceleration to obtain their initial values: v(0) 24(02 ) 66(0) 27 v (0) 27 ms-1 (b) (c)

a(0)

48(0) 66

a(0)

v(3)

24(32 ) 66(3) 27

a(3)

48(3) 66

a(3)

66 ms-2

v (3)

45 ms-1

78 ms-2

The values of t for which the object changes direction are obtained by solving the equation v(t ) 0 , as the object must first come to rest before changing direction: 0

24t 2 66t 27

3(4t 9)(2t 1)

0

As the equation v(t ) 0 is the same as s '(t )

9 1 ,t 4 2 0 , its two solutions represent

t

stationary points for the displacement.

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(d)

To find the required value for t, the equation v '(t ) 0 (or a(t ) 0 ) must be solved: 0

(e) 7.

48t 66

48t

66

11 1.375 8

t

The acceleration is 0.

Find the expression of D ' : 100 x2 Solve the equation D ' 0 to find the value of x which minimises the delivery cost: D

3 x 100 x

1

D ' 3 100( 1) x

2

D' 3

100 100 100 100 3 x2 x 2 2 x x 3 3 But x 0 x 5.773... x 5.77 tonnes The corresponding delivery cost is: 0 3

5.773...

100 D 20 3 34.64... thousand D $34, 600 100 3 To determine the nature of this stationary point, the second derivative is found: 200 D ' 3 100 x 2 D '' 100( 2) x 3 D '' x3 D 3

100 3

100 into the expression of D '' to analyse the concavity of the graph: 3

Substitute x D ''(

100 ) 3

200 3

D ''(

100 3 point is a minimum point.

8.

100 ) 1.039... 0 3

If the point ( 1, 8) is on the graph representing y satisfy the equation of the curve: 8 ( 1) 4 a( 1) 2 b( 1) c a b c

the curve is concave up, hence the

x4

ax 2 bx c , then its coordinates

9

To use the other given information, y ' and y '' must be found: y ' 4 x 3 2ax b, y '' 12 x 2 2a

Equate the expressions of y '( 1) and y ''( 1) to 6 to obtain two more equations: 6

4( 1)3 2a( 1) b

6 12( 1)

2

2a

2a

2a b 10 6

a

3

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Substitute the value of a in the remaining equations to find the values of b and c:

9.

2( 3) b 10

b

4

3 4 c

c

2

9

To find the stationary points, the equation y ' 0 must be solved. Find the expression of y ' : y

x3 3 x 1 x2

0 1

3 x2

x 1, x

y

x 3x

1

x

x 3 3x 2 x3

2 x3

2

y ' 1 3( 1) x

2

( 2) x

( x 1)( x 2 x 2) x3

0

0

3

y' 1

3 x2

2 x3

( x 1)( x 2)( x 1) x3

0

2

To determine the nature of the two stationary points, the second derivative is used: 6 6 y ' 1 3x 2 2 x 3 y '' 3( 2) x 3 2( 3) x 4 y '' x3 x 4 Substitute the x-coordinates of the stationary points into the expression of y '' , to determine the concavity of the function: 6 6 y ''(1) 3 4 0 the point is neither a minimum nor a maximum (there is an 1 1 inflection point at x 1 , and the tangent at this point is horizontal) 6 6 6 3 9 y ''(2) 0 the function is concave down, so there is a 3 4 ( 2) 2 8 8 8 maximum at x 2 .To find the y-coordinates, substitute the two x-values into the expression of the original function: 13 3 1 1 x 1 y y 3 12

x

2

y

( 2)3 3 ( 2) 1 ( 2)2

15 4

y

15 and a stationary point of inflection at (1,3) . 4 To determine the behavior of the function when x , the limit of the function must be evaluated:

There is a point of maximum at

x3 3 x 1 lim x x2

lim x

x

1 x

1 x2

2,

lim x

x

lim

x

1 x

lim

x

1 x2

0 0

This means the values of the function will increase towards when x and will decrease towards when x , in both cases without bound. It must be noticed that © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

the values of the given function will approach in the same manner as the values of the linear function y x (the graph has an oblique asymptote). The graph of the function is shown below:

10.

In each of the following questions, in order to be able to draw the graph of y the gradient of the curve y (a)

f '( x ) ,

f ( x) has to be analysed.

The graph has one maximum point at x 0 (0, 0) is a x-intercept for the graph of y

f '(0)

0 , which means that point

f '( x). Mark this point on a new set of

axes. Next, determine the gradient of the curve representing the function y

f ( x) ,

going from left to right, keeping in mind that the values of the gradient are y values for the graph of y f '( x) . At the left end the gradient is (because the tangent to the graph of the function at this point is vertical), then it decreases until it reaches 0, then it becomes negative and, at the right end, it approaches . The graph is shown below:

(b)

The graph has one maximum point at x 0

f '(0)

0

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This means that point (0, 0) is a x-intercept for the graph of y

f '( x ) . At the left

end the gradient has a positive value, then it decreases until it becomes 0, then it becomes negative. The graph is shown below:

(c)

The graph has two stationary points, a minimum at x x a. f '(a) f '( a) 0

a , and a maximum at

This means that points ( a, 0) and ( a, 0) are the x-intercepts of the graph of y

f '( x ) . At the left end of the given graph, the gradient is negative and

a , then it becomes positive and increases up increases until it becomes 0 at x to point (0, 0) , due to the symmetry of the graph of an odd function, after which it starts to decrease until it becomes 0 again at x a . Finally, past x a , the slope is negative. The graph is shown below:

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(d)

The graph has three stationary points, minimum points at x and a maximum at x 0 . f '(a) f '( a) f '(0) 0

a and x

a,

This means that points ( a, 0) , (0, 0) and ( a, 0) are the x-intercepts of the graph of y

f '( x ) . To start with, the gradient is negative and increases until it reaches 0 at

x

a, then it becomes positive and increases up to a point, say ( b, 0) , where b 0, b a , after which it starts to decrease until it reaches 0 again, at (0, 0) . Past x 0 , the gradient turns negative and decreases up to the point where x b , then it increases towards 0, which is reached at x a . Finally, past x a , the slope is positive. The graph is shown below:

(e)

As the graph represents a periodic function, the points of minimum and maximum occur following a pattern related to the period of the function. The repeating part of the graph has two sections, each of them a straight line, the first one with a constant positive gradient, the second one with a constant negative one. The graph of the derivative will also be periodic, and it will be represented by repeating horizontal line segments:

Notice that at the extrema the derivative is not defined because of the nature of the original function. © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

11.

A function f is increasing when its derivative, , is positive, and is decreasing when negative. (a)

is

By inspection, f is increasing for 1 x 5 and is decreasing for x 1 or x 5. The function has a minimum at x 1 , because f '(1) 0 , and f ' changes

(i) (ii)

its sign from negative to positive about x 1 . Similarly, the function has a maximum at x 5 because f '(1) 0 , and f ' changes its sign from (b)

12.

positive to negative about x 5 . f is increasing for x 1or 3 x 5 , and is decreasing for 1 x 3 or x 5 . The function has maximum points at x 1 and x 5 , and a minimum at x 3.

(i) (ii)

By inspection, there are three points on the given graph where the second derivative is 0, at x 0.5, x 4 and x 7.5 . Additionally, for a point of inflection to occur, there has to be a change in the sign of f '' about the point where f '' is 0. This only happens for x 0.5 and x 7.5 , the values of f '' about x 4 are always positive.

13.

(a)

f ( 2)

(b)

f (0)

4

y

4 is the y-intercept of the graph

(c)

f (2)

0

x

2 is an x-intercept of the graph

(d)

f '(2)

(e)

(g) (h)

f '( 2)

0

there are stationary points at x

The first derivative is positive for x

x (f)

( 2,8) is on the graph

8

2

2 , this means the function is increasing for

2 or x 2 .

The first derivative is negative for x

2 , this means the function is decreasing for

2 x 2. The second derivative is negative for x 0 , this means the graph is concave down on this interval. The second derivative is positive for x 0 , this means the graph is concave up on this interval. The graph is shown overleaf:

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14.

(a)

To determine the direction in which the object moves, its velocity has to be analysed. As the velocity is the rate of change of displacement, the first derivative of s has to be found:

v(t ) s '(t )

6t 2 30t 24

v(t )

The intervals where the velocity is positive will give the set of values for t when the object is moving to the right. Similarly, when the velocity is negative, the object moves to the left. To find when the change of direction occurs, the equation v(t ) 0 must be solved:

0

6t 2 30t 24

6(t 4)(t 1) 0

t 1, t 4

This means that the object is moving to the left when 0 t 1 or t 4 (velocity takes negative values), and it moves to the right for 1 t 4 (velocity is positive). (b)

(c)

The initial values are found by substituting t 0 into the expressions of the velocity and acceleration, respectively: v(0) 6 02 30 0 24 v(0) 24 (i) (ii)

a (t )

(i)

The displacement is at a maximum at either t 1 or t 4 . To decide which value will give the maximum point on the graph of the displacement, the second derivative, s '' , will be used: s ''(t ) s ''(1)

v '(t )

a (t )

a(t )

12t

s ''(t )

12 1 30 18

12t

30

a (0)

12 0 30

a(0)

30

30

the graph is concave up,

0

so t 1 is a minimum s ''(4)

12 4 30

18

0

the graph is concave down,

so t 4 is a maximum © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

The maximum displacement is:

2 43 15 42 24 4

s(4)

s(4) 16

The maximum velocity is obtained when v ' 0 :

(ii)

0

12t

30

5 2

t

The maximum velocity is: 5 v 2

6

5 2

2

5 2

30

24

(d)

t

v

5 2

13.5

5 (see part (c)(ii)). 2

At this point the object has achieved its greatest velocity. 15.

(a)

(b)

The minimum value of the function is y

f ( x) x

2 sin x

f '( x) 0

1

f

f

4 7 4

4 7 4

f '( x) 1

2 cos x 0

2 sin

f

4

2 sin

7 4

6.50

2 cos x

cos x

4 f

0.215 , the maximum is y

4 7 4

1 2

x

2

2 2

7 4

4 f

2

,x

7 4

4

4

2 2

f

1 7 4

7 4

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1

Exercise 12.4 1.

In each of the questions in this exercise, follow the steps listed below: i.

find the y-coordinate of the point of tangency question by substituting the given x-coordinate into the expression of the function find y ' , the first derivative of the given function

ii. iii.

find m, the gradient (slope) of the tangent line, by substituting the given x-coordinate into the expression of y '

iv.

substitute into the slope-intercept form of the equation of a line, y y1 m( x x1 )

v.

rearrange this equation into the required form, y mx c .

(a)

x

y ( 3)2 2( 3) 1

3

y x2 2x 1 m

y 4

the point of tangency is ( 3, 4).

y ' 2x 2

2( 3)

2

m

4

The equation of the tangent line is: y

(b)

4

4( x ( 3))

2 3

x

4 x 12 4

3

2 3

y

y x3 x 2

y

2 3

2

y

y

4 27

4x 8

the point of tangency is

y ' 3x 2 2 x 2

2 2 m 3 2 m 0 3 3 The equation of the tangent line is:

y (c)

4 27 2

x 1

4 27

y

y 3 02 0 1

x 0 y 3x

2 3

0 x

y 1

the point of tangency is (0, 1)

y ' 6x 1

m 6 0 1 m 1 The equation of the tangent line is: y 1

(d)

1( x

x

1 2

y

y

2x

x

0)

y

1 2

1 1 2

2

1

y'

x 1

y

the point of tangency is

3

2 ( 1) x

2

y'

2

1 ,3 . 2

1 x2

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2 4 , . 3 27

1

m 2

m 2 2 1 2 The equation of the tangent line is:

y 3 2.

1 2

y

2x 4

The gradient of the normal line is the negative reciprocal of the gradient of the tangent line. 3 1 19 4 y x 4 4 4 2 4 , The line tangent to the curve at is a horizontal line; this means the 3 27 2 normal line at the same point is a vertical line, its equation is x 3 1 y 1 ( x 0) y x 1 1 1 1 1 1 1 11 y 3 x y x 3 y x 2 2 2 4 2 4

(a)

y

(b)

(c) (d) 3.

2 x

1 ( x ( 3)) 4

4

y

1 x 4

The points of intersection between the curve and the x-axis have y 0 x3 3x 2 2 x

0 x( x 1)( x 2)

y x 3 3x 2 2 x

y ' 3x 2 6 x 2

0:

x 0, x 1, x 2

2 0 : m 3(0) 6 0 2 m 2 The equation of the tangent line at (0, 0) is:

When x y

0

2( x 0)

y

2x

2 When x 1 : m 3(1) 6 1 2

m

1

The equation of the tangent line at (1, 0) is: y

0

1( x 1)

y

x 1

2 2 : m 3(2) 6 2 2 m 2 The equation of the tangent line at (2, 0) is:

When x y

0

2( x

2)

y

2x

4

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4.

x 2y 1

2y

x 1

y

1 x 2

1 2

the gradient of the line is m 1

2

The gradient of the required tangent is the negative reciprocal of m, namely 2 . The first derivative of the given function is: y x2 2x

y ' 2x 2

The gradient of the curve is the same as the gradient of the tangent line, so an equation can be formed to find the x-coordinate of the tangency point: 2x 2 2 x 0 The y-coordinate is: y 02 2 0 y 0 the point of tangency is (0, 0) The equation of the tangent line at (0, 0) is: y 0

5.

2( x 0)

y

2x

The two graphs are shown in the diagram below:

(a)

Find the first derivative of both functions to obtain the expression of their gradients, then equate them and solve for x: y x 2 6x 20 y ' 2x 6 y x 3 3x 2 x 2x 6

But x

y ' 3x 2 6 x 1

3x2

6x 1

3x2

8x 5

0

(3 x 5)( x 1)

0

x

5 ,x 1 3

x 1

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2 3 2 Let f ( x) x 6 x 20 and g ( x) x 3x x

(b)

f (1) 12 6 1 20

y 15

g (1) 13 3 12 1

y

the point of tangency for y

f ( x) is (1,15)

3

the point of tangency for y g ( x ) is (1, 3) The gradient of both tangent lines is: m 21 6 m 4 It follows that the equation of the tangent line to the graph of y f ( x) at (1,15) is: y 15

4( x 1)

y

4 x 19

Similarly, the equation of the tangent line to the graph of y y

6.

x

( 3)

4( x 1)

y ( 3)2 4( 3) 2

3

y

y

4x 1

5

the point common to the normal and the curve is

( 3, 5). The gradient of the curve at x

y x2 4 x 2

y ' 2x 4

g ( x) at (1, 3) is:

3 is:

y '( 3) 2( 3) 4

y '( 3)

2

The gradient of the normal line is the negative reciprocal of the gradient of the curve, so the equation of the normal at ( 3, 5) is: 1 ( x ( 3)) 2

y ( 5)

y

1 x 2

3 2

5

y

1 x 2

7 2

To find where this line intersects the curve again, equate the expressions of the curve and the normal and solve for x: 1 x 2

7 2

x

x2

4x 2

1 ,x 2

3

x2

7 x 2

3 2

2x2

0

7x 3

0

(2 x 1)( x 3)

0

1 , the corresponding y-coordinate is: 2 1 2 1 1 15 y ( ) 4( ) 2 y 4 y 2 2 4 4 The other point of intersection is ( 1 , 15 ) 2 4

The required point is where x

7.

Find the expression of the first derivative: y

1 x3 x4

y

x

4

x

1

y'

4x

5

( 1) x

2

y'

4 x5

1 x2

The gradient of the tangent is found by substituting x 1 into the expression of y ' : m

4 15

1 12

m

3

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The equation of the tangent is: y

0

3( x 1)

y

3x

3

The slope of the normal line is the negative reciprocal of m, so the equation of the normal line is: 1 ( x 1) 3

y 0

8.

1 x 3

y

1 3

Based on the information given, the equation of the normal line is y

4.

x

1

x 1

y a 12 b 1

the point common to the normal and the curve is

y a b

(1, a b).

The coordinates of this point must satisfy the equation of the normal: y

x

4

a b

1 4

a b

3

a

3 b

The gradient of the curve at x 1 is: 1

y ax 2 bx

1 ax 2

y'

1 2

b

1 1 a12 b 2

y '(1)

y '(1)

a 2b 2

The slope of the normal line, n, is the negative reciprocal of the gradient of the curve, and is equal to 1: 2 a 2b

n

2 a 2b

1

a 2b

2

Substitute the expression of a into the last equation: a

9.

3 b

(a)

( 3 b)

2b

2

b

1

a

3 1

a

4

The first derivative of the function is: y

1 2 x 1 2

x3

3x 2

y'

x

The gradient of the tangent is found by substituting x m 3( 1)

y'

2

( 1)

1 into the expression of

m 2

The equation of the tangent is: y

(b)

1 2

2( x ( 1))

y

2x

5 2

The gradient of the other tangent is the same as the gradient of the tangent line found in (a). The gradient is given by the expression of y ' : 2

3x 2

x

3x2

x 2

0

The required x-coordinate is x y

2 3

3

1 2 2 3

2

1

y

(3 x 2)( x 1)

0

x

2 ,x 3

1

2 , the corresponding y-coordinate is: 3

41 27

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10.

the point of tangency is (4, 2). Find the expression of the first derivative: x 4

y

y

4 1

x 1

x

4

y

y

2

x x

1

y'

1

2 x

The gradient of the tangent, m, is found by substituting x m

1 2 4

1

4 into the expression of y ' :

3 4

m

The equation of the tangent is: 3 ( x 4) 4

y ( 2)

3 x 3 2 4

y

3 x 1 4

y

The slope of the normal line is the negative reciprocal of m, so the equation of the normal line is: 1 ( x 4) 3 4

y ( 2)

11.

(a)

x 1

4 16 x 2 3 3

y

y (1 x) 2 (5 x)

y

y (1 1) 2 (5 1)

4 x 3

22 3

y 16

the point of tangency is

(1,16). The first derivative is:

y (1 x)2 (5 x)

x3 3x 2 9 x 5

y

y'

3x 2 6 x 9

The gradient of the tangent, m, is found by substituting x 1 into the expression of y ' m 3 12 6 1 9 m 12 The equation of the tangent at point (1,16) is: y 16 12( x 1)

y

12 x

4

To show that the two graphs do not meet again, equate the expression of the function and the one of the tangent and solve for x the only solution for this equation should be the x-coordinate of the tangency point, x 1 : x3 3x 2 9 x 5 12 x 4

(b)

x 3 3x 2 3 x 1 0

( x 1)3 0

The gradient of the graph at the point where x y '(0)

30

2

60 9

x 1

0 is:

y '(0) 9

The equation of the tangent at point (0, 5) is: y 5

9( x 0)

y

9x 5

To show that the two graphs meet again, equate the expression of the function and the one of the tangent and solve for x there should be at least one more solution additional to the solution given by the x-coordinate of the tangency point, x 0 : x3 3 x 2 9 x 5 9 x 5

x3 3x 2 0

x2 ( x 3) 0

x 0, x 3

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The required x-coordinate is x y '(3)

where x (c)

12.

2

33

63 9

y '(3)

3 . The value of the gradient at this point is: 27 27 y '(3) 0 the point on the graph

3 is a turning point.

The three graphs are shown in the diagram below:

Let (a , b ) be a point of tangency. As this point is on the graph of the given function, it follows that its coordinates satisfy the equation of the curve. a2

b

a

2 the tangency point has coordinates (a, a a) .

The expression of the first derivative is: y x2 x y ' 2x 1 The gradient of the graph at the point where x a is: y '(a )

2a 1

2 The equation of the tangent at point (a, a a) is:

y (a 2 a) (2a 1)( x a)

y (2a 1) x a 2 a a(2a 1)

y (2a 1) x a 2 a 2a 2 a

y (2a 1) x a 2

The point (2, 3) lies on the tangent, so its coordinates must satisfy the equation of this line: 3 (2a 1)2 a 2 a 2 4a 5 0 (a 5)(a 1) 0 a 5, a 1 The two points of tangency are: a 5 b 52 5 b 30 (5,30) a

1

b ( 1)2 ( 1)

b 0

( 1,0)

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The gradients of the two lines are: a

5

a

y '(5) 1

2 5 1

y '(5)

y '(5)

2 ( 1) 1

11

y '( 1)

1

Consequently, the equations of the two tangent lines are: At (5, 30) : y 30 11( x 5) y 11x 25 At ( 1, 0) : y 0 13.

1( x

( 1))

y

x 1

Let (a , b ) be a point of tangency. b 1 (a 1)2

b a 2 2a 2 , the tangency point has coordinates (a, a 2 2a 2) .

The expression of the first derivative is: y x2 2x 2 y ' 2 x 2 The gradient of the graph at the point where x a is: y '(a )

2a

2

2 The equation of the tangent at point (a, a 2a 2) is:

y (a 2 2a 2) (2a 2)( x a)

y (2a 2) x a 2 2

The origin lies on the tangent, so its coordinates must satisfy the equation of this line: 0 (2a 2) 0 a 2 2

a2 2 0

a2

2

a

2

The two points of tangency are: 2

a

2

b

2

2 2 2

a

2

b (

2)2 2(

2) 2

b

4 2 2 b 4 2 2

( 2, 4 2 2) (

2,4 2 2)

The gradients of the two lines are: a

2

a

y '( 2) 2 2 2 2

y '(

2)

2 2 2

Consequently, the equations of the two tangent lines are: At ( 2, 4 2 2) : y (4 2 2) (2 2 2)( x At (

2)

2, 4 2 2) : y (4 2 2) ( 2 2 2)( x (

y (2 2 2) x 2))

y ( 2 2 2) x 4 2 2 4 2 2

y

(2 2 2) x

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14.

x 8

(a)

3

y

8

y 2

the point of tangency is (8, 2).

The first derivative is:

y

x

1 3

1 y' x 3

2 3

1

y'

2

3x 3 The gradient of the tangent, m, is found by substituting x 8 into the expression of y ' : 1

m

3 8

m

2 3

1 3 4

1 12

m

The equation of the tangent at point (8, 2) is: y

1 x 12

y

2 3

2

1 x 12

y

4 3

Substitute x 9 into the equation of the tangent:

(b)

1 9 12

y

15.

1 ( x 8) 12

2

x a

4 3

1 a

y

y

3 4

4 3

25 12

y

y

the point of tangency is a,

3

2.0833...

9

2.08

1 a

The first derivative is:

y

x

1 2

y'

1 x 2

3 2

1

y'

3 2

2x The gradient of the tangent, m, is found by substituting x a into the expression of y ' : m

1 2a

3 2

m

1 2a a

The equation of the tangent at point a, y

16.

1

1

a

2a a

( x a)

y

1 2a a

x

1 is: a

1

1

2 a

a

y

1 2a a

x

3 2 a

Let P (a, b) be the point of tangency. x a

y a3

3 the point of tangency is (a, a ).

2 The first derivative is y ' 3x .

The gradient of the tangent, m, is found by substituting x a into the expression of y ' m

3a 2

3 3 2 The equation of the tangent at point (a, a ) is: y a 3a ( x a)

y 3a 2 x 2a 3

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To find where the two graphs meet again, equate the expression of the function and the one of the tangent and solve for x: x3

3a 2 x 2a3

x3

3a 2 x

2a 3

0

As x a is a solution to this equation, it follows that ( x a ) is a factor. Use synthetic (or long) division to factorise the cubic

( x a)( x 2 ax 2a 2 ) 0

The required x-coordinates can be found by solving the equation: x 2 ax 2a 2 0 ( x a)( x 2a) 0 x a, x 2a Consequently, point Q has coordinates x

3 2a and y ( 2a)

8a3

Q( 2a, 8a3 ) or

Q( 2a, 8b) .

17.

Method 1 A geometric solution

A

r P

r

r

Q

B In triangle APQ with QA being the radius at the point of tangency and hence perpendicular to the tangent: AP

4r 2 r 2

r 3

Also, P

30 because the side opposite to it is half the hypotenuse. It can be easily shown that triangle APB is equilateral, and thus AB AP r 3. Method 2 Using calculus Consider a set of axes with the origin O(0, 0) at the center of the first circle. This means that the center of the second circle is at point (2 r , 0) , where r is the radius of the circles.

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Let point A have coordinates ( a, b) , it follows that B has coordinates (a, b) . The distance AB is vertical, so AB b ( b) AB 2b. 2 2 2 The equation of the circle centered at (2 r , 0) is ( x 2r ) y r

x

a

r 2 (a 2r )2

b

a2

b

4ar 3r

A(a,

a2

4ar 3r 2 )

a2 4ar 3r 2 ) are the two points of tangency.

and B(a,

To write the equation of the tangent at point A, start by finding the first derivative:

y

r y'

2

( x 2r )

2

y ( x

1 2 2

2

1 y' ( x 2 4 xr 3r 2 ) 2

4 xr 3r )

1 2

( 2 x 4r )

2r x x 2 4 xr 3r 2

The gradient of the tangent, m, is found by substituting x a into the expression of y ' :

m

2r a a 2 4ar 3r 2

The gradient of the curve representing the top semicircle is the same as the gradient of the tangent line. The gradient of the tangent line OA can be calculated as the gradient of the line joining

a2 4ar 3r 2 ) :

the origin, O(0, 0) , and A(a,

mOA But m

a2 4ar 3r 2 0 a 0

b

(

a2

2r a

mOA

2ar 3r 2

a 2 4ar 3r 2 a

mOA

a 0

2

4ar 3r

2ar

3r 2

2

a

3r 2 3r ) 4( )r 3r 2 2 2

a2

4ar 3r 2

2ar a 2

3r 2

9r 2 4

b

Consequently, the distance AB is: AB 2 18.

4ar 3r 2 a

r 3 2

3r 2

b

3r 2 4

b

r 3 2

AB r 3

Let P (a, b) be the point of tangency. x a

y 4 a2

2 the point of tangency is (a,4 a ).

The first derivative is y '

2 x.

The gradient of the tangent, m, is found by substituting x a into the expression of y ' : © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

m

2a

2 The equation of the tangent at point (a, 4 a ) is:

y (4 a 2 )

2 a( x a )

2ax 2a 2 4 a 2

y

2ax a 2 4

y

If point (1, 2) lies on any of the tangent lines, then its coordinates should satisfy the equation of the tangent: 2 2a 1 a 2 4 a 2 2a 2 0

(a 1) 2 1 0

(a 1) 2

1

a

This quadratic equation does not have any real solutions, so there is no line going through (1, 2) which is tangent to the graph of the given function.

Chapter 12 practice questions 1.

(a)

f ( x) x 2

f '( x) 2 x

The gradient of f at x 1.5 , m, is: m

(b)

f '(1.5)

f ( x) x 2

2 1.5

y 1.52

3

y 2.25

the tangency point is P(1.5, 2.25)

The equation of the tangent at this point is: y 2.25

3 x 4.5 2.25

y

(c)

The graph is shown below:

(d)

Point Q is the x-intercept of the line, so y

0:

0 3x

3( x 1.5)

9 4

3x

9 4

y

x

3 4

Q

3 x 2.25

y

3x

9 4

3 ,0 4

Point R is the y-intercept of the line, so x 0: y 30

9 4

y

9 4

R 0,

9 4

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(e)

Find the coordinates of M, the midpoint of line segment PR: xP

xM

xR 2

yP

yM

1.5 0 2

xM

xM

9 4

2.25

yR

yM

2

0.75

xM

3 4

0 2

yM

0

yM

2

3 ,0 is the midpoint between P and R. 4

M

This point has the same coordinates as point Q, so Q is the midpoint of [ PR] . (f)

The gradient of f at x = a is: f '(a )

2a

2 The equation of the tangent at point S (a, a ) is:

y a 2 2a( x a )

(g)

y 2ax 2a 2 a 2

y 2ax a 2

Point T is the x-intercept of the line, so y = 0.

0 2 ax a 2

2 ax a 2

a 2

x

a ,0 , a 0 2

T

Point U is the y-intercept of the line, so x 0: y 2a 0 a 2

(h)

a2

y

R(0, a 2 ), a 0

Find the coordinates of N, the midpoint of line segment SU: xS

xN

xU

xN

2

yS

yN

yU

yN

2 N

a 0 2

xN

a 2

yn

0 2

a2 ( a2 ) 2

yn

0

a ,0 is the midpoint between S and U. 2

This point has the same coordinates as point T, so T is the midpoint of [ SU ] for any a 2.

,a

0.

The first derivative needs to be found: y

Ax

B Cx

1

y'

A 0 C ( 1) x

2

y'

A

C x2

If the point (1, 4) is a stationary point for the graph, then the first derivative of the function takes value 0 when x 1 : y'

A

C x2

0

A

C 12

0

A C

The equation of the graph is now y

Ax

A B

C A . x

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Also, if points (1, 4) and ( 1, 0) are on the graph, then their coordinates must satisfy the equation of the curve: 4

A(1)

0

A( 1)

A 2A B 4 1 A B 2A B 0 1

B

Subtracting these two equations leads to: 4A

4

A 1

B

4 2(1)

B

2

The values of A, B and C are: A 1, B 3.

(a)

2, C

1

Find the expression of the first derivative and set it equal to 0, then solve for x: f ( x)

8x 8 x2

0

1

2x

2

f '( x)

8 x2

But x 0, so x

2

8( 1) x

2

4

x

x2

2

f '( x)

8 x2

2

2

2

To have a turning point at x

2 , there must be a change in the sign of f ' about

this point: f '( x )

8 x2

2

x2

0 8 x2

Similarly, f '( x)

2

4

x

0

x2

2 (recall that x

4

x

0)

2

Consequently, there is a change in the sign of f ' about x turning point for the graph of the function. (b)

2 , so this point is a

The limit when x 0 has to be evaluated when the values of x decrease towards 0 ( x 0 ): lim x 0 x 0

8 x

2x

lim x a x a

8 x

lim 2 x x a x a

0

Consequently, the graph has a vertical asymptote, x 0. To determine the existence of a horizontal asymptote, the end behavior of the function will be investigated, by considering the limit of the function when x : © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

lim x

8 2x x

8 x

lim

x

lim 2 x 0

x

This means the values of the function will increase towards when x , so there is no horizontal asymptote. It must be noticed that the values of the given function will approach in the same manner as the values of the linear function y 2 x , consequently, the graph has an oblique asymptote). The graph of the function is shown below:

4.

To find the stationary point, determine the expression of the first derivative and set it equal to 0, then solve for x: y ' 8x

1 x2

0

8 x3 1 0 x2

8x3 1 0

The corresponding y-coordinate is y 4 5.

6.

1 2

x 1 2

2

1 1 2

y

3

the stationary point is

1 ,3 . 2

2 Find the expression of the first derivative: y ' 3ax 4 x 1

y '(2)

3

(a)

The equation of the tangent to the graph of y y

(b)

3 12a 9

f (2)

f '(2)( x

a

1.

2)

y 3

5( x

2)

y

The equation of the normal to the graph of y y

f (2)

1 ( x 2) f '(2)

y 3

1 ( x 2) 5

f ( x) at the point where x

y

2 is:

5x 7 f ( x) at the point where x

1 17 x 5 5

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2 is:

7.

(a)

The maximum point is where x 1 , because g '(1) 0, g '( x) 0 when x 1, and g '( x) 0 , when x 1. That is, the values of the function increase for 2 x 1

(b)

and decrease for 1 x 3 . The function g is decreasing on the intervals where g '( x) 0 , so g is decreasing for 3 x 2 or 1 x 3 .

(c)

The inflection point is where x g ''(0)

(d)

8.

f (1) 0

1 , because g '' 2

1 2

0, g ''( 1) 0 and

0 , meaning there is a change in the concavity of the graph of g,

from concave up to concave down. The graph is shown below:

12 3b 1 (c 2) 0

3b c

3

Find the expression of the first derivative: f ( x) x 2 3bx (c 2) f '( x) 2 x 3b Use f '(3) 0 to form another equation: 0 2 3 3b 6 3b 0 3b 6 b 2 Substitute in the first equation to obtain the value of c: 3 2 c 9.

3

c 3

Graph (a) is the graph of the derivative of f 4 , because it represents a constant function. This type of function is the derivative of a linear function (in this case with a negative gradient as the horizontal line is below the x-axis). Graph (b) is the graph of the derivative of f 3 , because it has three x-intercepts, which correspond to three stationary points for the original graph. The only graph satisfying this condition is the graph of f 3 .

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Graph (c) is the graph of the derivative of f1 , because it has a x-intercept at the origin. Also, there is a change in the sign of the derivative about x 0 (before x 0 the values of the derivative are negative, while past x 0 the derivative is positive), which means the original graph has a minimum at x 0 . The only graph satisfying this condition is the graph of f1 . Graph (d) is the graph of the derivative of f 2 , because this graph has a x-intercept at the origin and negative derivative values elsewhere, meaning the original function is decreasing for all x except x 0 , where it has a horizontal tangent. The only graph displaying this behaviour is the graph of f1 . 10.

(a)

When x x

0

y

2

y

f

f (0)

1 sin

2

1 , when

1 sin 0

2

1 1 2

The average rate of change is the gradient of the line going through points (0,1) and (b)

2

average rate of change

,2

2 1

1

2

2

The first derivative of f has to be found: f ( x)

1 sin x

f '( x )

cos x

The instantaneous rate of change of f at x

11.

2

2

2

4

is f '

4

cos

4

2 2

(c)

cos x

(a)

(i)

To find the equation of the vertical asymptote, equate the denominator of the rational function to 0 and solve for x. It follows that x 0 is the equation of the vertical asymptote.

(ii)

Evaluate the limit of the function when x

x

arccos

lim f ( x)

x

lim

x

x

3x 2 x

0.88068...

lim

x

3x x

x

lim

x

0.881

2 x

: 3 0

3

As the limit is finite, there is a horizontal asymptote, its equation is y (b)

f ( x)

3x 2 x

3 2x

1

f '( x)

0 2( 1) x

2

2 x2

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3.

(c)

2 x2

f '( x)

0 for all x

increasing for all x (d)

,x

,x

0 , this means that the function is

0

The value of the first derivative at a stationary point is 0. As f '( x) 0 for all x , x 0 , it will never take value 0, so there are no stationary point on the graph of f.

12.

To find the stationary points, the equation y ' 0 must be solved. y ' 4 x 4 x3 0

4 x(1 x 2 ) 0

x 0, x

1

To determine the nature of the stationary points, the second derivative is used: y '' 4 12 x 2

Substitute the x-coordinates of the stationary points into the expression of y '' , to determine the concavity of the function: y ''(0) 4 12 02 y ''(0) 4 0 the function is concave up, so there is a minimum at the point where x 2

y ''(1) 4 12 1

0. 8 0 the function is concave down, so there is a maximum at the point

where x 1 . y ''( 1) 4 12 ( 1) 2

8 0

the function is concave down, so there is a maximum at the

1. point where x To find the y-coordinates, substitute the three x-values into the expression of the original function: x 0 y y 2 02 04 y 0 ; x 1 y 2 12 14 y 1 ; x

1

y 2( 1) 2 ( 1)4

y 1

There are two points of maximum, at ( 1,1) and (1,1) , and a minimum point at (0, 0) . 13.

The gradient of the curve at x 1 is: y

1

1

x2

x3

y'

1 x 2

1 2

1 x 3

2 3

y '(1)

1 1 2 3

5 6

The gradient of the normal line is the negative reciprocal of the gradient of the curve, so the equation of the normal at (1, 2) is: 6 6 6 6 16 ( x 1) y x 2 y x 5 5 5 5 5 Points ( a, 0) and (0, b) satisfy the equation of the normal: y 2

0

6 16 a 5 5

a

16 6

8 ;b 3

6 16 0 5 5

b

16 5

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14.

(a)

The velocity is the rate of change of displacement, so the first derivative of s has to be found: 1 2 t 2

s(t ) 10t

15.

v(t )

s '(t )

v(0)

10 t

10 0

v(0)

(b)

0 10 t

(c)

Substitute t 10 into the expression of s: s(10) 10 10 1 102

(a)

s(t ) 14t 4.9t 2

10 ms-1

t 10 s 2

s ''(t )

9.8 9.8

Solve the equation s '(t ) 0 14 9.8t

9.8t

0 to find the time needed to reach the maximum height:

14

14 9.8

t

The maximum height is: s(1.428...) 14 1.428... 4.9 1.428...2

t

1.428...

t

1.43 s

s(1.428...) 10 m

It takes approximately 1.43 seconds for the ball to reach the maximum height of 10 meters. s '(t ) 0 ,

(c)

which is the same as v(t ) 0 ). The acceleration is 9.8 ms-2, as it is constant at all times. 16.

To find the x-coordinate of the point of inflection, solve the equation y '' y x

3

12 x

2

x 12

y ' 3x

2

24 x 1

0.

y '' 6 x 24

0 6x 24 x 4 The corresponding y-coordinate is: y ( 4)3 12 ( 4)2 ( 4) 12 y 120 The point of inflection is at ( 4,120) 17.

(a)

50 m

s '(t ) 14 9.8t

v(t ) 14 9.8t , a(t )

(b)

s(10)

x

3

y

2 cos

3

3

y

2

1 2

3

y

2.

The point of tangency is y

2 cos x 3 m

2sin

3

, 2 . 3 2( sin x)

y' m

2

3 2

m

y'

2sin x

3

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The equation of the tangent line is:

y ( 2)

(b)

3 x

y

2 3 3 The gradient of the normal is the negative reciprocal of the gradient of the tangent, so the equation of the normal at y ( 2)

18.

(a)

3

3x

1 3

x

y

3

3

, 2 is:

3 x 3

3

2

9

The total surface area of a cylinder, A, is given by the sum of the curved surface area and the area of the two circular bases: A

2 rh

r2

2

54

2 rh

2 r2

Make h the subject: 2 rh 54

2 r2

h

2 (27 r 2 ) 2 r

h

27 r 2 r

The expression of the volume of the cylinder is: V

(b)

r2h

27 r 2 r

r (27 r 2 )

V

To find the value of the radius for which the maximum volume is obtained, solve for r the equation V ' 0 : V

27 r

0

27

But r 0 19.

r2

V

r3 3 r2

V' 27

27

3 r2

3 r2

r2

9

r

3

r 3

Point (0,10) is on the curve

10 a(0) 2 b(0) c

c 10

2 The equation of the function becomes y ax bx 10

Point (2,18) is also on the curve: 18 a 22 b 2 10 The curve has a maximum at (2,18) , so y '(2) 0 : y ax 2 bx 10

y ' 2ax b

0 2a 2 b

4a

2b

8

2a b

4a b 0

Solve the simultaneous equations to obtain the values of a and b: b 4 2a 4a 4 2a 0 2a 4 a 2 b 4 2( 2) b 8

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4

20.

(a)

1 2 1 x 5x 3 y ( 2) 2 2 2 tangency is ( 2,15). y' x 5 m 2 5 m 7 x

2

y

5( 2) 3

y

15

the point of

The equation of the tangent line is: y 15

(b)

7( x ( 2))

(a)

7 x 14 15

y

7x 1

The gradient of the normal is the negative reciprocal of the gradient of the tangent, so the equation of the normal at ( 2,15) is: 1 ( x ( 2)) 7

y 15

21.

y

y

1 x 7

2 15 7

y

1 107 x 7 7

Solve the equation y ' 0 to find the x-coordinates of the stationary points: 3 y ' 4 x3 3x 2 y ' x 2 (4 x 3) 0 x 0, x 4 To determine the nature of the two stationary points, the second derivative is used: y '' 12 x 2 6 x Substitute the x-coordinates of the stationary points into the expression of y '' , to determine the concavity of the function: y ''(0) 12 02 6 0 0 this is not a maximum nor minimum point (there is an inflection point at x 0 , and the tangent at this point is horizontal) 3 y '' 4

3 12 4

2

6

3 4

27 18 4 4

9 4

0

the function is concave up, so there

3 . 4 To find the y-coordinates, substitute the two x-values into the expression of the original function: x 0 y 04 03 0

is a minimum at x

4

3

3 3 3 81 27 27 x y y y 4 4 4 256 64 256 There is a point of minimum (relative, and absolute, as it is the only one) at M

3 27 , 4 256

and a stationary point of inflection at (0, 0) .

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(b)

The domain is x . For the range, the end behavior of the function has to be investigated: lim ( x 4 x3 ) lim x3 ( x 1) x

x

lim( x x

4

3

x ) lim x3 ( x 1) x

As the values of the function become infinitely large when x 3 27 , , it follows that the range is y 4 256

an absolute minimum at (c)

x

1 2

y

6 x (2 x 1)

1 2

4

1 2

0

x

y

1 16

3

1 2

0, x

There are two points of inflection: I1 (0, 0) and I 2 (d)

The graph is shown below:

(a)

2 3x 5 x 2 lim x 8 3x 2

(b)

27 . 256

To find the x-coordinate of the point of inflection, solve the equation y '' 0 . 0 12 x 2 6 x

22.

, and there is

x 4 2 lim x 0 x

2 x2

x2 lim x

x2

8 x2

3 5 x

x

0

x 4 4 x( x 4 2)

8 x2

x

3

x 4 2 lim x 0 x

lim

2 2 lim x

x 4 2 x 4 2

lim x

0

1 1 , . 2 16

3 5 x 3

0 0 5 0 3

5 3

( x 4) 2 22 lim x 0 x( x 4 2)

1 x 4 2

1 0 4 2

1 4

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(c)

lim

(d)

lim

x3 1 ( x 1)( x 2 x 1) lim x 1 x 1 x 1

x 1

h

( x h) 2 h

0

lim h

0

lim h

23.

0

f ( x)

(a)

( x h) 2 h

lim h

h( ( x h) 2 h h( x h 2

0

x

3 2

4x

1 2

( x h) 2

x 2

( x h) 2

x 2

( x h) 2 ( x 2) 0 h( ( x h) 2 x 2)

h

1 x h 2

lim h

0

3 12 x 2

f '( x)

x 2

lim

x 2) x 2)

x 1) 3

x 1

( ( x h) 2) 2 ( x 2) 2

f '( x )

24.

x 2

lim( x 2

1 4 x 2

1 2

1 2 x 2

x 2

f '( x)

3 x 2

2 x

3x 4 2 x

(b)

f ( x)

x 3 3sin x

(c)

f ( x)

1 x

(d)

f ( x)

7 3 x13

x 2

f '( x ) 3 x 2 3cos x

f ( x) f ( x)

1

x 7 x 3

1 x 2

13

f '( x )

( 1) x

7 ( 13) x 3

f '( x )

2

14

1 2

f '( x) f '( x )

1 x2 91 3 x14

1 2

The point ( p, q) is on the graph of the given function, it follows that its coordinates satisfy the equation of the curve: q p 3 p 2 9 p 9 the tangency point has coordinates ( p, p 3

p 2 9 p 9) .

The expression of the first derivative is: y x3 x 2 9 x 9 y ' 3x 2 2 x 9 The gradient of the graph at the point where x y '( p)

3p

2

2p 9

The equation of the tangent at point ( p, p 3 y ( p3

p 2 9 p 9) 2

y

(3 p

y

(3 p 2

p is:

2 p 9) x

(3 p 2 p (3 p

2 p 9) x 2 p 3

2 p 9)( x 2

p 2 9 p 9) is: p)

2 p 9) ( p 3

p 2 9 p 9)

p2 9

The given point (4, 1) lies on the tangent, so its coordinates must satisfy the equation of this line: 1 (3 p 2

2 p 9)4 2 p 3

p2 9

2 p 3 11 p 2 8 p 44 0

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The equation 2 p 3 11 p 2 8 p 44 0 has three solutions (GDC): x

11 ,x 2

2, x

2

The three points of tangency are: 11 p 2

q

p

b

2

p

25.

2

11 2

2

9

11 2

9

( 2)3 ( 2)2 9( 2) 9 3

2

2

2

9 2 9

b

15

b b 5

1105 8

11 1105 , 2 8

( 2, 5)

(2, 15)

1 , so the gradient of the tangent is 12. 12 The first derivative of the function is: 1 y x3 y ' 3x2 3 At the point of tangency, the gradient of the function and the gradient of the tangent are equal: 3x 2 12 x 2 4 x 2 The corresponding y-coordinates are: 1 1 23 x 2 y ( 2)3 y 8 y 3 3 3 1 1 25 x 2 y 23 y 8 y 3 3 3 The value of c can be found by substituting the coordinates of the tangency points into the equation of the normal line: 23 23 1 1 23 47 2, ( 2) c c c 0 3 3 12 6 3 6

The gradient of the normal is

2,

25 3

As c 0 26.

b

3

11 2

25 3 c

1 2 c 12

c

1 6

25 3

c

51 6

c

17 2

0

17 2

It is known that two parallel lines have the same gradient, this means that, at the point ( a, b) on the graph of the function, the gradient of the curve (which is the same as the gradient of the tangent line), must be the same as the gradient of the given line, namely 3. The gradient of the curve is given by the first derivative: 1 3 1 y x x y' 3x 2 1 y ' x 2 1 3 3

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The gradient of the curve at x a is 3: 3 a2 1 a2 4 a 2 The y-coordinates of the required points are: 1 3 8 2 b 2 2 b 2 b 3 3 3 1 8 2 b ( 2)3 ( 2) b 2 b 3 3 3 The points where the tangent lines are parallel to line y 27.

The gradient of the curve at x 1 is: y x x2 y ' 1 2x y '(1) 1 2 1

y '(1)

3 x are 2,

2 and 3

2,

2 . 3

1

The gradient of the normal line is the negative reciprocal of the gradient of the curve, so the equation of the normal at (1,0) is: 1 ( x 1) y x 1 1 To find where this line intersects the curve again, equate the expressions of the curve and the normal and solve for x: x 1 x x2 x2 1 x 1 The required point is where x 1 , the corresponding y-coordinate is: y 0

y

( 1) ( 1) 2

y

2

The coordinates of the point of intersection are ( 1, 2) . 28.

f '( x) lim h

0

f ( x h) h

lim h

0

lim h

0

( x h) 2 h

lim h

0

( x h) 2 h

x 2

h( ( x h) 2

x 2

( x h) 2

x 2

x 2) x 2)

x 2

( x h) 2

( ( x h) 2) 2 ( x 2) 2

h 0 h( x h 2

lim h

f ( x)

( x h) 2 ( x 2) 0 h( ( x h) 2 x 2)

lim h

1 x 0 2

x 2

1 2 x 2

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29.

v(t )

s '(t ), a(t )

s (t ) t

3

v(t )

(a)

9t 3t

2

24t

2

0 3t

s ''(t )

s '(t ) 3t 2 18t 24

18t 24 , a(t ) 2

18t 24

s ''(t )

6t 18

3(t 2)(t 4)

0

The displacement of the object is: t 2 s (2) 23 9 2 2 24 2 s (2) t

4

s (4)

4

3

9 4

6t 18

2

24 4

t

2, t

4

20

s (4) 16

The object has velocity 0 when is at (2, 20) or (4,16) (b)

0 6t 18 t 3 The displacement of the object is: t 3 s (3) 33 9 32 24 3 s (3) 18 The object has acceleration 0 when is at (3,18)

30.

(a),(b) In order for the particle to change direction, it must first come to rest, meaning its velocity should be 0. To determine when this happens, find the expression of velocity, v, by differentiating the expression of displacement, then set v equal to 0 and solve for t: s(t ) t sin t s '(t ) 1 cos t v(t ) 1 cos t

(c) (d)

0 1 cos t cos t 1 t The range for the velocity has to be considered: 0 t 2 1 cos t 1 0 1 cos t 2 0 v(t ) 2 This means that the velocity is always positive or 0, so the particle never changes its direction of movement, it merely stops at t , then it continues in the same direction. This also means that the particle is always on the same side of the origin. a(t ) s ''(t ) a(t ) sin t 0 sin t sin t 0 t 0, t ,t 2 The graph of the displacement is shown below:

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The maximum displacement is reached when t s(2 ) 2 sin 2 2 meters 31.

2 seconds and it is:

ax3 bx 2

If the point ( 1, 4) and (3, 7) are on the graph representing y then their coordinates satisfy the equation of the curve: ( 1, 4) 4 a( 1)3 b( 1) 2 c( 1) d a b c d (3, 7)

7

a 33 b 32 c 3 d

27 a 9b 3c d

cx d ,

4

(1)

7

(2)

To use the other given information, y ' and y '' must be found: y ' 3ax 2

2bx c, y '' 6ax 2b

The graph has a turning point where x

2 , so y '(2) 0 :

0 3a 22 2b 2 c 12a 4b c 0 The graph has an inflection point at ( 1, 4) , so y ''( 1)

0

6a ( 1) 2b

6a 2b

0

3a b

(3) 0:

(4)

0

We have a system of four equations, which you can use a GDC, or any other method of choice to solve. 1 3 5 a ;b ;c 6; d 4 4 2 1 3 3 2 5 The expression of the function is: y x x 6x 4 4 2 The y-coordinate of the turning point is: 1 3 3 2 5 5 19 y (2) 2 2 6 2 y (2) 2 3 12 y (2) 4 4 2 2 2 32.

Find the expression of y ' : y 1 9x

2

18 x

4

y'

9( 2) x

3

18( 4) x

5

y ' 18 x

3

72 x

5

y'

18 x3

To find the stationary points, the equation y ' 0 must be solved. 18x 2 72 18 x 2 72 0 x 2 4 x 2 5 x To determine the nature of the stationary points, the second derivative is used: 54 360 y ' 18 x 3 72 x 5 y '' 18( 3) x 4 72( 5) x 6 y '' x4 x6 0

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72 x5

Substitute the x-coordinates of the stationary points into the expression of y '' , to determine the concavity of the function: 54 360 9 the function is concave up, so there is a minimum y ''(2) y ''(2) 0 4 6 2 2 4 at the point where x 2 . 54 360 9 y ''( 2) y ''(2) 0 the function is concave up, so there is a 4 6 ( 2) ( 2) 4 minimum at the point where x 2. To find the y-coordinates, substitute the two x-values into the expression of the original function: 9 18 9 9 1 x 2 y 1 2 y 1 y 4 2 2 4 8 8 9 18 1 x 2 y 1 y 2 4 ( 2) ( 2) 8 There are two points of minimum, at 33.

(a)

2,

1 1 and 2, . 8 8

Find the expression of the first derivative: 1 y x1 y ( 1) x 2 y' x2 The gradient of the tangent is found by substituting x 1 into the expression of y ' : 1 m 1 12 The equation of the tangent is: y 1 1( x 1) y x 2 m

(b)

y

cos x

y'

m 1 2 The equation of the tangent is:

(c)

m

sin

y 0

1( x

sin x

) y x 2 2 1 1 y and y cos x are decreasing (their 0 x 0, sin x 0 2 x 2 x derivatives are always negative on the given interval). This means that the two tangent lines do not intersect again the graphs they are tangent to.

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A diagram helps clarify the argument.

1 2 is y '' , so the function is 0 on 0 x x 2 x3 concave up. It follows that the y-coordinates of the points lying on the tangent to 1 1 y are always less than the y-coordinates of the points lying on y . x x 1 x 2 x

The second derivative of y

Similarly, the second derivative of y cos x is y ''

, so the 2 function is concave down, so, for the same x-coordinate, the y-coordinates of the points lying on the tangent to y cos x are always greater than the y-coordinates cos x

0 on 0

x

of the points lying on y cos x . x

cos x cos x x 2 2 The tangent lines are parallel as they have the same gradient. The y-coordinates of the points on y x 2 are always greater than the y-coordinates of the points on

for the same x-coordinate, as 2 . 2 2 Combine the three inequalities to obtain the required inequality: 1 1 for all 0 x cos x x x 2 cos x 2 2 x x y

x

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34.

Let ( a, b) be the point of tangency. b

a3

the tangency point has coordinates (a, a 2 a 2)

a 2

The expression of the first derivative is: y x3 x 2 y ' 3x 2 1 y '(a )

3a 2 1

The equation of the tangent at point ( a, a 3 a 2) is: y ( a 3 a 2) y

(3a 2 1)( x a )

(3a 2 1) x a 2

y

a 2 2a 2

a

(3a 2 1) x a 3 a 2 a (3a 2 1) (3a 2 1) x 2a 3

y

2

The tangent should pass through the origin, (0, 0) : (3a 2 1) 0 2a 3 2

0

2a 3 2 0

a3 1

a 1

There is only one possible value for a, so there is only one tangent to y passing through the origin: y The point of tangency is: a 1 b 13 1 2 b 35.

(3a 2 1) x 2a 3 2

2

x 2

2x

(1, 2)

(a)

v(t )

(b)

The maximum displacement occurs when s '(t )

s '(t )

x3

v(t ) 50 20t

0:

5 s 2 The maximum displacement is: 0 50 20t

5 s 2

36.

t

5 50 10 2

5 2

2

1000

s

5 2

Let x1 and x 2 be the x-intercepts of the graph of y x

x3 and a minimum at x

1062.5 m f '( x) . The graph has a maximum at

x4 . This information is shown in the diagram below:

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As f '( x1 )

f '( x2 )

0 , it follows that the points where x

points for the graph of y

x1 and x

x2 are stationary

f ( x) .

To decide their nature, the sign of the first derivative about each point needs to be investigated. The values of f ' to the left of x x1 are positive ( f is increasing), after this point they become negative ( f is decreasing), this means that f has a maximum when x

x1 .

Similarly, the values of f ' to the left of x

x2 are negative ( f is decreasing), past this

point they are negative ( f is increasing), this means that f has a minimum when x At the points where x

x3 and x

means: ( f ') '( x3 ) ( f ') '( x4 ) 0

x4 , the gradient of the graph of y f ''( x3 )

points of inflection on the graph of y The graph of y

f ''( x4 )

x2 .

f '( x ) is 0, this

0 . It follows that there are two

f ( x) .

f ( x) is shown below:

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Exercise 13.1 1.

(a) y

3x 8

4

Let f (u ) u 4 , g ( x) 3 x 8, then f '(u )

4u 3

Using the chain rule

dy dx

4(3 x 8) 3 3 12(3 x 8)3

f '( g ( x )) g '( x)

(b) y

1 x 1

Let f (u )

u =u 2 , g ( x) 1 x,

then f '(u )

1 u 2

1 2

Using the chain rule

dy dx

1 1 x 2

f '( g ( x)) g '( x )

1 2

( 1)

1 2 1 x

(c) y sin x cos x Let f ( x ) sin x, g ( x) cos x Using the product rule dy dx

d f ( x ) g ( x) f ( x) g '( x) g ( x) f '( x) dx sin x ( sin x) cos x cos x sin 2 x cos 2 x cos 2 x sin 2 x

(d) y

2sin

x 2

Let f (u ) 2sin u , g ( x)

x , 2

then f '(u ) 2cos u Using the chain rule

dy dx

f '( g ( x)) g '( x) 2cos

x 2

1 2

cos

x 2

(e) f ( x)

x2

4

2

Let f (u ) u 2 , g ( x) then f '(u )

2x

x2

4,

3

Using the chain rule dy dx

2 x2

f '( g ( x )) g '( x) 4 x x2

4

4

3

2x

4x

3

x

2

4

3

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(f) x 1 x 1 Let f ( x) x 1, g ( x ) f ( x)

x 1,

Applying the quotient rule: x 1 1

x 1 1

x 1

dy dx

d f ( x) dx g ( x )

x 1 x 1

2

x 1

g ( x ) f '( x)

f ( x ) g '( x )

g ( x)

2

2

2

x 1

2

(g) y

1 x 2

x 2

1 2

1

Let f (u ) u 2 , g ( x ) 1 u 2

then f '(u )

x 2,

3 2

Applying the chain rule

dy dx

f '( g ( x )) g '( x )

1 x 2 2

3 2

1

1 2

x 2

3

(h) y

cos 2 x

Let f (u ) u 2 , g ( x ) cos x, then f '(u ) 2u dy Applying the chain rule dx

f '( g ( x)) g '( x) 2cos x

sin x

2sin x cos x (

sin 2 x )

(i) y

x 1 x

Let f ( x )

x, g ( x )

1 x

1 x

d f ( x ) g ( x) dx d applying also the chain rule for ( 1 x) dx 1 1 1 x 1 x 2 1 1 x 2 1 2 applying the power rule

Applying the product rule

dy dx

1 2

f ( x ) g '( x) g ( x) f '( x )

x 1 x 2 1 x finding common denominator x 21 x 2 1 x

2 3x 2 1 x

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1 1 3x 2 5x 7 3x 5 x 7 Let f (u) = u 1 , g ( x) 3x2 5 x 7,

(j) y

2

then f '(u) u2 Applying the chain rule: dy f '( g ( x)) g '( x) 3x 2 dx

3

(k) y

2x 5

2x 5

5x 7

2

6x 5

6x 5

3x

2

5x 7

2

1 3

1

Let f (u ) = u 3 , g ( x)

2 x 5,

2 3

1 u 3 Applying the chain rule: dy 1 f '( g ( x)) g '( x) 2x 5 dx 3 then f '(u )

(l)

y

2x 1

3

2

2

33 2x 5

2

x4 1 3

Let f ( x)

2 3

2 x 1 , g ( x)

x4 1

Applying the product rule: dy d 3 f ( x) g ( x) f ( x) g '( x) g ( x) f '( x) 2 x 1 4 x3 dx dx 2 Taking out the common factor of 2x 1 and simplifying:

2x 1

2

2 x 1 4 x3

6 x4 1

2x 1

sin x x Let f ( x) sin x, g ( x) x Applying the quotient rule: dy d f ( x) g ( x) f '( x) f ( x) g '( x) 2 dx dx g ( x) g ( x)

2

14 x 4

4 x3

6

x4 1 3 2 x 1

2 2x 1

2

7 x4

2

2 x3 3

(m) f ( x)

x cos x sin x x2

(n) f ( x)

x2

x 2 Let f ( x) x 2 , g ( x ) x 2 Applying the quotient rule: dy d f ( x) g ( x) f '( x ) f ( x ) g '( x ) 2 dx dx g ( x) g ( x)

x 2 2x x2 1 x 2

2

x2

4x

x 2

2

2

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(o) 2

y

3

x 2 cos x

x 3 cos x

2 3

Let f ( x) x , g ( x) cos x Applying the product rule: dy d f ( x ) g ( x) f ( x) g '( x) g ( x) f '( x) dx dx Applying the power rule: 2cos x 2cos x 3x sin x 3 2 x sin x 33 x 33 x 2. (a) y

3

2x2 1 , x

2

x3

sin x

cos x

2 x 3

1 3

1

Let f (u) = u3 , g ( x) 2 x 2 1, then f '(u) 3u 2 Applying the chain rule to the function: 2 dy f '( g ( x)) g '( x) 3 2 x2 1 4 x 12 x 2 x2 1 dx at x 1: 2 dy 12 ( 1) 2 ( 1)2 1 12 dx Hence the slope of the tangent is 12.

2

Finding y-coordinate of the tangency point: at x

1, y

2

1

2

1

3

1

So, the tangency point is: 1, 1 Using the point-slope form for a linear equation gives: 1 is the tangent to the graph at x (b) y

3x2

2, x 3 1

Let f (u )= u = u 2 , g ( x) 3x 2

2,

1 2

1 u 2 Applying the chain rule to the function: 1 dy 1 f '( g ( x)) g '( x) 3x 2 2 2 6 x 3x 3x 2 dx 2 at x 3 : 1 dy 9 3 3 3 32 2 2 dx 5 9 Hence the slope of the tangent is . 5 then f '(u )

2

1 2

Finding y-coordinate of the tangency point: at x 3 ,

y

y

3 32

2

5

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So, the tangency point is: 3,5 . Using the point-slope form for a linear equation gives: 9 9 2 is the tangent to the graph at x 3 . y 5 x 3 y x 5 5 5 (c) y sin 2 x, x Let f (u )= sin u, g ( x)

2 x,

then f '(u ) cos u Applying the chain rule to the function: dy f '( g ( x)) g '( x) 2cos 2 x dx at x : dy 2cos 2 2 dx Hence the slope of the tangent is 2.

Finding y-coordinate of the tangency point: at x , y sin 2 0 So, the tangency point is: , 0 Using the point-slope form for a linear equation gives: y 0 2 x y 2 x 2 is the tangent to the graph at x (d) x3 1 , x 1 2x Let f ( x ) x 3 +1, g ( x ) 2 x Applying the quotient rule: f ( x)

dy dx

d f ( x) dx g ( x)

at x 1 : dy 2 13 1 dx 2 12

g ( x) f '( x)

f ( x) g '( x)

g ( x)

2

2 x 3x2

x3 1 2 4 x2

2 x3 1 2x2

1 2

Hence the slope of the tangent is

1 . 2

Finding y-coordinate of the tangency point: 13 1 at x 1 , y 1 2 1 So, the tangency point is: 1, 1 Using the point-slope form for a linear equation gives: 1 1 1 is the tangent to the graph at x 1 y 1 x 1 y x 2 2 2

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3. (a) s(t ) cos t 2 1 d cos t 2 1 dt Let s(u)=cos u, g (t ) t 2 1,

v(t )

s '(t )

then s '(u) sin u Applying the chain rule: ds s '( g (t )) g '(t ) sin t 2 1 2t dt

2t sin t 2 1

(b) at t 0 v(0) 2 0 sin( 1) 0 (c) 0 t 2.5 Substituting 0 as v (t ) : v(t ) 0

2t sin t 2 1 sin t 2 1

0

0 or t

0 (outside the interval)

t 2 1 0 or t 2 1 1 (I choose positive solutions from the interval 0 t 2.5 ) t 1 or t

(d)

2t sin t 2 1

v t a t

v' t

(the blue graph)

2sin t 2 1

4t 2 cos t 2 1 (the red graph)

Accelerating to the right (both positive) then slowing down (positive velocity and negative acceleration), turning around, accelerating to the left (both negative), slowing down (negative velocity, positive acceleration), turning around again, then accelerating to the right (both positive), slowing down. 4. (a) y

2 2

2 x2 8

x 8 Let f (u )=2u 1 , g ( x) then f '(u )

2u

1

, 3, 2

x 2 8,

2

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Applying the chain rule: dy f '( g ( x)) g '( x) 2 x2 8 dx at x 3 : dy 4 3 2 2 dx 3 8

2

4x

2x

x

2

8

2

12

Hence the slope of the tangent is 12 ang the slope of the normal is

1 . 12

Using the point-slope form for a linear equation gives: 12 x 3 y 12 x 38 (i) tangent: y 2 (ii) normal: y 2 (b) y

1 4x ,

2,

1 x 3 12

1 x 12

y

7 4

2 3

1

Let f (u )= u = u 2 , g ( x) 1 4 x, 1 12 u 2 Applying the chain rule: dy 1 f '( g ( x)) g '( x) 1 4x dx 2 then f '(u )

at x 2 : dy 2 dx 1 4 2

1 2

4

2 1 4x

2 3

2 Hence the slope of the tangent is and the slope of the normal is 3 Using the point-slope form for a linear equation gives: 2 2 2 2 y x 2 y x (i) tangent: 3 3 3 3 2 3 3 11 (ii) normal: y x 2 y x 3 2 2 3

(c)

1 x 1 2 Let f ( x) x, g ( x) x 1 Applying the quotient rule: dy d f ( x) g ( x) f '( x) f ( x ) g '( x) 2 dx dx g ( x) g ( x) f ( x)

x

at x 1 : dy 1 dx 1 1

2

at

3 . 2

1,

x 1 1 x 1 x 1

2

1 x 1

2

1 4

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1 Hence the slope of the tangent is and the slope of the normal is 4 . 4 Using the point-slope form for a linear equation gives: 1 1 1 1 y x 1 y x (i) tangent: 2 4 4 4 1 9 (ii) normal: y 4 x 1 y 4x 2 2

5. y

sin 2 x

2

Let f (u )=sin u, g ( x) 2 x

2

,

then f '(u ) cos u Applying the chain rule: dy f '( g ( x)) g '( x ) 2cos 2 x dx d2 y dx 2

d dy dx dx

d 2cos 2 x dx

Let f (u ) = 2cos u, g ( x ) 2 x

2sin 2 x

2

2

2

,

then f '(u ) 2sin u Applying the chain rule: d2 y f '( g ( x)) g '( x) 2sin 2 x 2 dx 2 2 There is an inflection point at the point x0 2

2

d y dx 2 0

x 4sin 2 x

x

2 4

2

at x0 :

0

0 or 2 x

or x

2

3 4

4sin 2 x

2

2

d y dx 2 x 4

4cos 2 x

2

d2 y dx 2

at x =

2

d y d y is undefined and 2 changes sign at x0 2 dx dx 3 2x 2 2 2

0 or

2x

4sin 2 x

y = sin 2

4

2

0 at x 4

or x

or x

3 d2 y and 2 changes sign at 4 dx

4 3 (from the graph) 4

0

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at x =

3 4

y = sin 2

Inflection points: 6. y

4

3 4

0

2

3 ,0 4

,0 ,

2

x x 4 (a) (i) x-intercepts where y = 0 x x 4

2

0

x 0 or x 4 2

(ii) Let f ( x) x, g ( x) x 4 Applying the product rule: dy d f ( x) g ( x ) f ( x) g '( x ) g ( x ) f '( x ) dx dx

x 2 x 4

x 4

2

1

x 4 3x 4

dy 0 dx dy 4 0 x 4 3x 4 0 x 4 or x = dx 3 Using first derivative test to determine the nature of the stationary points:

Stationary points occur at the points where

dy dx

x 4 3x 4

First derivative changes sign from relative maximum at x

x

4 3

there is a

4 3

2

4 4 256 4 y 4 3 3 3 27 First derivative changes the sign from x 4. At x

at x 4

there is no relative maximum at

4 256 , 3 27 (ii) There is an inflection point at the point x0 at x0 :

The coordinates of the maximum point:

d2 y dx 2 dy dx d2 y dx2 d2 y dx 2

0 or is undefined and x 4 3x 4

d2 y changes the sign at x0 dx 2

3x2 16 x 16

6 x 16 2 3 x 8 2 3x 8

0

x

8 3

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The second derivative changes the sing at x inflection at x

8 (as a linear function) 3

there is a point of

8 3 2

8 8 128 8 At x : y 4 3 3 3 27 8 128 - the point of inflection , 3 27 (b) x 0 (double root) x 4 (double root; the graph reflects in the x-axis) 4 256 - relative maximum , 3 27 8 128 - the point of inflection , 3 27 Table: 0 4 8 4 4 8 8 0, , ,4 3 3 3 3 3 3

4

f (x)

+

+

0

-

-

-

0

f (x)

-

-

-

-

0

+

+

f(x)

0

7. f ( x)

x2

0

128 27

256 27

3x 4 x 1

(a) Let g ( x)

2

x2

3x + 4, h( x)

x 1

2

Applying the quotient rule: f '( x)

d g ( x) dx h( x)

h( x) g '( x) g ( x) h '( x) h( x )

2

x 1

2

2x 3

x2 x 1

3x 4

x 1

4

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x 1 5x 11 x 1

5x 11

4

x 1

3

(b) Let g ( x) 5x 11, h( x)

x 1

3

Applying the quotient rule: f ''( x) x 1

d g ( x) dx h ( x ) 2

h( x )

10 x 38 x 1

x 1

h( x ) g '( x) g ( x) h '( x)

6

3

5

5 x 11 3 x 1

2

x 1

2

6

10 x 38 x 1

4

38 38 0 4.83 (2) f '' changes the sign at x = 3.8 (denominator is always positive, linear function in the numerator changes the sign at x = 3.8)

(c) (1) f ''(3.8)

1) there is a point of inflection at x = 3.8 2) x a 8. f ( x) x a 1st derivative: Let g ( x) x a, h( x) x a Applying the quotient rule: x a 1 x a 1 d g ( x) h( x) g '( x ) g ( x )h '( x ) f '( x) 2 2 dx h( x) x a h( x ) 2nd derivative: 2 f '( x) 2a x a Applying the chain rule: d 2 3 f ''( x) 2a x a 4a x a dx 4a f ''( x) 3 x a 1 1 9. y x 1 1 x Derivatives: dy 2 2 1st: 1 x 1 dx d 2 y d dy 2 1 ( 1) 2 x 1 2nd: 2 dx dx dx 3rd:

d3 y dx3

d d2 y dx dx2

1

3

3

( 1) 2! 3 x 1

1 4

3

2! x 2

1

4

2a x a

2

3

3! x 2

4

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4th:

d4 y dx4

d d3 y dx dx3

1

4

( 1) 3! 4 x 1

4

1

5

4! x 2

5

Recognizing that the nth derivative of the function is formed by: multiplying the previous by n decreasing the power of x 1 by 1 we get the formula: dn y nth derivative: dx n

10. y

d dn 1 y dx dx n 1

1

n

( 1) ( n 1)! n x 1

n 1

1

n 1

x 1

n! n 1

8 4 x2

(a) Relative extrema occur at the points where

y 8 4 x2

dy dx

0 or is undefined and

dy changes sign dx

1

Let f (u ) 8u 1 , g ( x) 4 x 2 , then f '(u ) 8u 2 Applying the chain rule: dy f '( g ( x)) g '( x) 8 4 x2 dx

dy dx

0

16 x 4 x2

2

0

2

16 x 0

2x

16 x 4 x2

2

x 0

dy 0 16 x 0 x 0 dx 2 2 4 x 0 when x dy 0 16 x 0 x 0 dx At x = 0, the function changes its monotonicity from increasing to decreasing there is a relative maximum at x = 0 8 there is a relative maximum at 0, 2 f (0) 2 4 0 d2 y d2 y The inflection points occur at the points where 2 0 and 2 changes sign dx dx 2 dy 16 x 4 x 2 dx

Let f ( x)

16 x, g ( x)

4 x2

2

Applying the product rule: d 2 y d dy d f ( x) g ( x) f ( x) g '( x) g ( x) f '( x) 2 dx dx dx dx 2 dy Applying also the chain rule to 4 x2 dx © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

3

2 4 x2

16 x

d2 y dx 2

4 x2

48 x 2

0

3

4 x

4 x2

2x

64 2 3

0

2

48 x 2

16

64 0

the function changes its concavity at x

f

f

2 3 3

2 3 3

4

x2

4 3

2 2

4

x

4 x

x

2 3 or x 3

64 2 3

2 3 3

2 3 and at x 3

2 3 3

there are points of

2 3 2 3 and x 3 3 8 3 2 2 2 3 3 8

4

4 x

48 x 2

16

2 3

d2 y 2 3 2 3 0 48 x 2 64 0 x or x 2 dx 3 3 2 d y 2 3 2 3 0 48 x 2 64 0 x 2 dx 3 3

0 when x

inflection at x

64 x 2

2 3 3

Points of inflection:

2

3 2

2 3 3 2 3 3 , ; , 3 2 3 2

(b)

Since g(x) is the quotient of a positive constant and a positive expression, it will never be negative or zero. It is positive for all real numbers.

(c)

Since the numerator is constant, then as x

(d)

Below is a sketch of the graph of g.

11.

, 4 x2

8

lim

x

8 4 x2

0

d d cf ( x) c f ( x) dx dx Let g( x) c, then g '( x) 0 Applying the product rule: © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

dy d g ( x) f ( x) g ( x) f '( x) dx dx what was to be proved.

f ( x) g '( x) cf '( x)

12. y x 4 6 x 2 Consecutive derivatives: dy 4 x3 12 x 4 x x2 3 4 x x 3 x dx d 2 y d dy 12 x 2 12 12 x 1 x 1 dx 2 dx dx d3 y dx3

d d2 y dx dx 2

6

2) x

x

dy dx

0 or 0

0

x2 x

0

x

4x x

d2 y dx 2

0

1 x 1

4)

d3 y dx3 d3 y dx 3

6 x

3 x

3

in the interval 0 x 1

12 x 1 x 1

24 x 0

0

x 0

y 0

0 dy dx

0

0

in the interval 0 x 1

0

6

in the interval 0 x 1

6

3 or 0 < x < 3

3)

3

24 x

x4 6x2

1) y 0

f ( x) 0 cf '( x) RHS

d2 y dx 2

0

in the interval 0 x 1

0

Exercise 13.2 1. (a)

y

x 2e x

Let f ( x) x 2 , g ( x ) e x Applying the product rule: dy d f ( x) g ( x) f ( x) g '( x) g ( x) f '( x) dx dx (b)

x 2e x

e x 2x

x 2e x

2 xe x

y 8x Applying the formula of the derivative of an exponential function: dy 8x ln8 dx

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(c)

y

tan e x

Let f (u )= tan u, g ( x) e x ,

(d)

then f '(u) sec2 u Applying the chain rule: dy f '( g ( x)) g '( x) sec2 e x e x dx x y 1 cos x Let f ( x) x, g ( x) 1 cos x Applying the quotient rule:

dy dx

d f ( x) dx g ( x)

g ( x) f '( x)

1 cos x (e)

(f)

f ( x) g '( x)

g ( x)

1 cos x 1 x

sin x

e x sec2 e x

2

cos x x sin x 1

2

1 cos x

ex x Let f ( x) e x , g ( x) x Applying the quotient rule: dy d f ( x) g ( x) f '( x ) f ( x ) g '( x ) 2 dx dx g ( x) g ( x)

2

y

y

xe x

ex 1 x2

xe x e x x2

1 3 sec 2 x sec 2 x 3

1 Let f (u )= x 3 , g ( x) sec 2 x, 3 (1) then f '(u ) x 2 , g' x 2sec 2 x tan 2 x

Applying the chain rule: d 1 3 sec 2 x dx 3

(2)

f '( g ( x)) g '( x) sec 2 2 x 2sec 2 x tan 2 x

Let f (u )=sec u , g ( x )

2sec3 2 x tan 2 x

2x,

then f '(u ) sec u tan u Applying the chain rule: d sec 2 x f '( g ( x)) g '( x) dx

2sec 2 x tan 2 x

1, 2

dy dx

2sec3 2 x tan 2 x 2sec2 x tan 2 x 2sec 2 x tan 2 x sec 2 2 x 1

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dy dx

1 1 cos 2 2 x

2sec 2 x tan 2 x

2sec 2 x tan 2 x

sin 2 2 x cos 2 2 x

2sec 2 x tan 2 x

1 cos 2 2 x cos2 2 x

2 tan 3 2 x sec 2 x

x

(g)

(h)

(i)

1 y 4 4 Applying the formula of the derivative of an exponential function: x dy 1 1 ln dx 4 4 sin x y cos x tan x cos x sin x cos x Applying the formula of the derivative of a sine function: dy cos x dx x y ex 1 x

Let f ( x ) x, g ( x) e x 1 Applying the quotient rule:

(j)

dy dx

d f ( x) dx g ( x)

y

4cos sin 3 x

g ( x) f '( x)

f ( x) g '( x)

g ( x)

2

ex 1 1 x ex ex 1

xe x

2

Let f (u )=4cos u, g ( x) sin 3 x, then f '(u ) 4sin u g '( x) 3cos3 x (applying the chain rule) Applying the chain rule: dy f '( g ( x)) g '( x) 4sin sin 3x 3cos3x 12sin 3x cos3x dx

(k)

(l)

ex 1

ex 1

2

6sin 6 x

y 2x 1 2 2x Applying the formula of the derivative of an exponential function: dy 2 2 x ln 2 2ln 2 2 x dx

y

1 csc x sec x

csc x sec x

1

Let f (u )=u 1 , g ( x) csc x sec x, then f '(u) u 2, g ' x cot x csc x tan x sec x Applying the chain rule: dy 2 f '( g ( x)) g '( x) csc x sec x cot x csc x tan x sec x dx cot x csc x tan x sec x csc x sec x

2

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cos x 1 sin x 1 sin x sin x cos x cos x 2 1 1 cos x sin x

cos3 x sin 3 x sin 2 x cos 2 x

sin 2 x cos 2 x sin x cos x

2

cos3 x sin 3 x sin x cos x 2.

(a)

y sin x, x dy dx

at x dy dx

2

3

cos x

3

:

cos

3

1 2

Hence the slope of the tangent is

1 . 2

Finding the y-coordinate of the tangency point: 3 at x 1 y sin 3 2 3 , So, the tangency point is: . 3 2 Using the point-slope form for a linear equation gives: 3 1 1 3 3 y x y x is the tangent to the graph at x 2 2 3 2 6 (b)

y

x ex , x

1.

0

dy 1 ex dx at x 0 : dy 1 e0 2 dx Hence the slope of the tangent is 2.

Finding the y-coordinate of the tangency point: at x 0 y 0 e0 1 So, the tangency point is: 0,1 . Using the point-slope form for a linear equation gives: y 1 2 x 0 y 2 x 1 is the tangent to the graph at x 0

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(c)

y

4 tan 2 x, x

8 Let f (u )=4 tan u , g ( x)

2 x,

then f '(u ) 4sec 2 u Applying the chain rule to the function: dy f '( g ( x)) g '( x) 4sec2 2 x 2 8sec2 2 x dx

at x

8

dy dx

:

8sec2 2

1

8

8

16

cos 2

4 Hence the slope of the tangent is 16.

Finding the y-coordinate of the tangency point: at x

y

8

4 tan 2

4

8

So, the tangency point is:

,4 . 8 Using the point-slope form for a linear equation gives: y 4 16 x

3. g ( x)

8

x 2cos x, 0 x 2

(a)

Stationary points occur at the points where g ' x g' x

1 2sin x

g' x

0

1 2sin x 0

sin x

there are stationary points at x (b)

g '' y

1 2 6

x or x

6 5 6

0

or x

5 , in the interval 0 6

x 2

The nature of the stationary points: the 2nd derivative test: g ''( x) 2cos x Considering the sign of the 2nd derivative at the stationary points: g ''

4.

is the tangent to the graph at x

y 16 x 4 2

8

6 5 6

2cos

6

2cos

5 6

3 3

0 0

there is a relative maximum at x =

6 5 there is a relative minimum at x = 6

x ex

Stationary points occur at the points where

dy dx

0

dy 1 ex dx © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

dy dx

0

1 ex

0

ex

1

x 0

there is a stationary point at x 0 Applying the 2nd derivative test to determine the nature of the stationary point: d2 y ex dx 2 Considering the sign of the 2nd derivative at the stationary point: d2 y at x 0 : e0 1 0 there is a relative maximum at x =0 dx2 at x 0 : y 0 e0 1 The point 0, 1 is a relative maximum, the function is continuous in a set of real numbers and there are no other relative extremes 5.

(a)

f x

The point 0, 1 is an absolute maximum.

4sin x cos 2 x

Stationary points occur at the points where f ' x Applying the chain rule to

0

d cos 2 x : dx

Let h(u )= cos u , g ( x ) 2 x, then h '(u ) sin u dy h '( g ( x)) g '( x) 2sin 2 x dx f ' x 4cos x 2sin 2 x

f' x

0

4cos x 2sin 2 x 0

4cos x 1 sin x

4cos x 4sin x cos x 0

0

cos x 0 or sin x

1 and 0

x 2

3 2 2 Applying the 2nd derivative test to determine the nature of the stationary points: d Applying the chain rule to sin 2 x : dx Let h(u )= sin u, g ( x) 2 x, then h '(u ) cos u dy h '( g ( x)) g '( x) 2cos 2 x dx

There are stationary points at x

and x

f ''( x) 4sin x 4cos 2 x Considering the sign of the 2nd derivative at the stationary points: f ''

2

4sin

2

4 cos 2

2

3 3 3 4sin 4cos 2 2 2 2 nature using the 2nd derivative. f ''

8 0 0

there is a relative maximum at x = it is not possible to determine the

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2

1st derivative test at x f'

4cos

f' 2

3 : 2

2sin 2

4cos 2

4 0

2sin 4

4 0

1st derivative changes the sign from minimum at x = f

4sin

2 3 2

f

2

2

4sin

3 2

x=

3 2

there is a relative

3 . 2 cos 2

5

2

cos 2

3 2

3

,5 - relative maximum point

3 , 3 - relative minimum point 2

(b)

g x

tan x tan x 2

Stationary points occur at the points where g ' x 0 Let f ( x) tan x, h( x) tan x 2 Applying the product rule: d g' x f ( x ) h( x ) f ( x) h '( x) h( x) f '( x ) dx tan x sec2 x tan x 2 sec 2 x 2sec 2 x tan x 1 g' x

0

sec2 x 1 cos x

2sec 2 x tan x 1

0

0 or tan x 1 0 and 0 0 (contradiction) or tan x

x 1

2 x

3 or x 4

7 4

3 7 or x 4 4 Applying the 2nd derivative test to determine the nature of the stationary point:

There are stationary points at x

Let f ( x) 2sec2 x, h( x) tan x 1 Applying the product rule: d g '' x f ( x) h( x) f ( x) h '( x) h( x) f '( x) dx 2sec2 x sec 2 x tan x 1 4sec x sec x tan x 2sec2 x sec2 x 2 tan 2 x 2tan x

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Considering the sign of the 2nd derivative at the stationary points: 3 3 3 3 3 g '' 2sec 2 sec2 2 tan 2 2 tan 4 4 4 4 4 2

1 1 1 1 2 2

2 1 2

1

8 0 3 4 7 2 tan 2 4

there is a relative minimum at x = g ''

2

7 4

2sec 2

1 1 1 1 2 2

7 7 sec 2 4 4

2 1 2

1

2 tan

7 4

8 0

there is a relative minimum at x =

7 4

2nd coordinates of the points: 3 3 3 g tan tan 2 1 1 2 1 4 4 4 7 7 7 g tan tan 2 1 1 2 1 4 4 4 3 7 , 1 ; , 1 - relative minimum points 4 4 6.

y 3 sin x, x dy dx

2

cos x

at x

2

dy dx

:

cos

2

0

the tangent is a horizontal line

The vertical line going through the point with 7. f x

e

x

x

2

the normal is a vertical line.

as its first coordinate x

2

3

x 2 (a) 1st derivative: f ' x e 3 x x 2nd derivative: f '' x e 6 x

(b) f ' x

x

ex

0

3x 2

0

0.459 or x 0.910 or x 3.73

(c) Reading from the graph, considering the sign of the 1st derivative: f' x 0 0.459 x 0.910 or x 3.73 f is increasing on 0.459,0.910 ; 3.73, f' x

0

x

0.459 or 0.910 < x 3.73

f is decreasing on

, 0.459 ; 0.910,3.73

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(d)

Applying the 1st derivative test: y = f ´(x are relative minimums at these points. y = f ´(x maximum at this point.

x = 0.459 and x = 3.73, therefore there x = 0.910, therefore there is a relative

(e) Inflection points occur at the points where the second derivative is 0 or is undefined and the second derivative changes sign. Exploring the graph of f '' x f '' x

0

x

0.204 or x

1) at x 0.204 to

(f)

6 x , we get:

2.83

f '' x

0 and f '' changes sign from . 0 and f '' changes sign from .

2) at x 2.83 f '' x

1) 2)

ex

there are inflection points at x 0.204 or x 2.83 Reading from the graph, exploring the sign of the second derivative: f '' x 0 x ,0.204 2.83, f '' x

0

x

0.204, 2.83

the function is concave up on

,0.204 , 2.83,

the function is concave down on 0.204, 2.83

8.

y e x y e x cos x The common points are at the arguments where e x e x cos x e x 1 cos x 0 e x 0 (contradiction) or 1 cos x

0

cos x 1

x

2k , k

Z

Therefore, the common points are at x 2k , k Z At the points of intersection, the gradient of each curve are given by the value of each derivative. y e x Applying the chain rule to the function: dy 2k f '( g ( x)) g '( x) e x . At the points of intersection, the gradient is e dx y

e x cos x

Applying the product rule: dy d f ( x) g ( x ) f ( x) g '( x) g ( x) f '( x) e x ( sin x) cos x e x e x sin x cos x dx dx cos 2k e 2k 0 1 e 2k At the points of intersection, the gradient is e 2 k sin 2k At the intersection points, the gradients are the same and the graphs are tangent to each other.

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9.

s t

4cos t cos 2t

d2 s dt 2

(a) Applying the derivatives to find the acceleration: a Applying the chain rule to ds dt

v(t )

d ds dt dt

4cos t

v t

4sin t 2sin 2t 0 1 cos t

d sin 2t dt

2cos 2t

4cos 2t

The particle is at rest 4sin t

2sin 2t

4sin t 2sin 2t

Applying the chain rule to a

d cos 2t dt

0

ds dt

0

4sin t 4sin t cos t 0

0

sin t

0 or

1 cos t

0

t

k

or t

2k , k

Z

T

is the minimum positive argument for which the particle is at rest Acceleration at T :

m s2 (b) maximum speed is at the point where d2 s a t 0 0 dt 2 a

4cos

4cos 2

8

4cos t 4cos2t 0 cos t 2cos2 t 1 0 cos t 1 2cos t 1 cos t 1 or cos t

t

2k

0 t T

or t

0 1 2

2 3 t

2k , k

Z

2 3

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10.

v

2 3

y

ex

Let A

4sin

2 3

2sin

4 3

4

3 2

a, b be the tangency point, b ea

3 2

2

A

3 3 5.20

m s

a,ea

dy ex dx at x a : dy ea dx Hence the slope of the tangent is e a .

Using the point-slope form for a linear equation gives: y e a x ae a e a is the tangent to the graph at x (1) y e a e a x a

a .

Substituting the coordinates of the origin 0, 0 : 0 e a 0 ae a e

a

ea

ea 1 a

0 contradiction

0

or a 1

the tangency point is A 1,e substituting a=1 in equation (1):

y ex e e y ex is the tangent to the graph going through the origin. 11.

f x

(a) (b)

2x

Applying the formula of the derivative of an exponential function: f ' x 2 x ln 2 0,1 is the tangency point

f' 0

20 ln 2 ln 2

Hence the slope of the tangent is ln 2 . Using the point-slope form for a linear equation gives: y 1 ln 2 x 0 y x ln 2 1 is the tangent to the graph at 0,1 . (c)

Stationary points occur at the points where f ' x f' x

0

0

2 x ln 2 0

No solutions ( 2x ln 2 is greater than 0 for each real number x) therefore, there are no stationary points. 12.

h( x)

(a)

x2 3 ex Let f ( x )

x2

3, g ( x)

ex

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Applying the quotient rule: d f ( x) g ( x) f '( x) f ( x) g '( x) h' x 2 dx g ( x) g ( x) ex 2x

x2 e

3 ex

ex 2x x2

x 2

e

3

2x

h' x

x

0

h

(b)

0

ex 0

x 1 x 3

x

2x 3

x2

2x 3 ex

Stationary points occur at the points where h ' x 2

x2 0

2x 3 0

1 or x 3 2

1

1

h 3

32 e3

h' x

0

e 3

3

2e

1

6 e3 6 1, 2e ; 3, 3 are stationary points e Using 1st derivative test to explore the nature of the stationary points: e x 0 when x the expression x 2 2 x 3 determines the sign of the derivative. x2

x 1 x 3 h' x

0

0

2x 3 0

x

1,3

x

, 1

3,

y

h ' x changes the sign from 1, 2e is a relative 1 the point x

minimum y

h ' x changes the sign from

x 3

3,

a relative maximum.

lim h x

x2 3 0 (as e x tends to infinity faster than x 2 3 ) x ex lim x 2 3 e x

(d)

lim h x

0

(e)

x-intercepts: x 2

(c)

lim h x x

x

x

lim x

y-intercept: h 0

y 0 is a horizontal asymptote 3

x

3 e

0

3 or x

3

3

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6 is e3

13.

y sin x

(a)

dy dx

cos x

d2 y dx2

d sin x dx

d3 y dx3

d sin x dx

a

(b)

sin x

2

,b

,c

sin x

1 2

n

d y dx

(a)

cos x

2

sin x

2

cos x

n

sin x

2

sin x

sin x

2 2

sin x

3 2

3 2

The term which is added to x is increased by dy dx

14.

(using related angles)

2

d2 y dx 2

, n 2

sin x

2 2

2 d3 y , 3 dx

each step: sin x

3 2

,n

y xe x Applying the product rule: dy xe x e x 1 x 1 e x dx

xe x

ex

d 2 y d dy d x 1 ex 2 dx dx dx dx Applying the product rule again: d2 y x 1 e x e x 1 x 2 e x xe x dx 2 d3 y Similarly, x 3 e x xe x 3e x 3 dx

2e x

Hence, we may notice that the general formula is: dn y x n e x xe x ne x dx n

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xe x

y

(b)

dn y

x n ex , n

n

dx Proof:

dy x 1 ex dx Thus, the proposition is true for n 1 For n 1 , we just showed that

dk y Assume that the proposition is true for n k so dx k then

dk

1

dx

k 1

y

d dk y dx dx k

d dx

x k ex

ex 1

x k ex , k

x k ex x k 1 ex

xe x

k 1 ex

Given that the proposition is true for n = k, we have shown that the proposition is true for n = k + 1. Since we have shown that the proposition is true for n = 1, the proposition is true for all n .

Exercise 13.3 2 y 2 16 1. (a) x Differentiate both sides implicitly:

d 2 x dx

d 16 dx

y2

d 2 y dx dy x dx y

Applying the chain rule:

2x 2 y

dy dx

0

2y

dy dx

2 2 6 (b) x y xy Differentiate both sides implicitly:

d 2 x y xy 2 dx

d 6 dx

Applying the product rule and the chain rule:

d 2 d y y2 x 0 dx dx d 2 dy d y 2y , y Applying the chain rule: dx dx dx dy dy x2 2 xy 2 xy y2 0 dx dx d y x 2 2 xy 2 xy y 2 dx x2

d y dx

y

d 2 x dx

x

dy dy 1 dx dx

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2 xy y 2 x 2 2 xy

dy dx (c)

x tan y Differentiate both sides implicitly:

d x dx

d tan y dx

Applying the chain rule for:

1 dy dx (d)

sec2 y

sec2 y

dy , dx

dy dx

1 sec 2 y

x 2 3xy 2

d tan y dx

dy dx

y3 x y 2

cos 2 y

2

Differentiate both sides implicitly:

d 2 x 3xy 2 dx

d 2 dx d d 2 d 3xy 2 3x y y2 3x and Applying the product rule for: dx dx dx d 3 d d 3 y x y3 x x y : dx dx dx d 2 d d d 3 dy 2 x 3x y y2 3x y 3 x x y 2y 0 dx dx dx dx dx d 2 dy d 3 dy y 2y y 3y2 Applying the chain rule for: , : dx dx dx dx dy dy dy 2 x 3x 2 y 3 y 2 y 3 3xy 2 2y 0 dx dx dx dy 6 xy 3xy 2 2 y 2 x 3 y 2 y3 dx dy 2 x 3 y 2 y3 dx 6 xy 3xy 2 2 y (e)

x y

y x

y3 x y 2

1

Differentiate both sides implicitly:

d x dx y

y x

d 1 dx

Applying the quotient rule for

d x d y and : dx y dx x

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y

d x dx

y x y2

dy dx

x y2 x

d y dx

x

d y dx

dy y dx x2

0

y x2

d x dx

0

Multiplying both sides by x 2 y 2 :

dy dy xy 2 y3 dx dx dy xy 2 x2 y y3 dx 2 x y y3 x3 xy 2

x 2 y x3 x3

dy dx (f)

0

xy x y 1 Differentiate both sides implicitly:

d xy x y dx

d 1 dx

Applying the product rule:

xy

d dx

x

y

x

Applying the chain rule for

d xy dx

d xy dx

d dx

0

1 dy 1 and the product rule for dx 2 x y

x y

d dy x x y: dx dx 1 dy dy xy 1 x y x dx dx 2 x y x

d y dx

y

y

Multiply the whole equation with 2 x

dy 2 x dx dy 2 y 2 3xy dx 3xy 2 x2 x sin y xy xy 1

(g)

y

x

dy dx

y

y

0

y and simplify 0

Differentiate both sides implicitly:

d x sin y dx

d xy dx

d sin y dx d dy y x x dx dx

dy and the product rule for dx

Applying the chain rule for

cos y

d xy dx

y , we get:

x

dy y dx

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dy dy x y dx dx dy cos y x y 1 dx dy y 1 dx cos y x x2 y3 x4 y 4

1 cos y

(h)

Differentiate both sides implicitly:

d 2 3 x y dx

d 4 x dx

y4

d 2 3 d 3 d 2 x y x2 y y3 x : dx dx dx d 3 d 2 d 4 x2 y y3 x 4 x3 y dx dx dx d 3 dy d 4 dy y 3y2 y 4 y3 Applying the chain rule for and : dx dx dx dx dy dy 3x 2 y 2 2 xy 3 4 x3 4 y 3 dx dx dy 3x 2 y 2 4 y3 4 x3 2 xy 3 dx 3 dy 4 x 2 xy3 dx 3x 2 y 2 4 y 3 xy e y 0 Applying the product rule for

(i)

Differentiate both sides implicitly:

d xy e y dx

d 0 dx

Applying the product rule for

d xy dx

x

d y dx

y

d x dx

x

dy dx

y and the chain rule

d y dy e ey , dx dx dy dy x y ey 0 dx dx dy x ey y dx dy y dx x e y 2 2 x 2 y 3 25

for:

(j)

Differentiate both sides implicitly:

d dx

d 25 dx d x 2 Applying the chain rule for: dx x 2

2

y 3

2

2

2 x 2 1 and

d dx

y 3

2

2 y 3

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dy dx

2 x 2 1 2 y 3

(k)

dy dx

0

dy x 2 dx y 3 x arctan x y Differentiate both sides implicitly:

d x dx

d arctan x y dx 1 dy 1 2 1 x dx dy 1 1 dx 1 x 2 dy x2 dx 1 x2 (l)

y

xy

3x 3

Differentiate both sides implicitly:

d y dx

xy

d 3x3 dx

Applying the chain rule for:

dy dx

1 d xy 2 xy dx

d dx

1

1 d xy : 2 xy dx

9x2

Applying the product rule for:

dy dx dy dx

xy

d xy dx

x

d y dx

y

d x dx

x

dy dx

y:

1 dy x y 9 x2 2 xy dx x dy y 9 x2 2 xy dx 2 xy x dy 2 xy dx

9 x2

y 2 xy

Finding the common denominator:

2 xy

x dy dx 2 xy

dy dx

18 x 2 xy

18 x 2 xy

y

2 xy

y

x 2 xy

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2. (a) x3

xy 3 y 2

0 at

2, 2

Differentiate both sides implicitly:

d 3 x dx

d 0 dx

xy 3 y 2

d dy and the product rule for 3y2 6 y dx dx d d d dy xy x y y x x y: dx dx dx dx dy dy 3x 2 x y 6y 0 dx dx dy x 6y y 3x 2 dx 2 dy 3x y dx x 6 y at x 2, y 2:

Applying the chain rule for:

dy dx

3 22 2 2 6 2

7 5

Hence the slope of the tangent is

7 5 and the slope of the normal is 5 7

Using the point-slope form for a linear equation gives: 7 7 4 is the tangent to the graph at 2, 2 y 2 x 2 y x 5 5 5 5 5 24 is the normal to the graph at 2, 2 y 2 x 2 y x 7 7 7

(b)

16 x4

y4

32 at 1, 2

Differentiate both sides implicitly:

d 16 x 4 dx

y4

d 32 dx

Applying the chain rule for:

64 x 3 4 y 3

dy dx

d 4 y dx

4 y3

dy dx

0

dy 16 x3 dx y3 at x 1, y 2 : dy dx

16 13 23

2

Hence the slope of the tangent is

2 and the slope of the normal is

1 . 2

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Using the point-slope form for a linear equation gives: y 2 2 x 1 y 2 x 4 is the tangent to the graph at 1, 2 y 2

1 x 1 2

2 xy

(c)

1 x 2

y

sin y

2

3 is the normal to the graph at 1, 2 2

at

1,

2

Differentiate both sides implicitly:

d 2 xy dx

d 2 dx

sin y

d sin y dx d d d dy xy x y y x x dx dx dx dx dy dy 2x 2y cos y 0 dx dx dy 2x cos y 2y dx dy 2y dx 2 x cos y

cos y

Applying the chain rule for:

at x

1, y

2

2

dy dx

dy and the product rule for dx

y:

:

2

21

cos

2 2

Hence the slope of the tangent is

2

and the slope of the normal is

2

.

Using the point-slope form for a linear equation gives: y y

(d)

2

x 1

2 2

2

x 1

y y

2 2

is the tangent to the graph at 1,

x 2

x

4

2

2

is the normal to the graph at 1,

2

xy 14x y at 2, 32 Differentiate both sides implicitly: 3

d dx

3

xy

d 14 x y dx

Applying the chain rule for:

1 xy 3

2 3

d xy dx

14

d dx

3

xy

1 xy 3

2 3

d xy : dx

dy dx

Applying the product rule for

d xy dx

x

d y dx

y

d x dx

x

dy dx

y:

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1 xy 3 1 13 x y 3

dy dy y 14 dx dx 2 2 1 dy 1 3 3 dy 3 x y 14 dx 3 dx 1 2 2 1 dy dy x3 y 3 x 3 y 3 42 3 dx dx 1 2 2 1 dy x3 y 3 3 42 x 3 y 3 dx 1 3

x y

dy dx

2 3

x

2 3

2 3

dy 3 dx

42 3

42 x y

y x2

3

x y2

3

dy dx

42 3 x2

dy dx

42 3 x2 y 2

dy dx

42 3 22 322

3

1 3

3

y

x2

3 3

y2

x 3 3 y2

y

2

2

x 33 x y at x 2, y 32 : 3

2

2 3 2 32

32 2

42 16 32 2 3 16

Hence the slope of the tangent is

352 23

352 23 and the slope of the normal is . 23 352

Using the point-slope form for a linear equation gives: 352 352 32 is the tangent to the graph at 2, 32 y 32 x 2 y x 23 23 23 23 23 5655 is the normal to the graph at 2, 32 y 32 x 2 y x 352 352 176 3. x2 + y2 = r2 The center of the circle: (0, 0) Gradient of the line which passes through the points (x1, y1) and (0, 0):

m1

y1 0 x1 0

y1 x1

Gradient of the tangent to the circle at the point (x1, y1): Differentiate both sides implicitly:

d 2 x dx

d 2 r dx

y2

2x 2 y

dy dx

0

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dy dx

x y

At the point x1 , y1 :

4.

m1 m2

y1 x1

x2

y2

xy

(a)

x1 y1

x1 y1

dy dx

m2

the two lines are perpendicular

1

7 y

The curve intersects the x-axis

0 , we get: x

Substituting y

2

7

0

x

7 or x

7

7,0 - the points of intersection with the x-axis

7,0 ;

Differentiate both sides implicitly:

d 2 d x xy y 2 7 dx dx d d d xy x y y x dx dx dx dy dy 2x x y 2y 0 dx dx dy x 2y y 2x dx dy y 2x dx x 2y 0 2 7 7

7,0 :

(b)

0 2 7 7

m2

2 0 2

2 m2

the tangents are parallel

The tangent is parallel to the x-axis

y 2x x 2y y 2x

y:

2 m1

2 0

Gradient at dy dx m1

dy dx

7,0 :

Gradient at dy dx

x

dy dx

0

0

Substituting to the equation of the curve:

x2

x

2x

3x 2

7

x

7 or x 3

2x

2

7

7 3

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7 : y 3

For x

7 : y 3

For x

7 3

2

7 3

2 7 3

2

7 3

2

The points where the tangent to the curve is parallel to the x-axis:

7 7 , 2 , 3 3

7 7 ,2 3 3 (c)

The tangent is parallel to y-axis, so the gradient does not exist (the denominator is 0)

x 2y 0 x y 2 Substituting to the equation of the curve:

x2

x 2

x

3 2 x 4

x 2

2

7

7

7 7 or x 2 3 3 1 7 7 2 For x 2 : y 2 3 3 x

2

For x

2

7 : y 3

7 3

1 2 2

7 3

7 3

The points where the tangent to the curve is parallel to the y-axis: 2

2 5. (a)

x2

7 , 3

7 , 3

7 7 , 3 3 2 xy 3 y 2

0 at 1, 1

Differentiate both sides implicitly:

d 2 x 2 xy 3 y 2 dx

d 0 dx

Applying the chain rule for:

d xy dx

x

d y dx

dy dx dy 2x 6 y dx

2x 2x

d 2 y dx

d x dx dy 2y 6y 0 dx y

x

2y dy dx

dy and the product rule for dx

y:

2x 2 y

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dy 2 x 2 y dx 6 y 2 x at x 1, y 1 : dy 2 1 2 1 1 dx 6 1 2 1

1.

Hence the slope of the normal is

Using the point-slope form for a linear equation gives: y 1 1 x 1 y x 2 is the normal to the graph at 1, 1 Substituting y

x2

2x

x 2 to the equation of the curve:

x 2

3

2

x 2

0

x2 5x 4 0 x 1 - the tangency point or x 3 Substituting x 3 to the equation of the tangent, we get the y-coordinate of the point: y 3 2 1 The other point of intersection: 3, 1 . 6. (a)

4x2 9 y 2

36

Differentiate both sides implicitly: Applying the chain rule for: (1)

dy dx

d 2 y dx

d d 4x2 9 y 2 36 dx dx dy dy 2y 8x 18 y : dx dx

0

4x 9y

Differentiate both sides implicitly:

d dy dx dx

d dx

4x 9y ,

2

36 y 36 x

d y dx 2

81y 2

dy Substituting (1) as : dx

Applying the quotient rule to

d dx

4x : 9y

dy dx

d2 y dx 2

36 y 36 x 81y 2

4x 9y

16 x 2 36 y y 2 81y

Simplifying:

d2 y dx2

36 y 2 16 x2 81y 3

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(b)

xy

2x 3 y

d d xy 2x 3 y dx dx d dy 3y 3 Applying the chain rule for: and the product rule for dx dx d d d dy xy x y y x x y: dx dx dx dx dy dy dy x y 2 3 x 3 2 y dx dx dx dy 2 y (1) dx x 3 Differentiate both sides implicitly:

Differentiate both sides implicitly:

d dy dx dx

d 2 y dx x 3

Applying the quotient rule to

dy dx x 3

x 3

2

d y dx 2

d 2 y dx x 3

2 y 1

and the chain rule to

x 3

2

dy : dx 2 y x 3 2 x 3

dy dx x 3

d y dx

1

dy : dx

2 y 2

Substituting (1) as

x 3

d2 y dx 2

2

y

2 y 2 y x 3

2

Simplifying:

d2 y dx 2 7.

xy 3 (a)

2y 4 2

x 3

1 Expressing y in terms of x:

y

3

dy dx 2

d y dx2

1 x

1 3

x 1 x 3

d dx

4 3

1 x

1 x 3

4 3

4 3

4 x 9

7 3

4 7

9x 3

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(b)

Differentiate both sides implicitly:

d xy 3 dx

d 1 dx

The product rule for

x

d 3 y dx

y3

d xy 3 dx

d x dx

dy dx dy (1) dx

y3

d 3 y dx

d 3 y dx

3y2

y

d x dx

0

Applying the chain rule to

3 xy 2

x

dy : dx

0

y 3x

Differentiate both sides implicitly:

d dy dx dx

d dx

y 3x d y and the chain rule to y dx 3x dy 3x 3y dx 9x2

d dx

Applying the quotient rule to

2

d y dx 2

3x

dy dx 3x

y 3 2

2

d y dy : dx 2 dx d2 y 4 y Simplifying: (2) dx 2 9 x 2

3x

Substituting (1) as

8.

x2

y2

2 x2 2 y 2

2

x

at

and to

d 2 x2 2 y 2 dx

2x 2 y

at x

dy dx

0, y

d 2 y dx

x

2

y 3y 3x 9x2

1 2 d 2 x y2 dx dy 2y dx

2 2 x2 2 y 2

2 2 x2 2 y 2

x 4x 4 y

1 : 2

1 dy 2 dx

2 0 2

dy : dx

0,

Differentiate both sides implicitly: Applying the chain rule to

1

d 2 x2 dx

x 4x 4 y

2 y2

x

2

dy 1 : dx

dy 1 dx

2 2 02

2

1 2

2

0

4 0 4

1 dy 1 2 dx

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dy dx

2

dy 1 dx

dy 1 dx

Hence the slope of the tangent is 1.

Using the point-slope form for a linear equation gives:

1 1 x 0 y 2 (a) y ln x3 1 y

9.

(b)

1 1 is the tangent to the graph at 0, . 2 2

x

Applying the chain rule:

Let f (u )= ln u, g ( x)

dy f '( g ( x)) f '( x) dx y ln sin x

1 x3 1

dy dx

3x 2

3x 2 x3 1

Applying the chain rule: Let f (u )= ln u, g ( x)

dy dx (c)

dy dx

1 cos x sin x

f '( g ( x)) f '( x)

y log 5 x2 1 log 5 x 2 1

1 u

x3 1, then f '(u)

sin x, then f '(u )

1 u

cot x

1 2

Applying the chain rule: 1

dy dx

1

f '( g ( x)) f '( x)

1

x 2 1 2 ln 5

Applying the chain rule to

dy dx

f '( g ( x)) f '( x)

(d)

d 2 x 1 dx 1 x

Simplifying:

1 x ln 1 x

dy dx

x

1 2

1 2

2

1 ln 5

d 2 x 1 dx

f '( g ( x)) f '( x)

1 2

1 2 x 1 2 1 2 x 1 2

1 2

2x

2x

1 2

1 x 1 x

Applying the chain rule: Let f (u )= ln u, g ( x)

dy dx

1 2

x 1 ln 5

2

1 x ln 1 x

1 u ln 5

x 2 1 2 , then f '(u)

Let f (u)= log 5 u, g ( x)

1 1 x 1 x

1 2

d dx

1 x 1 x

1 2

, then f '(u)

1 u

1 2

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Applying the quotient rule to

d 1 x chain rule to dx 1 x dy dx

(e)

1

dy dx

1 x 1 x 1 2 x 1

y

log10 x

1 2

1 2

1 2

2 1 x

2

1 x

1 x 1 2

1 1 x 2 1 x

1 1 x 2 1 x

1 x 1

d 1 x dx 1 x

d 1 x dx 1 x

1 1 x 2 1 x

x 1 1 x 1 1 x

1

2

2 1 x

1 2

2

2 1 x

2

2

Applying the chain rule: 1 2

Let f (u )= u = u , g ( x) log10 x, then f '(u ) dy dx dy dx (f)

ln

1 u 2

1 2

f '( g ( x)) f '( x)

1 log10 x 2

1 2

1 x ln10

1 2 x ln10 log x a x a x

Applying the chain rule:

a x , then f '(u) a x 1 d a x f '( g ( x)) f '( x) a x dx a x a x

Let f (u)= ln u, g ( x)

dy dx

Applying the quotient rule to

dy dx

1 a x a x

Simplifying:

d a x dx a x

1 u

a x

1 a x

a x 2

1

2a a x

2a a x dy dx

2

2a x

2

a

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2

and

(g)

y

ln ecos x

Applying the chain rule: Let f (u )= ln u, g ( x)

dy dx

1

f '( g ( x)) f '( x)

e

Applying the chain rule to

dy dx (h)

y

1 e

cos x

1 log 3 x

d cos x e dx

ecos x

d cos x dx

dy dx

1

f '( g ( x)) f '( x)

Simplifying, (i)

dy dx

ecos x sin x

sin x

Applying the chain rule: Let f (u) = u 1, g ( x)

dy dx

1 u

d cos x e dx

cos x

ecos x sin x , simplifying

log3 x

ecos x , then f '(u )

2

log 3 x

log3 x, then f '(u)

u

2

1 x ln 3

1 x ln 3 log 3 x

y x ln x x Let f ( x) x, g ( x)

2

ln x

Applying the product rule:

dy dx

(j)

d f ( x ) g ( x) x dx dy ln x Simplifying: dx y ln ax ln b logb x Method 1 y

ln ax

f ( x) g '( x) g ( x) f '( x) 1 x

ln b log b x

ln a ln x ln b

ln x ln b

1 1 ln x 1 x

ln a , and the

derivative of a constant is zero. Method 2 Applying the chain rule: Let f (u )= ln u, g ( x)

dy dx

f '( g ( x)) f '( x) ln b

Simplifying: 10.

dy dx

1 x ln b

ax, then f '(u )

1 u

1 1 a ax x

0

y

log 2 x at x 8 dy dy 1 (a) , at x 8 : dx dx x ln 2

1 1 , hence, the slope of the tangent is . 8ln 2 8 ln 2 Finding y-coordinate of the tangency point: at x 8 , y log 2 8 3 . So the point of tangency 8,3 .

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Using the point-slope form for a linear equation gives:

y 3

(b)

1 x 8 8ln 2

11.

y

ln x and the tangent line: ln 2

We graph y

2

1 2

2

x 1 x2 1

1 1 x 3 is the tangent to the graph at x 8 . 8ln 2 ln 2

y

x 1 x2 1

Method 1 1

x2 1 , then f '(u) x2 1

Applying the chain rule: Let f (u )=u 2 , g ( x)

dy dx

1 2

1 x2 1 2 x2 1

f '( g ( x)) f '( x)

1 u 2

1 2

d x2 1 dx x 2 1

Applying the quotient rule to

x2 1 2x

d x2 1 dx x 2 1

x2 1

x2 1

2 x x2 1 x2 1

2x

2

x 2 12

4x x dy dx

2

1

2 1 2

1 x2 1 2 x2 1

f '( g ( x)) f '( x) 2x

4x x2 1

2

2x

x2 1 x2 1

3

x2 1

1 2

x2 1

3 2

Method 2

ln y

ln

x2 1 x2 1

2 ln y

ln

1 x2 1 ln 2 2 x 1 x2 1 x2 1

ln x 2 1

ln x 2 1

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Differentiate both sides implicitly:

d 2 ln y dx

d ln x 2 1 dx

2 dy 2x 2 y dx x 1 dy 2 xy dx x 4 1

dy dx 12.

x

1

x

4

1

4x x

4

1

x2 1 x2 1

Substituting: y

2x

2 x x2 1 x2 1

2x 2

ln x 2 1

x2 1 x2 1 x4 1

2 x x2 1 x

2

1 x

2

1

x

2

dy dx

1

2x x

2

1

1 2

x

2

1

3 2

x2 ln x 2

f x

Inflection point is where f '' x 1st derivative:

0 or is undefined and f '' x changes sign

dy dx

dy dx

2 ln x 2 2 x x

x2

2 x 1 ln x 2

2nd derivative:

d2 y dx 2 d2 y dx 2

d 2 x 1 ln x 2 dx 2 2x 1 ln x 2 x

f '' x

2 3 ln x 2

0

ln x2

3 1 e4

f ''

f '' 1 e g x (a)

e

2 3 ln e 2 3 ln1

0

3

1 or x e3

x

4

2 3 ln x 2

2 0

1 e3

and 0

x 1

there is an inflection point at x

x

1 e3

1 3

e2

6 0

1

x 13.

x2

d2 y dx 2

2

3 2

ln x x (1) Let f ( x) ln , h( x)

g' x

d f ( x) dx h( x)

x

Applying the quotient rule:

h( x) f '( x)

f ( x)h '( x)

h( x )

2

x

1 x

ln x 1 x

2

1 ln x x2

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1 3

e2

x2

(2) Let f ( x) 1 ln x, h( x) Applying the quotient rule:

d f ( x) dx h( x)

g '' x

h( x) f '( x)

1 ln x 2 x x

(b)

2

h( x )

1 x

x2

f ( x)h '( x)

3x 2 x ln x x4

4

3 2 ln x x3

The function is continuous on its domain 0,

g' x

0

g '' e x

1 ln x x2 3 2ln e e3

0

ln x 1

3 2ln e e3

x 1

e

e

0

3

there is an absolute maximum at

e (the only extreme value in the continuous domain) ln e 1 Maximum value is g e e e 14.

(a)

y arctan x 1 Applying the chain rule to the function:

dy dx (b)

y

f '( g ( x)) g '( x)

sin

1

1 1

x 1

2

1

x

2

1 2x 2

x 1 x2

Applying the chain rule to the function:

dy dx

1

f '( g ( x)) g '( x) 1

d dx

x 1 x2

x 1 x2

1 x2

d Applying the quotient rule to dx 1 x2 1 x

2

x

d dx

1

2 1 x2 1 x2

x 1 x 2x

2

and the chain rule to

1 x2

x2

1 x2 1 x

d dx

1 x2 2

1 1 x2

3 2

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dy dx

1

1 2

x

1

1 x2 1

1 1 2

1 x2

(c)

1 x

1 1 1 x2

3 2 2

1 x

3 x2

y arccos

3 2 2

1 3 2

1 x2

1 1 x2

2

arccos 3x

Applying the chain rule to the function:

dy dx

1

f '( g ( x)) g '( x) 1 6

ln y

2 2

6

3

x3 1 9 x

4

6

x x4 9x (d)

3x

6x

4

x4

x x4 9

ex tan

(1) y

x arctan x

1

x

Differentiate both sides implicitly:

d d ln y x arctan x , but dx dx d d d x arctan x x (arctan x) arctan x ( x) dx dx dx x arctan x 1 x2

x

1 1 x2

arctan x 1

Thus,

1 dy y dx

x 1 x2

Substituting y 15.

f x f' x

16.

(a)

ex tan

1

arcsin x arccos x 1 1 1 x y

2

arctan

dy dx

arctan x

1 x

2

x

(from (1)):

0

tan 1 x dy dx

x 1 x2

tan 1 x

y

x 1 x2

e x tan

1

x

the function f is constant.

x a

Applying the chain rule to the function:

dy dx

1

f '( g ( x)) g '( x) 1

x a

2

1 a

1 x a a

a 2

a

2

x2

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d x arctan dx a x arcsin a

y

(b)

a a

2

x2

Applying the chain rule to the function:

dy dx

1

f '( g ( x)) g '( x)

d x arcsin dx a

17.

y

dy dx at x

1 1

2x

2

a2

x2

1 a2

x2

d 2x dx

1 2 1 1 4 2

arctan 2 x 4

8

1 dy : 2 dx

2

1

1 2

4 x arctan 2 x, x

4x

x a

1

1 a

4 arctan 2

2

Hence the slope of the tangent is 2

1 2

8x 1 4x2

4arctan 2 x

2

.

The y-coordinate of the tangency point: at x

1 2

y

4

1 1 arctan 2 2 2

2

4

2

. So, the point of tangency is:

1 , 2 2

Using the point-slope form for a linear equation gives:

y 18.

2 f x

(a)

1 y 2 arcsin cos x , 0 x 2

x

x 1 is the tangent to the graph at x

2

1 . 2

1

Let f (u )= arcsin u, g ( x) cos x, then f '(u)

1 u2

Applying the chain rule:

dy dx (as 0 (b)

f '( g ( x)) g '( x)

x

)

1 1 cos2 x

sin x

sin x

sin 2 x

1

the function has a constant gradient and hence it is linear

f ( x) arcsin cos x Applying the related angles:

f ( x) arcsin sin

2

x

f ( x)

x

2

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.

19.

2 , tan x

tan

tan

5 x

tan tan 1 tan tan

tan

5 x

2 x 5 2 1 x x

3x x 10 2

3x is maximum x 10 Let f ( x) 3 x, g ( x) x 2 10

the function y

2

Applying the quotient rule:

dy dx

d f ( x) dx g ( x )

g ( x) f '( x )

g ( x)

x 2 10 3 3 x 2 x x 2 10

dy dx

x 2 10

x x 2 10

2

0

2

10 or x 2

3 x 2 30

3 x

10

0

10 dy the sign of depends on the sign of the numerator. dx x 10

the derivative changes its sign from (the only maximum) at x

2

3x 2 30 0

10 and x

0 when x

The numerator:

2

x 2 10

3x 2 30

0

f ( x ) g '( x)

x

x

10

there is a maximum value

10

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The distance between the observer and the base of the column should be x 20.

s t (a)

10 m 3.16 m

arctan t v t

ds dt

d arctan t dt

Applying the chain rule to the function:

v t

1

f '( g (t )) g '(t ) 1

v1

(b)

a t a t

1 2 1 21 1

dv dt

d dt

2t

1 2

3 2

2t

2t

1 2

t

2 t

1 m , v 4 4 s

1 2

2t

t

2 t

1 2 4 2 4

1 2

3t

1 2

1 31 1 2 1 21 1

(c) (d)

4

1 m 20 s

1 3t t 2 t

a 1

2t t

1

3 2

2

1

2

1 m , a 4 4 s2

2t t

2

1 3 4 4 2 4 2 4 4

2

The particle is moving fast to the right and then gradually slows down while continuing to move to the right. t , when t Let x then x

lim arctan x x

2

lim arctan t t

2

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13 m 800 s 2

Exercise 13.4 1.

(a)

We know the rate of change of the volume with respect dV to time: 2m3 min 1 , the height h 8m , the dt diameter of the top is 6m. We need to find the rate of change of the level of the dh water with respect to time: . dt The formula of the volume of a cone gives us an equation that relates the variables: V, r and h. 1 2 V r h 3 r 3 3 By using similar triangles, we get: r h and substituting the result into the h 8 8 2

(b)

2.

1 3 3 formula of the volume, we get: V h h h3 3 8 64 Differentiate both sides implicitly with respect to t: dV 9 dh dh 64 dV h2 dt 64 dt dt 9 h2 dt dV Substitute h 5, 2: dt dh 64 128 2 0.181083 m min 1 18.1 cmmin 1 2 dt 9 5 225 r 3 8 By using similar triangles, we get: h r and substituting the result into the h 8 3 1 8 8 3 formula of the volume, we get: V r2 r r 3 3 9 Differentiate both sides implicitly with respect to t: dV 8 2 dr dr 3 dV r dt 3 dt dt 8 r 2 dt r 3 15 We use similar triangles when h = 5: r 5 8 8 15 dV Substitute r , 2: 8 dt dr 3 48 2 0.0679 m min 1 6.79 cm min 1 2 dt 225 15 8 8

dV 240 cm3s dt dR We need to find the rate of change of the radius R respect to time: . dt The formula of the volume of the ball relates the variables V and R: 4 3 V R 3

We know the rate of change of the volume with respect to time:

1

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Differentiate both sides implicitly with respect to t: (a)

Substitute R 8 :

(b)

After 5 seconds: V Substitute R

3.

3

dR dt

1 4

8

2

240

15 16

240 5 1200 cm3

900 dR , dt

1 4

3

900

2

dV dt

dR dt cm 0.299 s

dR dt

4 R2

then 1200

4 R3 3

240 0.439

cm s

We know the rate of change of the radius with respect to time:

dr dt

1 dV 4 R 2 dt

R

1 cmh

3

1

We need to find the rate of change of the circumference with respect to time: (a)

(b)

4.

900

dl dt

We use the formula of the circumference of the circle l 2 r dl dr Differentiate both sides implicitly with respect to t: 2 dt dt d l cm dr Substitute: 2 1: dt h dt We use the formula of the area of the circle A r2 dA dr Differentiate both sides implicitly with respect to t: 2 r dt dt dl cm dr 2 4 8 Substitute: r 4, 1: dt h dt

We know the rate of change of the height with respect to time:

We need to find the rate of change of the angle

dh dt

50

with respect to time:

m min

d . dt

h 150 Differentiate both sides implicitly with respect to t: d 1 dh d cos 2 dh sec2 (1) dt 150 dt dt 150 dt 250 5 When h 250 tan 150 3

We use the trigonometric ratio: tan

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Using Pythagoras theorem and the trigonometric ratio: 25 9 sec2 1 tan 2 1 cos 2 9 34 9 d 9 d h 34 50 3 Substitute to (1): cos 2 , 50 34 dt dt 150 34 5.

0.0882

rad min

We know the rate of change of the horizontal distance x with respect to time:

dx dt

We need to find the rate of change of length of the string y with respect to time: Using Pythagoras theorem y 2

722

dy dt

2x

dx dt

(1)

dy dt

x dx y dt

1202 722

96 dx Substitute to (1): x 96, y 120 , dt

6.

dy . dt

x2

Differentiate both sides implicitly with respect to t: 2 y When y 120 , x

m s

6

6:

dy dt

96 6 120

24 5

4.8

m s

We know the rate of change of the horizontal distance x with respect to time:

dx dt

6

We need to find the rate of change of the horizontal distance y with respect to time: 20 3y y x 5 Differentiate both sides implicitly with respect to t: dy dx dy 4 dx 3 4 dt dt d t 3 dt dy 4 dx 4 ft dx 6 8 Substitute 6 dt 3 dt 3 s dt

By using similar triangles, we get:

y

ft s

dy . dt

4x

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7.

and the rate of change of the vertical distance y with respect to time v2

dy dt

We need to find the rate of change of the distance z with respect to time,

dz after 3 hours. dt

1802 1052

After 3 hours: x 60 3 180km , y 35 3 105 km , z Using Pythagoras theorem: z 2

x2

Substitute x 180, y 105, z 15 193 , dz dt

180 15 193

60

105 15 193

35

dz dt

dx dt

60 ,

69.5

dx 4 dt Square both sides of the equation: y 2 y

15 193

y2

Differentiate both sides implicitly with respect to t: 2 z

8.

dx km 60 dt h km 35 h

We know the rate of change of the horizontal distance x with respect to time v1

dy dt

dz dt

2x

dx dt

2y

dy dt

dz dt

x dx z dt

y dy z dt

35 :

km h

x2 1,

x2 1

Differentiate both sides implicitly with respect to t: 2 y When x 3, y

dy dt

2x

dx dt

dy dt

x dx y dt

10

Substitute x 3, y

10,

dx dt

4:

dy dt

3 10

4

12 10

3.79

9. We know the rate of change of the amount of water dV m3 with respect to time: We need to find 0.03 dt s the rate of change of the level of the water h with dh respect to time: . dt We use the formula of the volume of the prism: 1 V ah 4 2ah 2 a h By using similar triangles, we get: a 1.5h 1.5 1 © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

2 1.5h h V 3h2 dV dh dh 1 dV Differentiate both sides implicitly with respect to t: 6h (1) dt dt dt 6h dt 3 2 After 25 seconds: V 0.03 25 0.75 m 3h 0.75 h 0.5 m h 0

Substitute in the formula of the volume: V

Substitute h 0.5, 10.

dV dt

dh dt

0.03 in equation (1):

1 0.03 6 0.5

dh dt

m s mm 3 s

0.01

dR dt dV We need to find the rate of change of the volume V with respect to time . dt 4 3 We use the formula of the volume of the ball: V R 3 dV dR Differentiate both sides implicitly with respect to t: (1) 4 R2 dt dt We use the formula of the surface of a surface to find the radius when the surface is 10 mm 2 :

We know the rate of change of radius of a sphere with respect to time:

4 R2

10

Substitute R

R

5 2

R

5 dR and 2 dt

0 dV 3 in (1): dt

4

5 2

2

dV dt

3

30

mm3 s

11. dy km and the 40 dt h dz km velocity of the second car : v2 We need to find 50 dt h the rate of change of the distance between them with respect to dx time . dt

We know the velocity of the first car : v1

Applying the cosine rule: x 2 y 2 z 2 2 yz cos 60 Differentiate both sides implicitly with respect to t: 1 2 , the triangle is equilateral, so x 2 dy dz dx Substitute x y z 2 , 40 , 50 to (1) : 2 2 dt dt dt dx km 45 dt h

When y

x2

2x

dx dt

y2

z2

2y

dy dt

yz

2z

dz dt

y

dz dt

z

dy dt

z

2 2 40 2 2 50 2 50 2 40

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12. We know the rate of change of the diagonal of a cube with respect db cm to time: 1 . We need to find the rate of change of a 8 dt s da side a with respect to time dt Using Pythagoras theorem: a 2

b2

2

b2

b2

3a 2

Differentiate both sides implicitly with respect to t: db da da b db 2b 6a 2 dt dt dt 3a dt 3 b a 3 a, b 0

3a2

Substitute (1) and (3) to (2):

13.

a 2

da dt

a 3 8 3a

da dt

8 3 3

4.62

cm s

Let l be the distance travelled around the circle, then, l d 1 dl l r 0.3units s ec 10 dt 10 dt If the vertical distance to the x-axis is 5, we can calculate: 51 5 dx sin or , x 10 cos 10sin 6 6 d 10 2

dx dt

dx d

d dt

dx dt

6

sin

6

10 0.3 1.5units sec

14. We know the rate of change of the angle with d 1 rad respect to time: We need to find the dt 60 s dx speed of the jet v dt Applying trigonometric ratios: tan

10000 x

Differentiate both sides implicitly with respect to t: d 10000 dx dx x2 d 1 sec2 sec2 dt x2 dt dt 10000 dt 10000 10000 x When : tan 3 x 3 3

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Substitute

3

,x

10000 d , dt 3

1 to 1 : 60

2

dx dt

Speed:

15.

10000 1 3 sec 2 10000 3 60

dx dt

800

_

222.2

m s

800

km h

km h

(a)

We know the rate of change of the horizontal distance x with respect to time dx km m (the velocity of the car): 288 80 and the horizontal distance of the dt h s cameraman from the racing track: 40 m . We need to find the rate of change of the angle with respect to d 0 time when the car is directly in front of the camera dt x Applying the trigonometric ratios: tan 40 d 1 dx d cos 2 dx Differentiate both sides implicitly: sec 2 dt 40 dt dt 40 dt 2 d cos 0 rad degrees dx 80 2 115 80 , Substitute 0: dt 40 s s dt (b)

We know the rate of change of the vertical distance x with respect to time (the velocity dy km m 288 80 and the horizontal distance of the cameraman from of the car): dt h s the racing track: 40 m . We need to find the rate of change of the angle with respect d to time a half second later. dt The distance driven by the car during 1 second: 80 m the distance driven during 1 of a second: 40 m we need to find the rate of change of the angle with respect 2 d to time when x = 40 m the triangle is right-angled and isosceles dt 4 x Applying the same method as in (a): tan 40 d cos 2 dx Differentiate both sides implicitly and transform: dt 40 dt

dx Substitute dt

80 ,

d : 4 dt

cos 2 40

4 80 1 rad s

57

degrees s

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16. We know the rate of change of the horizontal distance dx km with respect to time: and the rate of change 640 dt h of the vertical distance with respect to time: dy m km We need to find the rate of change 180 10.8 dt h h of the distance between the tower and the airplane with dz respect to time: dt Applying Pythagoras theorem: x 2 y 2 z 2 dz dx dy dz x dx y dy Differentiate both sides implicitly with respect to t: 2 z 2x 2y dt dt dt dt z dt z dt 52

When x 6, y 5 : z

62

61

dy dx 640 , 10.8 : dt dt dz km 485 dt h

Substitute x 6, y 5 , z = 61 , dz dt

6 61

5

640

61

10.8

Exercise 13.5 1. The area of the rectangle shown on the picture is 1 A 2c b From Pythagoras theorem:

c2 b

2

1

1 c2

2 b

Substitute (2) into (1) : A c

b 0

2c 1 c2 , 0 c 1

Differentiate the function with respect to c:

A' c

2c

A' c

2c

A' c

0

d dc

1 c2

c 1 c2 4c2 2 1 c2

The critical point is c The denominator

1 c2

1 c2

1 c2 2 0

c

d 2c dc 2c2

2 2c2

4c2 2

1 c2

1 c2

1 c2

2 2

c

2 2

and 0 c 1

c

2 2

2 2

0 for each c

0,1

the sign of the first derivative depends

on the sign of the numerator.

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2 2

Numerator: 4 c

A' c A' c

0

0 c

0

2 2

Then b

2 2

1 c

there is a relative

c 1 2 2

2 A 2

maximum at c

2 2

c

1

2 2

2 by

2 2

2

The dimensions:

2 2 2 2

2 2

1

2 , 2c 2

2

2

1 2

1

2

2. Let the rectangle have dimensions x × y, where x is the fold into the base and y is the height of the cylinder. The radius of the base will be: 2r

x

x . 2

r

Therefore, the volume of the cylinder is:

V

r

2

h

V

x 2

2

y

x2 4 2

y

1 2 x y. 4

Since the perimeter is equal to 40 cm, we can express the volume in terms of x only:

2x 2 y

40

x

y

20

y

20 x, and so V x

1 2 x 20 x . To find the 4

maximum volume, we need to differentiate the volume with respect to x and find the zero of the derivative:

V' x x

1 2 x 20 x 4

0 or 40 3x

0

x2 x

1

1 x 40 3x 4

V' x

0

1 x 40 3 x 4

0

40 . 3

The first solution is not possible, so we take the second and calculate y:

y

20 x

y

20 . 3

So, the dimensions of the rectangle are 3.

40 20 cm and cm 3 3

Any point on the graph has coordinates x, x ; therefore, to find the distance to the given point, we will use the distance formula. To make the calculation simpler, we will look at the square of the distance and then, at the end, we will simply take the square root of the value we obtain. © Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free.

g x

3 x 2

g' x

0

2

2

x 0

2x 2 0

x 1

g 1

The volume of the box: 1000

4. (a)

9 x 4

x 2 3x

x 2 2x

9 4

12 2 1

2x x h

h

s x

9 4

5 4

g' x

d

2x 2

5 4

5 2

500 cm x2 4x 2 6hx

2 x 2x 2x h 2 2x h s x 500 Substitute h : x2 500 3000 s x 4x2 6x 2 s x 4x2 ,x 0 x x 3000 (c) Differentiate with respect to x: s ' x 8 x x2 3000 s ' x 0 8x 0 8 x3 3000 x 5 3 3 7.21 2 x 6000 2nd derivative test: s '' x 8 x3 6000 there is a relative minimum at x 5 3 3 s '' 5 3 3 8 24 0 3 53 3 (b)

Then 2 x

500

2 5 3 3 10 3 3 14.4 cm , h

53 3 Dimensions: 7.21 cm 5.

Let AD

14.4 cm

2

20 9

3

9.61 cm

9.61 cm

2y

The area of the rectangle: 100

2xy

y

50 - the radius of the semicircle x

The perimeter of the figure (two sides AB + the circle with the radius y (2 semicircles)):

100 ,x 0 x 100 2 Differentiate with respect to x: l ' x x2 100 l' x 0 2 0 x 2 50 x2 l x

2x 2

50 x

l x

x 5 2

x

5 2

200 5 2

3

and x

0

x 5 2

200 x3

2nd derivative test: l '' x

l '' 5 2

2x

0

there is a relative minimum at x

5 2

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The perimeter of the figure is minimum when x

6.

5 2

12.5 cm

Denote the angles at the vertical posts by and . Using the property that alternate interior angles have the same measure, we establish the relationship .

Now, using the right-angled triangle trigonometry formulae, both angles can be expressed in terms of x only: x x arctan tan 12 x 10 x 12 x arctan arctan 10 x 12 8 10 x tan arctan 8 8

x

x

12 8 . Simplifying this expression will give 2 x 144 x 20 x 164 4 x2 60 x 204 2

x2 144 x 2 20 x 164

At this point a GDC must be used because the work will require intensive symbolic manipulation. x 3.62 will give the maximum value for .

7.

Method 1 As shown below, let c be the straight distance from the wide hallway to the narrow one. The ladder will pass the corner when it matches the minimum such distance.

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Using similar triangles, we get:

2 x

y 3

6 x

y

c

6 2 x

2

2

2

3 x ,x 0

6 2 x

Consider the function f x

2 2

3 x , x 0 and try to see the minimum value of

this quantity.

f x

x2

6 x 13

24 x

36 x2

Differentiate with respect to x:

f' x

0

2x 6

24 x2

72 x3

f' x 0

Solution of polynomial equation 2 x 4

x 2.2894 or x

3 and x 0

24 72 x 2 x3 2 x 4 6 x3 24 x 72 0 x3 6 x3 24 x 72 0 will require a GDC: x 2.29 2x 6

Using the first derivative test, you will notice that f x is switching from negative values to positive values. This implies that f(x) itself has a minimum at this point. The value of this minimum is f 2.29 7.02. Thus, the longest ladder to pass this corner must be less than or at most equal to 7.02 m. Method 2 (Seen in chapter 7) Denote the angle between the 2-metre wide hallway and a straight line touching the corner by . Then the angle between the 3-metre wide hallway and this line is:

180

90

90

. We split the total length of the line segment a and b,

to correspond to the lengths in the 2-metre wide and 3-metre wide hallways respectively. Now, from the corresponding right-angled triangles, we get

sin

2 a

a

2 sin

and sin 90

cos

length of the segment can be expressed in terms of

3 b only: l

b

3 cos a b

. Therefore, the total

l

2 sin

Again, we will use a GDC to find the answer.

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3 cos

So, the longest ladder that can be carried around the corner is 7.02 m. Note: The minimum of this function is the maximum length of the ladder that can be carried around the corner of the hallway. 8.

We know the rate of change of the distance in the sandy terrain of Erica with respect to time dz km (the velocity v1 ): and the rate of change of the distance on road of Erica with 2 dt h respect to time (the velocity v 2 ):

dy dt

5

km h

We need to find the distance d, such that the

time T

T

t1

t1 t2 ( t1 - the time during walk on sandy terrain, t2 - the time during walk on the road) z z d d z d , t2 T v1 2 v2 5 2 5

Express z in terms of d: z

7 d

2

100

149 14d

d 2 , and substitute:

149 14 d d 2 d ,0 d 7 2 5 Differentiate with respect to d and find the value of d that minimises T. T

Calculations are long and it is better to use a GDC for such cases:

dT dd

0

d

2.64

So, point A is 2.64 km due west from the office.

9.

P

x,

8 x

2

, x 0 - is a point on the graph of the function y

4

The area of the rectangle: A

2 xy

2x

8 x

2

4

16 x ,x x2 4

8 x

2

4

0

Differentiate with respect to x:

x 2 4 16 16 x 2 x

dA dx

dA dx

x

0

2

4

64 16 x2 x

2

4

2

2

16 x 2 64 32 x 2 2

4

64 16 x2

0

x

0

2

64 16 x 2 x

2

4

2

,x

0

x 2 since we take the positive value.

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x2

4

2

0 for all values; thus, the sign of the expression 64 16x 2 determines the sign of

the derivative. The graph of 64 16x 2 is a parabola that has a maximum of 64 at x = 0. It will intersect the x-axis at x = 2 and it will become negative. Therefore, the function A has a maximum when x = 2. At x 10.

16 2 22 4

2: A

4 is therefore the maximum area of the rectangle.

Let t (in hours) be the time, x be the horizontal distance, y the vertical distance, z the distance between the ships

z2

x2

y2

Substitute: x 12t , y

z2

10 16t :

12t

2

2

10 16t

Differentiate both sides implicitly with respect to t:

2z

dz dt

2 12t 12 2 10 16t

Substitute z

dz dt

12t

2

10 16t

160 400t

0 12t

12t

2

2

2

10 16t

10 16t 2

dz dt

16 dz dt

:

12t

0

2

160 400t z 160 400t 2

10 16t

160 400t

0 for every t

0

t

2

2 5

the sign of the expression 160 400t

determines the sign of the derivative.

y

160 400t is a linear function changes sign from negative to positive at t =

2 12 5

Thus, the minimum distance is z

11.

2

2 10 16 5

2 . 5

2

6

r 2 h is the volume of the inscribed cylinder. If we look at the cross-section of the

V

sphere, we can find the relationship between R, r and h.

R

2

r

h 2

2

2

h

2 R2 r 2

V' r

4r

R2 r 2

2r 2

V' r

0

2 R 2 3r 2

0

V r

2r

2r 2 R2 r 2

r1,2

2r 2

2 2 R 3

R2 r 2 2 R 2 2r 2 R2 r 2

r2

2r

2 R 2 3r 2 R2

r2

6R 3

Since the radius r of the base of the cylinder cannot be negative, we have only one solution:

r

6R 3

h 2 R2

2 2 R 3

2 3R . 3

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12.

Let p be the distance between the points X and P, and t the time.

a2

AP

2 c a2

pr

c a2

p2 r 2 c2 p2

a 2c 2

b

p

c

r

p2

p2

pr c a 2

1 r

p2

pr c a 2

0

p2

2p

t' p t' p

a2

p2 , t p

cr a 2

p2

0 p 2r 2

c 2a2 c2 p 2

a2c2 r 2 c2

p2

p2

a2c2 r 2 c2

p

ac r 2 c2

, since p has to be

positive. 13.

This question is best done with a CAS. Since the circumference of the base is: 2

10 x

20

x

20 2

r

x

10

x , 2

we can also express the height h in terms of x.

h

100 r 10

2

100

x2 4 2

x

2

x 10 2

100 100 10

x2 4 2

x

40 x x 2 2

Now, the volume can be expressed in terms of x only.

V

1 2 r h 3

V' x

1 24

1 x 10 3 2

2

2 20

x

2

20 24

2

24

2

x

40 x x 2 20 x 40 x x 24

2 x 2 40 x

120

20

6 3

20

2 x 2 80 x 400

2

40 x x 2

2

20 x

2

x

2

2 20

x

2 40 x x 2

2

x2

40

2

40 x x 2

3 x 2 120 x 400

0 x1,2

x 20

2

x

24

40 x x 2

1

x 3x 2 120 x 400

20

V' x

2

20

40 x x 2 2

14 400 6

11.5 cm

2 2

V

0 4800

2

20

2000 3 27

20

6 3

403 cm3 , to 3 significant figures.

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Notice that the other possible solutions are discarded because the value of x exceeds the perimeter of the circle. 14.

Set up the coordinate system in such a way that the origin is at the point R. The distance P Q is a fixed positive number d and the distance P R is our variable x, x > 0. Then, the coordinates of the points are as follows: P x, a , R 0, 0 , Q d x, b .

tTotal

PR u

t1 t2

d

x2 a2 u

RQ v

x

2

b2

v dt dx d x

0.

Since the ray travels in such a way that the time is a minimum, we can deduce that

dt dx

2 d

2x u 2 x2 a 2

v 2

x d

1 x

2

b2

1 u

0

x x2 a2

1 u

d

x

2

b2

By looking at the triangles PP R and RQ Q, we can establish the following relationships.

sin

P'R PR

x x

2

a

and sin

2

RQ ' RQ

d d

x x

2

b

2

.

Therefore, we obtain the formula:

1 sin u

1 sin u

sin sin

u v

Exercise 13.6 1.

(a)

lim x

0

1 cos x x2

Substituting x

0 into the rational expression gives:

1 1 02

0 0

the limit is of the

0 and we can apply l 0 1 cos x sin x lim lim 2 x 0 x 0 x 2x 0 This is also of the indeterminate form and we can apply l 0 sin x cos x 1 lim lim x 0 x 0 2x 2 2 indeterminate form

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n:

(b)

x 1

lim x 1

x

2

Substituting x 1 into the rational expression gives:

3 2

1 1 2 2

0 0

and we can apply l

x 1

lim x

(c)

x2 3 2

1

1 ln x

lim x 1

1 2x

lim x

1

x

1

1 x 1

, as substituting x 1 we get the indeterminate form

x 1 ln x lim x 1 x 1 ln x

1 x lim x 1 1 x 1 ln x x 1

lim x 1

Applying l

x2 1 x2 4x 3

x 1

x 1 x 1 x ln x

0 : 0

0 0

:

x 1 lim x 1 x 1 x ln x

lim

2

2 x2 3 x 1 ln x lim x 1 x 1 ln x

Apply l

2. (a)

x2 3 x

lim

1 lim x 1 1 1 x ln x x

1 1 2 ln x

lim x

Substituting x 1 , we get

1 2

0 , so 0

Method 1 We may factorise the numerator and the denominator and reduce the fraction to find the limit:

x 1 x 1 x 1 x 3

lim x

1

Method 2 Apply l

x

lim

x

x 1

2

lim x

1

x 1 x 3

1

:

2

1 4x 3

lim x

1

2x 2x 4

1

3

(b)

(c)

1 x 1 Apply l x 0 x 0 form : 0 2 1 1 x 3 3 1 x 1 lim lim 3 x 0 x 0 x 1 x sin x Apply lim x 0 x3 0 indeterminate form : 0 lim

, as substituting x 0 we get the indeterminate

1

1 3

rule three times, as substituting x 0 we get the

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x sin x x 0 x3 1 1 lim x tan x x 0 x lim

(d)

1 cos x 3x 2

lim x

0

0

get the indeterminate form

tan x x lim 2 x 0 x tan x

x

0

sin x 6x

lim x

tan x x Apply x 2 tan x

lim x

lim

0

cos x 6

1 6

rule, as substituting x 0 we

0 : 0

1 1 cos 2 x lim x 0 1 x2 tan x 2 x cos 2 x

1 cos 2 x sin 2 x cos 2 x lim lim x 0 x 0 x2 x2 sin x 2 x sin x cos x 2 x 2 cos x cos x 0 This is still the indeterminate form . 0 Note that sin 2x 2sin x cos x , and apply rule again: 2 sin x 2sin x cos x lim 2 lim x 0 x x 0 2x x sin 2 x 2 x cos 2 x sin 2 x lim x

0

sin 2 x 2 x 2 x cos 2 x sin 2 x

2 cos 2 x 2 4 x sin 2 x 2 cos 2 x 2 cos 2 x 2 1 2 0 2 2 3

lim x

(e)

ln x 1 log 2 x

lim x

form

lim

ln x 1 log 2 x

we get the indeterminate

x ln 2 x 1

1 lim x 1 x 1 x ln 2

lim x

x ln 2 x 1

rule again as when x

Apply

x

rule, as when x

Apply

:

x

lim

0

lim x

ln 2 1

we get the indeterminate form

ln 2

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:

ln 1 x 2

lim

(f)

x

rule, as substituting x

Apply

ln 1 x 2

0

0 we get the

0 and transforming the expression: 0 2x 2 1 x2 1 x 2 lim x 1 lim 1 x 0 x 0 1 x2 2x 1 x2 1 x2 x 2 2cos x Apply rule twice, as substituting x 0 we get the e x 2 cos x

indeterminate form

ln

lim 0

ln

lim

2 ex

x

(g)

x

0

0 : 0 2 cos x 2 cos x

indeterminate form

2 x2 lim x x 0 e e x

lim x

0

2 x 2sin x e e x 2sin x

1

3.

lim 1 x x . as x

we get the indeterminate form

x

lim

x

x

0

0

2 2cos x e e x 2 cos x

4 1 4

x

.

We can symbolically manipulate the expression to reduce it to a form where can apply. 1 x

1 x

e

1 x

ln 1 x

e

lim e x

x

x

lim 1 x x

ex

x

, but,

lim

ln 1 x x 1 x

1 lim ln 1 x x

ln 1 x

ln 1 x , which when x is of the indeterminate form x x 1 1 lim 1 x lim 0 , and therefore, x x 1 1 x

1 lim ln 1 x x x

lim

1 ln 1 x x

1

1 x

Now, lim 1 x

e

lim

x

rule

1 ln 1 x x

e0

1

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.

Chapter 13 practice questions 1.

The important points of the first derivative are zeros, where the original function reaches minimum or maximum, x 4 and x 1, and the maximum point, at the midpoint. The intervals of the positive and negative values of the first derivative are to be established by the increasing/decreasing intervals of the original function. So:

f' x

0, 4 x 1, and f ' x

0, x

4 or x 1

y

x

2.

(a)

Given the product form of the function, the values a and b are zeroes that can be easily read from the graph. Therefore, (i) a 4 and (ii) b 2 .

(b)

(i)

We can use the product rule, but we have to be careful since there are three factors. f' x x 4 x 2 x x 2 x x 4 x2

(ii)

f' x

8 4 x 3x 2

2x

x2

3x 2

0

4x

8 4 x 3x 2

4x 8 0

16 96 2 2 7 2 2 7 2 2 7 x or x 6 3 3 3 D is relative maximum and the 1st derivative changes sign from x

(iii)

0

2x 8 x2

4

2 2 7 3 2 2 7 x-coordinate of the point D is 3

positive to negative at x

f

2 2 7 3

2 2 7 3

3

16

10 7 7 27

Note: This is very close to (1, 5) as it may look on the graph.

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(c)

m

(i) (ii)

f' 0

8

Equation of tangent: y 8 x

x x 4 x 2

8x

x2

x 8

2x 8

0

x2 x 2

0.

Since the point differs from the origin, we can conclude that the x-coordinate of the second point is x 2 . 3.

(a)

(b)

(c)

4.

(a)

66 66e

0.15 0

(i)

When t

0:v 0

(ii)

When t

10 : v 10

(i)

a t

v' t

(ii)

a 0

9.9e

(i)

lim 66 66e

(ii)

lim 9.9e

(iii)

Since the velocity is constant (66 m/s) the acceleration must be zero.

66e 0.15 0 0.15t

2 or x 3

x x

x