Collins Cambridge International AS & A Level Mathematics Pure Mathematics 1 Worked Solutions [First Edition] 9780008257736

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Cambridge International AS & A Level Mathematics

Pure Mathematics 1 STUDENT’S BOOK: Worked solutions

Helen Ball, Chris Pearce Series Editor: Dr Adam Boddison

Pure Mathematics 1 International Students Book Title page.indd 1 57736_Pi_viii.indd 1 Mathematics 1.indd 1 WS TITLE PAGE_Pure

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1 

WORKED SOLUTIONS

Worked solutions 1 Quadratics Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge

y 15

1 2x2 – 5x – 3 = 0 (2x + 1)(x – 3) = 0 x=−



10 5

1 or x = 3 2

0 –25 –20 –15 –10 –5 –5

2 3x2 + x – 7 = 0 a = 3, b = 1, c = −7 x=



−1 ± 1 + 84 x= 6



x=



−1 ± 85 6



x < 4 0

1

4 2

3

4

5

6

7

Exercise 1.1A b 2x2 – 8x = 2(x2 – 4x)

= 2[(x − 2)2 − 4]



= 2(x – 2)2 – 8 c x2 + 8x + 7 = (x + 4)2 – 16 + 7 = (x + 4)2 − 9

2 x2 – 10x + 11 = (x − 5)2 – 25 + 11 = (x − 5)2 − 14



2

4

2

2

2

2

2



1 a x2 + 4x = (x + 2)2 − 4



2

2

1 2

4 3x – 5 < 7 3x < 12

–1

( 32 ) − 94 − 7 3 37 = (x + ) − 2 4 3x 9 c 2x + 3x + 9 = 2 ( x + + 2 2)  3 9 9 = 2 ( x + ) − +  4 16 2   3 63  = 2 ( x + ) +  4 16   5 25 x – 5x + 7 = ( x − ) − +7 2 4 5 3 y = (x − ) + 2 4 2

x = 2, y = −3



= (x − 4)2 − 21



b x2 + 3x – 7 = x +

5x – y = 13 2x + y = 1 Add 1 and 2 . 7x = 14



x

3 a x2 – 8x – 5 = (x − 4)2 – 16 − 5

x = 1.37 or − 1.70 3

10 15 20 25

–10

−1 ± 12 − (4)(3)(–7) 6



5

At the turning point y has a minimum value 5 5 3 i.e. when x = . When x = , y = , the 2 2 4 coordinates of the turning point. y 25 20 15 10 5 0 –25 –20 –15 –10 –5 –5

5

10 15 20 25

x

1

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1 Quadratics

(

2 13 13 5 2 ( x + 2) −  = 2 x 2 + 4x + 4 −  2  2

)

5 3 9 4x 2 + 5x − 3 = 4  x 2 + x −   4 4 

= 2x2 + 8x + 8 − 13 = 2x2 + 8x − 5 1 x 3 − 6 a 5x 2 + x − 3 = 5 x 2 + 2 10 5

(

)

( )  1 241  = 5 ( x + − 20 ) 400   1 241 = 5( x + − 20 ) 80 18 b 5x – 18 = 5 ( x − 5 ) but can go no further. 7x 5 c 5 – 7x – 3x = −3 ( x + − 3 3)  7 49 5  = −3 ( x + ) − −  6 36 3   7 109  = −3  ( x + ) − 6 36   7 109 = −3( x + ) + 6 12 3x 11 2x – 3x + 11 = 2 ( x − + ) 2 2  3 9 11  = 2 ( x − ) − +  4 16 2   3 79  = 2 ( x − ) +  4 16    3 79  2 ( x − ) + >0 4 16    1 = 5 x + 20 



2

2

2

2

2

7

2

2

2

2

2

) − ( 85 ) − 34 

(

) − 6473 

 5 = 4 x + 8 

1 3 − − 400 5 

(

=4 x+

2

2

(

 5 = 4 x + 8 

2

2

2

Consequently b2 – 4ac < 0 because the curve of this equation will not intersect the x-axis and so there are no real roots. y 25 20

5 8

2

2

) − 1673 2

When 4x2 + 5x – 3 = 0

(

) − 1673 = 0 5 73 4(x + ) = 8 16 ( x + 85 ) = 6473

4 x+

5 8

2

2

2

x+

5 73 =± 8 64 x=

−5 ± 8

x=

−5 ± 73 8

73 64

2

 b  b2 10 ax 2 + bx + c =  ax + − +c  4a  2 a ax2 + bx + c = 0 2

 b  b2  ax +  − 4a + c = 0 2 a 2

 b  b2 b 2 − 4ac  ax +  = 4a − c = 4a 2 a ax +

b b 2 − 4ac =± 4a 2 a

ax +

b ± b 2 − 4ac = 2 a 2 a ax = −

15 10 5 –20 –15 –10 –5

0

2

5

10 15 20 25

b 2 − 4ac 2 a

ax =

−b ± b 2 − 4ac 2 a

x=

−b ± b 2 − 4ac a ×2 a

x=

−b ± b 2 − 4ac 2a

x

8 (1 – 2x) [(x + 4)2 − 1] = (1 – 2x)(x2 + 8x + 16 − 1) = (1 – 2x) (x2 + 8x + 15) = x2 + 8x + 15 – 2x3 – 16x2 – 30x = 15 – 22x – 15x2 – 2x3

b ± 2 a

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WORKED SOLUTIONS

Exercise 1.2A c

1 a x2 + 2x + 13 = 0



= −48

b2 – 4ac < 0 so no real roots. b2 – 4ac = (3)2 – (4)(2)(1)



=1

+ 12 – 5x = 0 b2 – 4ac = (−5)2 – (4)(3)(12)



= −119

b2 – 4ac < 0 so no real roots.

6 2x2 + 5x – c = 0

b 2x2 + 3x + 1 = 0

3x2

b2 – 4ac = (2)2 – (4)(1)(13)

b2 – 4ac = 0 so two equal real roots.



b2 – 4ac > 0 so two distinct real roots.

a No real roots:

b2 – 4ac < 0



(5)2 – (4)(2)(−c) < 0

c x2 + 4x + 4 = 0

b2



b2



– 4ac =

(4)2

– (4)(1)(4)

=0

2

– 4ac = 0 so two equal real roots.

2x2 – 5x + 3 = 0 b2 – 4ac = (−5)2 – (4)(2)(3) = 25 − 24 = 1

b2

3 a x2

– 4ac > 0 so has two distinct real roots. – 3x – 5 = 0



b2 – 4ac = (−3)2 – (4)(1)(−5)







b2 – 4ac > 0 so two distinct real roots.

= 29

b 3x2 – 2x + 7 = 0 b2 – 4ac = (−2)2 – (4)(3)(7) = −80

4

25 + 8c < 0

c 0



(5)2 – (4)(2)(−c) > 0

25 + 8c > 0 c>−



25 8

13 − 3x 3

7 x =

b2 – 4ac < 0 so no real roots. c x2 – 12x + 36 = 0 b2 – 4ac = (−12)2 – (4)(1)(36) =0 2 b – 4ac = 0 so two equal real roots.

3x2 = 13 – 3x

x2

As b2 – 4ac > 0 there are 2 distinct solutions.

+ (2k + 2)x + (5k − 1) = 0 b2 – 4ac = 0 for one repeated real roots. 2 (2k + 2) – (4)(1)(5k – 1) = 0 Expanding and simplifying. k2 – 3k + 2 = 0 or – k2 + 3k – 2 = 0 Both quadratic equations have the solutions k = 1 or k = 2. When k = 1, x2 + 4x + 4 = 0 and x = −2; when k = 2, x2 + 6x + 9 = 0 and x = −3

5 a 4 + 2x – 3x2 = 0 b2 – 4ac = (2)2 – (4)(–3)(4) = 52 b2 – 4ac > 0 so two distinct real roots. b 49 + x2 – 14x = 0 b2 – 4ac = (−14)2 – (4)(1)(49) =0

x2 =

13 − 3x 3

3x2 + 3x – 13 = 0 b2 – 4ac = 9 + 4 × 3 × 13 = 165 > 0 8 x2 – 3x + a2 = 0 No real roots when b2 – 4ac < 0 b2 – 4ac = 9 – 4a2 9 – 4a2 < 0 9 < 4a2 a2 > a
2 2

9 y = 3 – x meets y = x2 – 5x + 7 when 3 – x = x2 – 5x + 7 ie 0 = x2 – 4x + 4 b2 – 4ac = 16 – 16 = 0 One repeated root, so the line touches, but does not cross the curve.

3

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1 Quadratics

iii (x − 3)2 – 9 = 0

10 3x – xy + 1 = 0 3x + 1 x substitute into 3y – xy + 1 = 0 y=

( 3xx+ 1 ) − x ( 3xx+ 1 ) + 1 = 0 3x + 1 3( ) − 3x = 0 x



2 Factorisation: ›› can only be used on equations that factorise

›› sometimes spotting factors can be difficult ›› can solve: x2 + 3x + 2 = 0 ›› cannot solve: x2 − 5x – 7 = 0.

3



3x2 – 9x – 3 = 0 b2 – 4ac = 81 + 4 × 3 × 3 = 117 > 0 ∴ the curves intersect in two places.

mistake

›› can solve: x2 − 5x – 7 = 0 ›› cannot solve: x2 − 5x + 7 = 0.

Exercise 1.3A x2 – 9 = 0



(x + 3)(x − 3) = 0





x = 3 or −3

ii x=

0 ± 0 − ( 4 )(1)( −9) 2

x = 3 or −3

iii x2 = 9

x = 3 or −3

b i

x2

Quadratic formula: ›› can be used to solve any equation with real roots including ones that don’t factorise

›› cumbersome and consequently easy to make a

As b2 – 4ac > 0 there are two distinct real roots

1 a i

x = 0 or 6



Completing the square: ›› can be used to solve any equation with real roots including ones that don’t factorise

›› can be cumbersome manipulations if b is odd and a > 1, and consequently easy to make a mistake

›› can solve: x2 − 5x – 7 = 0 ›› cannot solve: x2 − 5x + 7 = 0. 3 a (x − 4)2 = 13

+ 3x + 2 = 0



x = 4 ± 13



(x + 1)(x + 2) = 0







Completing the square. The equation was already in form of a completed square.

x = −1 or −2

−3 ± 9 − ( 4 )(1)( 2) ii x= 2

x = −1 or −2

( 32 ) − 14 = 0

iii x+

2



x = −1 or −2

b 2x2 – 5x – 11 = 0

x=

5 ± 25 − ( 4 )( 2)( −11) 4



x=

5 ± 113 4



Does not factorise so used the quadratic formula. Alternatively, could have completed the square but chose not to as a > 1.

c i x2 – 5x – 7 = 0

Cannot be factorised.

ii x=



5 ± 25 − ( 4 )(1)( −7 ) 2

( 52 ) − 534 = 0 2



x=

5 ± 53 2

d i x(x − 6) = 0

x = 0 or 6

6 ± 36 − ( 4 )(1)( 0 ) ii x= 2

3x2 + 14x – 5 = 0

(3x − 1)(x + 5) = 0 1 x = or −5 3



Equation factorises so easy to do this. Alternatively, could have completed the square but chose not to as a > 1.

5 ± 53 x= 2

x− iii

c 3x2 = 5 – 14x

x = 0 or 6

4

d 4x2 – 16x + 7 = 0

(2x − 1)(2x − 7) = 0 1 7 x = or 2 2 Equation factorises so easy to do this. Alternatively, could have completed the square but chose not to as a > 1.

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WORKED SOLUTIONS

4 x2 + bx + c = 0

Subtract any extra values that completing the b2 square has produced, in this case− . 4



( )



Manipulate the equation to make x the subject.

x+

b 2

2

−

( ) b 2

x+





x+

b2 +c =0 4

2

=

b 2 − 4c 4

b ± b 2 − 4c = 2 2

−b ± b 2 − 4c 2 5 a  Cannot be factorised if b2 − 4ac is not a square number. x=



b2 – 4ac = 16 – (4)(3)(−11) = 148



148 not a square number so does not factorise.

b 3x2 + 4x – 11 = 0

x=

−4 ± 148 6



x=

−2 ± 37 3

6

(x − 3)(x − 5) = 8 x2 – 8x + 15 – 8 = 0 x2 – 8x + 7 = 0 (x − 7)(x − 1) = 0 x = 1 or 7 7 ax2 + bx + c = 0 bx c a x2 + + =0 a a

(





( ( ( (

)

b x+ 2a

) ) ) )

2

x+

b 2a

2

x+

b 2a

2

x+

b 2a

2

b2 c − 2+ =0 a 4a

x2 =

74x + 33 7

7x2 – 74x – 33 = 0 (7x + 3) (x – 11) = 0 The equation can be solved by factorising a quadratic expression. 9 a 9x2 + 55x – 56 = 0 b2 – 4ac = 552 + 4 × 9 × 56 = 5041 b 2 − 4ac = 71 so the quadratic can be factorised.



b (9x – 8)(x + 7) = 0 8 x = or x = – 7 9

Exercise 1.4A 1 a x2 +3x – 4 > 0

x2 + 3x – 4 = 0



(x + 4)(x − 1) = 0



x < −4 or x > 1

b x2 – 6x + 8  0

(x − 4)(x − 2) = 0



2x4

c x2 – 9  0

(x + 3)(x − 3)  0



x  −3 or x  3

d x2 – 6x < 0

x(x − 6) = 0



0 4

x2 − 4 > 0



(x − 2)(x + 2) > 0



x < −2 or x > 2

d 3x2  5x

3x2 – 5x  0



x(3x − 5)  0



0x

5 3

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1 Quadratics

3 a 6 – 5x – x2 > 0

c 12 + 9x +x2 > 3(x + 1)



(6 + x)(1 − x) = 0



12 + 9x + x2 > 3x + 3



−6 < x < 1



9 + 6x + x2 > 0



and – x + 6 > 0



(3 + x)2 > 0



6>x



Inequality true for all values of x.



x 3 – 4x – 3x2



{−5, −4, −3, −2, −1, 0}



0 > 2 – 5x – 3x2

b

2x2



3x2 + 5x – 2 > 0



(2x − 1)(x + 5) = 0 1 −5  x  2 or 2x < 5 5 x< 2 5 So x < 2



(3x – 1)(x + 2) > 0 x < –2 or x  1 [all integers excluding –2, –1 3 and 0]



+ 9x – 5  0

{all integers less than 2.5}

c (2x +

1)2

–90



6 a x2 + 5x – 6 < 0

(x + 6)(x – 1) = 0



–6 < x < 1 and x2 + 3x – 4 < 0



(x + 4)(x – 1) = 0



–4 < x < 1



2x + 1 = ±3



So –4 < x < 1



−2  x  1 and 2x < 6



{−3, −2, −1, 0}



x −3x



4x2 – 3x < 0



−6 < x < 1 and x < −4 or x > 1



x(4x −3) < 0 0 < x < 3 or 1 − 4x > 0 4 4 1 > 4x 1 x< 4 So x < 3 4 {all integers less than 0.75}



So −6 < x < −4



{−5}



πr 2 > 250



r > 8.9



So 8.9 < r < 10

5 a x2 + 4 < 7x − 2

x2



(x − 1)(x − 6) = 0



1 0 10 – 3x2  x – 5 3x2 + x – 5  0 3x2 + x – 5 = 0 when x =

−1 ± 1 + 60 −1 ± 61 = 6 6

−1 + 61 6

49 < 61 < 64 Consider



Multiply 2 by 2.



10x + 2y = 50



Add 1 and 3 .



13x = 52

3

−1 + 49 −1 + 7 6 = = =1 6 6 6

−1 + 61 ∴ >1 6

c x – 2y = 13 

1



−x + 3y = −15

2



Add 1 and 2 .

y = −2 and x = 9 2 5x + 6y = 6

–3

The higher value is

5x + y = 25

2

x = 4 and y = 5

3

–6

1





3x – y = 22 Multiply 2 by 6. 5x + 6y = 6



18x – 6y = 132  Add 1 and 3 . 23x = 138 x=6 Substitute x into 2 . 18 – y = 22

1 2 1 3

y = −4

3 If there are n unknowns then you need n distinct equations involving the n unknowns. 4 3x – 4y = 3

1



6x + 4y = 3

2



Add 1 and 2 . 9x = 6

5

x=

2 1 and y = − 4 3

x + y = 50 x − 1 = 15(y − 1) x − 15y = −14 Subtract 2 from 1 . 16y = 64

1 2

y = 4 and x = 46. Helen was 42 when her daughter was born.

No, the solution set for – 3x2  x – 5 is not a subset of x  1.

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1 Quadratics

6 2x − y = −14

1

1h + 1m = 33.9



3y − 2z = 16

2

Multiply A by 2



z−x=3 Multiply 1 by 3.

3

4m + 2p = 67.2

2A

2h + 2p = 42.6

B



6x − 3y = −42

4

Work out 2A – B



3y − 2z = 16 Add 4 and 2 .

2

4m – 2h = 24.6



C

D

Multiply C by 2

6x − 2z = −26 Multiply 3 by 2. 2z − 2x = 6  Add 5 and 6 . 4x = −20 x = −5 Substitute x into 3: z = −2 Substitute x into 1: y = 4

5 6

7 A sketch diagram helps see what is required. y = 2x + 8

y = 7 – 3x y

2h + 2m = 67.8

2C

Work out D + 2C 6m = 92.4 m=

92.4 = 15.4 6

1 maths textbook costs $15.40 (remember to add the ‘0’ to make the money notation correct).

Exercise 1.5B 1 a x + y = 3 

2x2



Add 1 and 2 .



2x2 + x − 28 = 0



(2x − 7)(x + 4) = 0

Find the coordinates of the point of intersection of the lines y = 2x + 8 and y = 7 – 3x



x=

7 – 3x = 2x + 8, so x = –0.2, y = 7.6



y=−

Required area x x-axis has equation y = 0

0

The intersections of the lines with the line y = 0 7 are when 0 = 2x + 8, x = –4 and 0 = 7 – 3x, x = 3 Area of the triangle =

( )

1 × base × height 2

1 7 1 = × 4+ × 7.6 = 24 square units 2 3 15 8 y = 4 – 5x meets y = 7 – 3x when 4 – 5x = 7 – 3x, so, x = –1.5, y = 11.5  (–1.5, 11.5) y = 4 – 5x meets y = x + 2 when 4 – 5x = x + 2, so, 7 1 1 7 x = ,y = , 3 3 3 3

( )

y = 7 – 3x meets y = x + 2 when 7 – 3x = x + 2, so x = 1.25, y = 3.25 The coordinates of the vertices are (–1.5, 11.5), 1 7 , , and (1.25, 3.25). 3 3

( )

− y = 25 

1 or 7 2



x2 + xy = −12



From 1 .



2x − 20 = y



Substitute into 2 .



x2 + x(2x − 20) = −12



3x2 − 20x + 12 = 0



(3x − 2)(x − 6) = 0



x=

2 or 6 3



y=

−56 or −8 3

c y = 4x

5 − x2 = y

1 2

1 2

1 = 2



5 − x2 = 4x

Let m be the cost of a maths textbook, h the cost of a history textbook, p the cost of a pen.



x2 + 4x − 5 = 0



(x + 5)(x − 1) = 0

2m + 1p = 33.6

A



x = −5 or 1

2h + 2p = 42.6

B



y = −20 or 4

9 Use the information to make 3 equations.

2

7 or −4 2

b 2x − y = 20



1

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WORKED SOLUTIONS

2 a Substitution or elimination: for elimination the first equation would need to be multiplied by 2 so that y can subsequently be eliminated.



b S  ubstitution only: neither addition or subtraction of the equations will eliminate a variable. c S  ubstitution only: neither addition or subtraction of the equations will eliminate a variable. 3 a 2x2 + y = 14

x − 2y = 11



Multiply 1 by 2.



4x2 + 2y = 28



Add 2 and 3 .



(4x + 13)(x − 3) = 0





x=

−13 or 3 4





y=

−57 or −4 8

b xy − x = −4 x + y = 1



From 2 .



y=1−x



Substitute into 1 .



xy = 140



From 1 .



x = y + 10



Substitute into 2 .

y2 + 10y − 140 = 0



y=



y = −10 ± 660 2



y = −5 ± 165 x = 5 ± 165

x2 − y = 7 From 2 . x2 − 7 = y Substitute into 1 .



)

x2

y2

+ = 64 From 1 . y=x+1 Substitute into 2 . x2 + (x + 1)2 = 64 2x2 + 2x − 63 = 0

1 2

x=



2

So the coordinates of the points of intersection of this line and this circle are (5.135, 6.135) and (−6.135, −5.135).

6 y − x = 10

1 2

y2

1



x2



The line is a tangent to the circle at (−5, 5).

+ = 50 From 1 . y = x + 10 Substitute into 2 . x2 + (x + 10)2 = 50 2x2 + 20x + 50 = 0 x2 + 10x + 25 = 0 (x + 5)2 = 0 x = −5, y = 5

2

7 Let the numbers be x and y. Then x + y = 129, x2 + y2 = 8433 The two numbers must satisfy both equations. x + y = 129 so y = 129 – x

−10 ± 100 − (4)(1)(−140) 2

4 2x − 3y = 13

5 y − x = 1

1

y(y + 10) = 140



(

4 47 this line and this curve are − 3 , − 9 and (2, −3)

−2 ± 4 − (4)(2)(−63) 4 x = 5.135 or −6.135 y = 6.135 or −5.135

x = ±2, y = −1 or 3



So the coordinates of the points of intersection of



x − x2 − x = −4

c x − y = 10



3

x(1 − x) − x = −4



2

4x2 + x − 39 = 0





1



2x − 3(x2 − 7) = 13 3x2 − 2x − 8 = 0 (3x + 4)(x − 2) = 0 −4 x= or 2 3 −47 or −3 y= 9

 Substitute into x2 + y2 = 8433   x2 + (129 – x)2 = 8433 x2

+ 16 641 – 258x + x2 = 8433 2x2 – 258x + 8208 = 0 x2 – 129x + 4104 = 0

1 2

 (x – 57)(x – 72) = 0 x = 57, x = 72 The two numbers are 57 and 72.

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1 Quadratics

8 The line y = 9 – 2x meets the circle (x – 3)2 + (y + 2)2 = 25 when (x – 3)2 + (9 – 2x + 2)2 = 25 (x – 3)2 + (11 – 2x)2 = 25

x = −5 square root sign missed



No real solutions.

3 x4 – 7x2 + 1 = 0 Let y = x2

x2 – 6x + 9 + 121 – 44x + 4x2 = 25

y2 – 7y + 1 = 0

5x2 – 50x + 105 = 0 x2 – 10x + 21 = 0

y=

7 ± 49 − 4 2

y=

7 ± 45 7 ± 3 5 = 2 2

(x – 3)(x – 7) = 0 x = 3, x = 7 y = 3, y = –5

y = x2

There are two intersections, and their coordinates are (3, 3) and (7, –5). 9 3x2 − 7x + 2 > 5x – 6 3x2 – 12x + 8 > 0 3x2 – 12x + 8 = 0 when 12 ± 144 − 96 12 ± 48 2 3 = =2± 6 6 3 2 When x = 2, 3x – 12x + 8 = –4, x=

∴ the solution set of 3x2 − 7x + 2 > 5x – 6 is x 0

x < −1.48 or x > 2.48

(

7x 1 − 4 2



= 4 x2 −



 7 = 4 x − 8 



=4 x−7 8

)

( ) − 6449 − 12 

(

2

) − 1681 2

b At the stationary point y will have a 7 minimum value i.e. when x = . 8

b 2x2 − 3x − 7 = 0

2

3 4

2

Let y = x3 −1 ± 12 + 8 −1 ± 3 y= = 4 4

2

2

19 ± 297 2

2y2 + y – 1 = 0

2

( 34 ) − 89 + 13 3 7 = 2 ( x + ) + 11 4 8 =2 x+

19 ± 361 − 4.1.16 2

x2 =

) − 169 + 132 





When x =

(

y=4 x− =−

7 . 8 7 8

) − 1681 2

81 16

So coordinates of stationary point are

6 a 3x − 2y = 7 x2 − y2 = 8 Rearrange 1 . 7 + 2y x= 3 Substitute into 2 .

( 78 ,− 1681 ).

1 2

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1 Quadratics

2



 7 + 2y  − y 2 = 8  3 

x 1− 2x = 1 + 21 × 2−2x = 1 + 9 4 = 1 + 2

4x = 22x

49 + 28y +4y2 − 9y2 = 72 5y2 − 28y + 23 = 0 (5y − 23)(y − 1) = 0 27 23 , and ( 3,1) 5 5

(

22x = 1 +

)



3x − 7 2 Substitute into x2 – y2 = c



x2 −



4x2

– (3x –



4x2

(9x2



– 5x2 + 42x – 49 – 4c = 0



5x2

b2 –

Let y = 22x y2 – y – 2 = 0

2

(y + 1) (y – 2) = 0 y = –1, y = 2

= 4c

22x = – 1 not possible, so 22x = 2

– 42x + 49) = 4c

– 42x + 49 + 4c = 0

4ac



= 1764 – 4 × 5 × (49 + 4c)



= 1764 – 980 – 80c



= 784 – 80c



= 8(98 – 10c)



For no intersection b2 – 4ac < 0



8(98 – 10c) < 0



98 – 10c < 0



98 < 10c

2x = 1, so x =

1 3 x < − or x > 3 2 8 a x2 + 10x + 21 > 0

(x + 3)(x + 7) > 0



x < −7 or x > −3 and 3x − 5 > 0 5 x> 3



5 3 b y = gf (x) = 3(x2 + 10x + 21) – a



x=



2 ± 4 − (4)(3)(−17) 6

2 ± 208 6 x = −2.070 or x = 2.737 =

x < −2.07 or x > 2.74 (3 s.f.) 11 a px2 + 5x – 6 = 0 b2 – 4ac < 0

25 – 4 × p × (–6) < 0



25 + 24p < 0

25 p



4p4 + 20p2 + 25 – 20p2 + 4pq = 0



4p4 + 4pq + 25 = 0 as required.

12 a

a 2x 2 + 3bx + c = 0  3bx c a2  x 2 + 2 + 2  = 0  a a 

= 3x2 + 30x + 63 – a 3(x2

= + 10x) + 63 – a 2 = 3 ( x + 5) − 75 + 63 − a 2 = 3 ( x + 5) − (12 + a )

1 2

10 3x2 + 5x − 2 > 7x + 15 3x2 − 2x − 17 > 0

c > 9.8 7 6x2 > 3 − 7x 6x2 + 7x − 3 > 0 (2x + 3)(3x − 1) > 0

– 22x – 2 = 0

∴ 42x – 4x – 2 = 0

( 3x2− 7 ) = c

–

2 22x

22x × 22x = 22x + 2 24x

b 3x – 2y = 7, so y =

7)2

2 22x

2





 x + 3b  − 9b 2 + c = 0   2a 2  4a 4 a 2

y = gf(x) has 2 real roots when 12 + a > 0, i.e. a > –12

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1 

WORKED SOLUTIONS

2

 x + 3b  = 9b 2 − c   2a 2  4a 4 a 2





=

9b 2 4a 2 c − 4a 4 4a 4

=

9b 2 − 4a 2 c 4a 4

x+



± 9b 2 − 4a 2 c 3b 2 = 2a 2a 2 −3b ± 9b 2 − 4a 2 c x= 2a 2



b  y = a2x2 + 3bx + c has a repeated root when c = 1

i.e. b2 – 4ac = 0



9b2 – 4a2 = 0



2 i.e. a 2 = 9b 4



x = −3b ±

16 hA = t + 4, t  0 hB = 8 + 3t – t2 hB > hA 8 + 3t – t2 > t + 4 4 + 2t – t2 > 0 0 > t2 – 2t – 4 t = 2 ± 4 + 16 = 2 ± 20 = 1 ± 5 2 2 When t = 0 hB > hA ∴ 0  t  1 + 5 seconds. 17 A sketch diagram often helps, it doesn’t need to be drawn accurately. A

x

B

a + 4x D

b 2 − 4ac = −3b = − 3b × 4 = − 2 2 9b 2 3b 2a 2 2a 2

13 a Pythagoras’ theorem AB2 = BD2 – AD2

C

Area = x(a + 4x) = 4x2 + ax When a = 3, Area  45 m2 4x2 + 3x  45

AB2 = (5x2 + 14xy + 10y2) – (2x + 3y)2

4x2 + 3x – 45  0

AB2 = 5x2 + 14xy + 10y2 – (4x2 + 12xy + 9y2)

Consider 4x2 + 3x – 45 = 0

AB2 = x2 + 2xy + y2 AB2 = (x + y)2 AB = x + y b So length = 19 + −12 = 7 or −1 + 3 = 2. 14 13πr  260 r  6.4 πr2 > 60 r > 4.4 4.4 < r  6.4 15 a











(

p + p −1

x=

−3 ± 32 − 4 × 4 × (−45) 2×4

−3 ± 27 −30 24 = or 8 8 8 24 = 3 is the only solution. As x > 0, x = 8 x=

The values of x for which the area is at least 45 m2 are x  3 m 18 6574t2 + 776t + 100 > 1 000 000

)

3

= 125

p + p −1 = 5 p +p−6=0 p+ p −6=0

6574t2 + 776t – 9 99 900 > 0 Consider 6574t2 + 776t – 999 900 = 0 −776 ± 776 2 + 4 × 6574 × 9 99 900 2 × 6574 −776 ± 162154.163 t= 13148 t=

b Let y = p

Reject the negative value, as t  0.

y2

t = 12.27°C

+y−6=0



(y + 3)(y − 2) = 0



So y = −3 or y = 2.



p = −3 or p = 2 So p = 4. (No solution from p = −3.)

The values of t for which the number of bacteria is great than 1 000 000 is t > 12.3° C (3 s.f.).

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1 Quadratics

Mathematics in life and work 3 =0 2

1 2x2 − 3x −

3 2



a = 2, b = −3, c = −



x=

3± 9+ 4×2×3 2 4



x=

3 ± 21 4

So x = 2

3 + 21 3 − 21 or . 4 4

3 − 21 3 + 21 −2, x ∈ } x+2 y

1 a g(1) = 12 = 1

4

b f(1) = (4 × 1) + 3 = 7

2

g(7) = 72 = 49 c g(−2) = (−2)2 = 4 g(4) = 42 = 16

1



1

2

3

4

5

6

7

8

9

10

x

d f(−2) = (4 × −2) + 3 = −5

–2 –3



Range {f(x): f(x) ⩾ 0, f(x) ∈ }

b gf(x) =

7 The relationship y =

1 isn’t currently a (x + a)2

function because there is a discontinuity at x = −a. The relationship could become a function if the domain was limited to either {x < −a, x ∈ − } or {x > −a, x ∈ + }

8 a { x: x > 3, x ∈ } As the curve continuous for x > 3. b { x: x < 0, x ∈ } Although the curve is

(

) − 94 − 15 = (2x + 32 ) 2

(a − 5)2 = 4



a2 − 10a + 25 = 4



a2 − 10a + 21 = 0



(a − 3)(a − 7) = 0



So a = 3 or a = 7.

4 fg(x) = 5(x2 + 3) – 2 = 5x2 + 13 gf(x) = (5x − 2)2 + 3 = 25x2 − 20x + 7

So fg ≠ gf. 5 a x + 2 = ff(x) b x2 + 2x − 6 = gf(x) 6 f(x) = 2x − 3, g(x) = x3 and h(x) =

hg(x) =

1 x3 + 1

}



fhg(x) =

2 −3 x3 + 1

10 f(x) = px5 + qx3 + 2



ffhg(x) =



f 2hg =

7

fg(x) = 2x − 3 gf(x) = 2x − 3 fg = gf 2x – 3 = 2x − 3

Hence f(x) is one-one for the set of values x: x ∈  , x  − 3 . 4

{

f(–1) = – p – q + 2 = 7 – p – q – 5 = 0 A f(2) = 32p + 8q + 2 = 10 32p + 8q – 8 = 0 4p + q – 1 = 0 B Work out A + B 3p – 6 = 0, so p = 2, q = – 7 11 2x2 – 4x + 13 = 2(x2 – 2x) + 13 = 2(x – 1)2 – 2 + 13 = 2(x – 1)2 + 11 This is a one-one function for values {x: x ∈ , x  1} , so p = 1.

2

2



2

69 – 4 3 This has a minimum value of – 69 when x = − . 4 4 3 f(x) is one-one for x  − . 4

9 4x2 + 6x – 15 = 2x + 3 2

2

3 fg(a) = 4



continuous for x < 3, the question specifies { x ∈ − }

( 1x ) − 1 = 1 −x x

c gg(x) = (x2 − 1)2 − 1 = x4 − 2x2 1 d ff(x) = 1 = x x

d=3



f(−5) = (4 × −5) + 3 = −17

2 a  Cannot be found as the range of g(x){x : x  – 1} is not fully included in the domain of f(x){x : x ≠ 0}.

6 h(2) = d − 2 j(2) = 2 + 2 = 1 4 d–2=1

f(1) = (4 × 1) + 3 = 7

3

0 –1 –1

2 



1 x +1

4 −6−3 x3 + 1

4 −9 x3 + 1

2x = 2x − 3 23 2x = (8)(2x) − 24 24 = (7)(2x)

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2 FUNCTIONS y

( )

ln 24 7 x= ln 2 = 1.78 (3 s.f.)

6 5 4

fg(x) can be found as the range of g is included in the domain of f, but gf(x) cannot be found as f(x) > 0 for all x and the domain of g is x: x < 0. 8

f(x) =

2 x −1

3 2 1 –4

2 2 = 2 −1 2 − x −1 x −1 x −1 x −1 2 2( x − 1) = = 3− x (3 − x) x −1 The domain is { x : x ∈  , x > 3} .

–3

–2

ff(x) = f ( f(x)) =

9

fg(x) = f(g(x)) = 4(x2 – 1)3 – 2(x2 – 1)2 + 7 =

–

3x4

+

3x2

– 1) –

2(x4

–

2x2

)

g(x) = x + 5 , or g(x) = x + 5 4 4 4

Exercise 2.3A a This is a linear function so it is one-one. f(x) = 5 − 4x {x ∈ } y = 5 − 4x 5− y x= 4

3

4

5

6

x

–3

b This is one-one for the given domain. f(x) = (x − 3)2 {x ∈ , x ⩾ 3} y = (x − 3)2

f−1 (x) = 3 + x {x ⩾ 0, x ∈ }

+ 1) + 7

10 f(x) = 4x – 5, for fg(x) = x ; g(x) must involve 1 x in 4 1 order for 4 × x = x . 4 5 5 It must also involve in order for 4 × = +5 . 4 4 x 5 Check: If g(x) = + : 4 4 x 5 –5=x+5–5=x fg(x) = 4 × + 4 4

1

2

–2

y

= 4x6 – 14x4 + 16x2 + 1

(

1

x=3+ y

f(x) = 4x3 – 2x2 + 7, g(x) = x2 – 1 4(x6

0 –1 –1

6 5 4 3 2 1 –2

0 –1 –1

1

2

3

4

5

6

x

c The cube of any number is unique so the function is one-one. f(x) = x3 {x ∈ } y = x3 x=

3

y

f−1 (x) = 3 x {x ∈ } y 4 3

5−x f−1(x) = {x ∈ } 4

2 1 –5

–4

–3

–2

0 –1 ––1

1

2

3

4

5

x

–2 –3 –4

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2

WORKED SOLUTIONS

2

f(x) = 7x − 2

5

Let y = 7x − 2. y+2 =x 7

a f(x) =

x+2 7 2+2 4 −1 f (2) = = 7 7

2 + 5 , where {x ∈ , x > 1}. x −1

Let y =

2 + 5. x −1

y−5=

2 x −1

f −1(x) =

3

Let y = 5 − 4x. x=

y

5−x {x ∈ } 4

6

6

5

4

4

2

3 2

–4

4

6

8

1 0 –1 –1

–8

b

f−1(x)

1

2

3

4

= x + 4 {x ∈ , x ⩾ 0}

4x2 − 24x + 12 = 0

f(x) = (x − 4)2 {x ∈ , x ⩾ 4}

x2 − 6x + 3 = 0

y 20

8

9

10 11

x

(Only 3 + 6 is valid as 3 − 6 is not included in the domain of either function.)

y = f(x)

15 10

y = f –1(x)

5 2

4

6

8 10 12 14 16 18 20 22 24

–10

x

6

f(x) = 2x2 + 5x − 13 Let y = 2x2 + 5x − 13.

 ( ) − 129 16 

 5 y = 2 x + 4  x=

 y= x+7 2 

f −1(x) =

( ) − 494 − 11   y = ( x + 7 ) − 93  2 4   2

2

ff −1(x) =

y + 93 − 7 4 2

{

8x + 129 5 − 16 4

 8x + 129 5   8x + 129 5  ff −1(x) = 2  −  + 5 −  − 13 16 4 16 4  

2

x + 93 − 7 4 2

2

8y + 129 5 − 16 4

f(x) = x2 + 7x − 11 Let y = x2 + 7x − 11.

f −1(x) =

7

x=3± 6

25

x=

6

2 2 +5= +1 x −1 x−5 2(x − 5) + 5(x − 1)(x − 5) = 2(x − 1) + (x − 1)(x − 5)

x = (y − 4)2

–6 –4 –2 0 –5

5

c f(x) = f−1(x)

Let y = x + 4.

4

y = f –1(x)

2

x

y = f –1(x)

–6

y = f(x)

7

y

0 –2 –2

–4

2 + 1, where {x ∈ , x > 5}. x−5

b

y = f (x)

–6

f−1(x) =

5− y 4

f(x) =

2 +1 y−5

x=

a f−1(x) = 5 − 4x {x ∈ }

8x + 129 8x + 129 25 8x + 129 25 52 −5 + +5 − − 8 16 8 16 4 4

ff −1(x) = x +

}

The domain is x : x ∈  , x > −93 . 4

129 25 50 104 + − − 8 8 8 8

ff−1(x) = x

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2 FUNCTIONS

7

3x − 1 f(x) = 3x − 1 Let y = x−2 x−2

3

= 4 x + 1 − 2x + 5 The domain is { x : x ∈  , x > −1}.

Then (x – 2) y = 3x – 1 xy – 2y = 3x – 1

10 f(x) = x6 – 2x3 – 7

xy – 3x = 2y – 1

Let y = x6 – 2x3 – 7

x(y – 3) = 2y – 1

Then y = (x3 – 1)2 – 1 – 7

2y − 1 x= y−3

y = (x3 – 1)2 – 8 y + 8 = (x3 – 1)2

f −1(x) = 2x − 1 x−3

x3 − 1 =

The domain is { x : x ∈  , x > 3}. 8

f(x) =

2x2

Let y =

x3 =

– 4x + 13

2x2

– 4x + 13

Then y = 2(x2 – 2x) + 13 y = 2[(x –

1)2

– 1] + 13

x=

3

y + 8 +1

f −1(x) =

3

x + 8 +1

Exercise 2.4A

y − 11 = (x − 1)2 2

1

y − 11 +1 2

f −1(x) =

y + 8 +1

The domain is { x : x ∈ , x  −8} .

y = 2(x – 1)2 + 11

x=

y+8

a f(x + 3) one solution y 25

x − 11 + 1 2

20 15 10

2

    ff −1(x) = 2 ×  x − 11 + 1 − 4  x − 11 + 1 + 13 2 2      x − 11   x − 11  x − 11 = 2 +2 + 1 − 4  + 1 + 13 2 2  2    = x − 11 + 4

x − 11 x − 11 +2−4 +9 2 2

5 –9 –8 –7 –6 –5 –4 –3 –2 –1

1

2

3

x

4

5

6

x

b f(x) + 3 no solutions y 25 20

=x

15

f −1f ( x ) =

10

2

2x − 4x + 13 − 11 +1 2

=

2x 2 − 4x + 2 +1 2

=

x 2 − 2x + 1 + 1

=

( x − 1)

2

5 –6 –5 –4 –3 –2 –1 0

3

y

+1

20 15 10 5 –4 –3 –2 –1 0

f(x) = 4x3 – 2x2 + 7, g(x) = x2 – 1 Let y = x2 – 1; then x = fg −1(x) = 4 x + 1 − 2

2

25

ff–1(x) = f–1f(x) = x as required.

3

1

c f(x − 4) one solution

=x–1+1=x 9

0

(

y + 1 , i.e. g −1 ( x ) = x +1

)

2

1

2

3

4

5

6

7

8

9 10 11

x

x + 1,

+7

3

= 4 x + 1 − 2x − 2 + 7

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2

WORKED SOLUTIONS

d f(x + 4) one solution

d y

y

3

25 20

2

15

1

10 5

–6

–11–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0

2

1

2

3

4

–5

–4

–3

0 –1 –1

–2

x

–3

–2

0 –1 –1

1

2

3

4

5

6

x

3

12

–3

10

–4

8

P (0, 3) P (0, 2)

3

–10 –8

2

–6

–4

–2

0 –1 –1

4 2 –2 0 –2

2

x

–4

1 –3

10

6

y

–4

8

x

y

–2

b

–5

6

6

14

One solution

–6

4

5

One solution

1 –4

4

–4

2

–5

3

–3

3

–6

2

–2

y

a

1

1

2

3

4

5

6

–6

x

–2

f(x + 1) = f(x) + 1 has one solution 4

–3 –4

a, b f(x − 180), where −180° ⩽ x ⩽ 180°; asymptotes at x = −90 and 90. Axis intercepts at (−180, 0), (0, 0) and (180, 0). y

One solution c y 3 2

–90°

1 –6

–5

–4

–3

–2

0 –1 –1 –2

1

2

3

4

5

6

0

x

90°

x c, d f(x) + 2; asymptotes at x = −90 and 90

–3 3

y

–4 4

One solution 2

–90° 90

0

90° 90

x

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2 FUNCTIONS

5

a, b f(x + 2) = 2x + 3

1  Translation, vector    2

f(x) + 2 = 2x + 1 + 2 9

f(x + 2)

y 25 20

f(x) = (x + 2)2 – 3 drawn in blue, g(x) = (x – 1)2 – 2 in red y 6

f(x)+ 2

15

4

10

2

5 –5

–4

–3

–2

–1

0

1

2

3

4

5

–8

x

–6

–4

0 –2 –2

2

4

6

x

–4 –6

c i If g(x) = 2x, then g(x + 3) = 2x + 3 = f(x + 2).

6

 3 a Translation vector   1

ii If g(x) = 2x, then g(x + 1) + 2 = 2x+1 + 2 = f(x) + 2. f(x) = x2 + 6x – 15 = (x + 3)2 – 24  − 3 f(x) is a translation of y = x2 with vector   −24

 −3 b Translation vector    −1  −2  c Translation vector    −3 

So, the transformation which maps f(x) to x2 is a  3 translation with vector   .  24 7

1  d Translation vector    −2 

f(x) = x2 – 2, g(x) = x – 1, ∴ g–1(x) = x + 1 fg–1(x) – 3 = (x + 1)2 – 2 – 3 = (x + 1)2 – 5 y

–8

–6

–4

10 f(x) = x3, ∴ f −1 ( x ) =

8

f(x + 2) = (x +

6

drawn in red

4

y

2

6

0 –2 –2

2

4

6

–8

–6

–4

0 –2 –2

= x2 – 2x + 1 – 2x + 2 + 3 = x2 – 4x + 6 = (x – 2)2 – 4 + 6 = (x – 2)2 + 2 y 10

2

4

6

x

Exercise 2.4B 1

a f(3x) one solution y 8

8

7

6

6 5

4

4

2 0 –2 –2

( x + 2)

–6

hfg(x) = hf(x – 1) = (x – 1)2 –2(x – 1) + 1 + 2

–4

3

–4

f(x) = x2 – 2x + 1, g(x) = x – 1, h(x) = x + 2

–6

drawn in blue ∴ f −1 ( x ) =

2

–6

–8

x

4

x

–4

8

2)3,

3

3

2

4

f(x) = x2 – 2x + 1 = (x – 1)2

6

2

x

1 –3

–2 –1 0 –1

1

2

x

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2

WORKED SOLUTIONS

b 3f(x)

b 2f(x)

one solution y

y 10

8

8

7

6

6 5

4

4

2

3

0 –4 –3 –2 –1 –2

2 1 –3

–2 –1 0 –1

1

2

1

2

3

4

5

x

–4

x

–6 –8

c f(2x)

c f(−x)

one solution y

y 3

8

2

7

1

6 5 4

–6

–5

–4

–3

–2

3 1 1

2

x

4

5

6

x

d −f(x − 2) y 10 8 6

y

1

4

–4 –3 –2 –1 0 –1

1

2

3

4

x

2

–2

–4

–3 –4

–2

0 –1 –2

–6

–6

–7

–8

a f(2x)

3 y

1

2

3

4

x

5

4

8

3

6

2

4

1

2 0 –1 –2 –4

( )

1 f(−2x); P (1, 0), R − ,0 , Q (0, −2) 2 f(−2x) = 0 has two solutions. y

10

–2

–3

–4

–5

–3

3

–8

one solution

–4

2

–3

–2 –1 0 –1

1 d − f(x) 2

2

1

–2

2

–3

0 –1 –1

1

2

3

4

5

x

0 –6 –5 –4 –3 –2 –1 –1

1

2

3

4

5

6

x

–2 –3

–6

−2f(x); P (−2, 0), R (1, 0), Q (0, 4)

–8

−2f(x) = 0 has two solutions.

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2 FUNCTIONS

6

y 5 4

a, b, d f(2x) = 22x + 1 no solutions and 2f(x) = (2)(2x + 1) no solutions y

3

7

2

6

1 0 –6 –5 –4 –3 –2 –1 –1

5 1

2

3

4

5

6

x

4

–2

4

y = 2f(x)

y = f(−x) and y = −f(x) in both cases asymptote at y = 0. For y = f(−x) P is at (0, 1). For y = −f(x) P is at (0, −1). f(−x) = −f(x) has no solutions (they do not intersect).

–7

–5

–4

–3

–2

0 –1 –1

1

2

3

x

7

f(x) = x2 – 2, g(x) = x – 5, ∴ g–1(x) = x + 5

2

g–1(–x) = – x + 5

1

2fg–1(– x) = 2[(5 – x)2 – 2] = 2(5 – x)2 – 4 This is the most useful form to use to produce a sketch graph. y

0 –6 –5 –4 –3 –2 –1 –1 –2 –3

1

2

3

4

5

6

x

y = –f(x)

10

–4

5

–6

y = f(2x)

c i If g(x) = 2x, g(2x + 1) = 22x + 1 and ii 2g(x + 1) = (2)(2x+1)

4 3

2 1

y y = f(–x)

3

a, b y = f(2x), where −180°  x  180°; asymptotes at x = −135°, −45°, 45° and 135°.

5

Axes intercepts at (−180°, 0), (−90°, 0), (0, 0), (90°, 0) and (180°, 0).

0

–5

y

5

x

–5

8

f(x) = x2 – 2x + 1, or f(x) = (x – 1)2 g(x) = 2x,

–180

0

–90

180 x

90

h(x) = – x hfg(x) = – [(2x)2 – 2(2x) + 1] = – 4x2 + 4x – 1 Or, using f(x) = (x – 1)2 hfg(x) = – (2x – 1)2

c, d y = −2f(x), where −180°  x  180°; asymptotes at x = −90° and 90°. Axes intercepts at (−180°, 0), (0, 0) and (180°, 0).

Horizontal stretch with scale factor 1 and a 2 reflection in the x-axis y 2

y –2

0

2

x

–2 –180

–90

0

90

180

x –4

–6

24

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2

WORKED SOLUTIONS

9

y = f(x) drawn in red, y = g(x) drawn in green. y

3

a

y

4

–5

–4

–3

–2

{f(x) ∈ , f(x) < 17 }

3

0.15

2

0.1

1

0.05

0 –1 –1

1

2

3

x

5

5.5

–0.05

–2

–0.1

–3

–0.15

–4

b

Let y =

1 . x+2

x+2=

1 y

–5 –6

a One-way stretch vertically with scale factor − 4 3 b One-way stretch vertically with scale factor − 3 4 c One-way stretch vertically with scale factor −3

x=

1 −2 y

=

1 − 2y y

f(−1)(x) =

1 − 2x x

d One-way stretch vertically with scale factor 4 10 f(x) = x3, so f −1(x) =

3

x

f(2x) = (2x)3 (red), f −1(2x) = 3 (2x) (green) y

4

4

2

–5

–4

–3

0 –1 –1

–2

1

2

3

x

–3 –4

b fg(x) =

Exam-style questions

gf(x) =

) − 474 47 1 y+ = 3( x − ) 4 2 2

f −1(x) = 2

5

y 47 + 3 12 1 + 2

x 47 + 3 12

 −2 a Translation with vector   0 b Reflection in the y-axis  0 c Translation with vector   1 

9.5

x

1 range of f(x) not all included x 2 + 5x − 13

within domain of g(x) so not an appropriate composition of functions.

2

1 x= + 2

9

2

b y = 3(x2 – x) – 11 1 2

8.5

( 1x ) + x5 − 13 range of g(x) included

within domain of f(x)

(

8

1 =x 1 x

a {f(x) : f(x)  –11.75, f(x) ∈ } y=3 x−

7.5

g2(x) = gg−1(x) The inverse of g(x) is the same as g(x). Consequently g2(x) = gg−1(x) become the same composition of functions.

–2

1

7

1 x

gg−1(x) =

1

6.5

a g2(x) = 1 = x 1 x g −1(x) =

3

6

Let y =

1 1 1 ; then 3 − x = and x = 3 − 3−x y y

f −1 ( x ) = 3 − 6

1 x

a fg(x) = 2(x − 3) − (x − 3)2 = 2x − 6 − (x2 − 6x + 9) = 2x − 6 − x2 + 6x − 9 = −x2 + 8x −15

25

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2 Functions

y

b

x + 7 = h2fg(x)

10 a

1

b x

0.5 –6

0 –2 –0.5

–4

2

4

6

8

10

x

1

11 a Let y =

7 a f(x) = x2 − 4x + 7

x2 − 4x + 7 = (x − 2)2 + 3



When x > 2, (x − 2)2 + 3 > 0.



=

(2x + 3)(x + 2) (x + 2)(x − 1)





=

2x + 3 x −1

6



y (x − 1) = 2x + 3

4



yx – 2x = y + 3

2



x (y – 2) = y + 3



x=



f −1(x) =

y 8

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

x

{f(x) ∈ , f(x) > 3}

c Let y = (x − 2)2 + 3.

x = 2+



f −1(x) = 2 + x − 3

y−3

2(x2

8 a f(x) = + 8x – 14 = + 4x) – 14    = 2[(x + 2)2 – 4] – 14 = 2(x + 2)2 – 22 f(x) = 2(x + 2)2 – 22 b f(x) = 2(x + 2)2 – 22 Let y = 2(x + 2)2 – 22. Then (x + 2)2 =



x=

y + 22 2

 − 2 with vector  .  −22

x + 3 = 7x − 14



6x − 17 = 0



x=

17 6

12 a f(x) = 3x2 + 15x – 10 = 3(x2 + 5x) – 10

(

) − 754 − 10 2

2

13 a

2 22x = 22x − x = gf(x) 2x

Let y = 2x. 2 × 2x × 2x – 2x = 2y2 – y = fg(x)

30



25

c 8x4 – 8x3 + x = 8x4 – 8x3 + 2x2 – 2x2 + x

20



= 8x4 – 8x3 + 2x2 – (2x2 – x)

15



= 2(4x4 – 4x3 + x2) – (2x2 – x)

10



= 2(2x2 – x)2 – (2x2 – x) = ff(x)

5 –4



b 2x(2x + 1 – 1) = 2x(2 × 2x – 1) = 2 × 2x × 2x – 2x

y

–6

x+3 =7 x−2

115 Minimum value of − 4 . b f(5) = 75 + 75 – 10 = 140, so range is f(x)  140.

c One-way stretch scale factor 2, translation

–10 –8





x + 22 − 2 2

9 a

x+3 x−2

= 3 x+5 2

y + 22 −2 2

f −1 ( x ) =

y+3 y−2

b f −1(x) = 7

2x2



2x 2 + 7x + 6 x2 + x − 2



2



+ 7 = ghf(x)

c x 2 + 7 = gh 2f(x)

–1

b

−1 2

0 –2 –5

2

4

6

8

x

b f(−x) reflection in y-axis

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WORKED SOLUTIONS

14 a gf(x) =

1 1 1 = = 1 − 3(2x − 5) 1 − 6x + 15 16 − 6x

c y = g(x) in red, y = h(x) in blue y 10

b 2g(x) = f(x) 2 = 2x − 5 1 − 3x 2 = (2x – 5)(1 – 3x)

= 2x – 5 – 6x2 + 15x



0 = – 6x2 + 17x – 7



x = −17 ± 121 = 1 or 2 1 −12 2 3

15 a f(x) = x3 + x2 + x + 1 g(x) = (x – 1)3 + (x – 1)2 + (x – 1) + 1 – 3 g(x) = (x – 1)3 + (x – 1)2 + x – 3 b h(x) = –

2x3

+

2x2

– 2x + 1



= 2(–x)3 + 2(–x)2 + 2(–x) + 1



= 2[(–x)3 + (–x)2 + (–x) + 1] – 1

5

–10

y(5 – x) = x 5y = x + xy = x(1 + y)

x=

5y 1+ y

f −1(x) = 5x 1+ x

b ff(x) = f(x), x x 5− x 5− x = 5− x = x × 5(5 − x) − x 5 − x 25 − 6x 5− x 5−x 5− x x x = 25 − 6x 5 − x

x(5 – x) = x(25 – 6x) 5x – x2 = 25x – 6x2

0

–5

x

5

–5

 −1 d Translation vector   0

(

18 a f(x) = ax2 + bx + c = a x 2 +



One-way stretch vertically scale factor 2, 0 reflection in y-axis, translation with vector   .  −1 x 16 a Let y = 5−x

2 



(

) − 4ba  + c

 b = a x + 2 a 

(

=a x+

b 2a

q c −

b2 4a

2

)

b x +c a

2

2

) − 4ba + c 2

2

b f(–1) = –9, f(1) = 1, f(2) = 15 a – b + c = –9

A



a + b + c = 1

B

4a + 2b + c = 15

C



A+B

2a + 2c = –8, ie a + c = –4

2A + C

6a + 3c = –3, ie 2a + c = –1

D E

E–D

a = 3 c = –7 b = 5 c

y 5

5x2 – 20x = 0 5x(x – 4) = 0 x = 0 or x = 4 17 a x2 + 6x + 4 = (x + 3)2 – 9 + 4 = (x + 3)2 – 5  −3 b Translation vector    −5

–5

0

5

x

–5

–10

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2 Functions

19

296 – 20x

296 – 20x 10x

10x

10x

10x

214 – 20x

214 – 20x 10x

10x

a v(x) =  (296 – 20x)(214 – 20x)10x = 40(148 – 10x)(107 – 10x)x b Domain is {x: x ∈  ,0 < x < 10.7}. Range is {v(x): v(x) ∈ , 0  v(x) < 1158}. c x = 3 cm v = 236 × 154 × 30 = 1 090 320 mm3, or 1090 cm3 (3 s.f.)

Mathematics in life and work 1 {t ⩾ 0, t ∈ } 2 f(t) 14 000 12 000 10 000 8000 6000 4000 2000 0

1

2

3

4

5

6

7

8

9

10

t

3 As the car ages its value depreciates.  t  4 fg(t ) = 15 000 − 1500  210    5 Car Model A has the greater rate of depreciation, as illustrated on the following graph. f(t) 14 000 12 000 10 000 8000 6000 4000 2000



0

1

2

3

4

5

6

7

8

9

10

t

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3 

WORKED SOLUTIONS

3 Coordinate geometry Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 Gradient =

3 a Multiply by 3: 3y = x − 21, and rearrange.

9−5 4 −2 . = = ( −2) − 8 −10 5



2 Gradient of first line = −3. 1 . 3 Equation of perpendicular line through (5, 8) is given by y − 8 = 13 (x − 5).

x − 3y + 19 = 0 3 Substitute y = 2 x − 2 into x + 2y = 17. 3

(



7 x = 21 3



7x = 63 x = 9 2 When x = 9, y = 3 × 9 − 2 = 4. Hence point of intersection = (9, 4). 4 Complete square on x2 − 14x + 23. x2 − 14x + 23 = (x − 7)2 − 49 + 23 (x − 7)2 − 26

Exercise 3.1A 1 The lines not in the form ax + by + c = 0, where a, b and c are integers are: c, because it does not equal zero d, because it does not equal zero e, because c is not an integer f, because a, b and c are not integers. 2 a Rearrange y = 4 + 5x.

5x − y + 4 = 0, where a = 5, b = −1 and c = 4

Multiply by 2: 6y = 8x + 21, and rearrange.

4 Rearrange: 8x + 3 = 2y

Rearrange and divide by 2: y = 4x +

3 2

gradient = 4 and the y-intercept

)

4 x − 4 = 17 3

21 . 2

8x – 6y + 21 = 0

( 32 )

2 x + 2 x − 2 = 17 3 x +

2x + 5y − 30 = 0

c Multiply by 3: 3y = 4x +

Perpendicular gradient =

3y − 24 = x − 5

x − 3y − 21 = 0

b Multiply by 5: 5y = − 2x + 30, and rearrange.

has coordinates 0,

5 The line in the form ax + by + c = 0, where a, b and c are integers is: 5y =x−4 3 5y = 3(x – 4) = 3x − 12 3x − 5y − 12 = 0 6 When the line intersects the x-axis, y = 0. Substituting y = 0 −3x + 2 = 0



2 = 3x 2 x= 3 So the coordinates of the x-intercept are 2 , 0 . 3 When the line intersects the y-axis, x = 0.

( )

Substituting x = 0 −5y + 2 = 0 2 = 5y 2 y= 5 So the coordinates of the y-intercept are 0, 2 . 5 y 1 7 x + − = 0 . 3 2 Multiply by 6: 6x + 2y  – 3 = 0, a = 6, b = 2, c = –3

( )

b Rearrange y = 3 − 2x.

2x + y – 3 = 0, where a = 2, b = 1 and c = −3

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3 Coordinate geometry

8

4x 7 5y + = 5 3 4 Multiply by 60: 48x + 140 = 75y, so 48x – 75y + 140 = 0 5 1 = − , 5 × 4y = – 1 × 7x, 20y = – 7x, 7x 4y

9

Numerator and denominator confused. Should be: m=

4 a m =

(x2, y2) = (4, −8)

1 5 − 3n = in the required form 2x 3y

multiply by the denominators, 2x, 3y. This gives a term 2x × 3y × ( – 3n) = – 18nxy, which is non-zero if n ≠ 0.

Exercise 3.1B y − y1 1 a m = 2 x 2 − x1

8−3 5 = =1 7−2 5 y − y1 b m = 2 x 2 − x1

m=

y-coordinate incorrect in (x2, y2). It should be (5, −8).

(x1, y1) = (1, −7)



(x2, y2) = (4, −16)



−16 − −7 = −9 = −3 3 4 −1 Lie on a straight line with a gradient of −3. m=

y 2 − y1 x 2 − x1

1 − 7 −6 = −3 = 3−1 2 Lie on a straight line with a gradient of −3. m=

5 Ascent: y − y1 m= 2 x 2 − x1 (x1, y1) = (−10, 0) ( x2, y 2 ) = 0, 12

( )

m=

−5a − a −6a = = −3 3a − a 2a y − y1 3 Should be m = 2 . x 2 − x1

y 2 − y1 x 2 − x1





m=

(x2, y2) = (3a, −5a)

m=

(x2, y2) = (3, 1)

(x2, y2) = (3, −9)

5a − 3a 2a 2 = = 8a − 3a 5a 5 y − y1 b m = 2 x 2 − x1 (x1, y1) = (a, a)

(x2, y2) = (4, 9)

(x1, y1) = (1, 7)

(x1, y1) = (1, −3)





d m =

y − y1 c m = 2 x 2 − x1

(x1, y1) = (3a, 3a)

(x1, y1) = (1, 3)



9−3 6 = =3 5−3 2

(x2, y2) = (8a, 5a)

y 2 − y1 x 2 − x1





(x2, y2) = (5, 9)

−9 − −3 −6 = −3 = 3−1 2 y − y1 2 a m = 2 x 2 − x1

Lie on a straight line with a gradient of −3.

c m =

m=

(x1, y1) = (3, 3)





9−3 6 =2 = 4 −1 3 Do not lie on a straight line with a gradient of −3.

(x2, y2) = (7, 8)

m=

m=



(x1, y1) = (2, 3)



−8 − 1 = −9 = −3 3 4 −1



b m =

n = 0 is the only value.



y 2 − y1 x 2 − x1

(x1, y1) = (1, 1)

20y + 7x = 0, a = 7, b = 20, c = 0 10 To write

−8 − 2 −10 −5 = = 5−1 4 2



1 1 −0 1 2 = 2 = 0 + 10 10 20 Descent: y − y1 m= 2 x 2 − x1

m=

( 12 ) ( x , y ) = ( 2, 83 )

(x1, y1) = 0, 2

2

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3 

WORKED SOLUTIONS

m=



Exercise 3.1C

− 12 − 18 1 = =− 16 2−0 2 3 8

The descent is the steepest because 1 > 1 . 16 20 6 m =

y 2 − y1 x 2 − x1

( 12 , 13 ) ( x , y ) = ( 34 , − 32 )

1 a m = 2 and (x1, y1) = (3, 0) y – y1 = m(x – x1) y − 0 = 2(x − 3) y = 2x − 6

( x1, y1 ) =



2

b m = 3 and (x1, y1) = (0, 3) y − y1 = m(x – x1) y − 3 = 3(x − 0) = 3x y = 3x + 3

2

−2 − 1 3 3 = −1 = −4 3−1 1 4 2 4



m=

7

1 5 − 3n = 2x 3y

c m = 2 and (x1, y1) = (3, 4) y − y1 = m(x − x1) y − 4 = 2(x − 3) = 2x − 6 y = 2x − 2

Multiply by 3y × 2x. 3y – 3n × 2x × 3y = 5 × 2x

d m = −5 and (x1, y1) = (2, 3) y − y1 = m(x − x1) y − 3 = −5(x − 2) = −5x + 10 y = −5x + 13

For this to be a straight line n = 0. 1 3y = 10x. Gradient = 3 . 3 8

1 1 y1 − y 2 a + 5 a + 5 1 = = = −6 a + = −6a − 1.2 5 x1 − x 2 1 − 1 −1 6 3 2

( )

Gradient = –2 ∴ – 6a – 1.2 = – 2, 6a = 0.8, 0.8 8 2 a= = = 6 60 15 9

4x 7 5y + = . Multiply by 60 : 48x + 140 = 75y, 5 3 4 so y =

2 a m = −4 and (x1, y1) = (−2, −5) y − y1 = m(x − x1) y − −5 = −4(x + 2) y + 5 = −4x − 8 y = −4x − 13 b m = −1 and (x1, y1) = (2, −2) y – y1 = m(x – x1) y − − 2 = −1(x – 2) y + 2 = −x + 2 y = −x

48x 140 + 75 75

Gradient =

48 48 , and 0 < < 1 as required. 75 75

10 For AB: 1+1 5 y1 − y 2 5 40 200 = 6 9 = 18 = − × =− x1 − x 2 − 1 − 7 − 43 18 43 774 5 8 40 −1 < −

200

> as 2 10 10 10

–1 –2 , 5 7

( 6445 ) + ( 569 ) 2

2

= 1.43 3 s.f.

Exercise 3.3A 1 a centre = (−5, 8), radius = 6 b centre = (19, 33), radius = 20 c centre = (0, −4), radius = 3 5 d centre = (−3, −10), radius = 2 7 2 a  (x + 5)2 + (y − 9)2 = 49 and x2 + y2 + 10x − 18y + 57 = 0 b (x + 11)2 + (y + 1)2 = 169 and x2 + y2 + 22x + 2y − 47 = 0 d  (x − 14)2 + (y − 6)2 = 44 and x2 + y2 − 28x − 12y + 188 = 0 b (x + 4)2 + (y − 2)2 = 9

3 1 is not the mid-point , Therefore, the point 5 2 of AB. b Gradient of AB is − 7 9 1 7 1 Equation is y − = − x − 3 9 2 18y − 6 = −14x + 7

( ) (

)

14x + 18y − 13 = 0.

c Gradient of perpendicular is 9 7 13 13 , Midpoint of AB is 20 60 Equation of perpendicular bisector is

(

(

)



13 9 13 y− = x− 60 7 20 420y − 91 = 540x − 351



540x − 420y = 260.



Length =

3 a (x − 5)2 + (y + 3)2 = 16

1 1 3 2 > as > 2 3 6 6



11 –1 , 9 8

c (x − 3)2 + y2 = 48 and x2 + y2 − 6x − 39 = 0

8 a Consider the y-coordinates of the points.

x

0

9 1 2 – = 7 8 56

( )

(

)

)

c (x + 1)2 + (y − 4)2 = 20 4 a (x − 9)2 − 81 + (y + 7)2 − 49 − 14 = 0

(x − 9)2 + (y + 7)2 = 144



centre = (9, −7), radius = 12

b (x + 4)2 − 16 + y2 = 9

(x + 4)2 + y2 = 25



centre = (−4, 0), radius = 5

c x2 + y2 + 10x + 18y + 79 = 0

(x + 5)2 − 25 + (y + 9)2 − 81 + 79 = 0



(x + 5)2 + (y + 9)2 = 27



centre = (−5, −9), radius = 3 3

d x2 + y2 + 30x − 6y +

855 =0 4

(x + 15)2 − 225 + (y − 3)2 − 9 +

855 =0 4

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3 

WORKED SOLUTIONS

5



81 (x + 15)2 + (y − 3)2 = 4



centre = (−15, 3), radius =

d2

( x + 76 ) + ( y − 65 ) = 366 + ( 76 ) + ( 65 ) 2

9 2

2

=

= (x1 − x2 + (y1 − y2)2 r2 = (−4 − 1)2 + (7 − 3)2

)2

2

6 49 25 80 + + = 36 36 36 36

80 2 Area = π r = 36 π

= (−5)2 + (4)2 = 41



2

The circle has the equation (x +

4)2

+ (y −

7)2

= 41.

10 Smaller circle radius 3 units centre (7, 3) has equation (x – 7)2 + (y – 3)2 = 9

6 (x − 9)2 + (y − 2)2 = 81

Larger circle radius 7 units centre (7, 3) has equation (x – 7)2 + (y – 3)2 = 49

7 d2 = (x1 − x2)2 + (y1 − y2)2

Area of path = area of larger circle – area smaller circle

1)2

+ (7 −

3)2



= (4 −



= (3)2 + (4)2 = 25

= 49π – 9π = 40π

The circle has the equation (x − 1)2 + (y − 3)2 = 25. The locations on the x-axis will lie on the circumference of the circle. When y = 0, (x − 1)2 + (−3)2 = 25 (x − 1)2 = 16

The required coordinates are (5, 0) and (−3, 0). 8 Rewrite the equation of the circle by completing the square.

( x + 12 p) − 14 p + (y + 3) − 9 = 96 ( x + 12 p) + (y + 3) = 14 p + 9 + 96 2

2

2

2

2

Since



1 2 p + 9 + 96 = 121 4

1 2 p = 16 4 2 p = 64 Since p is a positive constant, p = 64 = 8 . The centre of the circle has the coordinates 1 − p, − 3 . 2

)

Since p = 8, the centre has the coordinates (−4, −3). This is a distance of 5 units from the origin. 1 2 7 5 1 x2 + y2 + x − y = 3 3 6

Let the centre (8, 1) = (x1, y1) and D(2, 9) = (x2, y2).



d2 = (x1 − x2)2 + (y1 − y2)2



r2 = (8 − 2)2 + (1 − 9)2



= (6)2 + (−8)2 = 100 r = 100 = 10

c The circle has the equation (x − 8)2 + (y − 1)2 = 100. 3 a  The student has correctly worked out that AB2 = 1369, but the equation of the circle needs the square of the radius, not the square of the diameter. The diameter is 1369 = 37, so the radius is 37 . 2 b (x + 0.5)2 + (y − 17)2 =

2

( 372 )

2

x + 9 y + 11  , = (5, −3) 4 a i  2   2 x + 9 = 2 × 5 = 10

( x + 76 ) − ( 76 ) + ( y − 65 ) − ( 65 ) = 366 2

( 2 +214 , 9 −2 7 ) = (8, 1)





9 3x2 + 3y2 + 7x – 5y =

2

Centre =

b The radius is the distance between the centre and a point on the circumference.

2

1 = 121, p 2 + 9 + 96 = 121 4



(

= (−44)2 + (33)2 = 3025



x = 1 ± 4 = 5 or −3

r2

= (−17 − 27)2 + (25 − (−8))2

2 a The centre is the midpoint between D and E.

x − 1 = ±4



1 d2 = (x1 − x2)2 + (y1 − y2)2

d = 3025 = 55

(x − 1)2 + 9 = 25



Exercise 3.3B



x=1

y + 11 = 2 × −3 = −6

2



y = −17 G = (1, −17)

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3 Coordinate geometry

ii d2 = (x1 − x2)2 + (y1 − y2)2 r2



= (9 −

5)2

(4)2

+ (5 −

7 Centre of circle is the mid-point of AB and has  1 5 1 3 3 −1 coordinates  2 + 2 2 − 2  i.e. , 2 2  2 , 2  Radius is distance from centre to one of the points

( )

(−3))2

(8)2

 = + = 80 r = 80 = 4 5 iii (x − 5)2 + (y + 3)2 = 80 b Substitute the coordinates of H into the equation of the circle. 5)2

3)2

(−8)2

(4)2



(−3 −



Since both sides of the equation are satisfied, H lies on C.

+ (1 +

=

+

= 80

5 a The centre of the circle is the mid-point of −9 + 5 4 + 10 , ST = = (−2, 7). 2 2

(

)

The midpoint of UV will also be the centre of the circle, (−2, 7). Hence  

( 52 − 32 ) + ( 32 − 12 ) = 1 + 1 = 2 3 1 Equation of circle is ( x − ) + ( y + ) = 2 2 2 2

=

2

2

2

2 Substitute x = 2, y = − into the LHS of the 5 equation

( x − 32 ) + ( y + 12 ) = (2 − 32 ) + (− 52 + 12 ) 2

2

2

=

1 1 + = 0.26 ≠ 2 4 100

The point with coordinates x = 2, y = − lie on the circle.

1 + p 14 + q  = (−2, 7). , 2 2 

x2 – 10x + (x + 2)2 – 14(x + 2) + 73 = 0

p = −5

x2 – 10x + x2 + 4x + 4 – 14x – 28 + 73 = 0



2x2 – 20x + 49 = 0

14 + q = 2 × 7 = 14

q = 0 b

+ (4 −

x=

7)2

= (−7)2 + (−3)2 = 58



= (−9 −

(−2))2

The circle has the equation (x + 2)2 + (y − 7)2 = 58.

6 The equation of the circle can be rewritten as (x + 3)2 + (y − 8)2 = 125, so (−3, 8) is the centre of the circle. Substitute y = 2x + 14 into

x2

+

y2

20 ± 400 − 392 20 ± 8 2 = =5± 4 4 2

x =5−

2 2 , y =7− 2 2

x =5+

2 2 , y =7+ 2 2

A sketch diagram helps: 5 + √ 2 , 7 + √2 2 2

+ 6x − 16y = 52.

x2 + (2x + 14)2 + 6x − 16(2x + 14) = 52

y =x+2

Expand and simplify. x2 + 4x2 + 56x + 196 + 6x − 32x − 224 = 52 5x2 x2

2 does not 5

8 x2 – 10x + y2 – 14y + 73 = 0 meets y = x + 2 when

1 + p = 2 × −2 = −4

r2

2

2×

+ 30x − 80 = 0

+ 6x − 16 = 0

5–

Solve.

√2 , √2 7– 2 2

2×

(x + 8)(x − 2) = 0 x = −8 or 2

Use Pythagoras:

M and N are the two intersection points, it doesn’t matter which is which.

Diameter =

When x = −8, y = 2 × −8 + 14 = −2, so M = (−8, −2). When x = 2, y = 2 × 2 + 14 = 18, so N = (2, 18). −8 + 2 −2 + 18 , The midpoint of MN = = (−3, 8). 2 2 Since the midpoint of M and N is the centre of the circle, MN is a diameter of the circle.

(

)

√2 = √2 2

( 2) + ( 2) 2

√2 = √2 2

2

= 2 units.

9 A sketch diagram helps: –5 , 3 2 5 1 3 17 + = 4 5 20

3 5 + =4 2 2

3 , –1 2 4

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3 

WORKED SOLUTIONS

( ) 17 20

Diameter

2

+ 4 2 = 4.0893 ,

1  10 Centre of the circle has coordinates  3 + 2 −1   2 , 2   7 1 i.e. , − 6 2 2 2 1 7 Distance of centre from the origin is + 2 6 = 1.27 units (3 s.f.)

)

() ()

Exercise 3.4A

=



(−12)2

= (−1 − =

AC 2 + BC 2 = AB2

16x2 − 72x + 81 + 196 − 28x + x2 = 4x2 + 20x + 25 13x2 − 120x + 252 = 0

+

(−4)2

13)2

(−14)2

= 160

+ (11 −



9)2

+ (2)2 = 200

Since 40 + 160 = 200, DE 2 + EF 2 = DF 2.



Green area = (100π − 60) m2

(4x − 9)2 + (14 − x)2 = (2x + 5)2

EF 2 = (1 − 13)2 + (5 − 9)2 DF 2





= (−2)2 + (6)2 = 40





= (−8)2 + (−6)2 = 100 1 1 Area of triangle = bh = × 360 × 40 = 60. 2 2 Area of circle = πr2 = π × 102 = 100π.

3 Since AB is a diameter, ACB = 90° and the lengths of the sides satisfy Pythagoras’ theorem.

DE 2 = (−1 − 1)2 + (11 − 5)2



r2 = (3 − 11)2 + (2 − 8)2

c 100π − 60 = 254.16 m2 254.16 ÷ 5 = 50.8, so 51 tins required. 51 × 4 = $204

1 a  d2 = (x1 − x2)2 + (y1 − y2)2



Area = πr2 = 13.1 square units (3 s. f.)



d2 = (x1 − x2)2 + (y1 − y2)2



radius = 2.04466.

(



b Gradient of DE = 11 − 5 = −3. −1 − 1



When x =



Radius =



9−5 1 = . 13 − 1 3



Gradient of EF =



Since m1 × m2 = −1, DE and EF are perpendicular.

(13x − 42)(x − 6) = 0 42 x= or 6 13 42 42 149 ,d = 2× +5= 13 13 13

1 149 149 × = 2 13 26 When x = 6, d = 2 × 6 + 5 = 17 Radius = 1 × 17 = 17 2 2

4 a Substitute x = 1 and y = 0 into (x + 2)2 + (y − 1)2 = r2

(1 + 2)2 + (0 − 1)2 = r2

(3)2 + (−1)2 = r2

c DF is a diameter of the circle.

( −1 2+ 13 ,112+ 9 ) = (6, 10).



Centre =



For the radius.



d2



r2 = (6 − 1)2 + (10 − 5)2



)2

b Substitute x = −3 and y = q into (x + 2)2 + (y − 1)2 = 10

)2

= (x1 − x2 + (y1 − y2 =

(5)2

+

(5)2



PQ 2  = (−5 − 11)2 + (−4 − 8)2 = (−16)2 + (−12)2 = 400



PR2   = (−5 − 13)2 + (−4 − 2)2 = (−18)2 + (−6)2 = 360



QR2

QR2

+ (8 −

2)2

=

(−2)2

+

PQ2,

(6)2

)

−5 + 11 −4 + 8 , = (3, 2). b Centre = 2 2

For the radius.



q − 1 = ±3

q=1±3



q = −2 or 4

Since q > 0, q = 4

c Gradient of AB =

2−0 1 =− . −5 − 1 3



Gradient of AC =

2−4 = 1. −5 − (−3)



Gradient of BC =

0−4 = −1 . 1 − (−3)

= 40

Since + = Pythagoras’ theorem is satisfied and the triangle PQR is right-angled.

(

(q − 1)2 = 9

2 a  Proof using Pythagoras’ theorem (could compare gradients).

PR2

(−1)2 + (q − 1)2 = 10

The equation of the circle is (x − 6)2 + ( y − 10)2 = 50.

= (11 −

(−3 + 2)2 + (q − 1)2 = 10

= 50

13)2

r2 = 10



d m1 × m2 = −1 so AC and BC are perpendicular.

Since the angle in a semicircle is a right angle, AB is a diameter of the circle.

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3 Coordinate geometry

6 −1 1 5 a Gradient of L1 = 5 − (−10) = 3.

Gradient of L2 =



d2 = (x1 − x2)2 + (y1 − y2)2

4 − (−2) = −3. 9 − 11

r2 = (8 − 1)2 + (4 − 5)2

m1 × m2 = −1 so L1 and L2 are perpendicular.





At W, 1 (x − 5) + 6 = −3(x − 9) + 4. 3



x − 5 + 18 = −9(x − 9) + 12



x + 13 = −9x + 81 + 12



x=8



JK 2 = (−11 − (−1))2 + (6 − (p − 8))2 = (−10)2 + (14 − p)2 = p2 − 28p + 296

Because L1 and L2 are perpendicular and intersect at point W on the circumference, SV is a diameter.

( 5 +211 , 6 −2 2 ) = (8, 2).

Centre =



For the radius. d2 = (x1 − x2)2 + (y1 − y2)2 r2 = (5 − 8)2 + (6 − 2)2 = (−3)2 + (4)2 = 25



The equation of C is given by (x − 8)2 + (y − 2)2 = 25.



Expanding and simplifying: x2 + y2 − 16x − 4y + 43 = 0.

KL2 = (−1 − 7)2 + ((p − 8) − 22)2 = (−8)2 + (p − 30)2 = p2 − 60p + 964



Hence 580 = p2 − 28p + 296 + p2 − 60p + 964.



0 = 2p2 − 88p + 680



= p2 − 44p + 340



= (p − 10)(p − 34)



When x = 8, y = −3(8 − 9) + 4 = 7.



JL2 = (−11 − 7)2 + (6 − 22)2 = (−18)2 + (−16)2 = 580



Equation of L2 is given by y − 4 = −3(x − 9), from which y = −3(x − 9) + 4.

10x = 80

The equation of C is given by (x − 8)2 + (y − 4)2 = 50.

b JL2 = JK 2 + KL2

1 from which y = (x − 5)+ 6. 3



= (7)2 + (−1)2 = 50



1 b Equation of L1 is given by y – 6 = (x − 5), 3



For the radius.

7

p = 10 or 34

Gradient of AB =

y1 − y 2 2 − 1 1 = = x1 − x 2 6 + 1 7

23 − 13 y1 − y 2 2 − 5 = = 5 26 x1 − x 2 6− 4 5 5 13 5 1 =− × =− 5 26 2 23 − 1 18 y − y2 18 5 Gradient BC = 1 = 5 = 5 = × =2 x1 − x 2 4 +1 9 5 9 5 5 AC is perpendicular to BC (product of gradients is −1). Gradient AC =

∴ AB is the diameter. Mid-point of AB has coordinates i.e. (2.5, 1.5)

6 a PR2 = PQ2 + QR2

Length AB =

( 6 + 1 )2 + ( 2 − 1 )2

( 6 2− 1 , 2 2+ 1 )

= 49 + 1 = 50



PR2 = (1 − (t + 17))2 + ((t + 7) − 3)2 = (−16 − t)2 + (t + 4)2 = 2t2 + 40t + 272



PQ 2 = ( 1 − 3)2 + ((t + 7) − (t + 11))2 = (−2)2 + (−4)2 = 20



QR2 = (3 − (t + 17))2 + ((t + 11) − 3)2 = (−14 − t)2 + (t + 8)2 = 2t2 +44t + 260



Hence 2t2 + 40t + 272 = 2t2 + 44t + 260 + 20.

x2 – 5x + 6.25 + y2 – 3y + 2.25 – 12.5 = 0



−8 = 4t

x2 + y2 – 5x – 3y – 4 = 0



t = −2



Hence the coordinates of P and R are (1, 5) and (15, 3). 1 + 15 5 + 3 Centre of circle = , = (8, 4). 2 2



(

Radius of circle =

50 2

Equation of circle is (x – 2.5)2 + (y – 1.5)2 =

50 4

8 If PR is diameter and Q lies on the circle then PQ and QR must be perpendicular to each other.

)

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3 

WORKED SOLUTIONS

Gradient PQ =

Exercise 3.4B

17 1 y1 − y 2 6 − 3 = = 3 =1 x1 − x 2 −1 + 4 1 3 3

(

17 + 3 26 y1 − y 2 3 3 = − 26 × 3 = − 13 = = Gradient QR = x − x 4 3 4 2 1 2 − −0 −4 3 3 As −

13 × 1 ≠ −1 PQ is not perpendicular to QR 2

9 A sketch diagram helps. x=– 1 2 y

c Gradient of FG = − 1 . 4 d Equation is given by y − 1 = −

y = 3x + 7



d2 = (x1 − x2)2 + (y1 − y2)2 r2 = (9 − 4)2 + (5 − 2)2 = (5)2 + (3)2 = 34

x

B

3y = –x – 3

The diagram shows the intersection of the line 1 x = − with the two lines, these are labelled as 2 A and B. AB is the diameter of the circle. C, the centre of the circle, is the mid-point of AB.

( ) 1 1 1 5 Coordinates of B are x = − , y = − × ( − ) − 1 = − 2 3 2 6

1 1 11 Coordinates of A are x = − , y = 3 × − + 7 = 2 2 2

1 11 5 38 + = =6 6 3 2 6

y-coordinate of C is 11 − 3 1 = 7 6 3 2

(

)

10 If one of the line segments is a diameter then the other two line segments must be perpendicular to each other. y − y 2 −23 + 19 4 1 Gradient AB = 1 = =− =− x1 − x 2 −5 + 17 12 3 Gradient BC =

y1 − y 2 −3 + 19 16 = = =1 x1 − x 2 −1 + 17 16

y1 − y 2 −3 + 23 20 = = =5 x1 − x 2 −1 + 5 4 As none of the gradients multiply to give −1 none of the lines are perpendicular to each other. Gradient AC =

2 Gradient of JK =

7−2 5 =− . 3 − 10 7

7 Gradient of perpendicular bisector = . 5 Mid-point =

( 3 +210 , 7 +2 2 ) = (132 , 92 ) .

Equation of bisector is given by y − 9 = 7 x − 13 . 2 5 2

(

When y = 8, 8 −

)

(

)

9 7 13 = x− . 2 5 2

63 7  = x 5 5

1 7 1 Equation of the circle with centre − , , r = 3 is 2 3 6 2 2 2 1 7 19 x+ + y− = 2 3 6

) ( )

The equation is given by (x − 4)2 + (y − 2)2 = 34.

7 7 91  = x− 2 5 10

1 1 Radius = AB = 3 6 2

) (

y = 1 − 1 + 2 = 2 as required.

f For the radius. C

(

1 (x − 8) 4

from which 4y + x = 12 1 e When x = 4, y – 1 = − (4 − 8). 4

A

Distance AB =

)

1 a 9 + 7 , 5 − 3 = (8, 1) 2 2 5 − (−3) b Gradient of DE = 9 − 7 = 4 .

x = 9 Centre of circle = (9, 8). For the radius. d2 = (x1 − x2)2 + (y1 − y2)2 r 2 = (9 − 3)2 + (8 − 7)2 = (6)2 + (1)2 = 37 The equation is given by (x − 9)2 + (y − 8)2 = 37. 10 + x 11 + y  , = (6, 3). 3 a Mid-point =  2   2

10 + x = 2 × 6



x=2



11 + y = 2 × 3



y = −5



Coordinates of V are (2, −5).

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3 Coordinate geometry

11 − 3 b Gradient of UV = 10 − 6 = 2 .

d The quadrilateral MCNX has ∠XMC = ∠XNC = 90° because MX and NX are perpendicular bisectors. ∠MXN = 90° from c. Hence ∠ABC = ∠MBN = 360° − (3 × 90°) = 90°.



Gradient of perpendicular bisector = − 1 . 2





1 Equation of bisector is given by y − 3 = − (x − 6). 2 1 When x = 8, y − 3 = − (8 − 6). 2 y=2



Centre of circle = (8, 2).



For the radius.



d2 = (x1 − x2)2 + (y1 − y2)2





r2 = (8 − 10)2 + (2 − 11)2

6  A(12, 8), B(11, 1) and C(20, 4) 12 + 11 8 + 1 23 9 Mid-point of AB = , = , . 2 2 2 2

(



= (−2)2 + (−9)2 = 85

e AC is a diameter.

The equation is given by (x − 8)2 + (y − 2)2 = 85.

4 a  Equation of circle can be rewritten as (x + 3)2 + (y + 5)2 = 185.



) ( )

8 −1 = 7. 12 − 11 1 Gradient of perpendicular bisector = − . 7 9 1 23 . Equation of bisector is given by y − = − x − 2 7 2 1 43 y=− x+ 7 7 12 + 20 8 + 4 , = (16, 6). Mid-point of AC = 2 2

Gradient of AB =

(

(

)

8−4 1 =− . 12 − 20 2

Hence centre of circle = (−3, −5).



Gradient of AC =

b When y = 6, (x + 3)2 + (6 + 5)2 = 185.



Gradient of perpendicular bisector = 2. Equation of bisector is given by y − 6 = 2(x − 16). y = 2x − 26 1 43 For point of intersection, − x + = 2x − 26 . 7 7 225 15 = x 7 7 x = 15 When x = 15, y = 2 × 15 − 26 = 4. Centre of circle = (15, 4). For the radius:



(x + 3)2 + 121 = 185



(x + 3)2 = 64



x + 3 = ±8



x = 5 or −11



Length of chord = 5 − (−11) = 16.



c Height of triangle = 6 − (−5) = 11.

Base of triangle = PQ = 16.



Area of triangle = 1 bh = 1 × 16 × 11 = 88 . 2 2

(



)

5 a Mid-point of AB = −4 + 2 , 3 + 7 = (−1, 5). 2 2



d2 = (x1 − x2)2 + (y1 − y2)2





= (15 − 12)2 + (4 − 8)2

r2 = (3)2 + (−4)2 = 25 Final answer is x2 – 30x + 225 + y2 – 8y + 16 = 25



7−3 2 = . Gradient of AB = 2 − (−4) 3





3 Gradient of perpendicular bisector = − . 2

7 a (x + 6)2 + (y – 6)2 = 145 (x – 8)2 + (y + 1)2 = 40



3 Equation of bisector is given by y − 5 = − (x + 1). 2 2y − 10 = −3x − 3



3x + 2y = 7

b Mid-point of BC =

( 2 +210 , 7 −2 5 ) = (6, 1).

7 − (−5) 3 =− . 2 − 10 2

x2 + y2 – 30x – 8y + 216 = 0

Multiply out the brackets and collect terms.

x2 + 12x + 36 + y2 – 12y + 36 = 145 x2 + 12x + y2 – 12y = 73 x2

– 16x + 64 +

y2

A

+ 2y + 1 = 40

x2 – 16x + y2 + 2y = – 25

B

Find A – B



Gradient of BC =



2x – y = 7



2 Gradient of perpendicular bisector = . 3 2 Equation of bisector is given by y − 1 = (x − 6). 3 3y − 3 = 2x − 12



2x = 3y + 9

x2 – 16x + 4x2 – 28x + 49 + 4x – 14 = – 25



)

28x – 14y = 98 y = 2x – 7

Substitute this into equation B.

x2 – 16x + (2x – 7)2 + 2(2x – 7) = – 25

c 90° because gradients are perpendicular.

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3 

WORKED SOLUTIONS

5x2 – 40x + 60 = 0 x2 – 8x + 12 = 0 (x – 6) (x – 2) = 0 x = 2, y = – 3 i.e. (2, –3) x = 6, y = 5 i.e. (6, 5) b Length =

4 2 + 8 2 = 16 + 64 = 80 = 4 5

Mid-point of AB = Gradient of AB =

( 3 2+ 1 , 8 +2 4 ) = (2, 6).

8−4 = 2. 3−1



) (

)

8−5 1 = . Gradient of AC = 3 − (−6) 3 Gradient of perpendicular bisector = −3. Equation of bisector is given by 13 3 y− = −3  x − −  . 2 2  

( )



y = −3x + 2 For point of intersection, − 1 x + 7 = −3x + 2. 2 5 x = −5 2 x = −2 When x = −2, y = −3 × −2 + 2 = 8. Centre of circle = (−2, 8). For the radius: d2 = (x1 − x2)2 + (y1 − y2)2 r2 = (−2 − 1)2 + (8 − 4)2 = (−3)2 + (4)2 = 25 As all three points lie on a circle with centre (−2, 8) with radius 5 none of the line segments could be a radius, but all three line segments are chords. 9 A(19, 4), B(17, 0) and C(4, −1)

(

)



19 + 17 4 + 0 Mid-point of AB = , = (18, 2). 2 2



Gradient of AB =



1 Gradient of perpendicular bisector = − . 2



Equation of bisector is given by 1 y − 2 = − (x − 18). 2 1 y = − x + 11 2



Gradient of AC =



Gradient of perpendicular bisector = −3.



Equation of bisector is given by 3 23 . y − = −3 x − 2 2



1 Equation of bisector is given by y − 6 = − (x − 2). 2 1 y =− x+7 2 3−6 8+5 3 13 Mid-point of AC = 2 , 2 = − 2 , 2 .

(





1 Gradient of perpendicular bisector = − . 2



Mid-point of AC =



8 A(3, 8), B(1, 4) and C(−6, 5)

4−0 = 2. 19 − 17

(192+ 4 , 4 2− 1 ) = ( 232 , 32 ).



4 − (−1) 1 = . 19 − 4 3

(

)

y = −3x + 36 1 For point of intersection, − x + 11 = −3x + 36. 2 5 x = 25 2



x = 10



When x = 10, y = −3 × 10 + 36 = 6.



Centre of circle = (10, 6).



For the radius:



d2 = (x1 − x2)2 + (y1 − y2)2



r2 = (10 − 17)2 + (6 − 0)2



= (−7)2 + (6)2 = 85



The equation is given by (x − 10)2 + (y − 6)2 = 85.

Exercise 3.4C 1 Gradient of radius = 6 − 8 = 2 . 4−7 3 3 Gradient of the tangent = − . 2 The equation of the tangent is given by y − 8 = − 3 (x − 7). 2 2y − 16 = −3x + 21 3x + 2y = 37 2 a 22 + (−2 − 1)2 = 4 + 9 = 13

Since both sides of the equation agree, T lies on the circle.

b Centre of circle = (0, 1). 1 − (−2) 3 =− . 0−2 2



Gradient of radius =





2 . 3 The equation of the tangent is given by 2 y + 2 = (x − 2). 3 3y + 6 = 2x − 4



Hence 2x = 3y + 10.



Gradient of the tangent =

3 a Centre = (3, 6). b Radius = 8.

Let centre = X, point P = (15, 7) and point where tangent meets circle = T.

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3 Coordinate geometry

c The length of PT is required, where PT 2 + XT 2 = PX 2.

b AXBQ is a kite. Split into congruent triangles AXQ and BXQ.



For PX:

19 25 = . 4 4



Base of triangle AXQ = 11 −



PX 2

(x1 − x2)2 + (y1 − y2)2 (3 − 15)2 + (6 − 7)2



Perpendicular height of triangle AXQ = 3.



PX 2 = (−12)2 + (−1)2 = 145



Area of triangle AXQ =



Area of AXBQ = 2 ×



d2

= =

PT 2 + XT 2 = PX 2 PT 2 + 82 = 145



PT 2 = 81



75 75 . = 8 4

5 Rewrite the equation of the circle as (x + 4)2 + (y − 1)2 = 81.

PT = 81 = 9 m

Centre = (−4, 1); radius = 9.

4 a For A:

Let centre = X and point where tangent meets circle = T.

11 − 7 4 = . 7−4 3



Gradient of radius =



3 Gradient of the tangent = − . 4





The equation of the tangent is given by 3 y − 7 = − (x − 4). 4 Hence y = − 3 (x − 4) + 7. 4 For B:



Gradient of radius =



Gradient of the tangent =



The equation of the tangent is given by 3 y − 7 = (x − 10). 4



1 25 75 × ×3= . 2 4 8

PT 2 + XT 2 = PX 2. For PX:

d2 = (x1 − x2)2 + (y1 − y2)2



PX2 = (7 − (−4))2 + (4 − 1)2 = (11)2 + (3)2 = 130



PT 2 + XT 2 = PX 2 PT 2 + 92 = 130

11 − 7 4 =− . 7 − 10 3

PT 2 = 49



3 . 4

PT = 49 = 7

6 a  Rewrite the equation of the circle as (x − 15)2 + (y − 23)2 = 400.

3 (x − 10) + 7. 4



Centre = (15, 23), radius = 20.



Gradient of radius =

35 − 23 3 = . 31 − 15 4



Hence y =



Put tangents equal.



4 Gradient of the tangent = − . 3



3 3 − (x − 4) + 7 = (x − 10) + 7 4 4



The equation of the tangent is given by 4 y − 35 = − (x − 31). 3 3y − 105 = −4x + 124



3 3 − (x − 4) = (x − 10) 4 4



−(x − 4) = x − 10

4x + 3y = 229



−x + 4 = x − 10

When x = 10, y = q.





40 + 3q = 229



x = 7 (this could also be deduced by symmetry since AB is horizontal)



When x = 7, y =

14 = 2x

3 (7 − 10) + 7. 4



= 19 4

3q = 189



q = 63

b Use d2 = (x1 − x2)2 + (y1 − y2)2

3 y = (−3) + 7 4



For PT: PT 2 = (10 − 31)2 + (63 − 35)2



= (−21)2 + (28)2 = 1225



( )

19 Hence the coordinates of Q are 7, 4 .



For PX: PX 2 = (10 − 15)2 + (63 − 23)2

= (−5)2 + (40)2 = 1625

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WORKED SOLUTIONS



TX is the radius, so TX 2 = 400.

3y – 69 = – x – 8



Hence PT 2 + TX 2 = PX 2 because 1225 + 400 = 1625.

3y + x – 61 = 0

c Since triangle is right-angled, area = 1 × TX × PT. 2 TX = 20

PT = 1225 = 35



Area =

This line meets the tangent when 3(3x + 5) + x – 61 = 0 9x + 15 + x – 61 = 0

1 × 20 × 35 = 350. 2

7 Gradient of tangent = 1. Gradient of radius = −1.  quation of line with gradient −1 through E (5, 11) is given by y − 11 = −(x − 5).

10x – 46 = 0 x = 4.6, y = 18.8 Radius, r =

(x + 8)2 + (y – 23)2 = 176.4 10 Draw a sketch T1

y = 16 − x

X

16 − x = x + 2 14 = 2x



x=7



y=7+2=9

r T2

 ence the point (7, 9) lies on the H circumference. For the radius: d2 = (x1 − x2)2 + (y1 − y2)2   r2 = (7 − 5)2 + (9 − 11)2 = (2)2 + (−2)2 = 8



 he equation of the circle is given by T (x − 5)2 + (y − 11)2 = 8. 8 a Gradient of CT =

P

r

 ind point of intersection of y = 16 − x and F y = x + 2.

12.6 2 + 4.22 = 176.4

y1 − y 2 4 + 6 10 5 = = =− x1 − x 2 −3 − 1 −4 2

Let the tangents from P meet the circle at T1 and T2 and the centre of the circle be X. PXT1 is a right angled triangle (tangent meets radius at 90˚) PXT2 is a right angled triangle (tangent meets radius at 90˚) PX is common to both triangles and is the hypotenuse in each triangle. T1X = T2X = radius of the circle. In triangle PXT1, by Pythagoras’ theorem PT1 = PX 2 − r 2 In triangle PXT2, by Pythagoras’ theorem

Gradient of the tangent at T = 2 5 2 Equation of tangent is y + 6 = (x − 1) 5 5y + 30 = 2x – 2

PT2 =



so PT =  145 − 16 = 129 for both T1 and T2.



– 2x + 5y + 32 = 0

b Equation of diameter 5 Gradient − through the point (1, – 6) 2 5 y + 6 = − (x − 1) 2 2y + 12 = – 5x + 5

Therefore PT1 = PT2 In this question, r = 4, PX = 145 ,

Exercise 3.4D 1 a Substitute y = x − 10 into x2 + y2 = 50.

Equation of this radius is 1 y − 23 = − (x + 8) 3

x2 + (x − 10)2 = 50

x2 + x2 − 20x + 100 = 50

2y + 5x + 7 = 0 9 Gradient of tangent is 3, so gradient of radius to 1 point of contact is − . 3

PX 2 − r 2





2x2 − 20x + 50 = 0



x2 − 10x + 25 = 0



(x − 5)2 = 0

The equation has repeated roots so y = x − 10 is a tangent of the circle.

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3 Coordinate geometry

b Substitute x = 7y − 50 into x2 + y2 = 50. +

y2

= 50

− 700y + 2500 +

y2

= 50



(7y −

49y2



50y2

50)2



− 700y + 2450 = 0



y2 − 14y + 49 = 0



(y − 7)2 = 0

The equation has repeated roots so 7y = x + 50 is a tangent of the circle.

2 Substitute y = 4x − 5 into x2 + y2 + 10x − 18y + 38 = 0. x2 + (4x − 5)2 + 10x − 18(4x − 5) + 38 = 0 x2 + 16x2 − 40x + 25 + 10x − 72x + 90 + 38 = 0 17x2 − 102x + 153 = 0 x2 − 6x + 9 = 0 (x − 3)2 = 0 The equation has repeated roots so y = 4x − 5 is a tangent of the circle. x2

y2

3 Substitute y = 3x + c into + − 10x − 6y − 6 = 0. x2 + (3x + c)2 − 10x − 6(3x + c) − 6 = 0 2 x + 9x2 + 6cx + c2 − 10x − 18x − 6c − 6 = 0 10x2 + (6c − 28)x + (c2 − 6c − 6) = 0 b2 − 4ac = 0 2 2 (6c − 28) − 4 × 10 × (c − 6c − 6) = 0 36c2 − 336c + 784 − 40c2 + 240c + 240 = 0 −4c2 − 96c + 1024 = 0 c2 + 24c − 256 = 0 (c − 8)(c + 32) = 0 c = 8 or −32 1 4 Substitute y = − x + c into 2



(1 + m2)x2 + (28m − 36)x + 468 = 0 b2 − 4ac = 0 (28m − 36)2 − 4 × (1 + m2) × 468 = 0 784m2 − 2016m + 1296 − 1872 − 1872m2 = 0 −1088m2 − 2016m − 576 = 0 34m2 + 63m + 18 = 0 (17m + 6)(2m + 3) = 0 6 3 m=− or − 17 2 6 For the tangent y = − x + 24. 17 17y = −6x + 408 6x + 17y − 408 = 0 For the tangent y = − 3 x + 24. 2 2y = − 3x + 48 3x + 2y − 48 = 0

6 Substitute y = 3x – 2 into x2 + y2 + 10x – 18y + 38 = 0. x2 + (3x – 2)2 + 10x – 18(3x – 2) + 38 = 0 x2 + 9x2 – 12x + 4 + 10x – 54x + 36 + 38 = 0 10x2 – 56x + 78 = 0 5x2 – 28x + 39 = 0 b2 – 4ac = 784 – 780 = 4 As b2 – 4ac > 0 there are two intersections of the line with the circle, ∴ the line is not a tangent to the circle. 1 7 x2 + y2 – 8x – 4y – 6 = 0 has 2 tangents y = − x + c 2 Substitute the equation of the tangent into the circle.

(

x2 + y2 − 12x − 26y + 125

1 x2 + − x + c 2

x2

x2 +

= 0. 1 1 2 + ( − x + c) − 12x − 26(− x + c) + 125 = 0 2 2

x2 +

1 2 x − cx + c2 − 12x + 13x − 26c + 125 = 0 4 5 2 x + (1 − c)x + (c2 − 26c + 125) = 0 4



5x2 + (4 − 4c)x + (4c2 − 104c + 500) = 0 b2 − 4ac = 0 2 2 (4 − 4c) − 4 × 5 × (4c − 104c + 500) = 0 16 − 32c +16c2 − 80c2 + 2080c – 10 000 = 0 −64c2 + 2048c − 9984 = 0 c2 − 32c + 156 = 0 (c − 6)(c − 26) = 0 c = 6 or 26 1 1 The tangents are y = − x + 6 and y = − x + 26. 2 2

5 Substitute y = mx + 24 into (x − 18)2 + (y − 10)2 = 52. (x − 18)2 + (mx + 24 − 10)2 = 52 (x − 18)2 + (mx + 14)2 = 52 x2 − 36x + 324 + m2x2 + 28mx + 196 = 52

) − 8x − 4(− 12 x + c ) − 6 = 0 2

1 2 x – cx + c2 – 8x + 2x – 4c – 6 = 0 4

5 2 x – (c + 6)x + c2 – 4c – 6 = 0 4 For the line to be a tangent b2 – 4ac = 0 5 b2 – 4ac = (c + 6)2 – 4 × × (c2 – 4c –6) = c2 + 12c + 36 4 – 5c2 + 20c + 30 = –4c2 + 32c + 66 –4c2 + 32c + 66 = 0 −b ± b 2 − 4ac −32 ± 1024 + 1056 = 2a −8 −32 ± 45.607 = −8

c=

The values of c are –1.70 and 9.70. 8 x2 + y2 – 16x – 22y + 115 = 0 or (x – 8)2 + (y – 11)2 = 70 Circle has centre (8, 11). 1 so gradient of the 3 diameter between the points of contact is –3.

Gradient of tangent(s) is

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WORKED SOLUTIONS

b

Equation of diameter:

y

y – 11 = –3(x – 8), so y = – 3x + 35

10

Substitute this in the equation of the circle:

8

(x – 8)2 + (–3x + 35 –11)2 = 70

6

Multiply out the brackets and simplify:

4

10x2

2

– 160x + 570 = 0

–2 –1 0 –2

x2 – 16x + 57 = 0 x = 5.35, y = 18.9

–4

x = 10.6, y = 3.06

–6

For the tangent through x = 5.35, y = 18.9 1 y =  x + 17.2 3 For the tangent through x = 10.6, y = 3.06 1 y =  x – 0.0486 3 9 (x – 7)2 + (y + 4)2 = 57; tangents have gradient –2, and circle has centre (7, –4).

x

(4, –9)

7 6 5 4

4)2

3

= 57

2

2

−b ± b − 4ac 17.5 ± 306.25 − 21.25 = 2a 2.5 17.5 ± 16.882 = 2.5

1



–1

3

x

1

2

3

x

2 1 –1

1 a

0 –1 –2

y

–3

10

–4

8 6



4

4

2

–6

2

y

Exercise 3.5A

–4

1

3

x = 0.246, y = – 7.38, the point is (0.246, –7.38)

0 –7 –6 –5 –4 –3 –2 –1 –2

0

b

x = 13.8, y = –0.624, the point is (13.8, –0.624)



8

8

– 17.5x + 4.25 = 0

(–3, –4)

7

y

x2 – 14x + 49 + 0.25x2 – 3.5x + 12.25 – 57 = 0

x=

6

9

Diameter meets the circle when

1.25x2

5

3 a

1 (x – 7) 2

+ (0.5x – 7.5 +

4

2 Roots at x = 1 and x = 3 so factors are (x − 1) and (x − 3). (x − 1)(x − 3) = x2 − 4x + 3, so the equation is y = x2 − 4x + 3

y = 0.5x – 7.5 (x –

3

–8

Equation of diameter

7)2

2

–10

Gradient of the diameter joining the points of 1 intersection is . 2

y+4=

1

1

2

x

–5

3 , –9 2 2

The sketch is not correct. x2 − 4x − 21 = (x − 7)(x + 3) So should cut x-axis at (−3, 0) and (7, 0). a > 0 so graph should be U-shaped. When x = 0, y = −21 so y-intercept at (0, −21). x2 − 4x − 21 = (x − 2)2 − 25 So turning point at (2, −25).

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3 Coordinate geometry

5 a

8 Red graph is of y = – 2x2 – 9x + 7.

y

Blue graph is of y = 2x2 – 9x + 7. y

24 20

20

16 12

15

8 4 –1

10

0

1

2

3

4

5

6

7

8

x 5

b y

–5 , 51 4 8

–10

0

–5

5

3

9 a

2

y 4

1 –3

–2

0

–1

–1

x

–5

4

–4

5

2

1

2

(–6.416, 0)

x

–8

–6

–4

(0.416, 0)

0

–2 –2



2

4

6

(0, –1.333)

8

x

–4

6 a 4 m c Roots at t = −1 and t = 3 so factors are (t + 1) and (t − 3).  ∩ – t2

–(t + 1)(t − 3) = –t2 + 2t + 3, the equation is h = –t2 + 2t + 3

(

y = 3x 2 − 7x + 11 = 3 x −

7



–6

(–3, –5.833)

b 3 s

(

7 6

=3 x−

7 6

(–2.416, 0) (–6.416, 0)

4 2

(0.416, 0)

–8

2

–6

–4

0 –2 –2

(–1, –3.883)

–4

) + 1283 2

(–3, –5.833)

–6

(4.416, 0) 6 8 x (0, –1.333) 2

4

(0, –5.333) (1, –5.833)

10 Let the equation of the curve be y = a(x – b)2 + c

12

Minimum point is at

10

(

8

y=a x−

6 4

5 2

( 52 , 72 ) so b = 52 , c = 72

) + 72 2

When x = 0, y = 16

2 0

y

) − 1249 + 11

y

–2

b

2

4

x

( 52 ) + 72

16 = a × Intersects the y-axis at y = 11. Turning point is  7 , 83  .  6 12 

2

16 = a 0 −

25 7 + 4 2

a = 12.5 ×

(

4 =2 25

∴y=2 x−

5 2

) + 72 2

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3 

WORKED SOLUTIONS

Exercise 3.5B

13 21 c x + 4 = − x + 4

1 a x2 + x − 2 = x + 2



2x = 2



x = 1, y =



x2 − 4 = 0



x = ± 2 so two points of intersection.

b x2 + x − 2 = x + k

x2 – 2 – k = 0



x2 – (2 + k) = 0



b2 − 4ac = 0 – 4.1. – (2 + k) = 8 + 4k

i intersects: b2 – 4ac > 0 8 + 4k > 0

k > −2

ii is a tangent to: b2 − 4ac = 0 8 + 4k = 0

k = −2

iii does not touch: b2 – 4ac < 0 8 + 4k < 0

k < −2

2 x2 + x − 12 = x − 13 x2 = −1 No real solutions to this equation consequently the line and the curve do not intersect. 3 a x2 − 2x + 5 = 5 − x

x2 − x = 0



b2 − 4ac = 1 − 0 = 1 > 0 so the curve and the line intersect

b x2 − 2x + 5 = x +

5 2

5 − 3x + = 0 2 4.1.5 2 = −1 so the curve and the b − 4ac = 9 − 2 line do not intersect x2

c x2 − 2x + 5 = x + 3

x2 – 3x + 2 = 0



b2 – 4ac = 9 − 4.1.2 = 1 so the curve and the line intersect

4 a −x2

+ 2x + 3 = x + k

−x2 + x + 3 – k = 0

b2 – 4ac = 1 – 4. – 1(3 – k) = 0 13 k= 4 b −x2 + 2x + 3 = −x + k

−x2



b2



k=

+ 3x + 3 – k = 0

– 4ac = 9 – 4. – 1(3 – k) = 0 21 4

( )

17 , the point is 1, 17 4 4

5 2x2 + 5x – 3 = x + k 2x2 + 4x – (3 + k) = 0 b2 – 4ac = 16 – 4.2. – (3 + k) = 0 k=−5 Tangent: y = x − 5 Point where curve meets tangent: 2x2 + 5x – 3 = x − 5 x2 + 2x + 1 = 0 (x + 1)2 = 0 x = − 1, y = −6 Equation of perpendicular: y + 6 = −1(x + 1) y = −x − 7 Points where curve meets perpendicular to tangent: 2x2 + 5x – 3 = − x − 7 2x2 + 6x + 4 = 0 x2 + 3x + 2 = 0 (x + 2)(x + 1) = 0 Intersect at x = − 1, y = − 6 and x = − 2, y = − 5 6 a For y = x + k: 3x2 – 2x + 5 = x + k

3x2 – 3x + (5 – k) = 0





b2 – 4ac = 9 – 4.3.(5 – k) = −51 + 12k

i intersects: b2 – 4ac > 0 −51 + 12k > 0

k>

51 12

ii is a tangent to: b2 – 4ac = 0 −51 + 12k = 0

k=

51 12

iii does not touch: b2 – 4ac < 0 −51 + 12k < 0 51 12



k
0 −59 + 12k > 0

k > 59 12

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3 Coordinate geometry

ii is a tangent to: b2 – 4ac = 0

8 The line y = 2x + k meets the curve

(

−59 + 12k = 0 k=



y = 2x −

59 12

4x 2 − 28 x + 49 + 5 = 2x + k 3 9

59 12

0 = 4x 2 −

b For line of symmetry: x+



2x =



x=

3x 2 −

( 343 ) − 4 × 4 × ( 949 − k )

b 2 − 4ac = −

8 12

=

1 3

7 y = 3x 2 −

3 x − (7 + m) = 0 2 For the line to be a tangent b2 – 4ac = 0

k=

Meets the curve y = – 2x2 – 7x + 3 when – 2x2 – 7x + 3 = 3x + 15.5 – 2x2 – 10x – 12.5 = 0 2x2 + 10x +12.5 = 0 A tangent when b2 – 4ac = 0

1 x − 7 meets the line y = – x + n when 2

b2 – 4ac = 100 – 4 × 2 × 12.5 = 100 – 100 = 0

1 x − 7 = −x + n 2

The line is a tangent to the curve. 10 y = 2x2 – 7x + 11 meets the line y = –2x + k

1 x − (7 + n) = 0 2

when 2x2 – 7x + 11 = – 2x + k 2x2 – 5x + 11 – k = 0

For the line to be a tangent b 2 − 4ac = 0

()

Tangent when b2 – 4ac = 0

2

1 1 b − 4ac = + 4 × 3 × (7 + n) = + 84 + 12n 4 2

b2 – 4ac = 25 – 4 × 2 × (11 – k) = – 63 + 8k = 0

= 84.25 + 12n = 0 337 n=− 48

k=

The tangents meet when x − 115 = −x − 337 16 48 337 115 1 −2x = − =− 48 16 6

63 8

To find the coordinates: 2x2 – 7x + 11 = –2x + 63 8 25 2x2 – 5x + =0 8

337 y = −x − 48

x=

29 12

9 2y – 6x – 31 = 0, so y = 3x + 15.5

115 y=x− 16

2

29 12

2

= 86.25 + 12m = 0 115 m=− 16

3x 2 +

348 9

Line is not a tangent when k ≠

( 32 ) + 4 × 3 × (7 + m) = 94 + 84 + 12m

b 2 − 4ac = −

3x 2 −

348 + 16k = 0 9

16k =

3x 2 −

2

1156 1504 − + 16k 9 9

=−

1 x − 7 meets the line y = x + m when 2

1 x −7= x +m 2

y = 3x 2 −

34 94 x+ −k 3 9

Tangent when b2 – 4ac = 0

51 59 = −x + 12 12



2

2

−59 + 12k < 0 k
sin θ so cos θ

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4 Circular measure and trigonometry

12 Write t for tan θ to get t (t – 2) = 3; rearrange as t2 – 2t – 3 = 0. Factorise: (t – 3) (t + 1) = 0; tan θ = 3 or –1; hence θ = 1.25 or 2.36.

y

b

1

Exercise 4.4 A

0.5

–90

0

–45

45

90

q

1 a

1 2 3 2

b sin 120° = sin 60° =

–0.5

c sin 225° = −sin 45° = − –1

1 ; 3θ = 18.4° or 198.4° or 378.4° or… 3 Hence θ = 6.1° or 66.1° or 126.1° or 186.1° … The only obtuse angle is 126.1°

8 a tan 3θ =

b 4θ + 30° = 78.7° or 258.7 ° or 438.7° or 618.7° or … Hence θ = 12.2° or 57.2° or 102.2° or 147.2° or … The two obtuse solutions are 102.2° and 147.2°. 9

( )



If tan x = 2, then x = 1.11 or 1.11 − π = −2.03

sin x cos x

2

= 4; tan x = ± 2

If tan x = – 2, then x = −1.11 or −1.11 + π = 2.03. sin x 10 a 2  sin  x = tan x ⇒ 2sin x = cos x ⇒ 2 sin x cos x – sin x ⇒ sin x (2 cos x – 1) = 0 Either sin x = 0 or cos x = 1 . Hence x = 0°, 2 180° 360°, 60° or 300° b Either sin (2y + 35°) = 0 or cos ( 2y + 35° ) = 1 . 2 Either 2y + 35° = 0° or 180° or 360° or 540° or 720°… so y = 72.5° or 162.5° or 252.5° or 342.5° are the values in the interval given.



Either 2y + 35° = 60° or 300° or 420° or 660° or 780° or …



Possible values are y = 12.5° or 132.5° or 192.5° or 312.5° or 372.5°.

c In this case sin z (cos z – 2) = 0 and hence either sin z = 0 or cos z = 2.

The second equation has no solution. Hence z = 0° or 180° or 360° are the only solutions.

11 a x2 – 2x + 1 = 0; (x – 1)(x – 1) = 0; x = 1 is the only solution. b This factorises to (tan θ – and so θ = 45° or 225°. c Now

((tan θ)2

–

3 2

d

1)2

= 0 so

1)2

e cos 135° = −cos 45° = −

b tan120° = − tan 60° = − 3 c tan 225° = tan 45° = 1 3 a sin 3π = sin π = 1 4 4 2 b cos 3π = − cos π = − 1 4 4 2 3π = − tan π = −1 c tan 4 4 4 a

3 1 b − 3 − c 2 2 π 2π π 7π , b  θ = , 3 3 6 6

5 a θ =

(

6 a sin

3π 4 2

) + (sin 43π ) + (cos 54π ) 2

2

2

  =  1  +  − 3  +  − 1   2   2  2

2

=

2

1 3 1 + + = 3 2 4 2 14

b sin

π 2π 4π 5π − sin + sin π − sin + sin  − sin 2π 3 3 3 3



 3  3 3 3 + − − + (0) −  − −0 =0 2 2  2   2 

=

7 a tan 210° + tan 240° = tan30° + tan60° 1 3 4 3 = + 3= + 3= 3 3 3 b cos30° + cos45° + cos60° = =

= 1.

1 2

f cos 300° = cos 60° = 1 2 1 2 a tan 30° = 3

= 0 so tan θ = 1

(tan θ)2

1 2

(

3 1 1 + + 2 2 2

3 2 1 1 + + = 1+ 2 + 3 2 2 2 2

)

tan θ = 1 or −1 and so θ = 45°, 135°, 225° or 315°.

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4 

WORKED SOLUTIONS

b

1+ 3 2 = 2 1 + 3  c sin 30° + sin 60° = 2  2 sin 45° 1 2  2 1 1 2+ 6 = 2+ 2× 3 = 2 2

(

8



)

(

y 10

)

0

5π 9π π 2θ + 3π = π or 4 or 4 or … ⇒ 2θ = − or 7π 8 4 8 8 π 7π 15π or 15π or … ⇒ θ = − 16 or 16 or 16 or … 8 The two solutions in the interval are 7π or 15π . 16 16

c period = 2π = 4π; amplitude = 2 0.5 d

y 6

4

    = a  3 − 1  = a  3 − 1 = a × 2 = a 2 3 2 3  2 3 3

2

10 a In triangle ABC, angle BAC is 30°.

0

The length of AC = 2 × AB cos30°

π

2π

3π

4π

x

5 It is a sine wave of amplitude 4 and period 360

3 = 3a . 2

° = 2a × 2 × AB cos30

x

–10

a 9 BD = a sin 60° = 3a; AB = a cos60° = ; 2 2 1 a a = BC = AB tan 30° = × 2 3 2 3 Hence CD = BD – BC = 3a − a 2 2 3



0.5π

−45 translated by  .  0 

b The area of triangle ABC is

y

1 1 1 3 2. × AB × AC sin 30° = a × 3a × = a 2 2 2 4

The area of triangle ACE is 1 × AC × AE sin 60° 2



= 1 × 3a × 3a × 3 = 3 3 a 2 . 2 2 4



The area of the kite is 3 a 2 + 3 3 a 2 = 4 4

4 2

–180

–135

–90

–45

0

45

90

135

180 x

–2 2

3a .

–4

Exercise 4.5A 1 a y = 15 sin x 2 a y = −2 cos x 3 a period = b

b y = tan 2x b y = 10 + 5 sin x

360 = 180, amplitude = 4 2

6 a y = 2 + tan (x − 1) b y 4 2

y

0 –2

4

π

2π

3π

4π

5π

6π

x

–4 0

90

180

270

360

x

–4

4 a period = 2π = π , amplitude = 10 4 2

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4 Circular measure and trigonometry

Exercise 4.6A

7 a and b y

y = 2 – 2cosx

y = 2 – 2cos0.5x

4

1 Calculator operation 2 a cos θ = ± 1 − 0.68 2 = ± 0.733 two possible values

3

b tan θ =

2

3 a sin θ = ± 1 − 0.44 2 = ± 0.898

1

–360

–270

–180

–90

0

sin θ 0.68 =± = ± 0.927 cosθ 0.733

90

180

270

360

b tan θ =

x

sin θ 0.898 =± ± 2.041 cosθ 0.44

5 5 4 so cos2 x = 1 – sin2 x = 1 − = ; 9 9 9 2 cos x = 3

4 a sin2 x = 8 a Start with the graph of y = cos x.





Reflect it in the y-axis. The equation is y = cos (–x) but the graph is symmetrical and unchanged. 90 Translate by   so that it is now identical  0  to the graph of y = sin x. The equation is y = cos (–(x – 90°)) or y = cos (90° – x).

b cos (90° – x) = 2 cos x ⇒ sin x = 2 cos x ⇒ tan x = 2. Hence x = 63.4° or 243.4°.





The value is positive because x is acute.

b tan x =

sin x 5 2 5 = ÷ = cos x 3 3 2

1 . 2 2  2  6 (cos  x + 1)(cos x − 1) = cos x – 1 = −(1 – cos x) = −sin2 x 7 1 – sin θ cos θ tan θ = 1 – sin θ × cos θ × sin θ cosθ = 1 – sin θ × sin θ 5 One is a reflection of the other in the line y =

= 1 – sin2 θ = cos2 θ  9 The amplitude is 30 so start with y = 30 sin x 1 1 sin x cos x sin 2 x + cos2 x = + = = 8 a tan x + Stretch by 4 parallel to the x-axis to get the correct cos x sin x tan x cos x sin x cos x sin x 1 1 1 sin x cos x sin 2 x + cos2 x period. The equation is y = 30sin x . tan x + = + = = 4 cos sin x tan x cos x sin x cos x sin x x π   Finally translate by   . 1 1 2 2  0 b tan x + tan x = sin x cos x = 2sin x cos x = sin 2x . 1 The equation becomes y = 30sin ( x − π ) or 4 The range of values for sin 2x is – 1 ⩽ sin 2x ⩽ 1. 1 π . y = 30sin x − Hence 1  1 or ⩽ – 1 and hence 4 4 sin 2x π Hence a = 30, b = 1 and c = − . 4 tan x + 1  2 or ⩽ −2. 4 tan x Another possible value for c is 7π corresponding 9 (cos x + sin x)2 + (cos x – sin x)2 4  −7π to a final translation of  . = cos2 x + 2 cos x sin x + sin2 x + cos2 x – 2 cos x sin x +  0  sin2 x = 2(cos2 x + sin2 x) = 2 π 10 a The period is and so b = 2. 2 10 sin2 x + cos2 x = 1 so(sin2 x + cos2 x)2 = 1 The graph crosses the y-axis at 0.4 and so c = 0.4. Therefore sin4 x + 2 sin2 x cos2 x + cos4 x = 1 and so The equation is y = a tan 2x + 0.4. sin4 x + cos4 x = 1 – 2 sin2 x cos2 x. Substitute the coordinates π , 0.5 ; 8 1 sin 2 x cos2 x π ≡ 11 tan 2 x + 2 + 0.5 = a tan + 0.4. 2 2 +2+ 4 tan x cos x sin 2 x Hence 0.5 = a + 0.4 and so a = 0.1. sin 4 x + 2sin 2 x cos2 x + cos4 x ≡ b The curve will look the same but cross the sin 2 x cos2 x y-axis at (0, −0.4). 2 sin 2 x + cos2 x 1 The equation is y = 0.1 tan 2x – 0.4. ≡ ≡ sin 2 x cos2 x sin 2 x cos2 x

(

)

(

)

(

)

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4 

WORKED SOLUTIONS

(

)(

)



1 × (sin θ + cosθ )(sin θ − cosθ ) ≡ sin θ + cos θ sin θ − cosθ



sin 2 θ + cos2 θ sin 2 θ − cos2 θ sin 4 θ − cos4 θ ≡ 12 sin θ − cosθ sin θ − cosθ ≡

13 cos2 A = 1 − sin 2 A = 1 − 9 = 16 ; hence 25 25 4 cos A = ± 5 25 144 2 cos B = 1 − sin 2 B = 1 − = ; hence 169 169 12 cos A = ± 13 4 12 3 5 + × Hence cos A cos B + sin A sin B = ± × 5 13 5 13 63 33 48 15 + = 65 or − 65 = ± 65 65 14 sin θ −

1 sin 2 θ − 1 cos2 θ = =− sin θ sin θ sin θ



cosθ −

1 cos2 θ − 1 sin 2 θ = =− cosθ cosθ cosθ



tan θ +

1 sin θ cosθ sin 2 θ + cos2 θ = + = tan θ cosθ sin θ cosθ sin θ



1 cosθ sin θ cos2 θ  sin 2 θ  1 The product is − × − × sin θ  cosθ  cosθ sin θ = cosθ sin θ = 1 cosθ sin θ =

Exercise 4.7A 3 . One solution is 2x = π . Other 2 3 π π π π solutions are π − ,2π + ,3π − ,4π + , etc. 3 3 3 3 π 2 π 7 π 8 π 13 π That is 2x = , , , , etc. 3 3 3 3 3

1 a sin  2x =



So x = π , π , 7π and 4π . There are four solutions. 6 3 6 3

The next value, 13π is outside the interval. 6 b 4 tan  0.5x = −3 and so tan 0.5x = −0.75 tan−1 (−0.75) = −36.9°.





Therefore x = –73.7° or 286.3°. These are the only values in the interval −360° ⩽ x ⩽ 360°.

Therefore 0.5x = –36.9° , –216.9° , 143.1°, etc.

2 a cos (x + 75)° = 0.5 and cos−1 0.5 = 60°

So x + 75 = 60, 300, 420, 660, … and x = –15, 225, 345, 585, ….



The solutions in the interval 0° ⩽ x ⩽ 360° are x = 225° or 345°.

b Rearrange as 50 sin 2x = −35 and sin 2x = −0.7; sin−1 (−0.7) = −0.775.



2x = −0.775, −π + 0.775, π + 0.775, 2π – 0.775 … π x = –0.388, − π + 0.388, 2 + 0.388, π – 0.388 … 2 The four solutions in the interval –π ⩽ x ⩽ π are x = –0.388, x = –1.18, 1.96 and 2.75.

3 3 cos2 x – 2 cos x = 0 → cos x (3 cos x − 2) = 0

Either cos x = 0 → x = 90° or 270°



or 3 cos x – 2 = 0 → cos x =



There are four solutions.

2 → x = 48.2° or 311.8°. 3

4 cos2 x = 0.25 → cos x = ±0.5

π 5π or 3 3 2π If cos x= −0.5, x = or 4π . These are the four 3 3 solutions. If cos x = 0.5, x =

5 (3 sin x − 1)(2 sin x − 1) = 0

Either 3 sin x – 1 = 0 → sin x =



or 2 sin x – 1 = 0 → sin x =

1 → x = 19.5°, 160.5° 3

1 → x = 30°, 150°. 2

6 4 sin2 x – 7 sin x – 2 = 0 → (4 sin x + 1) (sin x − 2) = 0 1 Either 4 sin x + 1 = 0 → sin x = − → x = –0.253 4 which is not in the range so x = π + 0.253 and 2π − 0.253 = 3.394and 6.03.

or sin x – 2 = 0 → sin x = 2.



This has no solution.

7 sin x + 1 = 1 – sin2 x → sin x + sin2 x = 0 → sin x (1 + sin x) = 0

Either sin x = 0 → x = 0°, 180°, 360°



or 1 + sin x = 0 → sin x = −1 → x = 270°.



There are four solutions.

8 a Diameter = 67.5 – 67.5 cos180

= 67.5 – 67.5(−1)



= 135 m

b First solve 67.5 – 67.5 cos θ = 100



67.5 cos θ = −32.5 cos θ = −0.481









θ = 180 – 61.2, 180 + 61.2 = 118.8°, 241.2°

 The wheel is above the ground for 122.4° out of 360°. Length of time =

122.4 × 30 mins 360

= 10.2 mins

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4 Circular measure and trigonometry

9 a The graphs cross twice between 0° and 360° so there are 2 solutions. sin x b cos x = tan x → cos x = cos x  Multiply by cos x: cos2 x = sin x → 1 − sin2 x = sin x.



Rearrange: sin2 x + sin x − 1 = 0





Use the quadratic formula.



sin  x=



sin  x = −1.618 has no solution.





−1 ± 1 + 4 = 0.618 or −1.618 2

Exam-style questions 1 If θ is the angle of the sector, 16 θ = 36 so θ = 2.25

b The area of the sector containing the 1 triangle is a 2θ and the area of the triangle 2 1 2 1 1 is a sin θ and so a 2θ = 2 × a 2 sin θ and 2 2 2 hence θ = 2 sin θ.

10 20(1 − sin2 x) + 27 sin x = 29 Rearrange as 20 sin2 x − 27 sin x + 9 = 0.



This can be factorised as (5 sin x − 3)(4 sin x − 3) = 0



Hence sin x = 3 or sin x = 3 4 5



solutions to 3 s.f. are x = 0.644, 2.50, 0.848 or 2.29.

3 a sin2 x + cos2 x = 1 therefore 2

 3 3 13 sin 2 x = 1 −  − =1− = . 16 16  4  b sin x = ±

11 8(1 − cos² x) = 2cos x + 5 8cos²x + 2cos x – 3 = 0 (4cos x + 3)(2cos x – 1) = 0 cos x = −

3 1 or cos x = 4 2

x = −138.6°, 138.6°, −60°, 60° 12 cos x = 1 − cos² x + 0.3 cos² x + cos x – 1.3 = 0 cos x =

−1 ± 1 + 5.2 = 0.745 or −1.745 2

cos x = −1.745 has no solution. x = 0.730, 5.55 13 a 6(1 – sin2 x) + sin x = 5 ⇒ 6 sin2 x – sin x – 1 = 0 1 1 ⇒ (2 sin x – 1)(3 sin x + 1) = 0 ⇒ sin x = or − 3 2

Hence x = 30°, 150°, 199.5° or 340.5°.

b If you write sin2 y = sin x then you find 1 1 or − sin 2 y = 3 2 1 2 sin y = − has no solution. 3 1 1 1 2 then sin y = If sin y = or − 2 2 2 Hence y = 45°, 135°, 225° or 315° 14 a tan2 x + k = 2 tan x; tan2 x – 2 tan x + k = 0

This is a quadratic in tan x and the discriminant is (–2)2 –4 × 1 × k = 4 – 4k.



There is no solution if 4 – 4k  < 0 ⇒ k > 1.

b tan2 x – 2 tan x – 8 = 0; (tan x – 4)(tan x + 2) = 0

tan x = 4 or –2; x = 76.0°, –104.0°, –63.4 or 116.6°

1 × 2.25 × 162 = 288 cm2. 2

2 a The angle of the sector is 2π – θ so the area is 1 a2 (2π − θ) or πa2 − 1 a2 θ. 2 2

The smallest solution of sin x = 0.618 is 38.2°.



Area =



13 4 sin x 13 3 13 =± ÷− =± . cos x 4 4 3



Then tan x =



This could be written as ± 39 . 3

4 a 2 sin2 x = 3 cos x; 2(1 – cos2 x ) = 3 cos x ; 2 cos2 x + 3 cos x – 2 = 0 1 (2 cos x – 1)(cos x + 2) = 0; cos x = or −2 2 1 cos x = −2 has no solution. If cos x = then 2 π. x= 3 b Use the solution to part a

cos2θ =

π 1 π so 2θ = or 5π ; θ = or 5π 3 6 2 3 6

are the two solutions in the domain 0 ⩽ θ ⩽ π. 5 a The amplitude is a = 5; b = 7π as only the 4 positive value is required. b Again the amplitude is c = 5 ; d =

5π as only 4

the positive value is required. 6 a 2 sin2 x − sin x = 0

Factorise: sin x (2 sin x − 1) = 0.



Either sin x = 0 or 2 sin x − 1 = 0.



If sin x = 0 then x = 0 or π.



If 2 sin x − 1 = 0 then sin x = 1 so x = π or 5π . 2 6 6

There are 4 values for x. θ b Using the solution to part a, either sin 2 = 0 θ 1 or sin = 2 2

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4 

WORKED SOLUTIONS

θ θ = 0 , then = 0 and θ = 0 is the only 2 2 solution in the domain 0 ⩽ θ ⩽ π. θ 1 If sin = , then θ = π or 5π or … and 2 2 2 6 6 the only solution is θ = π . 3 1 2 7 Area of triangle = a sin 60° . 2 1 Area of sector = a 2θ so θ = sin 60° = 0.866. 2 The perimeter of the triangle is 3a and of the sector is 0.866a + 2a = 2.866a. The sector perimeter is 2.866 × 100 = 95.53% of the 3 triangle’s, about 4.5% less.

b sin θ tan θ = 4 sin θ In the domain, sin θ ≠ 0 so tan θ = 4 and θ = 1.33 to 3 s.f.

If sin

8 a 2y + 40 = 30° or 180 − 30 = 150° or 360 + 30 = 390°, etc

2y + 40 = 30 → y = − 5° which is outside the interval.



2y + 40 = 150 → y = 55°; 2y + 40 = 390 → y = 175°;

2y + 40 = 510 → y = 235°; 2y + 40 = 750 → y = 355°. There are four solutions. 1 b sin ( 2y + 40° ) = ; cos2 (2y + 40°) 2 1 3 = 1 – sin2 (2y + 40°) = 1 − = 4 4

So cos ( 2y + 40° ) = ±



sin ( 2y + 40° ) 1 3 1 c tan ( 2y + 40° ) = = ÷± =± 2 cos ( 2y + 40° ) 2 3 9 a cos (x + 30°) = − 0.8 x + 30° = 143.1° or 360 − 143.1 = 216.9°



x = 113.1° or 186.9°

b This is the equation in part a with 2x + 30° = θ.

So tan (2x + 30°) = 4; 2x + 30° = 76° or 256° or −104° or −284°.



The solutions are x = 23° or 113° or −67° or −157°.

12 cos A = 6 = 0.6 ; A = 0.9273 and the reflex angle at 10

A is 2π – 0.9273. The length of the major arc on the left is 6(2π – 0.9273) = 32.135.



cos B =



The length of the major arc on the right is 8(2π – 0.6435) = 45.117



The total length is 77.3 cm.



Rearrange: 2 sin θ cos θ − sin θ = 0. sin  θ (2cos θ − 1) = 0



Either sin θ = 0 so θ = 0, π or 2π



π or cosθ = 1 so θ = or 5π . 3 3 2



The solutions are 0, π , π, 5π and 2π. 3 3



y 9



5

60

240

x

–5

10 a

1 1 − cos2 θ sin 2 θ sin θ 1 − cosθ = = = × cosθ cosθ cosθ cosθ cosθ

1 − cos2 θ sin 2 θ sin θ − cosθ = = = sin θ × = sin θ tan θ cosθ cos θ = cosθ

sin θ = 2 sin θ → sin θ = 2 sin θ cos θ cosθ

b Using the answer to part a, sin (4x – π) = 0 or 1 cos(4x − π ) = 2 If sin(4x – π) = 0 then 4x – π = 0, ± π, ± 2π, ± 3π …

b

0 –60 –1

8 = 0.8 ; A = 0.6435 and the reflex angle at 10

B is 2π – 0.6435.

13 a

3 2



11 a Divide by cos θ: tan θ = 4 → θ = 76.0° or 76.0 − 180 = −104.0°.

So 4x = π, 0, 2π, −π, 3π, −2π, 4π, …

3π Hence x = 0, π , π , , π, 5π , 3π , 7π or 2π. 4 2 4 4 2 4 These are the values in the given domain. If cos(4x − π ) =

1 π π then 4x – π = − , , 3 3 2

5π , 7π , 11π , … 3 3 3 So 4x = 2π , 4π , 8π , 10π , 14π , … 3 3 3 3 3 π π 2 π 5 π 7 π and x = , , , , , 4π , 5π 6 3 3 6 6 3 3 = sin θ tan θ

or 11π . 6

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4 Circular measure and trigonometry

17 a The perimeter is 2r + r θ cm and hence 2r + r θ = 50. 50 − 2r Rearrange: r θ = 50 – 2r; θ = r

14 a y 1

0

(

)

1 2 r θ = 150 ; 1 r 2 50 − 2r = 150 ; 2 2 r 2 2 25r – r = 150; r – 25r + 150 = 0.

b The area is

0.5

90

180

270

360

x

c (r – 10)(r – 15) = 0; Hence r = 10 or 15 and 1 50 − 2r = 50 − 20 50 − 30 = 3 or =1 . 3 r 10 15 θ θ 2r = 2 ; = 1.107 18 a If angle AOB = θ, then tan = 2 2 r and θ = 2.21 to 3 s.f. 1 b Area of sector AOB = r 2 × θ = 1.107r 2 ; area 2 of ACBO = r × 2r = 2r 2.

θ=

 The graphs of y = cos x and y = tan x cross twice between 0° and 360°. b tan x = cos x therefore



therefore Therefore sin x = sin x = 1 − sin2 x. Rearrange: sin2 x + sin x − 1 = 0 Use the quadratic formula:

−1 ± 1 + 4 −1 ± 5 = 2 2 = 0.618 or −1.618 sinx = −1.618 has no solutions x = 38.2° or 141.8° 1 = 5. 15 a tan x + tan x sin x =



Multiply by tan x: tan2 x + 1 = 5 tan x.



Rearrange: tan2 x − 5 tan x + 1 = 0.

19 a 6 cos2 x + sin x = 5; 6(1 – sin2 x) + sin x = 5; 6 – 6 sin2 x = – sin x + 5;

c 6 sin4 y + 5(1 – sin2 x) = 6

16 a 6(1 − cos2 x ) + cos x = 5; 6 − 6cos2 x + cos x = 5; 6cos2 x − cos x − 1 = 0 1 b Factorise: (2 cos x − 1)(3 cos x + 1) = 0 so cos x = 2 1 or − 3 . 1 π If cos x = then x = or 5π ; if cos x = − 1 2 3 3 3 then x = 1.91 or 4.37.

6 sin4 y + 5 – 5 sin2 y = 6



6 sin4 y – 5 sin2 y – 1 = 0 (6 sin2 y + 1)(sin2 y – 1) = 0



only has a solution if k2 − 4 ⩾ 0. Hence k2 ⩾ 4 so that k ⩾ 2 or k ⩽ − 2.



d This is a quadratic in sin2 y:

5 ± 25 − 4 = 4.791 or 0.2087. Therefore  tan x = 2 x = 1.37 or 0.206. 1 c tan x + = k ; tan2 x − k tan x + 1 = 0. This tan x

6 sin2 x – sin x – 1 = 0

b Factorise: (3 sin x + 1)(2 sin x – 1) = 0; 1 1 sin x = − or . 2 3 x = 30°, 150°, 199.5° or 340.5°

b Use the quadratic formula.

Area of ACBO outside the circle is

2r2 – 1.107r 2 = 0.893r2. 0.893r 2 × 100 = 44.6%. c Percentage = 2r 2

cos2 x





sin x = cos x cos x

1 or 1 6 1 sin 2 y = − has no solution. 6 If sin2 y = 1 then sin y = 1 or −1; y = 90° or 270°. sin 2 y = −

20 cos AXY = 0.75; Angle AXB = 2 × 0.7227 = 1.4455 1 The area of sector AXB = × 1.4455 × r 2 = 0.7227r2. 2 1 The area of triangle AXB = × sin AXB × r 2 = 2 0.4961r2.

The difference is 0.2267 r2. The area in common is 2 × 0.2267 r² = 0.4533 r2. 0.4533 The percentage of one circle is × 100 = 14.4%. π

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4 

WORKED SOLUTIONS

Mathematics in life and work 1 The first oscillation is completed when 512πt = 2π Hence t = 2 = 1 . There are 256 cycles in one 512 256 second so the frequency is 256 Hz. 2 The frequency is 256 ×

3 = 384. 2

1 If the equation is y = a sin kt , then kt = 2π when 2 1 t= so k = 2π × 384 = 768π. 384 1 The equation is = a sin 768πt. 2



3

y a 0.5a 0

1 256

t

–2

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5 Series

5 Series Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 a

1 × 5 × 6 = 15 2

b If

 5  5 c (x + 4)5 = x5 + 5 × x4 × 4 +   × x3 × 42 +   ×  2  3

1 n(n + 1) = 54, then n2 + n − 108 = 0. 2

20x4 + 160x3 + 640x2 + 1280x + 1024 4 a (x − y)4 = x4 + 4x3(−y) + 6x2(−y)2 + 4x(−y)3 + (−y)4 b (1 − 2y)4 = 1 + 4(−2y) + 6(−2y)2 + 4(−2y)3 + (−2y)4

=x  3 +3x2 + 3x + 1 − x3 + 3x2 − 3x + 1 = 6x2 + 2 or 2(3x2 + 1)



$24 000 ×

1.055



 10 10! 10 × 9 b   = = = 5 × 9 = 45 2×1  2  2!8!

 12 12! 12 × 11 × 10 d   = = = 2 × 11 × 10 = 220 3× 2×1  3  3!9!

( )

3 a (2 + x)4 = 24 + 4 × 23x + 6 × 22x2 + 4 × 2x3 + x4 = 16 + 32x + 24x2 + 8x3 + x4

b (3 + 5x)3 = 33 + 3 × 32 × 5x + 3 × 3 × (5x)2 + (5x)3

= 32 − 80y + 80y2 − 40y3 + 10y4 − y5

also be found directly with a calculator.  6 b The x3 term is   23(4x)3 = 20 × 8 × 64x3 =  3  6 6 × 5 × 4 . 10 240x3. Here   =  3 3 × 2 × 1

= 27 + 135x + 225x2 + 125x3

(2 + x2 ) = 2 + 3 × 2 × x2 + 3 × 2( x2 ) + ( x2 ) = 8 + 6x + 32 x + 18 x 3

6 a

 8 8! 8 × 7 × 6 × 53 2 e   = = x = 7 3× 2 × 5 =2 70x x  4 4!4! 4 × 3 2× +2 2× 1 = 2 + 3 × 2 × 2 + 3 × 2 2 +  10 10 × 9 × 8 × 7  10 = 210 2 The coefficient is   or   = 4 × 3× 2×1  6  4 0 10 × 9 × 8 × 7 = 210. This can also be found directly with a = 4 × 3× 2×1 6  calculator.





 5 5 a The x3 term is   (3x)3 = 5 × 4 × 27x 3 = 10 × 27x 3 = 270x 3 2×1  3  5  5   5 5×4 3 3 3 3  3 (3x) = 2 × 1 × 27x = 10 × 27x = 270x . The value of  3 =  2 could

 12 12! 12 × 11 c   = = = 6 × 11 = 66 2  10 10!2!



× 22(−y)3 + 5 × 2(−y)4 + (−y)5



 6 6! 6 × 5! 1 a   = = =6  1  5!1! 5!× 1

= 1 − 8y + 24y2 − 32y3 + 16y4



c (2 − y)5 = 25 + 5 × 24(−y) + 10 × 23(−y)2 + 10

= $30 630.76.

Exercise 5.1A



= x4 − 4x3y + 6x2y2 − 4xy3 + y4



3 The multiplier for a 5% increase is 1.05.

 5   5 5 × 4 Now   =   = = 10 so (x + 4)5 = x5 +  2  3 2 × 1



This cannot be factorised with whole numbers so 54 is not a triangular number.

2 (x + 1)3 − (x − 1)3 = x3 + 3x2 + 3x + 1 − (x3 − 3x2 + 3x − 1)

x2 × 43 + 5x × 44 + 45



3

2

( ) ( x2 ) = 8 + 6x + 32 x + 18 x 3

2

2

3

2

3

b (5x − 4y)4 = (5x)4 + 4(5x)3(−4y) + 6(5x)2(−4y)2 + 4(5x)(−4y)3 + (−4y)4

= 625x4 − 2000x3y + 2400x2y2 − 1280xy3 + 256y4  10 7 a The term involving x3 is   × 27 x 3, where 3   10 10! 10 × 9 × 8  3  = 7!3! = 3 × 2 × 1 = 120, so the coefficient is 120 × 27 = 15 360.



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5 

WORKED SOLUTIONS

14 (1 + x2)3 (1 – x2)3 = {(1 + x2) (1 – x2)}3 = (1 – x4)3 

 9 b The term involving x3 is   × 26(3x)3, where  3  9 9! 9×8×7  3 = 6!3! = 3 × 2 × 1 = 84, so the coefficient

= 1 – 3x4 + 3x8 – x12 15 a (1 + x)3 + (1 – x)3 = 1 + 3x + 3x2 + x3 + 1 – 3x 

is 84 × 64 × 27 = 145 152.

b (1 + 3y)3 + (1 – 3y)3 = 2 + 6(3y)2 = 2 + 54y2

 7  7 7 × 6 × 5 = 35 8 The x3 term is   × 24(bx)3 and   =  3  3 3 × 2 × 1 so the term is 35 × 16 × b3x3 = 560b3x3. Hence 560b3 = 70 000 and b3 = 125 so b = 3 125 = 5 . 9 a (x2 + 2)4 = (x2)4 + 4 × (x2)3 × 2 + 6(x2)2 × 22 + 4x2 × 23 + 24 = x8 + 8x6 +24x4 + 32x2 + 16



c 2 + 54y2 = 11 ⇒ 54y2 = 9 ⇒  y 2 = 1 and so 6 1 y=± 6  8 2 16 The terms in x2 and x3 are   a 6 ( bx ) and  2  8 5 3 a ( bx ) .  3

b (1 − y3)5 = 1 + 5(−y3) + 10(−y3)2 + 10(−y3)3 +



=1

+

10y6

−

10y9

+

5y12

−

y15

 8  8 10 (1 + 2x)8 = 1 + 8(2x) +   (2x)2 +   (2x)3 +…, where  2  3

1 a 58 – 25 = 33; 91 – 58 = 33; 124 – 91 = 33; an arithmetic progression with d = 33. b The 15th term is 25 + 14 × 33 = 487.

 8 8 × 7  8 8 × 7 × 6  2 = 2 = 28 and  3 = 3 × 2 × 1 = 56 .

c The 30th term is 25 + 29 × 33 = 982.



So (1 + 2x)8 = 1 + 16x + 28 × 4x2 + 56 × 8x3 + … = 1 + 16x + 112x2 + 448x3 + …



So (1 + x)(1 + 2x)8 = (1 + x)(1 + 16x + 112x2 + 448x3 + …). x3



448x3

112x2

The term is 1 × +x× = 560x3. The coefficient is 560.

2 a d = 57 – 45 = 12; u12 = 45 + 11 × 12 = 177 b d = 191 – 200 = –9; u14 = 200 – 9 × 13 = 83 c d = 55.1 – 48.7 = 6.4; u20 = 48.7 + 19 × 6.4 = 170.3 d d = 192 – 215 = – 23; u15= 215 – 14 × 23 = –107

= (448 + 112)

x3

11 (x + 1)(x − 1) = x2 − 1 so (x + 1)8(x − 1)8 = (x2 − 1)8  8 The x10 term is   (x 2)5(−1)3 and  5 8   8×7×6  5 = 3 × 2 × 1 = 56 . The x10 term is 56x10 × (−1) = −56x10.



12 a (1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4 b If x = 2 , then (1 + 2)4 = 1 + 4 2 + 6

( )

2

(1 + 2)4 = 1 + 4 2 + 6 2 + 4



( 2) + ( 2) 3

( 2)

4

= 1 + 4 2 + 12 + 8 2 + 4 = 17 + 12 2

Hence a = 17 and b = 12. 13 a The constant term is

( ) = 20 × 8 × 1 = 160  4 3 b   ( x ) ( a ) = 6 × 1 a hence a 2 4  2 2 x  6 3 1  3 ( 2x ) x 2

3

2

a2 = 64; a = ±8

2

Hence 28a6 b2 = 56a5 b3; hence 28a = 56b and a 56 2 and so a : b = 2 : 1. = = b 28 1

Exercise 5.2A

5(−y3)4 + (−y3)5 −5y3

+ 3x2 – x3 = 2 + 6x2

2

= 96 ;

2

+4

3 a S10 =

10 (2 × 20 + 9 × 5) = 425 2

12 (2 × 120 + 11 × (−9)) = 846 2 15 c S15 = (2 × 9 + 14 × 2.5) = 397.5 2 b S12 =

4 a a = 20 and d = 7; 10th term = 20 + 9 × 7 = 83 10 {2 × 20 + 9 × 7} = 515. 2 180 so n = 26.7. c If 20 + 7(n − 1) = 200 then n − 1 = 7 3 4 2 + 2 So the term required is the 26th. 20 5 The series is 2 + 6 + 10 +... (20 terms). S20 = (2 × 2 2 + 19 × 4) = 800. 6 a a = 15, and d = 6  ⇒  un = 6n + 9 b S10 =

( ) ( )









On Saturday, n = 7  ⇒  u7 = 6 × 7 + 9 = 51

Alex read 51 pages on Saturday. b Sn = n (30 + 6(n – 1)) 2 S7 = 7 (30 + 6 × 6) = 7 × 66 = 231 pages 2 2

7 a = 50, d = 5 and n = 24; S24 =

24 (2 × 50 + 23 × 5) = 2

2580. So she will save $2580.

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8 u5 = a + 4d = 27.6 and u10 = a + 9d = 24.1.

14 The 10th term is twice the 4th term so a + 9d = 2(a + 3d); a + 9d = 2a + 6d; a = 3d.



Subtract to get 5d = −3.5 so d = −0.7.



Substitute into the first equation: a − 2.8 = 27.6 so a = 30.4.



The 18th term is 50 so a + 17d = 50; 3d + 17d = 50; 20d = 50; d = 2.5.



Therefore u20 = 30.4 − 19 × 0.7 = 17.1.



Then the first term a = 3d = 7.5.

9 a The multiples form an arithmetic progression with a = 6 and d = 6.

S10 =

10 (2 × 6 + 9 × 6) = 330 2

n n (12 + (n − 1) × 6) = (12 + 6n − 6) 2 2 n = (6n + 6) = 3n(n + 1) 2 10 a The odd numbers form an arithmetic progression with a = 1 and d = 2 so Sn = n (2 + (n − 1) × 2) 2 n n = (2 + 2n − 2) = × 2n = n2. 2 2 b Sn =

15 If n = 1, then 1 + 4 = 5 = a.

If n = 2, then 4 + 8 = 12 = a + a + d = 2a + d = 10 + d and so d = 2.



The nth term is 5 + 2(n – 1) = 2n + 3.



Hence the 100th term is 203

16 If n = 1, then the first term is 6 + 8 = 14.

If n = 2, then the first + the second term is 24 + 16 = 40 so the second term is 40 – 14 = 26.



If an arithmetic sequence starts 14, 26 … , then a = 14 and d = 12 and the sum of n terms is 1 1 n 28 + 12(n − 1)} = n(28 + 12n − 12) 2 { 2 1 2 = n(12n + 16) = 6n + 8n, so the sequence is 2 arithmetic.



b Each even number is 1 more than the corresponding odd number so Sn = n2 + n. n (20 + 4(n − 1)) = 792 2 n Therefore (16 + 4n) = 792. 2

11 Sn =

8n + 2n2 = 792



n2 + 4n − 396 = 0



(n + 22)(n − 18) = 0



n = –22 or 18



You want the positive root; there are 18 terms.

12 a The differences between terms are 1, 3, 1, 3, 1, 3, … They are not constant so it is not an arithmetic sequence. b Here are two possible methods:

Method 1: Add pairs of terms to get this sequence: 21, 29, 37, …



This is an arithmetic progression with a = 21 and d = 8. The sum required is 50 terms of this sequence. S50 = 25 {42 + 49 × 8} = 10 850.



Exercise 5.3A 1 a a = 0.04 and = b a = 20 and r =

b x = a + 7d and hence x = a + 7(y – x); x = a + 7y – 7x; a = 8x – 7y.

0.01 = 2; 0.005

u15 = 0.005 × 214 = 81.92 d a = 3 and r =

−6 = −2; u10 = 3 × (−2)9 = −1536 3

2 a a = 5 and r = 10 ÷ 5 = 2; u15 = 5 × 214 = 81 920 b S15 = 3 a S10 = b S8 =

Each is an arithmetic progression with d = 4. Add 50 terms of each one.

13 a The difference is y – x and the next term is y + (y – x) = 2y – x.

24 = 1.2; u6 = 20 × 1.25 = 49.7664 20

c a = 0.005 and r =

Method 2: Split the sequence into two: 10, 14, 18, … and 11, 15, 19, …

Sum = 25 {20 + 49 × 4} + 25 {22 + 49 × 4} = 5400 + 5450 = 10 850.

0.2 = 5; u8 = 0.04 × 57 = 3125 0.04

5(215 − 1) = 163 835 2−1 2(310 − 1) = 59 048 3−1

100(1 − 0.98) = 569.533 570 (3 s.f.) 1 − 0.9

c a = 4 and r = –2 so S12 =

4((−2)12 − 1) = −5460 −2 − 1

d a = 20 and r = 1.1 so S20 =

20(1.120 − 1) = 1150 1145.5 (3 s.f.) 1.1 − 1

4 The total amount is a geometric series with a = 12 000, r = 1.03 and n = 10.

S10 =

12 000(1.0310 − 1) = $137 566.55 (3 s.f.) 0.03

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WORKED SOLUTIONS

1 1 ;    S∞ = 2 1− 1 2 1 2 a = 2 and r = ; S∞ = 3 1− 1 3 5 1 S∞ = a = 5 and r = − ; 3 1+ 1 3 3 1 a = 1 and r = ; S∞ = 4 1− 3 4 2 −4 6 a = 6 and r = = − ; S∞ = 6 3 1+ 2 3

5 a a = 1 and r = b c d e

b The distance walked is 800 + 400 + 200 + ... m.

=2

=3 =



15 4

c Distance north = 800 + 200 + 50 + …

18 5

6 a The amounts are a geometric progression with a = 0.01 and r = 2.

0.01(221 − 1) = $20 971.51 2−1

c On his 65th birthday he would receive $0.01 × 264 which is more than $1.8 × 1017. This is an absurdly large amount. 7 a The amounts each year are a geometric progression with a = 2000 and r = 1.25.

The amount in the third year is $2000 × 1.252 = $3125.

2000(1.259 − 1) = $51 605 After 9 years she has S9 = 1.25 − 1 9 2000(1.25 − 1) S9 = = $51 605 so it does take 9 years. 1.25 − 1

b

( ) = 165

1 1 1 1 1 + of = + 4 4 4 4 4

2

( ) = 6421

5 1 + 4 16

3

1 c This is a geometric series with a = r = so 4 1 1 2 3 1 1 1 1 S∞ = + + +…= 4 = 4 = . 1 3 4 4 4 3 1− 4 4

() ()



One-third of the square is shaded.

9 a The distance north is 800 – 400 + 200 – 100 + ... m.

This a geometric series with a = 800 and r = − 1 . 2 800 2 1600 1 = 533  m north = 800 × = So S∞ = 3 3 3 1 + 12

1600 + ( 3200 3 ) ( 3 ) 2

2

 = 1193 m

10 This is a geometric series with a = 0.45 and r = 0.01. S∞ =

0.45 0.45 45 5 = = = 1 − 0.01 0.99 99 11

11 a = 80 and

80 a = 200 so 1 − r = 200. 1−r

80 = 0.4 and therefore r = 0.6. 200



1−r =



S11 =

80(1 − 0.611) = 199.27 but 1 − 0.6

S10 =

80(1 − 0.610) = 198.79 so 11 terms are needed 1 − 0.6

2000(1.258 − 1) = 39 684 $ b After 8 years she has S8 = 1.25 − 1 $39 684 which is less than $50 000.

8 a



= 400 = 400 = 1600 3 3 1− 1 4 4 By Pythagoras, distance from start =

The 21st term is 0.01 × 220 = $10 485.76.

b S21 =



= 800 = 800 = 3200 3 3 1− 1 4 4 Distance east = 400 + 100 + 25 + …

=4 =

This is a geometric series with a = 800 and r = 1 . 2 800 S∞ = = 1600 so she walks 1600 m. 1 − 12

before the total is greater than 199. 12 Split the sum into two separate series: 1 + 2 1 1 1 1 + 3 + 4 + … and + + + +… 2 4 8 16

The first is an arithmetic series, a = d = 1 and n 1 Sn = {2 + (n – 1)} = n(n + 1). 2 2



1 The second is a geometric series with a = r = 2 n 0.5(1 − 0.5 ) n . = 1 − 0.5 and S n = 1 − 0.5 So for the series given Sn = 1 n(n + 1) + 1 − 0.5n . 2





a 20 = 200 ; 1 − r = = 0.1 ; r = 0.9 1−r 200 1% of 200 = 2 so the total must be at least 198.



S43 =



20 1 − 0.944  = 198.06 1 − 0.9 This shows that the n must be at least 44

13 a = 20 and

S44 =

(

)

(

)

20 1 − 0.943  = 197.84; 1 − 0.9

of the starting point.

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14 The second term is ar and the sum to infinity is a . 1−r a = 4ar ; 1 = 4r(1 – r); 4r2 – 4r + 1 = 0; Hence 1−r (2r – 1)(2r – 1) = 0. 1 The only solution is r = . 2 15 a From her aunt she receives amounts in an arithmetic progression, a = 30 and d = 10.

On her 10th birthday, n = 6 and she receives 30 + 5 × 10 = $80.



From her uncle she receives amounts in a geometric progression, a = 0.1 and r = 2.



On her 10th birthday, n = 6 and she receives 0.1 × 25= $3.20.



The total is $83.20.

b Anna has had 17 birthday presents so n = 17. 17 From her aunt 2 {2 × 30 + 16 × 10} = $1870.

(

17

)

From her uncle 0.1 2 − 1 = $13107.10. 2−1 The total is $14977.10. 16 a S∞ = a = x = 8 ; x = 8(1 – 3x); x = 8 – 24x; 1 − r 1 − 3x 8 25x = 8; x = or 0.32 25 4 2 2 b ar = 4; r = ; r = 4 ; r = ± 2 a a a



S∞ =

a a = = 1−r 1± 2 a

 10 1 a (1 + 2x)10 = 110 + 10 × 2x +   × (2x)2 + … 2  = 1 + 20x + 45 × 4x2 + …





= 1 + 20x + 180x2 + …

b (3x − 4)(1 + 2x)10 = (3x − 4)(1 + 20x + 180x2 + …) =

The term x2 is 3x × 20x − 4 × 180x2 = 60x2 − 720x2



The coefficient is −660.

2 a

= −660x2.

 (1 − 3x)4 = 1 + 4(−3x) +

4 2 3 4  2 (−3x) + 4(−3x) + (−3x)





= 1 − 12x + 6 × 9x2 − 108x3 + 81x4





= 1 − 12x + 54x2 − 108x3 + 81x4

b (2 + ax2)(1 − 3x)4 = (2 + ax2)(1 −12x + 54x2 − 108x3 + 81x4)

4 a S20 = 10(2a + 19d ) and S30 = 15(2a + 29d ) Hence 15(2a + 29d ) = 20(2a + 19d ); 

3(2a + 29d ) = 4(2a + 19d );

6a + 87d = 8a + 76d; 2a = 11d; a = 11d 2 b  15(2a + 29d) = 400; substitute to get



15(11d + 29d) = 400; 15 × 40d = 400; 10 2 d = 15 = 3 Then a = 11 × 2 = 11 2 3 3 5 a An arithmetic sequence with a = 1200 and d = 70. 12 S12 = (2 × 1200 + 11 × 70) = 19 020. So he 2 earns $19 020. b In the second month he earns 1200 × 1.05 = $1260 which is less than 1200 + 70 = $1270. 1200(1.0512 − 1) = $19 100.55, 1.05 − 1 which is more than $19 020. 8 32 6 a ar 2 = and ar 4 = 3 27

In the year he earns



Hence r 2 =

32 8 32 3 4 ÷ = × = 27 3 27 8 9

8 4 ÷ = 6. 3 9

4 2 , then r = ± . 9 3 2 6 If r = , then S∞ = = 18. 3 1− 2 3 2 3 If r = − , then S∞ = 6 = 6 ÷ 5 = 18 or 3 . 3 5 2 3 5 1+ 3 6 7 a The term in x3 is   ( 2x )3 ( −k )3  3 = – 20 × 8x3 k3 = – 160k3 x3. b If r 2 =

Exam-style questions



( )

and a =

a a a ±2



3 The term independent of x, where the x terms 4  6 cancel out is   (x 2)2 × 1 = 6 × 5 x 4 × 1 4 = 15 2 x 2×1 16 16x  2



 6 5 The term in x5 is   ( 2x ) (−k) =  – 6 × 32x5 k  5 = – 192k x5.

The coefficients are equal so – 160k3 = – 192k; 192 = 1.2 ; k = 1.2 . hence k 2 = 160  6 4 2 b The term in x2 is   ( 2x ) ( −k )  2

= 15 × 4x2 × 1.22 = 86.4 x2.

The coefficient is 86.4.

x3

The coefficient of is 2 × − 108 + a × −12 = −216 − 12a. If −216 − 12a = −132 then −12a = 84 and a = −7.

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WORKED SOLUTIONS

8 a a = 24 and a + d = 18 so d = – 6 n n 2a + ( n − 1) d } = {48 − 6 ( n − 1)} 2 2{ n = ( 48 − 6n + 6 ) = 27n – 3n2 2 If Sn < – 30 then 27n – 3n2 < – 30; 3n2 – 27n – 30 > 0; n2 – 9n – 10 > 0

Sn =



(n – 10)(n + 1) > 0; hence n > 10 or n < – 1.



Since n is positive, n > 10. b a = 24 and ar = 18; r = 18 = 3 ; 24 4 a 24 = 96 S∞ = = 1−r 1− 3 4 9 a a + 2d = 51 and a + 10d = 187; hence 8d = 187 – 51 = 136; d = 136 ÷ 8 = 17.



If Sk = 504, then 2k2 + 18k = 504; k2 + 9k – 252 = 0; (k + 21)(k – 12) = 0.



Hence k = −21 or 12. The value of k is positive. Hence k = 12.

14 a The total length is 20 + 10 + 5 + ... which is a geometric series with a = 20 and r = 0.5 so the 20 = 40 . sum S∞ = 1 − 0.5 b The x-coordinate is 20 – 5 + 1.25 − ... which is a geometric series with a = 20 and r = −0.25 20 so the x-coordinate is S∞ = = 16 . 1 + 0.25



Hence a + 34 = 51 and a = 17; the terms of the sequence are 17, 34, 51, … the multiples of 17. b 1000 = 58.8, so there are 58 multiples of 17 17 between 0 and 1000.

 7 10 a The term in x3 is 3x ×   × 2x 2 × (−b)6 1

= 3 × 7 × 2b6 x3 = 42b6 x3.



Hence 42b6 = 2688; b6 = 64; b = ± 2.

 7 b The term in x2 is a ×   × 2x 2 × (−b)6 1 = 14ab6 x2. 2688 Hence 14ab6 = 2688; a = = 3. 14 × 64 11 a $400 has 1 month’s interest, another $400 has 2  month’s interest and so on.

Total = 400 × 1.005 + 400 × 1.0052 + … + 400 × 1.00512.



This is a geometric series with a = 400 × 1.005 = 402 and r = 1.005.

12 b  S12 = 402(1.005 − 1) = 4958.90 , she has 1.005 − 1 $4960 (3 s.f.).

12 a (a + 5d) + (a + 6d) + (a + 7d) = 12; hence 3a + 18d = 12; a + 6d = 4; a = 4 – 6d b a + 5d = 12; hence (4 – 6d) + 5d = 12; 4 – d = 12; d = –8 Hence a = 4 – 6d = 4 + 48 = 52.

13 a = 20, a + d = 24, d = 4

k k 2a + ( k − 1) d } = {40 + 4 ( k − 1)} 2{ 2 k = {4k + 36} = 2k2 + 18k 2

Sk =



15 a The ratios of successive terms are equal so 6p + 2 4p + 4 = . 4p + 4 3p + 3

With a = 17 and d = 17 then

S58 = 26(34 + 57 × 17) = 26 078





The y-coordinate is 10 – 2.5 + 0.625 – ... which is a geometric series with a = 10 and r = −0.25 so 10 = 8. the y-coordinate is S∞ = 1 + 0.25 The coordinates are (16, 8).



6p + 2 4 Hence 4p + 4 = 3 and 3(6p + 2) = 4(4p + 4);



18p + 6 = 16p + 16;



2p = 10 and so p = 5.

b The first term is 6 × 5 + 2 = 32, the second is 24 3 4 × 5 + 4 = 24 and r = = . 32 4

S∞ = 323 = 32 1 = 128 1− 4 4

16 a If the common difference is d then b = a + 2d 1 and d = (b − a). 2 5 The sixth term is a + 5d = a + (b − a) 2 5 5 3 1 5 = a + b − a = b − a or (5b − 3a). 2 2 2 2 2 b b If the common ratio is r, then ar2 = b; r 2 = ; a b r=± a .

The fifth term is ar4 = a ×

() b a

2

=

b2 ab 2 2 = a . a

63 17 a 64 n b The nth term is 2 n− 1 . 2 n 2 − 1 1 c =1− n . 2n 2

( )( )( )

Hence Sn = 1 − 1 + 1 − 1 + 1 − 1 + ... 2 4 8 1 +… + 1 − n  =  n −  1 + 1 + 1 + ... 1n  .  2 4 8 2  2 

73

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5 Series

The terms in the bracket form a geometric sequence with a = r =

1 and n terms. 2

()

n 1 1 − 1    2 2 1 1 1 1   =  1 − 1 + + + ... n =  2 1 2 4 8 2 1− 2

1 The value of the first payment after 1 year is $

()

18 If the common ratio is r, then S∞ =

1 1 1 Therefore = k; = 1 − r ; r = 1 − . 1−r k k



n

a = ka . 1−r

( k1 ) = a − ka .



12 a 12 S∞ = = r = = 50 1 − r 1 − r r(1 − r )



Hence 12 = 50r(1 – r); 6 = 25r – 25r2;



20 a If the shortest piece is a then the longest is a + 9d = 4a. 1 Hence 3a = 9d and d = a 3 The sum is 100 so 10 ( 2a + 9d ) = 100 ; 2 1 5 2a + 9 × a = 100; 25a = 100 so a = 4. 3 The shortest piece is 4 cm.

(

5000 1.02

((

)

1 15 1.02

−1

) = 5000 ((

S15

n = a(r − 1) = r −1



= 64 246. The value is $64 200 (3 s.f.)

1 1.02

−1

)

1 15 1.02

−0.02

−1

)

R , then in this case it is a 100 5000 1 geometric progression with a = and r = . c c Then 5000 5000 c (1 − c −15) × S15 = 1 − c −15 1 − 1 = c c c c −1

2 If you write c = 1 +



25r2 – 25r + 6 = 0 3 2 b (5r – 2)(5r – 3) = 0; r = or = 0.4 or 0.6 5 5 12 12 12 = 30 or = 20 so a = c a = r 0.4 0.6 3 3 The 4th term is ar = 30 × 0.4 = 1.92 or 20 × 0.63 = 4.32.

Total value = 5000 + 50002 + … + 5000 . 1.02 1.02 1.0215

This is a geometric progression with a = 5000 ,r = 1 1.02 1.02 1 and n = 15. 5000 a= ,r = 1.02 1.02

The second term is ar = a 1 −

19 a The first term is 12 ; r

5000 , 1.02

5000 the value of the second after 2 years is $ and 1.022 so on.

n

()

()

n  1 So Sn = n − 1 − 1   =  n − 1 + 2 2  

Mathematics in life and work

(

=

(

)

( )

)

5000 500000 × 1 − c −15 = × (1 − c −15) . c −1 R

3 If the annuity is for n years, then 500 000 Sn = × (1 − c −n). R

Now limn→∞ c−n = 0 so the value of a perpetuity, S∞ =

500 000 . R

)

b If the shortest piece is a, then the longest is ar9 = 4a; r9 = 4 ;  r = 9 4 = 1.1665

(

)

a r 10 − 1 100 = 100 ;  a =  = r −1 22.015 4.54 cm to 3 s.f.

The sum is

21 The fifth term is r4. The sum of the subsequent 5 terms is ar5 + ar6 + ar7+ ... =  ar . 1−r 5 Hence  ar = 9ar 4 ;  r = 9 ; r = 9(1 – r); 10r = 9; 1−r 1−r r = 0.9. a a = = 10a . Then S∞ = 1 − r 0.1

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6

WORKED SOLUTIONS

6 Differentiation Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1

2

a 7 1 c 2

6

a f(x) = x2 − 6x + 5 so f ′(x) = 2x − 6 b f ′(1) = −4 and f ′(11) = 16

b −0.3

c

d −2

y 8

a 2x2 − 8x b

x3

+

6

3x2

4

c x2 + 4x − 5x − 20 = x2 − x − 20 d 3

9x2

a

x−3

d

1 x3

+ 3x + 3x + 1 =

9x2

b

1 x2

e

3 x2

2

+ 6x + 1 c

–4

−1 x 2

0 –2 –2

3

4

b 6x b 2x − 4

2

4

6

x

4

c 6x + 4 c 4 − 2x

2

dy = 2x + 6 dx dy b y = x2 − 2x − 3, = 2x − c2 dx dy c y = 4x2 + 12x + 9, = 8x + 12 dx

–6

–4

0 –2 –2 –4 –6 –8 8 –10

dy = 2x + 1 dx ii If x = 0,

dy = 1. dx

7

a f ′(x) = − 2x

8

b f ′(−1) = 2 and f ′(3) = −6 c y 3

dy = 30x − 30 dx a If x = 1, then

c At (2, 0) and (−3, 0), the gradients are 5 and −5.

–2

x

–6

a y = x2 + 6x,

–3

8

y

b i at (0, −6)

5

6

6

a 3 a 2x + 4

a

4

–4

Exercise 6.1A 1 2

2

dy =0 dx

2k × 4 − 4k = 2

1

8k − 4k = 2

–2

1

2

3

x

dy = 120 dx

y = kx(x − 4) = kx2 − 4kx dy = 2kx − 4k dx dy When x = 4, = 2. dx

2

0 –1 –1

b If x = 5, then

4k = 2 1 k = or 0.5 2

–3 –4

75

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6 Differentiation

dy 9 a  = 2x − 6 ; where the curve crosses the y-axis dx

13 a

x = 0 and then the gradient is −6. b The curve crosses the x-axis where x2 – 6x – 16 = 0; (x – 8)(x + 2) = 0; x = 8 or −2. dy = 16 − 6 = 10; If x = 8, dx

10 a P is (1, 1) and A is is

The gradient of PB is

11 a  A is (1, 2) and B is (3, 12). The gradient of 12 − 2 AB =  =5 3−1 b The y-coordinate of Q is (2 – h)2 + (2 – h) = 4 – 4h + h2 + 2 – h = 6 – 5h + h2.

The y-coordinate of P



is (2 + h)2 + (2 + h) =  4 + 4h + h2 + 2 + h = 6 + 5h + h2.



The gradient of PQ = 

(6 + 5h + h ) − (6 − 5h + h ) = 10h = 5. 2

(2 + h) − (2 − h)

2h

dy = 2x + 1 so the gradient at (2, 6) is 2 × 2 + 1 dx = 5 = the gradient of PQ.

12 a If y = 0, then (x + 3)(x – 4) = 0 and so x = −3 or 4 y =

x2 –

dy x – 12 and so = 2x − 1. dx

At (−3, 0), 

dy = ( 2 × −3) − 1 = −7 ; at (4, 0),  dx

dy = ( 2 × 4 ) − 1 = 7. dx b At (5, 8) 

dy = 2 × 5 − 1 = 9 so the equation of dx

the tangent is y – 8 = 9(x – 5) or y = 9x – 37. dy 3 c If = 2, then 2x – 1 = 2 and x = = 1.5 ; 2 dx y = (1.5 + 3)(1.5 – 4) = – 11.25.

)

The gradient of PQ = 

1.1 − 1 = 3.31. 1.1 − 1

c As h approaches 0, 3 + 3h + h2 approaches 3 so the gradient = 3.



(

so the gradient of PA

1 + 3h + 3h 2 + h3 − 1 3h + 3h 2 + h3 = = 3 + 3h + h 2. 1+h −1 h



( h8 )  = 2 + h4 .

3

2

h , if h is 4

small then 4 + h ≈ 2 1 +



b (1 + h)3 = 1 + 3h + 3h2 + h3 so B is (1 + h, 1 + 3h + 3h2 + h3)

Using the result given with a =

b P and Q are (4, 2) and 4 + h, 4 + h .

if x = −2, dy = −4 − 6 = −10. dx (1.1, 1.13)



( h4 )  =  4 1 + h4  = 2 1 + h4

4+h = 4 1+

h 4+h −2 2+ 4 −2 1 ≈ = ; the gradient 4 4+h−4 h

1 of the curve at P is  . 4 dy 14 a  = 18 − 4x ; at A 18 – 4x = – 2; 4x = 20; x = 5 and dx y = 45; A is (5, 45) b At A the equation of the tangent is y – 45 = – 2(x – 5) or y + 2x = 55.

This crosses the y-axis at (0, 55) and the x-axis at (27.5, 0).



The area of the triangle is 55 × 27.5 ÷ 2 = 756.25.

Exercise 6.2A 1 a 2 × 3x2 = 6x2 b 0.5 × 4x3 = 2x3 4 4 c 0.1 × 5x = 0.5x d 50 × 3x2 = 150x2 2 a 3x2 + 8x − 8

b 6x2 − 10x + 6

3 a 4x3 + 16x

b 5x4 − 30x2 + 2

4 a 3x2 − 4x b  i When x = 2, 3x2 − 4x = 12 − 8 = 4.   ii When x = −1, 3x2 − 4x = 3 + 4 = 7. 5 a f ′(x) = 0.5 × 4x3 − 2 × 2x + 1 = 2x3 − 4x + 1 b  i When x = −1, 2x3 − 4x + 1 = −2 + 4 + 1 = 3.   ii When x = 1, 2x3 − 4x + 1 = 2 − 4 + 1 = −1. iii When x = 2, 2x3 − 4x + 1 = 16 − 8 + 1 = 9. 6 a y = 2x3 + 5x2

dy = 6x2 + 10x dx

b y = x(x2 − 8x + 16) = x3 − 8x2 + 16x

dy = 3x2 − 16x + 16 dx

c y = x3 + x2 + x + 1

dy = 3x2 + 2x + 1 dx

The point is (1.5, −11.25).

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6 

WORKED SOLUTIONS

7 a f(x) = x3 + 3x2

13

f ′(x) = 3x2 + 6x

b  i f ′(−3) = 27 − 18 = 9

  ii f  ′(−2) = 12 − 12 = 0



iii f  ′(1) = 3 + 6 = 9 8 y = x(x2 − 6x + 8) = x3 − 6x2 + 8x dy = 3x2 − 12x + 8 dx dy If x = 0, = 8. dx dy If x = 2, = 12 − 24 + 8 = −4. dx dy If x = 4, = 48 − 48 + 8 = 8. dx



dy dy = 1; = 3x 2 − 4x + 1; at A x = 0, y = 3 and  dx dx the equation of the tangent is y – 3 = x or y = x + 3. b If x = 2, then x3 – 2x2 + x + 3 = 5 and so (2, 5) is on the curve. Also if x = 2, then x + 3 = 5 so the point (2, 5) is also on the tangent.

dy 10 a  y = 8x2 – x4 ⇒ = 16x − 4x 3 ; if x = −3, then dx dy = −48 + 108 = 60 dx b If 16x – 4x3 = 0, then 4x – x3 = 0; x(4 – x2) = 0; x(2 + x)(2 – x) = 0

So x = 0, −2 or 2; The three points are (0, 0), (−2, 16) and (2, 16)

b If f ′(x) = 0, then 4x3 – 4cx = 0; x(x2 – c) = 0; x = 0 or x2 = c



If c  0 the only solution is x = 0 and there is one point where the gradient is 0. If c > 0, there are three solutions, 0,

c and

− c and hence there are three points. 12 y = x(x2 – 6x + 9) – 8 = x3 – 6x2 + 9x – 8 dy Hence  = 3x 2 − 12x + 9 dx If the gradient is −3, then 3x2 – 12x + 9 = – 3; 3x2 – 12x + 12 = 0 x2 – 4x + 4 = 0; (x – 2)2 = 0; x = 2 If x = 2, then y = 2(2 – 3)2 – 8 = – 6 so P is (2, −6) The equation of the tangent is y + 6 = – 3(x − 2) or y = – 3x and this passes through the origin.

b If f' (x) = 0, then 4x3 – 4a2 x = 0; x3 – a2 x = 0; x(x2 – a2) = 0; x(x + a)(x – a) = 0; x = 0, a or –a; then y = a4, 0 or 0

The points are (0, a4), (a, 0) and (−a, 0)

Exercise 6.3A dy 2 = −2x−3 = − 3 dx x dy 6 b y = 2x−3 and so 2 × (−3)x−4 = − 4 dx x

1 a y = x−2 and so

1

c y = x 3 ; d

dy 1 − 32 = x dx 3

3 dy 5 5 −1 = 4 × x 2 = 10x 2 2 dx 1

2 a s = 10t 2 so

1 −1 5 ds = 10 × t 2 = 2 dt t

b s = 50t–1 + 10 so

11 a f(x) = x4 – 2cx2 + c2 so f ′(x) = 4x3 – 4cx



The second of these is on the straight line and so the line is the tangent at that point.

14 a (x + a)(x – a) = x2 – a2 so f(x) = (x2 – a2)2 (x2 – a2)2 = x4 – 2a2 x2 + a4 hence f' (x) = 4x3 – 4a2 x

9 a



dy = 3x 2 − 4x − 5 . The gradient of the straight line dx is 10. dy If = 10, then 3x2 – 4x – 5 = 10; dx 3x2 – 4x – 15 = 0; (3x + 5)(x – 3) = 0 5 So x = −  or 3 and the points on the curve are 3 (−1.67, 4.15) and (3, 0)

50 ds = 50 × −1 × t −2 = − 2 dt t

c s = 10t2 – 10t–2 so

20 ds = 20t + 20t −3 = 20t + 3 dt t

3 a f(x) = 24x−1

f ′(x) = −24x−2 = −

24 x2

b  i f ′(6) = − 242 = − 2 3 6 24   ii f ′(4) = − 2 = − 3 2 4 24 iii f ′(−2) = − = −6 (−2)2 iv f ′(24) = − 242 = − 1 24 24 c If x ≠ 0, then x² is positive and the gradient 24 − 2 is negative. x

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6 Differentiation

5 5 =− 4 22 1 5 = −0.05 ii f ′(10) = − 2 = − 20 10

1

b   i f ′(2) = −

4 a y = x 2

dy 1 − 12 1 = x = dx 2 2 x



  i If x = 4,

dy 1 1 = = . dx 2 4 4

c f ′(x) = −

  ii If x = 9,

dy 1 1 = = . dx 2 9 6



iii If x = 100, b

1 2 x

=

1 =1 2 x







1 x = and the coordinates are 1 , 1 . 4 4 2





1 2

1 x + 2x−1 2 dy 1 1 2 = − 2x −2 = − 2 2 x dx 2 dy 1 2 1 1 = − = − = 0. dx 2 22 2 2



dy 1 2 1 1 = − = − = 0. If x = −2, dx 2 (−2)2 2 2

dy 1 2 1 1 3 = − = − = . dx 2 4 2 2 8 8



If x = 4,

c

dy 1 2 = − dx 2 x 2 If x is large, 22 is a small positive number x 1 and the gradient is close to . 2 1 The larger x is, the closer the gradient is to . 2 5 = 2 + 5x−1 x 5 f ′(x) = −5x−2 = − 2 x



1 3 x 8

dy 1 3 = × 3x 2 = x 2 ; Q is 8 dx 8

the point (10, 125).

The gradient is of

2 . 75

3 75 × 100 =  = reciprocal 8 2

8 a f(x) = x4 + 6x2 + 9 so f'(x) = 4x3 + 12x b h(x) = x3 (x4 + 6x2 + 9) = x7 + 6x5 + 9x3

So h'(x) = 7x6 + 30x4 + 27x2 y=

9

dy 1 1 2 1 b If x = 0.5, = − = − 8 = −7 . 2 dx 2 0.52 2

6 a f(x) = 2 +

−2 dy 2 2 1 2 = 125 3 = × 2 = 3 5 75 dx 3

c For the reflection

( )

If x = 2,



dy 2 − 32 = x ; if x = 125 then dx 3

b The equation is x = 3 8y ; 8y = x3; y =





Points are (1, 7) and (−1, −3).

7 a y = 2x 3  ⇒

c





1

1 2

x = 1 and the coordinates are (1, 1).

5 a y =

x2 = 1

x = 1 or −1

dy 1 1 . = = dx 2 100 20



x =

5 = −5 x2

x 2 + 2ax + a 2  = x + 2a + a2 x–1; hence x

dy a2 = 1 − a 2x −2 = 1 − 2 . dx x

a2  = 0, then x2 = a2 and x = ±a x2 (2a)2 = 4a. If x = −a, then y = 0. The If x = a, then y = a points are (a, 4a) and (–a, 0). If 1 −

( 44 ) = 5 = 25 so (4, 25) is on the

f (4) = 4 + 10 a 

2

2

curve. b f(x) = x2 + 8 + 16x–2 so

32 x3 32 1 15 c f ′ ( 4 ) = 8 − =8− = 64 2 2 f'(x) = 2x – 32x–3 = 2x −



The equation of the tangent is 15 y  − 25 = (x − 4) ; 2y – 50 = 15x – 60; 2 2y = 15x – 10.

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6 

WORKED SOLUTIONS

Exercise 6.4A



dy 1 a = 2(4x + 2) × 4 = 8(4x + 2) or 16(2x + 1) dx b

dy = 3(4x + 2)2 × 4 = 12(4x + 2)2 dx

c

dy = 5(4x + 2)4 × 4 = 20(4x + 2)4 dx

1

6 a y = (2x + 5)2 so

dy = 10(1 + 3x2)9 × 6x = 60x(1 + 3x2)9 dx

c

dy = 10(6x − 3x2)9 × (6 − 6x) = 60(6x − 3x2)9(1 − x) dx 1 ( x − 3) 2

(

)

−1 2

(

1 2

)

1 2 c f(x) = x 2 − 3 so f'(x) = 2 x − 3 1 − 1 2 x . x − 3 2 × 2x = 2 x2 − 3

(

=



)

1 . 2 x−3

−1 2

× 2x =



dy 20 = 20 × −1 × (2 + x)−2 × 1 = − dx ( 2 + x )2



If x = 2, the gradient

dy 20 =− = −1.25 16 dx

b Where the gradient is –0.2, − x)2

20 = −0.2. 2 + ( x )2



Therefore (2 +



Take the positive root: 2 + x = 10.



Therefore x = 8 and y = 2. The point is (8, 2).



b y = 600(x2 + 50)–1

x x2 − 3

= 100.

f ′ ( x ) = 20 × ( −1)( x − 2)

−2

×1 = −



=−

1200x (x 2 + 50)2

g′ ( x ) = 20 × ( −1)( x + 3) 20 = −5 22

×1 = −

20 (x + 3)2

 −5 c A translation of y = f(x) by   is 0 20 20 = = g(x). (x + 5) − 2 x − 3

So y − 4 = −

(

f′(x ) =

8 (x − 10), 8x + 15y = 140 15

)

1

9

(

1 100 − x 2 2

)

−1 2

× (−2x)  = 

−x 100 − x 2



So the gradient of the tangent = f ′ ( −6 ) =



The gradient of OP =

−2

dy 12000 12000 24 8 =− =− =− =− 22500 45 15 dx 150 2

b f ( x ) = 100 − x 2 2 ;

20 . (x − 2)2

b g(x) = 20(x + 3)–1 so



dy = –600(x2 + 50)–2 × 2x dx

8 a  f ( −6 ) = 100 − (−6)2 = 64 = 8 so (−6, 8) is on the curve.

20 = −5 22

g′ ( −1) = −





5 a f(x) = 20(x – 2)–1 so

f ′ (4) = −

600 600 = =4 10 2 + 50 150 so (10, 4) is on the curve.

c At x = 10,

4 a y = 20(2 + x)−1



dy 1 = and the equation of the dx 3 1 tangent is y − 3 = (x − 2). 3 1 On the x-axis y = 0 and −3 = (x − 2); 3 x – 2 = – 9; x = – 7. The point is (−7, 0).

7 a When x = 10, y =

1 2x − 3 − 12 × 2 = 1 . ) 2( 2x − 3

1

b f(x) = ( 2x − 3) 2 so f'(x) =

1 1 = 2x + 5 y

c At (2, 3)

b

3 a f(x) = ( x − 3) so f'(x) =

−1 dy 1 = × (2x + 5) 2 × 2 = dx 2

b If x = 2, then y = 4 + 5 = 3.

dy 2 a = 10(8 − x)9 × −1 = −10(8 − x)9 dx

1 2

The gradient at x = 4 on the first is the gradient at x = 4 – 5 = −1 on the second.

−

6 3 = . 8 4

8 4 =− −6 3

4 3 × = −1 so the lines are perpendicular. 3 4

dy dy = 3(2x − 3)2 × 2 = 6(2x − 3)2 ; if = 24, then dx dx 6(2x – 3)2 = 24; (2x – 3)2 = 4; 2x – 3 = 2 or −2; x = 2.5 or 0.5; then y = 8 or – 8; points (2.5, 8) and (0.5, −8)

10 a 1 −

1 x +1 1 x +1−1 x = − = = x +1 x +1 x +1 x +1 x +1

b f'(x) is the derivative of 1 − ( x + 1)

−1

= ( x + 1)

−2

=

1 . (x + 1)2

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6 DIFFERENTIATION

x 1 = + 1 and the derivative x −1 x −1 −2 1 is −1 × ( x − 1) = − . (x − 1)2 −2x 11 a f(x) = (x2 + 4)–1; f'(x) = –1 × (x2 + 4)–2 × 2x = x2 + 4 c g(x ) =

(

b g(x ) =

2

dy = 6 − 4x dx

a

At a stationary point,

)

6 − 4x = 0

2

x =1

x2 x2 + 4 − 4 4 = =1− 2 = 1 − 4f(x) x +4 x2 + 4 x +4 2

So g′ ( x ) = −4f ′ ( x ) = 12 a If x = 2, then y =

(x

8x 2

+4

)

2

.

(in fact it is always negative) so the point is a maximum point. c y 9.5

dy 160 = − 3 = −2.5. If x = 2, then dx 4

5

The equation of the tangent is y – 5 = –2.5(x – 2) or y + 2.5x = 10. 10 = 4; the area 2.5

0

3

a

1 × 10 × 4 = 20. 2

10

dy dx

= 6 – 2x

dy = 2x − 12 dx

8

y = x (6 – x)

6 4

At a stationary point,

2

dy = 0. dx

–4

2x − 12 = 0

b

x

1.5

a and b

Exercise 6.5A 1

2

d 2y = −4 which is negative when x = 1.5 dx 2

b

b y = 80(x + 2)–2; then dy −3 160 = 80 × ( −2) × ( x + 2) = − . dx (x + 2)3

of the triangle is

( ) = 9.5.

1 1 1 and y = 5 + 6 × 1 − 2 × 1 2 2 2

Coordinates are (1.5, 9.5).

80 = 5. 42

If x = 0, y = 10; if y = 0, x =

dy = 0. dx

–3

–2

0 –1 –2

x = 6 and y = 62 − 12 × 6 − 20 = −56

–4

Coordinates are (6, −56).

–6

1

2

3

4

5

6

7

8

x

–8

d 2y = 2 which is positive when x = 6 dx 2

y = 6x − x2 dy = 6 − 2x dx

(in fact it is always positive) so the point is a minimum point.

dy crosses the x-axis dx gives the x-coordinate of a turning point.

c

c Where the graph of y d 0

6

x

d 2y = −2, a constant value dx 2 y

–20 0

–56

x

–2

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6

WORKED SOLUTIONS

4

a f ′(x) = 6x2 − 18x + 12 At a stationary point: f'(x) = 0

6

6x2 − 18x + 12 = 0

4x3 − 4x = 0

Divide by 6 and factorise: x2 – 3x + 2 = 0

4x(x − 1)(x + 1) = 0

(x − 1)(x − 2) = 0;

x = 0, 1 or −1

x = 1 or 2.

The stationary points are (0, 0), (1, −1) and (−1, −1).

f(1) = 2 – 9 + 12 + 8 = 13 and f(2) = 16 – 36 + 24 + 8 = 12.

d 2y = 12x2 − 4 dx 2

The stationary points are (1, 13) and (2, 12).

d 2y = −4 and is negative so (0, 0) is a dx 2 maximum point.

If x = 0,

b f ″(x) = 12x − 18 f ″(1) = 12 – 18 = −6

d 2y = 8 and is positive so (1, −1) is a dx 2 minimum point.

(1, 13) is a maximum point.

If x = 1,

f ″(2) = 24 – 18 = 6 (2, 12) is a minimum point. c

13 12 8

5

a

2 If x = −1, d y2 = 8 and is positive so (−1, −1) is a dx minimum point.

y

0

7

1

2

dy = 4x3 − 4x dx At a stationary point:

b f ′(x) = 3(x2 + 4)2 × 2x = 6x(x2 + 4)2. This is negative if x is negative and positive if x is positive. The function is neither increasing nor decreasing. 300 c f(x) = 100x−3 so f'(x) = 100 × −3x−4 = − 4 . Now x 300 x ≠ 0 so x4 is always positive and so − 4 is x always negative. The function is decreasing.

x

dy = 3x2 − 12x − 180 dx At a stationary point,

dy = 0. dx

3x2 − 12x − 180 = 0 Divide by 3 and factorise. x2 − 4x − 60 = 0

a f ′(x) = 3x2 + 6x + 3 = 3(x2 + 2x + 1) = 3(x + 1)2 which, because of the square, is  0 for all values of x. The function is increasing.

8

y = 10x−1 +

1 x 4

dy 1 = −10x−2 + 4 dx

(x − 10)(x + 6) = 0 x = 10 or −6 (10, −1400)

At a stationary point: 1 −10x−2 + = 0 4

x = −6, y = (−6)3 − 6 × (−6)2 − 180 × (−6) = 648

Rearrange.

The stationary points are (10, −1400) and (− 6, 648).

10x−2 =

If x = 10, y = 103 − 6 × 102 − 180 × 10 = −1400.

d 2y b = 6x − 12 dx 2 d 2y = 48 > 0, so (10, −1400) is a If x = 10, dx 2 minimum point. d 2y = −48 < 0, so (− 6, 648) is a If x = − 6, dx 2 maximum point.

1 4

40 = x2 x = 40 = 2 10 9

dv =6−t dt At a maximum point:

dv = 0. dt

6−t=0 t=6

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6 DIFFERENTIATION

d 2v = −1 which is negative for any value of t so dt 2 this is a maximum point.

Exercise 6.6A 1

a This parabola crosses the x-axis at 1 and −3. y

If t = 6, v = 10 + 6 × 6 − 0.5 × 62 = 10 + 36 − 18 = 28. The maximum speed is 28 m s−1. 10 a

dy = 23 − 10t dt

0

–3

When the height is a maximum: dy =0 dt

–4 4

t = 2.3

b y = x2 + 2x − 3

d 2y = −10 < 0 so this is a maximum point. dt 2

dy = 2x + 2 dx

When x = 2.3, y = 23 × 2.3 − 5 × 2.32

c If x = 3,

= 26.45. The maximum height is 26.45 m.

dy = 2 × 3 + 2 = 8. dx

The equation of the tangent is y − 12 = 8(x − 3)

b Air resistance will reduce the maximum height of the ball.

y − 12 = 8x − 24 y = 8x − 12

dy 11 a = 3x 2 + 12x − 15 . At a stationary point dx

d The gradient of the normal is −

dy = 0 so 3x2 + 12x – 15 = 0; dx

1 8

The equation of the normal is y − 12 = − 1 (x − 3). 8 8y − 96 = −x + 3

x2 + 4x – 5 = 0; (x + 5)(x – 1) = 0; x = −5 or 1. If x = −5, then y = – 125 + 150 + 75 – 90 = 10 so (−5, 10) is a stationary point.

x + 8y = 99 dy = 2 × −3 + 2 = −4, the tangent is dx y − 0 = −4(x + 3)

2

d y d y = 6x + 12, and if x = −5 then = −18 dx 2 dx 2 and so (−5, 10) is a maximum point. b If y = x3 + ax2 + bx + c, then

x

–3

23 − 10t = 0

2

1

e If x = −3,

y + 4x + 12 = 0

dy = 3x 2 + 2ax + b . dx

1 The gradient of the normal is and the 4 equation is y = 1 (x + 3) or 4y = x + 3. 4

There are two stationary points if the quadratic equation has two distinct roots. That is when the determinant ‘b2 – 4ac’ > 0; (2a)2 – 4 × 3b > 0; 4a2 – 12b > 0; 4a2 > 12b; a2 > 3b. 12 a f′(x) = 3x2 + 2ax, and if f′(x) = 0 then 3x2 + 2ax = 0; x(3x + 2a) = 0; 2a x = 0 or − . There is a stationary point if x = 0. 3 b f ″ (x) = 6x + 2a; if x = 0, then f ″ (x) = 2a; if a > 0 then f ″ (x) > 0 and the point is a minimum point. c There is another stationary point where 2a 24 x=− =− = −8. 3 3 Then f(–8) = (–8)3 + 12 × (–8)2 + 6 = 262; the point is (−8, 262).

2

a If x = −1, y = (−1)3 − 4 × (−1)2 = −5 so (−1, −5) is on the curve. dy = 3x2 − 8x dx dy c If x = −1, = 3 × (−1)2 − 8 × (−1) = 3 + 8 = 11. dx The equation of the tangent is y + 5 = 11(x + 1). b

y + 5 = 11x + 11 y = 11x + 6 d The equation of the normal is y + 5 = − 1 (x + 1). 11 11y + 55 = −x − 1 11y +x = −56

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6

WORKED SOLUTIONS

3

y = 12x−2 dy = −24x−3 dx If x = 1, dy = −24. dx

The tangent crosses the x-axis at 3.5 and the 1 area of the triangle is × 14 × 3.5 = 24.5. 2 6

a

The gradient of the normal is 1 . 24

y

The equation of the normal is y − 12 = 1 (x − 1). 24 24y − 288 = x − 1

2

24y = x + 287 4

1

a y = 12x 2

1

−1 dy 1 = (x + 4) 2 . dx 2 dy 1 1 1 = × = . At P, x = 0 and so y = 2 and dx 2 2 4

b y = (x + 4)2 so

dy −1 6 2 = dx 6x = x 6 dy If x = 4, = 3. = dx 4 1 Gradient of normal = − . 3 1 Equation of normal is y − 24 = − (x − 4). 3 3y − 72 = −x + 4

The equation of the tangent at P is 1 y − 2 = (x − 0) or just y − 2 = 1 x. 4 4 At Q, y = 0 and so x = –8 and the coordinates of Q are (–8, 0).

x + 3y = 76 dy 6 3 b If x = 100, = . = dx 100 5 3 Gradient of tangent = . 5 3 Equation of tangent is y − 120 = (x − 100). 5 5y − 600 = 3x − 300

7

a

a

dy 2 − 13 = x dx 3 −1 dy 2 2 1 1 If x = 8, = ×8 3 = × = . 3 3 2 3 dx The equation of the tangent is y − 4 = 3y − 12 = x − 8

1 (x − 8). 3

3y = x + 4

5y = 3x + 300 5

x

0

–4

dy = −2x dx dy If x = 2, = −4 and the equation of the dx tangent is y − 6 = −4(x − 2).

b You need to find the coordinates of the point. If

dy −1 = 1, 2x 3 = 1. dx 6 3 6

4x

y − 6 = −4x + 8

−1 3

=1 1

4 = x3

y + 4x = 14

x = 43 = 64

b Draw a diagram. Only the tangent is required.

2

y = 64 3 = 42 = 16

y

Equation of tangent is y − 16 =

1 (x − 64). 6

6y − 96 = x − 64 6y = x + 32

10

8

a y = x(10 − x) so it crosses the x-axis at 0 and 10. y

–10

10

x 0

10

x

10

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6 DIFFERENTIATION

b

Where it meets the x-axis y = 0 100 100 = − 2 × ( x − a ) so a = x – a and x = 2a and − a a 1 200 × 2a = 200. The area of the triangle is × 2 a

dy = 10 − 2x dx dy If x = 3, = 10 − 2 × 3 = 4. dx Equation of tangent is y − 21 = 4(x − 3). y − 21 = 4x − 12

11 a

y = 4x + 9 1 Equation of normal is y − 21 = − (x − 3). 4 4y − 84 = −x + 3

The equation of the tangent is y – a2 = 2a(x – a); at the point T, x = 0 and y – a2 = – 2a2; y = – a2; so OT = a²

4y + x = 87

A is the point (0, a²) and so OA = a² = OT. 1 b The gradient of the normal is − and the 2a 1 equation is y − a 2 = − (x − a) 2a

Sketch the tangent and the normal (the curve is not necessary). y

1 At N, x = 0 and y − a 2 = − 1 × (−a); y = a 2 + ; 2 2a

(3, 21)

21.75

dy dy = 2x ; at P = 2a and the gradient of the dx dx tangent is 2a.

hence ON = a 2 + 9

12 The normal is y = 0

The tangent meets the y-axis at 9. The normal meets the y-axis at

87 3 = 21 . 4 4

The area of the triangle is

(

)

1 3 1 21 − 9 × 3 = 19 or 19.125. 2 4 8 9

Then y = 9 – 12 = – 3; P is (−3, −3). 13 a If x = 16 then y = 16 + 9 = 25 = 5 so P is on the curve. 1

−1 dy 1 1 = (x + 9) 2 = ; if dx 2 2 x+9 dy 1 = . x = 16, then dx 10

b y = (x + 9)2 so

y = a(1 + x2)−1 Using the chain rule,

dy = a × −1(1 + x2)−2 × 2x dx −2ax . = (1 + x 2)2

The equation of the tangent is y − 5 =

dy −2a −a a and = 2 = . 2 2 dx 2 a a The equation of the tangent is y − = − (x − 1). 2 2 If x = 1 then y =

Where this meets the y-axis, x = 0 and then a a y − = so y = a and this is the point where the 2 2 curve crosses the y-axis. 100 10 f(x) = 100x–1 so f ′ ( x ) = −100x −2 = − 2 ; x 100 f ′ (a ) = − 2 a The equation of the tangent is 100 100 y− = − 2 (x − a) a a Where it meets the y-axis, x = 0 and y−

84

1 3 x − which has a gradient of 2 2

1 so the gradient of the tangent at P is −2 2 dy = 2x + 4; if 2x + 4 = – 2, If y = x2 + 4x, then dx then 2x = – 6 and x = –3.

x

3

1 1 and AN = ON – OA = . 2 2

200 100 100 100 = − 2 × ( −a ) = so y = . a a a a

at A y = 0 and −5 =

1 (x − 16); 10

1 (x − 16); – 50 = x – 16; x = −34 10

and A is (−34, 0). The gradient of the normal is −10; The equation of the normal is y – 5 = –10(x – 16); at B y = 0 and 1 – 5 = –10(x – 16); = x − 16; x = 16.5 and A is 2 (16.5, 0). The length of AB is 34 + 16.6 = 50.5.

Exercise 6.7A 1

When the depth is h cm, the volume V = π × 302 × h = 900πh cm3. dV dV = −400 cm3 s−1 and = 900π dt dh

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6 

WORKED SOLUTIONS



dV dV dh dh = × so −400 = 900π × dt dh dt dt



50 dV dV dx dx dx = × = so 50 = 3x2 × and dt 3x 2 dt dx dt dt



Therefore

dh 400 =− = −0.141. 900π dt The depth is decreasing at a rate of 0.141 cm s−1.



The rate of increase of the surface area is



2 The area is A = πr2 and

dA = 2πr. dr

dr dA dA dr = 0.3 and = × = 2πr × 0.3 = 0.6πr dt dt dr dt



a When r = 5,

dA 200 = = 40 cm2 s−1. 5 dt



When x = 5,

7

dx = 0.2 and the volume is x2h = 1000 and dt

dA = 0.6π × 5 = 9.42 cm2 s−1. dt

dh dh dx = × . dt dx dt

b When r = 10,

dA = 0.6π × 10 = 18.8 cm2 s−1. dt



h=

c When r = 15,

dA = 0.6π × 15 = 28.3 cm2 s−1. dt



Therefore



When the shape is a cube, x = h and x3 = 1000 so dh 400 =− = −0.4. 1000 dt



The height is decreasing at a rate of 0.4 cm s−1

3 If the radius is r then the surface area A = 4πr2. dr dA = 0.5 and = 8πr dt dr dA dA dr = × = 8πr × 0.5 = 4πr dt dr dt dA When r = 2.8, = 4π × 2.8 = 35.2 cm2 s−1. dt 4 The volume V =

4 3 dV dV = 4πr2. πr , = 30 and 3 dr dt

dV dV dr dr = × so 30 = 4πr2 × and dt dr dt dt



dA dA dx 50 200 = × = 12x × 2 = . x dt dx dt 3x

dr = 30 = 7.5 . dt 4πr 2 πr 2 a When r = 4,

dr = 7.5 = 0.0746 cm s−1. dt 2π × 16

b When r = 8,

dr = 7.5 = 0.0187 cm s−1. dt 2π × 64

5 When the depth is h the radius of the circle 8 h = 0.4h. is 20 1 The volume of water V = π × (0.4h)2 × h = 0.1676h3 3 dV = 0.5027h2. and dh dV dV dV dh dh = 25 and = × so 25 = 0.5027h2 × dt dt dh dt dt dh 49.74 = . and dt h2 dh 49.74 When h = 12, = = 0.345 cm s−1. dt 122

1000 2000 dh = 1000x−2 so = −2000x−3 = − 3 . dx x2 x dh = − 2000 × 0.2 = − 400 . dt x3 x3

2

2 − 1 100 du = 150 × w 3 = 3 ; if w = 8, 3 dw w du 100 = 3 = 50 dw 8

8 a  u = 150w 3 so

b

dw = 0.05 dt

c

du du dw 100 dw 5 = × = 3 × = 3 ; when w = 64, dt dw dt dt w w du = dt

3

5 5 = = 1.25 64 4

dV dA dV dV dA 9 a  × = ; = 4πr 2 and = 2πr so dA dr dr dr dr dV × 2πr = 4πr 2 ; dV = 2r dA dA 1000 = 100; the volume is 100 cm³ 5+5 dV = −1000(t + 5)−2 ; when t = 5, c V = 1000(t + 5)–1; dt dV 1000 =− = 10 cm³s-1 dt 10 2

b V =

1

10 a  10 + 0.4t 2;

−1 dp 0.2 = 0.2t 2 = ; when t = 100, dt t

dp 0.2 = = 0.02 10 dt

6

If the side of the cube is x cm then the volume V = x3 and the surface area A = 6x2.

b

dv dv dp 1200 dv 1200 ;v= = × so =− 2 p dt dp dt dp p



dV dA = 50, dV = 3x2 and = 12x dt dx dx



When t = 100 p  = 10 + 0.4 100 = 14; dv dv dp 1200 = × =− × 0.02 = −0.122 to 3d.p. dt dp dt 14 2

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6 DIFFERENTIATION

Exam-style questions 1

5

1

a y = (a + bx 2)2 and

a y = 2x + 8x−1

−1 dy 1 = (a + bx 2) 2 × 2bx = dx 2

dy 8 = 2 − 8x−2 = 2 − 2 dx x

− 10x + 8 = 0.

6

Divide by 2: x2 − 5x + 4 = 0; factorise: (x − 1)(x − 4) = 0.

a y = x4 – px2;

If x = 1, then dy = 2 − 8 = −6; if x = 4, then dx dy = 2 − 0.5 = 1.5. dx 2

x (a +

1 bx 2)2

=

1 dy . × b dx

dy = 4x 3 − 2px dx dy = 4x 3 − 16x dx

b If p = 8, then

So x = 1 or 4

1

(a + bx 2)2

1

b Since y = (a + bx 2)2 , x = y

8 b If y = 10, then 2x + = 10; 2x2 + 8 = 10x; x 2x2

bx

At a stationary point 4x3 − 16x = 0. Divide by 4 and factorise. x(x2 − 4) = 0

The gradient is 0 when x = −1 or 4.

x(x − 2)(x + 2) = 0

y

x = 0, 2 or −2 If x = 0, y = 0. If x = 2, y = 24 − 8 × 22 = −16.

–1

3

0

If x = −2, y = (−24) − 8 × (−22) = −16.

x

4

Stationary points are (0, 0), (2, −16) and (−2, −16). c

a y = 3x2 – 3x + ax – a dy = 6x − 3 + a dx b 6x – 3 + a = a

d 2y = 48 − 16 = 32 > 0 so (2, −16) is dx 2 a minimum point.

If x = 2, then

6x = 3

d 2y = 32 > 0 so (−2, −16) is a dx 2 minimum point. If x = – 2, then

1 x= 2 y=

( 32 + a )(− 12 )

=−

3 a − 4 2

( 4

1 3 a ,− − 2 4 2

a y=

1 x 2(x

d 2y = 0 and this does not dx 2 determine the type of stationary point at (0, 0). If x = 0, then

However the other two stationary points are below (0, 0) so it is a maximum point.

)

+ 4) =

3 x2

+

b At a stationary point Multiply by

1 x2:

3 12 x 2

y

d

1 4x 2

−1 dy 3 12 = x + 2x 2 dx 2

x=−

d 2y = 12x 2 − 16 dx 2

+

−1 2x 2

–2

= 0.

0

2

x

–16

3 3 x + 2 = 0; x = −2; 2 2

4 1 or −1 . 3 3

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6 

WORKED SOLUTIONS

7

dy = 6x2 − 12x − 12 dx





If the gradient is 6, then 6x2 − 12x − 12 = 6.





x2



(x − 3)(x + 1) = 0

− 2x − 3 = 0

x = 3 or −1

If x = 3, y = 2 × 27 − 6 × 9 − 12 × 3 + 4 = −32.



If x = −1, y = − 2 − 6 + 12 + 4 = 8.



The points are (3, −32) and (−1, 8).



dh When t = 10, dt = 4 − 3 = 1. Speed is

1 m s−1

dh = 0. dt



0.4t − 0.03t2 = 0



t(0.4 − 0.03t) = 0



Either t = 0, which is when the drone starts, or 0.4 − 0.03t = 0.



t=

0.4 1 = 13 0.03 3

2 of 20 seconds, showing the ascent 3 is twice as long as the descent. This is

9 a y = x2 + 250x−1 dy = 2x − 250x−2 dx b At a stationary point, 2x − 250x−2 = 0.

2x3 = 250



x3 = 125

11 a f(x) = (5x +

d 2y = 2 + 500x−3 dx 2 d 2y If x = 5, = 2 + 500 × 5−3 = 6 > 0 so this is a dx 2 minimum point. 250 = 75. If x = 5, y = 52 + 5 The minimum is at (5, 75). dy 10 a y = px3 – 4px2 + 5x – 11 so = 3px 2 − 8px + 5 dx

At a stationary point 3px2 – 8px + 5 = 0. This quadratic in x has two distinct solutions if b2 – 4ac > 0. That is (– 8p)2 – 4 × 3p × 5 > 0; 64p2 – 60p > 0; 16p2 – 15p > 0; p(16p – 15) > 0

)

and so using the chain rule 1



If x  0, then 5x + 4 is always positive. So is f ′(x) and this means the function is increasing.

5 5 b When x = 1, f ′(x) = = The gradient of 2×3 6 5 the tangent is . 6 Hence the gradient of the normal is − 6 and 5 the equation is y − 3 = − 6 (x − 1) which can be 5 rearranged as 5y − 15 = −6x + 6 or 5y + 6x = 21. c f ′ ( x ) =

5 5 so f ′ (1) = and the 6 2 5x + 4

tangent at (1, 3) is y − 3 =





5 (x − 1). 6

5 Where it meets the y-axis, x = 0 and y  − 3 = − , 6 y=

x = 5

1 4)2

− 5 1 f ′(x) = × (5x + 4) 2 × 5 = . 2 2 5x + 4

upwards.

b At the highest point,



b If p = 1 the stationary points are given by 3x2 − 8x + 5 = 0. x = 1 or 5 . 3 If x = 1, y = –9 and if x = 5 , y = – 247 or –9.148. 27 3 The coordinates of the stationary points are (1, –9) and 5 , − 247 3 27

(

8 a dh = 0.4t − 0.03t2 dt

The roots of the equation p(16p – 15) = 0 are 15 p = 0 or 16 There are two distinct stationary points if 15 p  16

13 . 6

Where it meets the x-axis, y = 0 and 5 18 13 −3 = (x − 1), x − 1 = − ; x = − . 6 5 5 1 13 13 169 The area of the triangle is × × = 2 6 5 60 or 2.82 to 3 s.f.

12 a The other side is = 240 − 2x.

The area A = x(240 − 2x).

b A = 240x − 2x2

dA = 240 − 4x dx For maximum area, dA = 240 − 4x = 0. dx x = 60 d2A = −4 < 0 so this will be a maximum. dx 2 Maximum area = 60 × 120 = 7200 m².

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6 Differentiation

13 a

b

dy = 2x + 4 dx



dy = 2 × 2 + 4 = 8. dx The equation of the tangent is y − 7 = 8(x − 2); y = 8x − 9.

If x = 2,

dy dy dy dx = 2.4; = × so at A, dt dt dx dt

dx dy dy = ÷ = 2.4 ÷ 8 = 0.3 units s−1. dt dt dx dy c The gradient is so call this z. dx



Then the rate of change of the gradient is



Then

dz dz dx = × . dt dx dt



But

2 dz dz d y = = 2 and so = 2 × 0.3 = 0.6 units s−1. dt dx dx 2

h = 60 − r so V =

1 2 πr h. 3

1 2 πr (60 − r). 3



When the volume is a maximum, dV = 0 . dr 1 π(120r − 3r2) = 0 3 120r − 3r2 = 0



Divide by 3 and factorise.



r(40 − r) = 0



r = 0 or 40



If y  = 0, x = p and this is the intercept on the x-axis. p2 p 1 ×p = . The area of the triangle is × 2 p +1 2(p + 1) 4 16 When the radius is r, the volume V = πr3 and the 3 surface area A = 4πr2. dV dA dV = 4πr2, = 8πr and = 36 dr dr dt dV dV dr dr = × so 36 = 4πr2 × and dr = 9 2 dt dr dt dt dt πr

1 b V =  π (60r2 − r3) 3 dV 1 =  π (120r − 3r2) 3 dr





14 a Volume of a cone, V =

p If x = 0, y = and this is the intercept on the p +1   y-axis.

d 2V = 1 π (120 − 6r) dr 2 3 d 2V = 1 π (120 − 240) < 0 so When r = 40, dr 2 3 this is a maximum point.

c The maximum volume is

1 π 402(60 − 40) 3 

= 32 000π cm 3 . 3 15 y = (x + 1)(x – p) = x2 + x – px – p and hence dy = 2x + 1 − p . dx dy At (p, 0), x = p and = 2p + 1 − p = p + 1. dx 1 The gradient of the normal is − and the p +1 1 equation is y = − (x − p). p +1



dA dA dr 9 72 = × = 8πr × 2 = . r dt dr dt πr 1500 3 4 When V = 2000, πr 3 = 2000 and r = π and 3 r = 7.816. Then dA = 72 = 9.21. dt 7.816 The area is increasing at a rate of 9.21 cm2 s−1.

17 a f(x) = 10x – x2; f(3) = 21; f(3.1) = 21.39; f(3.5) = 22.75; f(4) = 24 24 − 21 The gradient of AD is = 3. 1 22.75 − 21 The gradient of AC is = 3.5. 0.5 21.39 − 21 The gradient of AB is = 3.9. 0.1 b The answers suggest that f'(3) is greater than 3.9 and could be 4. 18 a Where the curve crosses the x-axis, x(x + 2)(x – 5) = 0 so x = 0, −2 or 5; at P, x = −2. dy y = x3 – 3x2 – 10x and so = 3x 2 − 6x − 10; if dx dy x = −2, then = 12 + 12 − 10 = 14. dx

The gradient at P is 14.

b Where the gradient is 14, 3x2 – 6x – 10 = 14; 3x2 – 6x – 24 = 0; x2 – 2x – 8 = 0;

(x – 4)(x + 2) = 0; x = 4 or −2. Q is the point where x = 4 and then y = 4 × 6 × −1 = −24



Q is (4, −24).

19 a The height of the box is x cm, the length is 30 – 2x cm and the width is 20 – 2x cm.

Multiply these three to get v = x(30 – 2x)(20 – 2x).

b v = x(600 – 60x – 40x + 4x2) = 600x –100x2 + 4x3



dv = 600 − 200x + 12x 2; where the volume dx dv has a maximum value, = 0; dx 2 600 – 200x + 12x = 0; divide by 4: 3x2 – 50x + 150 = 0

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WORKED SOLUTIONS





50 ± 2500 − 1800 50 ± 700 = ; 6 6 x = 12.74 or 3.924. The first value is impossible; it would give a negative volume. The volume is a maximum when x = 3.92.

2 If the side of the triangle is x cm the area is 1 x2 sin 60° = 0.433x2. 2    Then the volume is 0.433x2l = 1000 and so

d 2v = −200 + 24x = −105.8 < 0 when x = 3.92 dx 2 which confirms that the volume is a maximum.



2309 . x2 The surface area is A = 2 × 0.433x2 + 3xl = 0.866x2 + 6928 . x



, and if dA = 0, then Then dA = 1.732x − 6928 dx dx x2

x=

a a ; At P, f ′ ( p ) = − a and the ; x f′(x ) = − x 2 p2 a a tangent is y − = − 2 (x − p). p p

20 f ( x ) =





a a = − 2 (x − p); p = x – p; x = 2p. p p a a a a At R, x = 0 and y  − = − 2 × (−p); y − = ; p p p p 2a y= . p 1 2a The area of triangle OQR = × 2p × = 2a. 2 p a A is (p, 0) so OA = p; B is  0, a  so OB = ; the area p  p a of OAPB is p × = a. This proves the result. p At Q, y = 0 and −

6 

 l =



x3 = 6928 = 4000. 1.732



Hence x = 3 4000 = 15.9.



The length is l =

2309 = 9.2 cm. 15.9 × 15.9

Mathematics in life and work 1 a Volume = x2h = 1000

Rearrange: h =

1000 x2

b The total surface area is 2x2 + 4xh = 2x2 + 4x × c If a = 2x2 +

4000 1000 . = 2x2 + x x2

4000 = 2x2 + 4000x−1, then x

da = 4x − 4000x−2 = 4x − 4000 . dx x2

When the surface area has a minimum value, da = 0. dx



=0 4x − 4000 x2 x3 = 1000







x = 10 2

d a = 4 − 8000x−3 dx 2 2 8000 = 12 which is When x = 10, d a2 = 4 + 103 dx positive, so the surface area has a minimum value. In fact, in this case the cuboid is a cube.

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7 Integration

7 Integration Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Prerequisite knowledge 1 4x

2 5 5 2 b 5 x − 2 x + 10x + c

3 a 2x3 − 2x2 + c

−1 2

c 4x2 −

2 The graph crosses the x-axis at 3 and −5.

10 3 x +c 3

1 2 3 4 a ∫ x 2 dx = x 2 + c 3

y

1

b 2 ×

4 5 2 52 x + c = x2 + c 5 5

1

c 3 × 2x 2 + c = 6x 2 + c

16

−1 2

d  ∫ 10x

1

dx = 20x 2 + c or 20 x + c

5 a ∫ (x 2 − 4x) dx = 1 x 3 − 2x2 + c 3 b ∫ (x 3 − 8x 2) dx = c

∫ 10x

−3

dx =

5 10 −2 x + c = −5x−2 + c or − 2 + c −2 x

1

–5

–1

0

3

6 a ∫ 4x 2 dx = 4 ×

x

3 a 12x2 − 6 b y =

1 x + 2x−1 2

7

dy 1 1 2 = − 2x−2 or − 2 2 x dx 2

∫

c

∫ 4x

8 3 2 32 x + c = x2 + c 3 3

4x dx = ∫ 2 x dx = 2 × −1 2

4 3 2 32 x + c = x2 + c 3 3

1

1

dx = 4 × 2x 2 + c = 8x 2 + c or 8 x + c

f(x) = ∫ (2x 3 − 2x)dx = 2 x 4 − x 2 + c = 1 x 4 − x 2 + c 4 2

8 a  You cannot integrate a product by integrating each term separately and multiplying the results.

1

c y = 6x 3

b

1 4 8 3 x − x +c 4 3

b First multiply out the brackets:

−2 dy = 2x 3 dx

∫ (2x

2

− x − 1) dx =

2 3 1 2 x − x −x+c 3 2

9 a ∫ (4 − 2x −3)dx = 4x + x−2 + c

Exercise 7.1A

b

6 1 a x 2 + c = 3x2 + c 2

b 2x2 + 2x + c

7 c x 2 − 5x + c 2

d 3x − 2x2 + c

2 2 a x 5 + c 5

2 −3 2 b x = − x −3 + c −3 3

1

∫ (2x 2 + 8x

−1 2 )d x

−1 2 )d x

= 2×

1 1 2 32 4 3 x + 16x 2 + c = x 2 + 16x 2 + c 3 3

1 1 2 3 4 3 = 2 × x 2 + 16x 2 + c = x 2 + 16x 2 + c 3 3

c

5 −2 5 x + c = − x −2 + c c 5 x 4 + c d −2 2 4

1

∫ (2x 2 + 8x ∫ (4x

−2

−3

+ 2x 2 ) dx= − 4x−1 − 4x

−1 2

+c

10 The derivatives of f(x) and g(x) are the same so the derivative of f(x) − g(x) is 0. This means that f(x) − g(x) is ∫ 0 dx = a constant.

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7 

WORKED SOLUTIONS

∫ ( x − 1)

11 a

2

dx = ∫(x 2 − 2x + 1) =

1 3 x − x2 + x + c 3

(

)

1 1 3 3 2 b 3 (x − 1) + c = 3 x − 3x + 3x − 1 + c = 1 x3 − x2 + x − 1 + c 3 3 12

∫ ( x +1)( x – 3)( x + 5) dx = ∫ ( x =

∫ (x

3

2

)

– 2x – 3 ( x + 5) dx

)

+ 3x 2 –13x –15 dx

= 1 x 4 + x 3 − 13 x 2 − 15x + c 4 2 2 13 a  x − 1 dx = (x + 1)(x − 1) dx ∫ x −1 ∫ x −1 = ∫ ( x + 1) dx = 1 x 2 + x + c 2

b

∫

)

(



c 15 a

(

2

)

2

+ 1 dx =



)

3

(

)

3 1 2 1 x + 1 + c or just 6 6

)

1 x3

8 2  5 −1  dx = ∫  x 3 − x 3  dx = 3 x 3 − 3 x 3 + c 8 2  

dy = 3 × (x + 7)2 × 1 = 3(x + 7)2. dx 1 Hence ∫ (x + 7)2 dx = (x + 7)3 + c . 3 dy = 4 × (x + 7)3 × 1 = 4(x + 7)3. dx 1 Hence ∫ (x + 7)3 dx = (x + 7)4 + c . 4

b If y = (x + 7)4, then

c If y = (x + 7)6, then

arbitrary constant.



1 2 (x + 1)6 + c 12

∫

1 x2 + 1 dx = ∫  1 + 2  dx =  x2 x  1 −1 x +c =x+ −1 =x− 1 +c x

−2 ∫ (1 + x ) dx

(

16 a y  = x 2 + a

)

using the chain rule for

differentiation,

(

dy 1 2 = x + a2 dx 2

=

∫

x dx = x 2 + a2

2

x . 2 x + a2 2

x +a +c





)

1 2 2;

dy = 3(2x − 3)2 × 2 = 6(2x − 3)2. dx 1 Hence ∫ (2x − 3)2 dx = (2x − 3)3 + c . 6 dy = 4(6x + 1)3 × 6 = 24(6x + 1)3. dx 1 (6x + 1)4 + c . Hence ∫ (6x + 1)3 dx = 24

b If y = (6x + 1)4, then

2

∫ 

dy = 6 × (x + 7)5 × 1 = 6(x + 7)5. dx Hence ∫ (x + 7)5 dx = 1 (x + 7)6 + c . 6

2 a If y = (2x − 3)3, then

(

b

x2 − 1

3 1 2 1 x + 1 + c , including the in the 6 6

2  x 2 + 1  1 + 1  dx  2  dx = ∫  2  x  x  2 −1 1 −3 x + x +c = ∫ 1 + 2x −2 + x −4 dx = x + −1 −3 = x − 2 − 1 3 + c x 3x

b

∫

1 a If y = (x + 7)3, then

b ∫ 6x x 2 + 1 dx = x 2 + 1 + c and hence

∫x (x

same as Ari’s answer except for the constant. This means that 1 (x + 2)3 + c is the same solution written 3 in a different way. Both students are correct. 1 3 2 5 18 a ∫ x × x 2dx = ∫ x 2dx = x 2 + c 5

Exercise 7.2A

= ∫(2 − x)dx = 2x − 1 x2 + c 2 14 a f(x) = 3(x2 + 1)2 × 2x = 6x(x2 + 1)2

(

However, (x + 2)3 = x3 + 3x2 × 2 + 3x × 22 + 23 = x3 + 6x2 + 12x + 8 Hence 1 (x + 2)3 = 1 x 3 + 2x 2 + 4x + 8 and this is the 3 3 3

b

4 − x 2 dx = (2 − x)(2 + x) dx ∫ 2+ x 2+ x

2

17 ∫(x + 2)2 dx = ∫(x 2 + 4x + 4) dx = 1 x 3 + 2x 2 + 4x + c 3 so Ari is correct.

)

−1 2

dy = 5(0.5x − 4)4 × 0.5 dx    = 2.5(0.5x − 4)4.

c If y = (0.5x − 4)5, then



Hence

4 1 5 ∫ (0.5x − 4 ) dx = 2.5 (0.5x − 4) + c

= 0.4(0.5x − 4)5 + c.



dy = −1 × (10x + 1)−2 × 10 dx −10 = (10x + 1)2 .

3 a  y = (10x + 1)−1 thus

× 2x b 4 a

3

3

∫ (10x + 1)2 dx = −10 (10x + 1) ∫

−1

+c =−

0.3 +c 10x + 1

1

x + 1 dx = ∫ (x + 1)2 dx

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7 Integration



3

d

If y = (x + 1)2 , then dy 3 = × (x dx 2

∫

1 + 1)2

×1 =

x + 1 dx =



Hence

b

∫



If y = (2x + 1)2 , then

1 + 1)2

3 (x 2

3

3 1 2x + 1 dx = (2x + 1)2 + c . 3

∫ 3 4x − 2 dx = ∫ 3(4x − 2)2 dx



If y = (4x − 2)2 , then

6

}

− 2 ( x + 2) dx 5

5

1 x x +1−1 x +1 = = − (x + 1)3 (x + 1)3 (x + 1)3 7 a (x + 1)3

1 1 dy 3 = × (2x + 1)2 × 2 = 3(2x + 1)2 . dx 2

c

∫ {( x + 2)

7 6 1 1 = 7 ( x + 2) − 2 × 6 ( x + 2) + c = 1 ( x + 2)7 − 1 ( x + 2)6 + c 7 3

1

∫

dx =

6

2x + 1 dx = ∫ (2x + 1)2 dx

Hence  

5

= ∫ ( x + 2) dx − 2∫ ( x + 2) dx

3 2 (x + 1)2 + c . 3



∫ x ( x + 2)

= b

1

3

1 1 − (x + 1)2 (x + 1)3 x

1

1

∫ (x + 1)3 dx = ∫ (x + 1)2 dx − ∫ (x + 1)3 dx −2 −3 = ∫(x + 1) dx − ∫(x + 1) dx

1 1 dy 3 = × (4x − 2)2 × 4 = 6(4x − 2)2 . dx 2

= 1 (x + 1)−1 − 1 ( x + 1)−2 + c −1 −2 3 3 1 1 1 1 Hence   ∫ 3 4x − 2 dx = 3 × (4x − 2)2 + c = (4x − 2)2 + c = − + +c 2 6 x + 1 2 ( x + 1)2 3 3 1 1 1 ∫ 3 4x − 2 d x = 3 × 6 (4x − 2)2 + c = 2 (4x − 2)2 + c . 8 a ∫ 10x + 5 dx = ∫(10x + 5)2 dx 5 a

1

∫ 3 0.6x + 5 dx = ∫ (0.6x + 5)3 dx

If y = (0.6x +

4 5)3 ,

3 3 = 1 × 2 (10x + 5)2 + c = 1 (10x + 5)2 + c 15 10 3

then

1 1 dy 4 = × (0.6x + 5)3 × 0.6 = 0.8(0.6x + 5)3 . dx 3



Hence 1

4

1

b 4

∫ 3 0.6x + 5 dx = 0.8 (0.6x + 5)3 + c = 1.25(0.6x + 5)3 + c 4

4

∫ 3 0.6x + 5 d x = 0.8 (0.6x + 5)3 + c = 1.25(0.6x + 5)3 + c . b

1

∫

−1 3

∫ 3 0.6x + 5 dx = ∫ (0.6x + 5)

−1 1 dx = ∫(10x + 5) 2 dx 10x + 5 1 1 × 2(10x + 5)2 + c 10 1 = 10x + 5 + c 5

=

dx

1 10x + 5 − 1 ( ) 2 x 10 = 1 2 + + 10 x 5 10 x 5 − +1 = 3 3 1 1 1 = 10x + 5 − × 2 2 10 10 x+5 3 If y = (0.6x + 5) , then Hence −1 −1 dy 2 = × (0.6x + 5) 3 × 0.6 = 0.4(0.6x + 5) 3 . x 1 1 1 dx 3 ∫ 10x + 5 dx = 10 ∫ 10x + 5 dx − 2 ∫ 10x + 5 dx 2 2 1 1 3 dx = (0.6x + 5)3 + c = 2.5(0.6x + 5)3 + c 1 Hence ∫ 3 1 1 1 0.4 0.6x + 5 = 10x + 5 + c × (10x + 5)2 − × 2 5 10 15 2 2 1 1 3 ∫ 3 0.6x + 5 d x = 0.4 (0.6x + 5)3 + c = 2.5(0.6x + 5)3 + c . 1 1 = (10x + 5)2 − 10x + 5 + c 5 6 1 150 10 6 a ∫ ( x + 2) dx = ( x + 2) + c 6 9 f ′ ( x ) = ∫(ax + b)3 dx = 1 (ax + b)4 + c b ∫ ( x + 2)6 dx = 1 ( x + 2)7 + c 4a 7 where c is a constant. 1 c (x + 2)6 = (x + 2) (x + 2)5 = x (x + 2)5 + 2 (x + 2)5 f ( x ) = ∫ f'(x)dx = ∫ (ax + b)4 + c 4a Hence x(x + 2)5 ≡ (x + 2)6 – 2(x + 2)5. 1 1 × (ax + b)5 + cx + d    = 4a 5a c

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WORKED SOLUTIONS

=

1 (ax + b)5 + cx + d, where d is a constant. 20a 2

4

∫

There are two arbitrary constants.

∫ ( 3 2x + 5 )

10 a

2

2

dx = ∫(2x + 5)3 dx =

5 1 3 × (2x + 5)3 + c 2 5

∫

b

2x + 5 dx = ∫(2x +

1 5)3

4 3 = 1 × 3 (2x + 5)3 + c = 8 2 4

=

(

5

a If x ≠ 0, then x2 is positive. This means that

dx 3

2x + 5

always positive.

)

4

c=5 The equation is y = − 2 + 5 = 5x − 2 . x x 6

− 2x + c

a y = ∫ (3x 2 − 3) dx = x3 − 3x + c

b When x = 0, y = 3.

2=c

3 = 2 × 02 − 2 × 0 + c

The curve is y = x3 – 3x + 2.

=c

b The turning points are where 3x2 – 3 = 0

The equation is y = 2x2 − 2x + 3. 2

a y= =

∫

1 x2

x2 = 1.

dx

x = 1 or −1 There are 2 stationary points.

2 32 x +c 3

If x = 1, y = 1 – 3 + 2 = 0 so one t.p. is (1, 0). If x = −1, y = − 1 + 3 + 2 = 4 so the other t.p. is (−1, 4).

b If x = 9, y = 25 25 =

3 2 × 92 + c 3

=

2 × 27 + c 3

7

a f (x ) =

4=−

= 18 + c The equation is y =

=

2 32 x + 7. 3

+ 3x + c

a 0=0+c Hence the equation is y = 0.2x2 + 3x. b 5=c y = 0.2x2 + 3x + 5 c 0 = 0.2 × 52 + 3 × 5 + c c = −20 y = 0.2x2 + 3x − 20

−3

dx = −10x −2 + c

10 +c 4

c = 4 + 2.5 = 6.5

y = ∫ (0.4x + 3)dx 0.2x2

20

∫ x 3 dx = ∫ 20x

f(2) = 4

c=7

3

dx = ∫ 2x −2 dx

4=−1+c

a y = ∫ (4x − 2) dx =

2 is x2

= −2x−1 + c

3 4 y +c 8

2x2

2

∫ x2

b y=

+c

Exercise 7.3A 1

10 +c 5

c = −1 10 The equation is y = x − − 1. x

5

3

x 2 + 10 dx = ∫ 1 + 10x −2 dx x2 = x − 10x−1 + c 2=5−

3 = (2x + 5)3 + c 10

7

f(x) = −10x−2 + 6.5 or f ( x ) = 6.5 −

10 x2

b y 8 6 4 2 0 –2

2

4

6

8

10 12

x

–4 –6 –8

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7 Integration

(

)

8 a ∫ 0.03x 2 ( x − 10 )2 dx = ∫ 0.03x 2 x 2 − 20x + 100 dx

dy 12 dy = 4 − 100 ; at a stationary point =0 2 d x dx x 4 3 2 5 4 3 = ∫ (0.03x − 0.6x + 3x )dx = 0.006x − 0.15x + x + c so 4 − 100 = 0; a2 = 25; a = 5. a2 x 4 − 0.6x 3 + 3x 2)dx = 0.006x 5 − 0.15x 4 + x 3 + c = ∫ (0.03  dx The equation of the curve is y = ∫  4 − 100  When x = 0, y = 100, so c = 100.  x2  5 4 3 y = 0.006x – 0.15x + x + 100               = 4x + 100 + c . x b After 10 seconds, x = 10 and 100 (5, 10) is on the curve so 10 = 20 + + c; c = −30. y = 0.006 × 105 – 0.15 × 104 + 103 + 100 5 = 600 – 1500 + 1000 + 100 = 200 The curve is y = 4x + 100 − 30 . x The distance from A is 200 m. 13 f′(x) = mx + c; f′(–2) = – 2m + c = 7 1 2 3 2 f ( x ) = mx 2 + cx + d; f ( 0 ) = d = 18; f(– 2) = 2m – 2c 9 y = ∫ (3x − 12x + 8)dx = x − 6x + 8x + c 2 + 18 = 8; 2m – 2c = – 10 When x = 0, y = 0 so c = 0. Add the two equations: –c = – 3 so c = 3

y = x3 – 6x2 + 8x







Where the curve crosses the x-axis, x3 – 6x2 + 8x = 0.

The equation is y = – x2 + 3x + 18.

x(x2 – 6x + 8) = 0;



x(x − 2)(x − 4) = 0







The curve crosses the axes at (0, 0), (2, 0) and (4, 0).

x = 0, 2 or 4.

−2

1

10 a y = ∫ 4x 3dx = 4 × 3x 3 + c 1



y = 12x 3 + c



When x = 8, y = 30.



30 = 12 × 8 3 + c



30 = 12 × 2 + c



  c = 6

y=

dy dy = −0.5 14 a  = k2 for some k. When x = 4, dx dx x so k = −0.5 and k = −8. 16

8 If x = 2, y = 8 + 4 = 8; if x = −2, y = + 4 = 0. 2 −2

+6

The gradient is −2 at (2, 8) and (−2, 0).

1

b When x = 20, y = 12 × 20 3 + 6 = 38.57…

dy 8 = − 2 ; y = −∫ 8x −2 dx; y = 8 + c x dx x

When x = 4, y = 6 so 6 = 8 + c and c = 4; 4 8 y = + 4. x dy 8 If the gradient is −2, = − 2 = −2; x2 = 4; dx x x = 2 or −2.

1

1 12x 3

Then – 2m + 3 = 7 so m = −2.

b

y

The radius is 38.6 cm to 3 s.f.

8

11 a f′(2) = a × 22 + b × 2 = 4a + 2b = – 0.8  ---------(1)

6

f′(5) = a × 52 + b × 5 = 25a + 5b = 2.5  ----------(2)

(1) × 5: 20a + 10b = – 4



(2) × 2: 50a + 10b = 5

4

Subtract: 30a = 9 so a = 0.3; then 1.2 + 2b = – 0.8 and b = −1. b f′(x) = 0.3x2 – x so f(x) = 0.1x3 – 0.5x2 + c

f(2) = 4 so 0.1 × 8 – 0.5 × 4 + c = 4; c = 5.2



The equation of the curve is y = 0.1x3 – 0.5x2 + 5.2.

2

–6

–4

–2

0

2

4

6

x

–2

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7

WORKED SOLUTIONS

Exercise 7.4A 1

2

a

∫1 3x

b

4 x3   2

2

(

dy 3 2 = x +9 dx 2

2

)

1 2

(

× 2x = 3x x 2 + 9

)

1 2

= 3x x 2 + 9 . 4

3 4   Hence ∫ x x 2 + 9 dx =  1 x 2 + 9 2  =  125  − [ 9] = 32 2  3  3 3 0  0 

dx =  x 3  = 8 − 1 = 7 1

(

= 64 − 8 = 56

)

4

3 1 2  2 2  125  2 x x + 9 d x = x + 9   =  3  − [ 9] = 32 3 . ∫0 3  0

(

4

3

c  x 3  = [ 27 ] −  −1 = 28 −1

6

Area =

2

)

2

∫−2(2 + 0.1x

4

2

)dx =  2x + 0.02x 5  = [ 4.64 ] − [ −4.64 ] = 9.28 −2

a  1 x 5  =  32  −  1  = 31 = 6.2  5 1  5   5  5 2 2 (2 + 0.1x 4)dx =  2x + 0.02x 5  = [ 4.64 ] − [ −4.64 ] = 9.28 ∫ 4 − − 2 2 b  2.5x 4  = [ 640 ] − [ 40 ] = 600

2

2

10 −2 −1 5 c ∫1 10x dx =  −10x 1 =  − 5  − [ −10 ] = −2 + 10 = 8 7   5 10 −2 −1  5    ∫1 10x dx =  −10x 1 =  − 5  − [ −10 ] = −2 + 10 = 8 5

8

4

 2x 3 − 2x 2  = [128 − 32] − [ 54 − 18 ] = 96 − 36 = 60  3

5

∫2 3x

c

2

2

=

5

+ 6x dx =  x 3 + 3x 2  2

4

∫1

a

∫1 20x

b

4

4

1  =  40x 2  = [ 80 ] − [ 40 ] = 40  1 4  3  = 6 × 4  − [ 6 ] = 32.10 1  

4

4 1   4 8x 3 dx = 6x 3  = 6 × 4 3  − [ 6 ] = 32.10   1 

(

x2 + 9 = x2 + 9

a Try y =

5

Then

4

∫0

(

dy 1 2 = x +9 dx 2

Hence

4

∫0

)

−

1 2

)

9

3

∫−3(9 − x

3

∫−3(9 − x

a

−0=

3

3

3

1 )dx = 9x − x 3  = [ 27 − 9] − [ −27 + 9] = 36 3  −3  6

∫0 (6x − x

b 6

∫0 (6x − x

6

1 )dx = 3x 2 − x 3  = [108 − 72] − [ 0 ] = 36  3 0 c y 10

x 2

x +9

y = 9 – x2

.

4

dx =  x 2 + 9  =  25  −  9  = 2. 0  x +9 3 1 dy 3 2 b Try y = x 2 + 9 2 . Then = x + 9 2 × 2x = 3x x 2 + 9 dx 2

(

)

y = x(6 – x)

4

dx =  x 2 + 9  =  25  −  9  = 2 2 0  x +9

)

8 6

4

x

x

6

1 )dx = 3x 2 − x 3  = [108 − 72] − [ 0 ] = 36 3 0 

2

2

1 2.

× 2x =

3

1 )dx = 9x − x 3  = [ 27 − 9] − [ −27 + 9] = 36  3  −3

2

2

2 –4

2

(

3

0

2 32 a .

This shows that area of OAP is 2 of the area of OAPB. 3

4  43  6x

1

2 32 a

= 180

∫1 8x 3dx = 

c

∫1

1 2 dx



1

4

−

3 a

The area of OAPB = OA × OB = a × a = a × a 2 = a 2.

1 3  2x 2 dx =  4 x 2  =  4 × 8  −  4  = 28 or 9 1   3  3 3 3  1  3

4

2

1

a

∫0 x 2 dx =  3 x 2 

= [125 + 75] − [ 8 + 12] = [ 200 ] − [ 20 ]

4

Suppose the coordinates of P are (a, a ). Then the area of OPA =

2

b  x + 2x + 3x  = [ 8 + 8 + 6 ] − [ 0 ] = 22 0 3

4

1  16 3 2 ∫0 (8 + 8x 2 − 6x) dx = 8x + 3 x 2 − 3x  0 4

= 32 + 42 2 − 48  = 26 2 m 2 3 3  

4

a  2x 3 − 2x 2  = [128 − 32] − [ 54 − 18 ] = 96 − 36 = 60 3

3

4

Area =

(

)

1 2

–3

–2

0 –1 –2

1

2

3

4

5

6

7

8

x

One graph is a translation of the other and =the 3xareas x 2 + are 9 the same.

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7 Integration

1 10 a If y = 2x − 2 , then y2 = 2x – 2 and x = y 2 + 1 . 2 4 4 4 1 2 The area is ∫ x d y = ∫ y + 1 dy =  1 y3 + y  0 2 0  6 0 64 2 = + 4 = 14 . 6 3 b The area is the area of OAPB − 14 2 = 36 − 14 2 3 3 1 = 21 . 3

(

11 a

∫0 x

1

n

b

∫0 x

1

n

)

1

–a

0

The area is

1

n

(1 − x 2)dx =

1

∫0 x

n

2

)

a

− x 2 d x =  a 2x − 1 x 3   3  −a

= 2 a3 + 2 a3 = 4 a3. 3 3 3 3

(2x + 16)4 = 0 so 2x + 16 = 0 and x = −8;

1

the coordinates are (−8, 0).

2 = 1 − 1 = n + 3 − n − 1 = n + 1 n + 3 (n + 1)(n + 3) (n + 1)(n + 3) 12 a  v = 6t2 – t3; dv = 12t − 3t 2 ; when the speed is a dt

Where it meets the y-axis, x = 0 and 3

y = 16 4 = 23 = 8 so the coordinates are (0, 8). b The area is

3

0

∫−8(2x + 16)4 dx 0

7 7 2    =  1 × 4 × (2x + 16)4  =  × 16 4  − [ 0 ] 7 2 7    −8 

maximum, dv = 0 . dt 12t – 3t2 = 0; 4t – t2 = 0; t(4 – t) = 0; t = 0 or 4 When t = 0, v = 0 and that will be a minimum. When t = 4, v = 32. b

a

14 a Where the curve meets the x-axis,

− x n + 2 dx

=  1 x n +1 − 1 x n + 3   n + 1 0 n+3



∫−a ( a

t

=  a 3 − 1 a 3  −  −a 3 + 1 a 3  3   3  

1 1 = n + 2− n −1 1 − n + 1 n + 2 (n + 1)(n + 2) = (n + 1)(n + 2)

∫0 x

a

1

(1 − x)dx = ∫ x n − x n +1 dx 1

c

v

1 1 −0= d x =  1 x n +1  = n +1  n + 1 0 n + 1

=  1 x n +1 − 1 x n + 2  n+2  n + 1 0 =

Sketch of the curve:

The maximum speed is 32 ms–1. v

= 2 × 128 = 256 or 36 4 . 7 7 7

Exercise 7.5A 1 a

32



2

1 8 4 − 2x)dx =  x 3 − x 2  =  − 4  − 0  = −  0  3  3 3 4 The area is . 3 2

∫0 (x

2

3

3 2 1 3 8 4 4 2 b Area = ∫2 (x − 2x) dx =  3 x − x  = [ 9 − 9] −  3 − 4  = 0 −  − 3  = 3 2     

0

c The distance is

4

∫0 (6t 6

2

6

t3

∫2 (x

2

)

6

1 − t 3 dt =  2t 3 − t 4  4 0 

= [ 432 − 324 ] − 0 = 108.

3

1 8 4 4 − 2x) dx =  x 3 − x 2  = [ 9 − 9] −  − 4  = 0 −  −  = .  2   3  3  3  3

The distance is 108  m.

13 Where the curve crosses the x-axis, a2 – x2 = 0 so x = ± a.

c Since the curve is symmetrical, this area is 4 also , the same as part b. 3 3

3 1 81 1 2 a  i ∫ x 3 dx =  x 4  =   − [ 4 ] = 16  4  2  4  4 2

  ii

3

1 4 81 1 3 ∫−2 x dx =  4 x −2 =  4  − [ 4 ] = 16 4 3

2

2 3 1 4 iii ∫−2 x dx =  4 x −2 = [ 4 ] − [ 4 ] = 0

96

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7

WORKED SOLUTIONS

b

4

4 1 64  − Area under curve = ∫ (4x − x 2) dx =  2x 2 − x 3  = 32 − 3 1  3  1 

y 40 4

∫1 (4x − x

30 20

4

1 64   1 − 2−  ) dx =  2x 2 − x 3  = 32 − 3 1  3   3  

2

2 2 = 10  − 1  = 9 .  3   3 

10 –4

–3

–2

0 –1 –10

1

2

3

4

x

Area of triangle =

–20

Area between the curve and the straight line 1 1 =9−4 = 4 . 2 2

–30 –40

The integral from −2 to 2 in part iii above is zero because the graph is symmetrical and the areas above and below the x-axis are the same. 3

1 1 ×3×3= 4 . 2 2

5

a The graphs look like this and cross at (0, 0) and (1, 1). y 6

a Where they intersect, x2 = 2 – x.

5

x2 + x – 2 = 0

4

(x + 2)(x − 1) = 0

3

x = −2 or 1

2

If x = –2, then y = 4

1

If x = 1, then y = 1 –3

The points are (–2, 4) and (1, 1).

–2

0 –1 –1

1

2

3

4

5

6

7

x

1

1 1 2 1 2 1 3 –2 b The area between them is ∫ (2 − x) dx − ∫ x dx =  2x − 2 x − 3 x  −2 −2  −–3  2 1 1 1 2 1 1 2 3  ∫−2(2 − x) dx − ∫−2 x dx =  2x − 2 x − 3 x −2 The area between them is

4

a Where the lines cross, 4 – x = x(4 − x). 4 – x = 4x – x2. x2 – 5x + 4 = 0 Factorise: (x − 1)(x − 4) = 0. x = 1 or 4 Points are (1, 3) and (4, 0).

1

n +1 1 1  n  1 1 n x n − x n +1  = − b Area = ∫ (x n − x n) dx =  1 1 n + 1 n + + +1 n n 0  0 1 n +1 1 1  n 1 n 1 n −1 n n +1  ∫0 (x n − x ) dx =  n + 1 x n − n + 1 x  = n + 1 − n + 1 = n + 1 . 0 c When n is large the area is close to, but less than, one.

b

6

y A

3

x2 – 4x + 3 = 0. Factorise: (x − 3)(x – 1) = 0.

2 1 0 –1 –1 –2 2

Where they cross, 4x – x2 = x2 – 4x + 6. Therefore 2x2 – 8x + 6 = 0.

4

–2

1

1 2 3 1 3 2 1 1 2 ∫0 (x 2 − x ) dx =  3 x 2 − 3 x  = 3 − 3 = 3 . 0 1

=  2 − 1 − 1  −  −4 − 2 + 8  = 7 + 10 = 4 1 . 2 3   3  6 3 2 

C 1

The curves cross where x = 1 or 3.

B 2

3

4

5

6

7

x

Area between =

2 2 2 ∫1 (( 4x − x ) − ( x − 4x + 6 )) dx = ∫1 8x − 2x − 6)dx 3

3

2 2 2 ∫1 (( 4x − x ) − ( x − 4x + 6 )) dx = ∫1 8x − 2x − 6)dx 3

–3 3

The area required is the difference between the area under the curve between A and B, and the area of triangle ABC.

3

3

2 =  4x 2 − x 3 − 6x  = [ 36 − 18 − 18 ] 3 1  2 8 8 2 −  4 − − 6  = [ 0 ] −  −  = or 2 .  3 3   3  3

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7 Integration

7 Where the curves cross, (x – 2)2 = (x – 10)(2 – x). Hence



Hence

2x2



Divide by 2: x2 – 8x + 12 = 0



Factorise: (x – 2)(x – 6) = 0.



Hence x = 2 or 6.

8



The required area is

6



+ 12x – 20.

If x = 2 or −2, y = 8; the curves cross at (2, 8) and (−2, 8).

– 16x + 24 = 0.

b

6

2

dx +

6

2

3 1 64 1 dx =  ( x − 2)  =   − [ 0 ] = 21  2  3   3 3

∫2 ( x − 2)

– 4x + 4 = −

x2

9 a  Where they cross 0.5x2 + 6 = 2x2; 1.5x2 = 6; x2 = 4; x = 2 or −2.

x2

∫2 ( x − 2)

y

10

∫6 ( x − 10 )( 2 − x ) dx . 6

4x

0

–4 2

2

2 2 c Area 10 = ∫ (0.5x + 6)d x − ∫ 2x d x −2 −2 1 3 2   ∫6 ( x − 10 )( 2 − x ) dx = ∫6 −x + 12x − 20 dx =  − 3 x + 6x − 20x 6 2 2 =  1 x 3 + 6x  −  2 x 3  10 10  6  −2  3  −2 1 x − 10 )( 2 − x ) dx = ∫ −x 2 + 12x − 20 dx =  − x 3 + 6x 2 − 20x  6  3 6   40 40   −  −  −  16  −  − 16  = 16.  3   3   3   3  2   1000  2 = 42 y = 0, then x3 – 5x2 + 6x = 0; x(x2 – 5x + 6) = 0; =  − 3 + 600 − 200  − [ −72 + 216 − 120 ] = 66 3  − [ 24 ]10 a If 3 x(x – 2)(x – 3) = 0; 2 2 − [ −72 + 216 − 120 ] = 66  − [ 24 ] = 42   3 3 x = 0, 2 or 3 and the coordinates are (0, 0), (2, 0)   and (3, 0). 1 2 The required area = 21 + 42 = 64. 3 3 y b 10

10

2

{

}

8 a Where they cross, x2 + 1 = 2x + 9; x2 – 2x – 8 = 0; (x – 4)(x + 2) = 0. x = 4 or −2. If x = 4, then y = 17; if x = −2, then y = 5. The points are (4, 17) and (−2, 5). b Here is a sketch: y

0

2

3

x

c Area between 0 and 2

∫0 ( x 2

= 0

–2



x

∫−2 ( 2x + 9) dx =  x

2

+ 9x 

−2

= [52] – [–14] = 66



(Alternatively, this is a trapezium and the area is 1 5 + 17 ) × 6 = 66 , the same answer.) 2(



Area under curve =

∫−2 ( x 4

2

∫2 ( x 3

4

)

+ 1 dx

2

)

− 5x 2 + 6x dx =  1 x 4 − 5 x 3 + 3x 2  3  4 0

=  4 − 40 + 12  − [ 0 ] = 8   3 3

Area under straight line = 4



4

3



3

)

3

− 5x 2 + 6x dx =  1 x 4 − 5 x 3 + 3x 2  3  4  2

=  81 − 45 + 27  −  8  = 9 − 8 = − 5 12  4   3  4 3

1 The total area is 8 + 5 = 37 or 3 . 12 3 12 12

4



=  1 x 3 + x  =  76  −  − 14  = 30.  3  −2  3   3  The area between is 66 – 30 = 36.

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7 

WORKED SOLUTIONS

11 a  y = x(x – 4) is a parabola that crosses the x-axis at 0 and 4.

At P, x =



( 72 , 174 ) or (3 12 , 4 14 ) .

The straight lines cross where 2x = 8 – 2x; 4x = 8; x = 2; cross at (2, 4).

7 1 7 5 17 and y = × + = ; P is 2 2 2 2 4

The lines look like this:

c The lines look like this: y

y 4

4

3

0

2

4

2

x

2

1

b

∫0 x(x − 4)dx = ∫0 ( x 4

4

2

)

− 4x dx



Area under curve =



The area below the x-axis is 32 . 3

3

4

x

2 ∫1 ( x − 4x + 6 ) dx 3.5

= [10.792] – [4.333] = 6.4583. Area of trapezium = 1 × ( 3 + 4.25) × 2.5 = 9.0625 2



(or by integrating, area =

3.5

∫1 (0.5x + 2.5) dx

3.5

= 0.25x 2 + 2.5x  = [11.8125] – [2.75] = 9.0625  1

The total area is 8 + 32 = 18 2 . 3 3

12 a dy = 2x − 4 ; at a stationary point 2x – 4 = 0; dx x = 2 and then y = 4 – 8 + 6 = 2. d 2y = 2 > 0 so (2, 2) is a minimum point. dx 2 dy b At (1, 3) = 2 − 4 = −2 ; the gradient of the dx tangent is −2; the gradient of the normal is 1 . 2 1 The equation of the normal is y − 3 = (x − 1) or 2 1 5 y= x+ . 2 2 Where the curve and the normal meet, 5 1 x 2 − 4x + 6 = x + ; 2x2 – 8x + 12 = x + 5; 2 2 ∞ 8 = 7 2 dx = 2x2 – 9x + 7 = 0; (x – 1) (2x – 7) = 0; x∫0= (1xor + 2) 2

2

3.5

The area of the triangle above the x-axis is 1 × 4 × 4 = 8. 2

1

=  1 x 3 − 2x 2 + 6x  1  3

4

=  1 x 3 − 2x 2  =  64 − 32  − [ 0 ] = − 32  3 0  3  3

0

–1



which is the same answer).

Area enclosed = 9.0625 – 6.4583 = 2.60 to 3 s.f.

Exercise 7.6A 1 Area =



25

∫0

10 dx = x

−1 2

25

1   dx = 10 × 2x 2  = [100 ] − [ 0 ] = 100  0

25

∞ 20

∞

−2

∫2

∞

dx = ∫ 20x −3 dx =  −10x −2  2 2 x3 . = 0 − − 2.5 = 2.5 ] [ ] [

3 The area = ∞

∫0

10x

=  20 x  = [100 ] − [ 0 ] = 100 . 0

2 The area =

25

∞

∞

8

∫0 (x + 2)2 dx = ∫0 8 ( x + 2)

+ 2) dx =  −8 ( x + 2) ∫0 8 ( x

−2

∞

−1 dx =  −8 ( x + 2)  = [ 0 ] − 0

−1 ∞

 = [0 ] −  − 8  = 4 . 0  2 

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7 INTEGRATION

4

a

y

b y 10

0 5

x

2

The integral is represented by the unshaded area. 2

c To evaluate  − 502  is impossible because  x 0 0

dx =

11

∫2 10 ( x − 2)

−1 2

−3 2

4

4 20 20 20 20 a11 ∫ 2 dx =  −  =  −  −  −  10 dx = 1110 x − 2 − 12 dx = 10 × 2 x − 2 12  =2x20 x − 211x  2  4   2  ( )  ∫2 ( )  x−2 = – 52 + 10 = 5 2 

8

11

∫2

11

1 11   dx = 10 × 2 ( x − 2) 2  =  20 x − 2  2  2 =  20 9  − ( 0 ) = 60.

The area is =

5

x

2

b Area =

you cannot divide by 0. The area between 0 and 2 is infinite.

∞

∞

∫a

24x

The area is 1 and so

dx =  − 20  = [ 0 ] −  − 20  = 5  x  4 x  4  2

∞

2

20  20   48   48  2 48  20   20   =  − x  = [ 0 ] −  − a  ∫=−2 xa2 dx =  − x  −2 =  − 2  −  − −2   0  a  = – 10 – 10 = – 20

dx =   48  48 . = a a 

∞

∞

∞ 20

∫4

c The student calculated the answer like this:

∞ −1   −48x 2

−3 2

−1    48   dx =  −48x 2  =  − = [0 ] −  − x 0   a 

b

However, the graph looks like this: y

48 = 1. a

Hence a = 48 and a = 48² = 2304. 6

Area =

∞

x

∫0

2 2

2 −1

2

Then

∫0

(

–2

.

dy = −1 × 1 + x 2 dx

Hence ∞

dx .

(1 + x ) 1 = (1 + x ) Try 1+ x

∞

∫0

(

)

−2

× 2x = −

(1 + x ) 2

because  − 20  cannot be evaluated.  x 0

2.

9

a If x = 0, then y = 242 and the point is  0, 242  .  a  a y

∞

1  1 1  1 = [0 ] −  −  = . 2 dx =  − 2 ×  2  2 1 + x 2 0  1 + x2

(

x

7

)

a

∞

∫2 100x

−3

x

2

2x ∞

)

2

The area between x = 0 and x = 2 is infinite

1  1 1  1 = [0 ] −  −  = 2 dx =  − 2 ×  2  2 1 + x 2 0  1 + x2 x

0

∞

∞

100 −2  50 = −  dx =  x  2  x 2  2  −2 = 0 − [ −12.5] = 12.5

x = –a

–a

0

x

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dx =

7

WORKED SOLUTIONS

∞

∞

∫0 24(x + a)

dx =  −24(x + a)−1  0 ∞ 24  24  24   = [0] − − = . = − 1  a  a  x + a 0 = ∫ πy 2dx =

b The area is

−2

−1

Volume =

4

∫−1 π ( x 1

c If a is negative the graph looks like this: y

2

)

2

+ 3 dx =

∫−1 πy dx = ∫−1 π ( x 1

1

2

∫−1 π ( x 1

2

)

2

+ 3 dx =

∫−1 π ( x 1

4

)

+ 6x 2 + 9 dx

)

4

+ 6x 2 + 9 dx

( ) ( ) ( =  π ( 1 x + 2x + 9x )  =  π ( 1 + 2 + 9)  −  π ( − 1 − 2 − 9)  = 11.2 π + 11.2π = 22.4π. 5  5    5   5

)

1

=  π 1 x 5 + 2x 3 + 9x  =  π 1 + 2 + 9  −  π − 1 − 2 − 9  = 11.2 π + 5    5  −1  5 

1

3

−1

5

x=a

If y = 0.4x2, then x2 = 2.5y. Volume =

4

∫2 πx

a

0

2

4

∞

∫4 x d y and

∞

∞

∫4 x d y = ∫4 64y

−2

2

dy =

4

∫2 π × 2.5y d y = 1.25πy

2 4  2

4

d y = x∫ π × 2.5y d y = 1.25πy 2  = [ 20π ] − [ 5π ] = 15π. 2 2 6

In this case the area is infinite because 24 x+a cannot be evaluated when x = – a. 10 a Area =

4

∫2 πx

The curve is a parabola that looks like this: y

8 64 x = 2; y so y

x =

dy

0

x

4

∞

=  − 64  = [0] – [–16] = 16  y  4 b If the area if finite, it is ∞

1  ∞ = 16x 2  = 16 x 0 . 0

∞

∫4

y dx =

∞

∫4 8x

−1 2

dx The curve meets the x-axis at 0 and 4. Volume =

However, this cannot be evaluated because ∞ 4 4 2 cannot be found. The area is infinite.∫ πy 2dx = ∫ π ( x ( x − 4 )) dx = 0

0

Exercise 7.7A 1

Volume = 4

4

4

∫0

4

4

2

2 5

∫2 π × 0.09x

y 2dx =

2

Volume =

( )

2

dx =

5

∫2 π × 0.09x

2

5

dx = 0.03π x 3  2

7

2

1 3 ∫0 π 8 x dx =

2

∫0

2

∫0

πy 2dx = 2

( )

2

1 3 ∫0 π 8 x dx = 2

π x 6dx =  π × x 7   64 7  64  0 128 π  − 0 = 2π. = × 7  [ ] 7  64

4

2

)

2

dx =

∫0 π ( x 4

2

)

2

− 4x dx

2

− 4x dx

(

(

)

)

a Volume =

5

Volume =

∫0 π ( x

(

)

4

∞

∫a πy

2

dx =

∞

∫a πx

−2

∞

dx =  −πx −1  a

∞

dx = 0.03π x 3  = [ 3.75π ] − [ 0.24π ] = 3.51π. 2 3

2

5

∫2 πy

4

2

=  π 1024 − 512 + 1024  − [ 0 ] = 34 2 π or 34.13π. 5 3  15 

π 2 ∫0 πy dx = ∫0 πx dx =  2 x 0 = [8π ] − [0 ] = 8π . 4

4

= ∫ π x 4 − 8x 3 + 16x 2 dx =  π 1 x 5 − 2x 4 + 16 x 3  3  5 0 0

π 2 ∫0 πx dx =  2 x 0 = [8π ] − [0 ] = 8π 4

πy 2dx =

∫0 πy dx = ∫0 π ( x ( x − 4 ))

2

∫0

2

π π π = −  = [0] − −  = .  a  a  x  a

π x 6dx =  π × x 7   64 7  b Volume = 64  0

a

∫0 πy

2

a

π dx =  −  .  x 0

π However, if x → 0, then → ∞ and the x volume is infinite.

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7 Integration

8

b The volume formed by the curved line is

y2 x2 + = 1. 225 100

4.5

∫0

y2 x2 4 =1− and y 2 = 100 − x 2 . 100 225 9

Hence

dx =

15

∫−15

4.5

0

4.5

2 π × 2y d y =  πy 0 = 20.25π.

The volume formed by the straight line is

∫

4.5

∫3

πx 2 d y 4 4  3  2 Volume = ∫ πy dx = ∫ π 100 − x dx =  π 100x − x  y = 0.5x + 3 so 2y = x + 6; x = 2y – 6 and 9 27 where −15 −15   −15 2 = 4y2 – 24y + 36. hence x 15 4.5 4 4  3  2 π 100 − x dx =  π 100x − x  The volume is ∫ π x 2 d y 9 27 3   −15 15

2



π x 2 dy = ∫

(

)



(

15

2

(

)

(

)

)



(

)

= [π(1500 − 500)] − [π(−1500 + 500)]

=



= 1000π + 1000π = 2000π.

= π  4 y 3 − 12y 2 + 36y   3 3

The volume of the ball is 6280 cm³ to 3 s.f.

π 4y 2 − 24y + 36 d y 4.5

= 40.5π – 36π = 4.5π (or use the fact that it is a cone with radius = 3 and height = 1.5 so the volume is 1 πr 2h The volume of the ring is the difference between 3 = 1 π × 32 × 1.5 = 4.5π). the volumes of the shapes formed by rotating each 3 of the lines. The volume of the shape is 20.25π – 4.5π = 3 3 Volume formed by the arc of the circle = ∫ πy 2 dx = ∫ π 25 − 15.75π x 2 dx −3 −3

(

3

∫−3 πy

2

2 ∫−3 π ( 25 − x ) dx 3



=



=  π 25x − 1 x 3  =  π ( 75 − 9)  −  π ( −75 + 9)  3   −3



= 66π + 66π = 132π.



The volume formed by the straight line is

dx =

(

3

∫−3 πy

2

)

dx =

3

3

∫−316π dx

)

12 a If x = 2 then 1 x 3 = 4 and (x – 4)2 = 4 so both 2 curves pass through (2, 4). b The volume is in two parts. For the curve y = 1 x 3 the volume 2 2 2 2 1 2 6 1 = ∫ πy dx = π ∫ 4 x dx = π  x 7  0 0  28 0 = 128 π = 32 π. 28 7



= [16π x ]−3 = [48π] − [−48π] = 96π.



(Alternatively, as this is a cylinder, the volume is πr2 h = π × 42 × 6 = 96π.)



4.5



9 Where the circle and the ring cross, y = 4 and x2 + 42 = 25 hence x2 = 9 and x = ±3.

150

3

The volume of the ring is 132π – 96π = 36π.



The curve y = (x – 4)2 meets the x-axis at 4 so for this part the volume is



∫2 πy

10 a The cylinder has a radius of 5 and height of 6. Volume = πr2h = π × 52 × 6 = 150π b When the curve is rotated the volume is 6

∫2

2

2

πy dx .

2

4

4 1 dx = ∫ π ( x − 4 )4 dx = π  (x − 4)5  2 5 2

= π [ 0 ] − π  − 32  = 32 π .  5  5

The total volume is

32 π + 32 π = 10 34 π or 7 5 35

11.0π (3 s.f.).

(

y = 2 x−2 Volume =

4

6

)

2

= 4(x – 2) = 4x – 8

∫2 π ( 4x − 8 ) dx

Exam-style questions 6

2 = π  2x − 8x  2

= π[24] – π[–8] = 32π. The volume of the shape is the difference between these answers = 150π – 32π = 118π. 11 a Where they cross 0.5x2 = 0.5x + 3; 0.5x2 – 0.5x – 3 = 0; x2 – x – 6 = 0 (x – 3) (x + 2) = 0; x = 3 or −2; the coordinates are (3, 4.5) and (−2, 2)

1 y = ∫ (8x 3 − 12x 2)dx = 2x4 – 4x3 + c When x = 2, y = –3 32 – 32 + c = −3 So y = 2x4 – 4x3 – 3.

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7 

WORKED SOLUTIONS

10

2 Distance = ∫

0

0.1 ∫ (15 − 0.1x dx ) = 15 x − 3 x 10

2

10 3

0

0

x (15 − 0.1x dx ) = 15x − 0.1 3 2

10 3

dV b When t = 20, dt = 0.6 × 20 – 10 = 2.

0



= 150 − 33 1  − [ 0 ] = 116 2 m.  3  3 

⌠ −3 −1 3 a   5x −2 + x 2  dx = −5x −1 − 2x 2 + c  ⌡ b

∞5

∫1

7 Area =

∞



16 = +c x



When x = 4, y = 6.



6=



c=6−4=2



The equation is y = 16 + 2 . x

16 +c 4

∞

∞ −1 24  = 0 − −12] dx =  −24 ( x + 2)  =  −  x + 2 0 [ ] [ 0

y = ∫ (2x + 6)dx





= x 2 + 6x + c At a turning point, dy = 0 dx 2x + 6 = 0







The turning point is on the x-axis so the coordinates are (−3, 0).



Put these into the equation for y.

x = −3



0 = 9 – 18 + c

x

0

1

4

9



c=9

y

−2

−1

0

1



The equation of the curve is y = x2 + 6x + 9.

(

2

–1

(

1

2

3

4

5

6

7

8

9

(

x



 2 32 x

 − 2 d x =  3  



4

 − 2x  =  16 − 8  − [ 0 ]  0  3

= − 8 so the area below the x-axis is 8 . 3 3 9

9

)

1 2

× 8x

= 12x 4x 2 + 5

)

3

b If y = 4x 2 + 5 2 , then

–2

4 1 x2 0 

3



1 0

)

9 a If f ( x ) = 4x 2 + 5 2 , then f ′ ( x ) = 3 4x 2 + 5 2

y

∫

−2

8

5 a Find some points on the curve:

b

∞

∫0 24 ( x + 2)

 5 2  = 0 − −5 − = + x 2] 7 [ ] [ dx =  − x − ∞ ∞ x 1 ∞ x2  −2  = [ 0 ] − [ −12] = 12 .   24 x + 2 ) dx =  −24 ( x + 2) −1  =  − x24 ∫0 ( + 2 0 0 

4 y = ∫ −16x −2 dx = 16x −1 + c



This is positive and implies that the volume is increasing. This cannot be the case. If dV will be negative. there is a leak, dt

(

dy 3 = 4x 2 + 5 dx 2

)

1 2

(

dy 3 = 4x 2 + 5 dx 2

)

1 2

× 8x = 12x 4x 2 + 5

× 8x = 12x 4x 2 + 5 .

(

Hence  ∫ x 4x 2 + 5 dx = 1 4x 2 + 5 12

)

3 2

+ c.

10 Where the curve crosses the x-axis, x(4 − x) = 0. x = 0 or 4

The area =

4

1 3 2 2 ∫0 ( 4kx − kx ) dx =  2kx − 3 kx 0 4

1 4  2 3  4 1 kx 3  = 32k − 21 1 k  10 2 k = 2 −     . ∫4  x 2 − 2 dx =  3 x 2 − 2x  ∫0 4kx − kx 2 dx =  2kx  3 3  3 0  4 2 = [18 − 18 ] −  − 8  = 8 so the area above the 10 k = 32 3  3  3 x-axis is 8 . k=3 3 11 Area beneath the curve between A and B is: Hence the total area is 8 + 8 = 16 or 5 1 . 3 3 3 3 a a 1 1 1 2 ∫ x 2 dx =  x 3  =  a 3  −  − a 3  = a 3 6 a V = ∫ (0.6t − 10) dt = 0.3t 2 − 10t + c   3 −a  −a  3   3  3 



(

)



When t = 0, V = 100 so c = 100.



The area of rectangle ABCD = 2a × a2 = 2a3.



V = 0.3t2 – 10t + 100



The area between AB and the curve is 2 4 2 2a 3 − a 3 = a 3 and this is of the area of the 3 3 3 rectangle.

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7 INTEGRATION

12 a First, find the coordinates of the points where the lines cross.

14 Volume =

– 3x – 4 = 0

Thus y 2 = 4x + 1.

(x − 4)(x + 1) = 0

Volume =

x = −1 or 4 If x = −1, y = 1 + 1 + 4 = 6. The points are (−1, 6) and (4, 16). y B

0

C −1

4

∫−1( x 4

2

dy 2 1 = = ; the gradient of dx 16 2

A

4

x

P

)

4

At P 4x – 8 = 0 so x = 2; at R –2x + 16 = 0 so x = 8. Volume formed by rotating curve = = π ∫ (4x − 8)dx = π  2x 2 − 8x  2 2

6 + 16 × 5 = 55. 2 1 5 The required area is 55 − 34 = 20 . 6 6

= [24π] – [– 8π] = 32π.

Area of trapezium =

1 1 3 x = 2x 2 and either x = 0 or 13 Where they cross, 16 5

x 2 = 32 so x = 4.

3

1 3 x and the x-axis is 16 4

1 dx =  x 4  = 4 − 0 = 4 .  64 0

Area between y = 2 x and the x-axis is 4

4 3  32 2 ∫0 dx =  3 x 2  = 3 − 0 = 10 3 . 0 Area between curves = 10 2 − 4 = 6 2 . 3 3 1 2x 2

x

R

6

64 1 1 1 5 1 =  − 8 + 16  −  − − − 4  = 29 + 4 = 34 .  3 6 6  3   3 2

1

2 ; 4x − 8

− x + 4 dx

1 1 =  x 3 − x 2 + 4x   −1  3 2

∫0 16 x

5

y

b You want the difference between the area of the trapezium ABCD and the area under the curve between A and B.

Area between y =

)

+ x  2

6

D

4

2

the normal is −2 and the equation is y – 4 = – 2(x – 6) or y = –2x + 16. b

Area under curve =

5

if x = 6, then

16

A

∫2 π( 4x + 1) dx =  π ( 2x

= [ 55π ] − [10π ] = 45π . 1 dy −1 1 15 a y = (4x − 8)2; = (4x − 8) 2 × 4 = dx 2

If x = 4, y = 16 − 4 + 4 = 16.

4

2

y = 4x + 1

Where x2 – x + 4 = 2x + 8 x2

5

∫2 πy dx.

6

6

∫2 πy

2

dx

6

Volume formed by rotating the normal is a cone with volume 1 π × 4 2 × 2 = 32 π 3 3 (or)

8

∫6 π(16 − 2x)

2

8

1 dx = π  − (16 − 2x)3   6 6

= [ 0 ] − π  − 64  = 32 π .  6  3 32 π = 42 2 π. The total volume is 32π + 3 3 16 a Where the curves cross, x2 – 10x + 25 = 5 + 4x – x2; 2x2 – 14x + 20 = 0; (x – 2)(x – 5) = 0; x = 2 or 5. When x = 2, y = 9 so A = (2, 9). When x = 5, y = 0 so B = (5, 0). b The volume is

2 ∫2 π (5 + 4x − x ) 5

2

5

(

)

2

dx − ∫ π x 2 − 10x + 25 dx 2

= 129.6π – 48.6π = 81π = 254.

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7 

WORKED SOLUTIONS

17 a Where they cross, x4 + 2 = x2 + 14; x4 – x2 – 12 = 0; (x2 – 4)(x2 + 3) = 0

b The areas are given by the formula in part a with a = 1, 2, 4, 8, etc.

x2 = 4 or −3



x2 = − 3 has no solution; if x2 = 4, then x = 2 or −2; in either case, y = 24 + 2 = 18 and the coordinates are B (2, 18) and A (−2, 18).

b The minimum points are where the curves cross the y-axis. y = x4 + 2 is at (0, 2) and y = x2 + 14 is at (0, 14) c If y = x4 + 2, then



=

18

∫2

1 This is a geometric series with a = 50 and r = . 2 a 50 = = 100 . Then S∞ = 1−r 1− 1 2 2 20 a Where it crosses the x-axis, y = 0; x 2 = 1; x2 = a2; a

y − 2 = x 2 and the volume

generated by y = x4 + 2 is

18

∫2

πx 2 d y



x = ± a; points (a, 0) and (– a, 0).



Where it crosses the y-axis, x = 0;

b Volume =



=

1

∫14 π ( y − 14 ) d y = π  2 y 18

2

18

∫14 πx

2



dy

18

− 14y  14



∫

1 dx = ax + 3

f (x ) =



−1 2 dx

1 2 = (ax + 3)2 + c = 2 ax + 3 + c a a

b f ( 0 ) = 0 →



∫ ( ax + 3)

2 × 3 + c = 0; 2 3 and c=− a a

2 2 3 ax + 3 − a a

f (a ) = 2 2 − 2 →

2 2 3 a2 + 3 − = 2 2 − 2; a a

From which

( 2 − 1) + 3 (3 − 2 2 ) + 2a 3 (

a2 + 3 = a a2 + 3 = a2

)

2 −1 + 3

2a − 2a 2 + 2 6 − 2 3 = 0

a=

a

So volume =

a

∫−a πb

2

x2   1 − a 2  dx

a a 2 2 = πb  a − 3  − πb  −a + 3     

= πb 2 × 2 a + πb 2 × 2 a = 4 πab 2 . 3 3 3

Mathematics in life and work

Assume that the segment is from x = −a to x = a.



The volume is =



=



= π 64x − 0.048x 3 + 0.000 016 2x 5  −a



= 2π 64a − 0.048a 3 + 0.000 016 2a 5  .



If the length is 15.6 cm then a = 7.8 and



a volume of 2996 cm³ is just under 3 litres.

2 ∫−a π (8 − 0.009x ) a

∫−a π (64 − 0.144x a

2

2

dx

)

+ 0.000 081x 4 dx a

2 3−2 6 = 3 2−2 2

2a 100

19 a  ∫

2 2 dx , where y 2 = 1 − x 2 and b a

a

128 104 The volume required = 3 π − 8π = 3 π or 34 2 π. 3

18 a f ( x ) =

2

x3  2 = πb  x − 2  3a  −a 

= [–90π] – [–98π] = 8π.

a

∫−a πy

 x2  y 2 = b2 1 − 2  .  a 

If y = x2 + 14, then y – 14 = x2 and the volume generated by y = x2 + 14 is

y2 = 1; b2

y2 = b2; y = ± b; points (0, b) and (0, – b).

18

1 3   π( y − 2)2 d y = π  2 ( y − 2)2  3  2

=  128 π  − [ 0 ] = 128 π. 3  3 

Total area = 50 + 50 + 50 + 50 + .... 1 2 4 8

2a

dx =  − 100  =  − 100  −  − 100   2a   a   x  a x 2

= − 50 + 100 = 50 a a a

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Summary REVIEW

Summary Review Please note: Full worked solutions are provided as an aid to learning, and represent one approach to answering the question. In some cases, alternative methods are shown for contrast. All sample answers have been written by the authors. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers, which are contained in this publication. Non-exact numerical answers should be given correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Warm-up Questions

 7 3 For ′x′:   26(ax)1 = 448ax.  1

1 a i 27, 38 ii  Add consecutive odd numbers to get the next term. b 1, 5, 9 2 5x + 2y = 16 3x − 4y = 7 ① × 2: 10x + 4y = 32 ② + ③: 13x = 39 x=3 In ①: 15 + 2y = 16 2y = 1

① ②



 7 For ′x2′:   25(ax)2 = 672a 2x 2 .  2

448a = 672a2   0 = 672a2 − 448a    = 3a2 − 2a    = a(3a − 2)

③  

− π 4 i A translation  2  and a stretch in the  0  y-direction, scale factor 3. ii The minimum point of f(x) is

y = 0.5 Area of circle = π × 7.52 = 177 cm2.

1 f′(x) = 3x2 − 7 Integrating: f(x) = x3 − 7x + c Substitute f(3) = 5  ⇒  27 − 21 + c = 5 ⇒ c = −1 ∴ f(x) = x3 − 7x − 1 2 i sin   = k  ⇒ sin2  = k2 2 sin   + cos2  = 1 2 2 k + cos θ = 1 2 2     cos θ = 1 − k









2     cosθ = − 1 − k Negative because  is obtuse.

ii tan θ =

sin θ k =− cosθ 1 − k2

iii sin ( + π) = −k

106

( 32π ,−1). After

the translation, this becomes (π, –1). After the stretch, this becomes (π, –3). So the minimum point of g(x) is (π, –3).

3 r = 7.5 cm

A Level Questions

2 a= 3  

5 tan 2x = 2 2x = 63.4…, 243.4… x = 31.7°, 121.7° 6 i Area OAB =

1 × 82a = 32a. 2



2 Area OCA = π × 4 = 8π 2





Therefore 64a  = 8π





P = 8 × π + 4π + 8 8

α=π          8 ii BA = 8a and OCA = 4π P = 8a + 4π + 8

P = 5π + 8 dy = −8x−3 − 1 dx  y = 4x−2 − x + c Substitute (2, 4) 4 = 1 − 2 + c           c=5 4           y= 2 −x+5 x 7

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WORKED solutions

8 i Let C be the centre of the circle. So A( –7, –7) and C( –1, 6). B is at the opposite side of the circle, so you can conclude B(5, 19). ii AC is a radius of the circle. By Pythagoras,





When x = 2,

AC = r = 6 2 + 132 = 205 iii Gradient of AC is 13 ⇒ gradient of 6 perpendicular bisector is − 6 and it passes 13







y − 6 = − 6 (x + 1) 13



13y – 78 = – 6x – 6



6x + 13y – 72 = 0





4 − 2 + 2a + 2 + 4tan a   cos α 4 = cos α + 4 tan a + 2a

Perimeter =



10 i y − 2t = −2(x − 3t) y = −2x + 8t At A, y = 0  ⇒  x = 4t. At B, x = 0  ⇒  y = 8t. 4t × 8t Area of triangle OAB = = 16t 2 . 2







1 ii  Equation of perpendicular line is y − 2t = (x − 3t ). 2 2y = x + t At C, y = 0  ⇒  x = −t. So C(−t, 0) and P(3t, 2t).

Mid-point of CP is

( 3t 2− t , 22t ) = (t, t), which

is on the line y = x. 11 i y = 8x −1 + 2x dy = −8x −2 + 2 = − 82 + 2 dx x d 2y −3 = 16 2 = 16x x d x −3



ii 0 = −82 + 2 x = −8 + 2x2 x=±2

(x + 2)2 = 0

x = –2  ⇒  y = 2

Arc length = 2a



d 2y = −2 < 0  ⇒ maximum d 2x

x2 + 4x + 4 = 0

CB ⇒  CB = 4  tan a 4   1 Area of triangle = × 4 × 4 tan a = 8 tan a. 2 1 Sector area = × 22a = 2a. 2 Shaded area = 8 tan a − 2a. 4 AC = 4 DC = 4 − 2 ii cosα = cosα cosα AC



When x = −2,

x2 + (x + 4)2 = 8

The line and the circle intersect once only at P( –2, 2) so the line is a tangent.

ii The second tangent is at the opposite side of the circle and passes through (2, –2). The equation is

9 i tan α =



d 2y = 2 > 0  ⇒ minimum. d 2x

12 i The centre of the circle is the origin O(0, 0). The line and the circle intersect when

through C.



Substituting into original equation gives y = ±8. So the stationary points are (2,8) and (−2, −8).



y + 2 = 1 (x – 2)  ⇒  y = x – 4 13 4 sin2 x + 8 cos x − 7 = 0 4(1 − cos2 x) + 8  cos x − 7 = 0 4 cos2 x − 8 cos x + 3 = 0 (2 cos x − 1)(2 cos x − 3) = 0 3 cos x = ⇒  no solutions 2  1 ⇒  x = 60°, 300° cos x = 2  14 i f(x) = x2 + 1 y = x2 + 1 y − 1 = x2 x = y − 1

f−1(x) = x − 1 (x > 1) ii f(x2 + 1) = (x2 + 1)2 + 1 =



(x2 + 1)2 =

185 16

169 16

13 4 3 x= 2

x2 + 1 = ±

Only one solution, since x  0

 6  6  6 15 i  (2x − x2)6 =   (2x)6 +   (2x)5(−x2) +    0 1   2 (2x)4(−x2)2 + ... = 64x6 − 192x7 + 240x8 + ...

ii (2 + x)(2x − x2)6 For ‘x8’: 2 × 240x8 = 480x8 and x × −192x7 = −192x8

So the coefficient of x8 is 288.

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Summary REVIEW

16 Completing the square: x2 + y2 – 2x – 16y + 40 = 0 becomes (x – 1)2 + (y – 8)2 = 25

20 i f(x) = 4(x – 3)2 + 2 ii

y 50

This is a circle with centre (1, 8) and radius 5.  −6 The translation   moves the centre to ( –5, 3)  −5 and the radius remains as 5.

40

17 i −x2 + 6x − 5 = −1(x − 3)2 + 4 by completing the square 30

ii m = 3

iii      y = −(x − 3)2 + 4 (x − 3)2 = 4 − y      x=3+ 4− y

   f−1(x) = 3 + 4 − x



The domain is x ⩽ 0

18 i S =

20

10

a a and 3S = 1−r 1 − 2r

3a a =   1 − r 1 − 2r 3(1 − 2r) = 1 − r 2 r= 5     

0

77 n −1

①



245 = 7 + (3n − 1)d  ⇒  d =

238 3n − 1

②



Equating ① and ②: 238(n − 1) = 77(3n − 1) 238n − 238 = 231n − 77 7n = 161

5

x

The minimum point is (3, 2)

iii f(2x) moves (3, 2) to (1.5, 2)

ii 84 = 7 + (n − 1)d    ⇒  d =

n = 23

19 i Perimeter = 400  ⇒ 400 = 2πr + 2x  ⇒ 2x = 400 − 2πr ① A = πr2 + 2xr Substituting ①: A = πr2 + r(400 − 2πr) A = πr2 + 400r − 2πr2 A = 400r − πr2 dA = 400 − 2πr dr  0 = 400 − 2πr 2πr = 400 Substituting into ①: 2x = 400 − 400  x = 0 Therefore there are no straight sections. ii



(0, 38)

d2A = −2π < 0  ⇒ maximum dr 2

f(2x – 3) moves (3, 2) to (4.5, 2)

–f(2x – 3) moves (3, 2) to (4.5, –2)

So the maximum point of – f(2x – 3) is (4.5, –2)

21 i ar2 = −108 ① ar5 = 32 ② ② ÷ ①: 8 r3 = − 27 2  r = − 3 ii Substituting value of r into ① gives: 4 a = −108 9    a = −243 −243 iii S∞ = 1− −2 3 = −145.8 22 i x = x2 − 4x + 4 0 = x2 − 5x + 4 0 = (x − 1)(x −4) So x = 1, y = 1 A(1, 1). And x = 4, y = 4 B(4, 4). Therefore, mid-point of AB is (2.5, 2.5). dy = 2x − 4 dx 2 x − 4x + 4 = (2x − 4)x x2 = 4 ii

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WORKED SOLUTIONS

x = ±2 x = 2 ⇒ y = 0 ⇒ m = 0 Ignore this solution since m = 0. x = −2 ⇒ y = 16 ⇒ m = −8 So the tangent meets the curve at (−2, 16).

ii

y 8

2

23 i 4x2 − 24x + 11 = 4(x − 3)2 − 25 Vertex is at (3, −25). ii g(1) = 4 − 24 + 11 = −9 Therefore g(x)  −9. iii y = 4(x − 3)2 − 25 y + 25 (x − 3)2 = 4 x =3+

y + 25 4

=3+

1 y + 25 2

g−1(x) = 3 +

1 x + 25 2

The domain is x  −9. 24 By Pythagoras: i

AB = 122 + 10 2 = 244 Similarly, AC = 12 + 112 = 122 and BC = 112 + 12 = 122 Therefore, (AC)2 + (BC)2 = (AB)2 ⇔ ABC is a right-angled triangle.

ii (AC)2 + (BC)2 = (AB)2 ⇒ AB is the hypotenuse ⇒ AB is the diameter of the circle ∴ The centre of the circle is the midpoint of AB. The centre is (–4, 2) and the radius is 244 = 61 2 So the equation of the circle is (x + 4)2 + (y – 2)2 = 61 4 − 1.75 = 4.5 and gradient of 25 i Gradient of CE = 4 − 3.5 DE = 4 − 3.51 = 4.9 4 − 3.9 ii The sequence of chords have gradients: 3, 4, 4.5, 4.9. This indicates that as the chords approach E, the gradients are converging towards 5. It indicates that f’(4) = 5. x 26 i 5 + 3 cos 2 = 7

() 2 x cos ( ) = 3 2

x = 0.841… 2 x = 1.68

2π

x

iii f has an inverse because it is a one−one function (or it has no turning point). x iv y = 5 + 3 cos 2

()

()

y−5 x = cos 3 2 2 cos−1  

y − 5 =x 3  f−1(x) = 2 cos−1

( x 3− 5 )

27 i y = 8 −2x − x2 dy = −2 − 2x dx 2 = −2 − 2x x = −2 y=8 Therefore 8 = −4 + c c = 12 ii 8 − 2x − x2 = 2x + 11 0 = x2 + 4x + 3 = (x + 1)(x + 3) x = −1 or x = −3 Area =

−1

∫−3 (8 − 2x − x

2

)dx −

−1

∫−3 (2x + 11)dx .

−1

−1  x3  4 Area = 8x − x 2 − −  x 2 + 11x  = . 3  3 −3  −3 28 i Completing the square: x2 + y2 – 8x – 6y + 21 = 0 ⇒ (x – 4)2 + (y – 3)2 = 4.

Therefore, the centre is (4, 3). ii The radius is 2. iii When x = 0 ⇒ (y – 3)2 = –12 ⇒ no real solutions ⇒ does not intersect y-axis. When y = 0 ⇒ (x – 4)2 = – 5 ⇒ no real solutions ⇒ does not intersect x-axis. iv The diameter lies on the line AB. The gradient of the tangent is 1 , so the 3 gradient of the diameter (and hence AB) is –3. AB also passes through the centre. So the equation is:

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Summary REVIEW

y – 3 = – 3(x – 4) y – 3 = – 3x + 12 3x + y – 15 = 0

1 2

ii Area = ∫ (3 − 2x)3 dx − ∫ (−24x + 20)dx 0

29 i 0 = 3u + − 10 0 = 3u2 − 10u + 3 0 = (3u − 1)(u − 3) u=1 x=1 9 3   u = 3   x = 9 f ′′(x) = ii f ′′

−

2

iii f(x) =

+

3 2(4)2

 −7 =  c = 5

+

3

− 10x +c

1 6(4)2

− 10(4) + c

So x + 2 = ± 4  ⇒  x = 2 or x = – 6



Therefore, P(2, 7).

The tangent is perpendicular to CP, so it has 4 gradient − . 3 The equation of the tangent is y − 7 = − 4 (x − 2) ⇒ 4x + 3y – 29 = 0 3    

iii For the tangent line, y = 0  ⇒  4x = 29  ⇒  x = 7.25  ⇒  Q(7.25, 0)

By Pythagoras, PQ = 5.252 + 7 2 = 8.75



So, the area of triangle CPQ is 5 × 8.75 = 21.875 units2 2

31 i y = (3 − dy = 3(3 − 2x)2(−2) dx = −6(3 − 2x)2 dy When = 1 , = −6(3 − 1)2 = −24. 2 dx 2x)3

So the equation of the tangent is: 1 y − 8 = −24 x − 2

( )

y − 8 = −24x + 12 y = −24x + 20

110

4

4

2

2

4

4

V = π∫(36x 2 − 12x 3 + x 4) dx − π∫64 dx 2

1

f(x) = 2x 2 + 6x 2 − 10x + 5



(x – 2)(x – 4) = 0

V = π∫ (6x − x 2)2 dx − π∫ 8 2 dx

ii Let C be the centre of the circle. The gradient of CP is 7 − 4 = 3 . 2 − −2 4



So x2 – 6x + 8 = 0 x = 2 or x = 4





32 The graphs intersect when 6x – x2 = 8

⇒ minimum

30 i y = 7  ⇒ (x + 2)2 + 9 = 25  ⇒ (x + 2)2 = 16



=9 8



2

1 6x 2

= 81 − 9 8



( 19 ) = −36 < 0 ⇒ maximum 3 2x 2

1

1 −  −12x 2 + 20x  2 =  − (3 − 2x) 0   8 0

3 − 32 x

4 f′′(9) = > 0 9

0

1 42

3u−1

3 − 12 x

1 2

2

4

V = π 12x 3 − 3x 4 + 1 x 5  − π [ 64x ]2  5  2 4

V = π[[204.8] – [54.4]] – 128π 112π V = 22.4π or 5

(

)

Extension Questions 1 i Radius and tangent meet at B, so angle ABC is a right-angle and ABC is a right-angled triangle.

Area of triangle ABD = 3(BD) 2   ⇒  Area of kite ABDC = 3 (BD)



Therefore, 12 = 3 (BD)  ⇒  BD = 4 cm



By Pythagoras: (AD)2 = 32 + 42  ⇒  AD = 5 cm

ii The locus of points is a circle, centre (5, 8) and radius 5 cm. So the equation of the locus is:

(x – 5)2 + (y – 8)2 = 5

2 i Sometimes. The result is not valid for a = 0. ii Sometimes. The result is only true when a = b. The circle is in the positive quadrant and touches both the x-axis and the y-axis.

If a > b, the circle extends beneath the x-axis.



If a < b, the circle touches the y-axis only and does not reach the x-axis.

iii Sometimes. Solving the equation:

tan θ = 1 ⇒ θ = π , 5π , … 4 4 In general, the result is true when θ = (4n − 3)π , where n ∈ Z . 4

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WORKED solutions

iv Always. This is an identity and should be written as sin2 θ + cos2 θ ≡ 1.



3 Solving the quadratic: (3x + 5)(x – 2) = 0 5 ⇒ x = − or x = 2. 3

For the cubic:



dy = 3x 2 − 8x + 6 . The discriminant dx



is 82 – 4(3)(6) = – 8  ⇒  no turning points. At the intersection of the two curves: x3 – 4x2 + 6x + 5 = 3x2 – x – 10  ⇒  x3 – 7x2 + 7x + 15 = 0 x = – 1 is a root  ⇔ (x + 1) is a factor By inspection: (x + 1)(x2 – 8x + 15) = 0  ⇔ (x + 1) (x – 3)(x – 5) = 0 This information allows a sketch to be drawn. y (5, 60)

20

(0,5) (−1.667, 0) 0

(2, 0)

5

(−1, −6)

The area between the curves from (3, 14) to (5, 60) is the area we need to find.



5



))

) (

2

(



3

3

5

3

 4 ( 2x )3 ( −3) × ( −4 )6 = −12582912x 3  3 

( 2)5 × 



( 2)5 × 



( 2)5 × 



( 2)5 × ( −3)4 × 



Summing: the coefficient of x3 is 11 802 624.

()

 4  6 ( 2x )2 ( −3)2 ×   x2 ( −4 )5 = −21233664x 3  2 1 

( ) (−4) = −6635520x

 4  6 ( 2x )( −3)3 ×   x2  1   2

2

4

( ) (−4) = −414720x

 6 x  3 2

3

3

3

3



sin2 x (which loses some solutions) and sinx = 1 2 has more solutions in the given range than shown. The correct solutions is: sin2x – cos2x = 4 sin3x – 1 sin2x – (1 – sin2x) = 4sin3x – 1 2sin2x – 1 = 4sin3x – 1 2sin2x = 4sin3x sin2x = 2sin3x 2sin3x – sin2x = 0 sin2x (2sinx – 1) = 0 1 sin x = 0 or sin x = 2

π 5π x = 0, π, 2π  x = , 6 6

)

4 All of the different combinations that contribute to the coefficient of x3 are:  5 3 2 4 6   ( x ) ( 2) × ( −3) × ( −4 ) = 13271040x 3  3

4

− x − 10 − x 3 − 4x 2 + 6x + 5 dx 6 i ax2 + bx + c = 0 3 b c x2 + x + = 0 5 a a Shaded area = ∫ −x 3 + 7x 2 − 7x − 15 dx x2 + b x = − c 3 a a 5 1 7 7 20 2 4 3 2 2   = − x + x − x − 15x = units b − b2 = − c  4 3 3 2 3 ii x + 2a a 4a 2 Shaded area =



∫ ((3x

4

2



x

(3, 14)



( ) (−4) = 6 220800x  5  4  6 x  1  ( x )( 2) ×  1  ( 2x )( −3) ×  1  ( 2 )( −4 ) = 53084160x  5 4 4  6 x  1  ( x )( 2) × ( −3) ×  2 2

5 The errors in this solution include division by

40

−5

 5 4  4 2 2 6 3  1  ( x )( 2) ×  2 ( 2x ) ( −3) × ( −4 ) = 70778880x

 5 2 3  4 3 6 3  2 ( x ) ( 2) ×  1  ( 2x )( −3) × ( −4 ) = −70778880x

()

 5 2 3 4  6 x 5 3  2 ( x ) ( 2) × ( −3) ×  1  2 ( −4 ) = −19906560x



( ) ( x + 2ba ) = 4ba − ac ( x + 2ba ) = 4ba − 44aca ( x + 2ba ) = b 4−a4ac 2

2

2

2

2

2

2

2

2

2

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Summary REVIEW



2 iii x + b = ± b − 42ac 2a 4a

b 2 − 4ac x+ b = ± 2a 2a



b 2 − 4ac x=− b ± 2a 2a



2 x = −b ± b − 4ac 2a

7 i gh(x) = g(sin x) = sin2 x, 0  x  π and 2

iii Not always. A point of inflection occurs when the sign of the curvature changes. This can occur without the need for a stationary point.

()

)

ii fgh ( x ) = f sin 2x = 1 − sin 2x = cos x 0 ⩽ x ⩽1 and 0 ⩽ fgh(x) ⩽ cos(1) 8 i Area of circle 1 is 100π and area of circle 2 is 25π. So a = 100π and r = 1 . The area is reduced by 4 a scale factor of 4 for each consecutive circle. The radius is halved for each consecutive circle. Therefore, the 4th circle has area 25π cm2 from 16 which its radius is 5 cm. 4 The distance of the centre of the 4th circle from O is the radii of circles 1 and 4 and the diameters of circles 2 and 3 ⇒ 10 + (2 × 5) + (2 × 2.5) + 1.25 = 26.25.

So the centre of circle 4 is (26.25, 0) with radius 5 cm. 4 100 π 400π cm2 ii S∞ = = 1 3 1− 4 iii The total area is found by doubling the sum to positive infinity and subtracting the area of the original circle as this will have been double-counted. 400π 500π Total area = 2 − 100π = units cm2 2 3 3

( )



dy d 2y = 4x 3 − 9x 2 = 12x 2 − 18x dx dx 2 At stationary points: 0 = 4x3 – 9x2 ⇒ x = 0 or x = 9. 4



2 9 d y 81 > 0 ⇒ minimum When x = , 2 = 4 dx 4

9 i

d 2y ii When x = 0, = 0 ⇒ may be a point of dx 2 inflection. dy = −0.094 and when When x = – 0.1, dx x = 0.1,

)

()

10 cm

8 cm

(

)

3 = π−x 2 2 π x= − 3 2 2

0 ⩽ gh(x) ⩽1

(

(

10 cosx = sin π − x ⇒ sin  cos π  = sin π − x 2 6  2  cos π = π − x 6 2

x = π− 3 2 11

5 cm



The gradient either side of zero is negative and the curve is continuous. This suggests that x = 0 is a point of inflection.

a 10 cm

b 8 cm

Use the cosine rule to find α and β. 52 = 102 + 102 – 2(10)(10) cos α 25 = 200 – 200 cos α cos α =

175 200

α = 0.505 radians 52 = 82 + 82 – 2(8)(8) cos β 25 = 128 – 128 cos β cos β =

103 128

β = 0.636 radians The shaded area can be found in two parts. In each case, subtract the area of the isosceles triangle from the area of the sector.

( 12 × 10 α − 12 × 10 sinα ) 1 1 + ( × 8 β − × 8 sinβ ) = 2.404… 2 2 2

Shaded area = 2

2

2

The required area is the area of each circle minus twice the shaded area. Required area = 100π + 64π − 2 × 2.404 … = 510.41 cm2

dy = −0.086 dx

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WORKED solutions

12 i The terms can be written as: 2, 2 + 2d, 2 + 5d.

Consecutive terms in a geometric sequence 2 2 ⇒ + d = 2 + 5d 2 2 + 2d



Solving: (2 + 2d)2 = 2(2 + 5d)



4 + 8d + 4d2 = 4 + 10d



4d2 – 2d = 0



2d2 – d = 0

d (2d – 1) = 0

The non-trivial value of d = 1 2 We know that r = 2 + 2d = 2 + 1 = 3 2 2 2

ii For the arithmetic progressions, the 4th term 1 = 7. is 2 + 3 2 2

()



For the geometric progression, the 4th term

( ) = 278 a

is a 3 2

3



Equating the terms:



27 a = 7 8 2



54a = 56



a = 28 27

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