Comprehensive Remedial Mathematics for B. Pharmacy 9781282805255, 9788188449776

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Comprehensive Remedial Mathematics for B. Pharmacy

"This page is Intentionally Left Blank"

Comprehensive Remedial Mathematics for B. Pharmacy

Dr. Shyam Patkar M.Sc. , Ph.D.

Professor & Head

Ramakant Bhardwaj

Sarvesh Agrawal

M.Sc, Ph.D.

M.Se , MPhil.

Truba Institute of Engineering & Information Technology, Bhopal (India)

•

til

..•••• ... _....... . PharmaMed Press ........... -....... :::!!~: An imprint of Pharma Book Syndicate '

,

4-4-316, Giriraj Lane, Sultan Bazar, Hyderabad - 500 095 .

Copyright © 2009, by Publisher

All rights reserved. No part of this book or parts thereof may be reproduced, stored in a retrieval system or transmitted in any language or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of the publishers.

Published by

•

PharmaMed Press •••• ...... .... .. An imprint of Pharma Book Syndicate ....... . · ...... .......... I'II········ .. " ... ... . -~

',

"

,

,

Au {'"print 0/

Ph.,",. Book Syndicate

4-4-316, Giriraj lane, Sultan Bazar, Hyderabad - 500 095. Phone: 040-23445605, 23445688; Fax: 91+40-23445611 E-mail: [email protected] WWIN. pharmamedpress.com/pharmamedpress .net

Printed at Adithya Art Printers Hyderabad

ISBN: 978-81-88449-77-6 (HB)

Dedicated to

Our Parents

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Preface Comprehensive Remedial Mathematics for B .. Pharmacy is specially designed for undergraduates.- The authors have evolved this from lecture notes and related material developed while teaching various courses in Remedial Mathematics during the last sixteen years. The authors have given special stress on basic fundamentals, solved examples and oriented problems of various universities in the book. This structure will provide students a platform for building confidence and sharpen their skills in Pharmaceutical Mathematics. The comprehensive edition will surely serve its purpose to help to achieve handsome marks in the examinations. We hope the students and teachers of Mathematics of various universities will appreciate our efforts and receive the book with enthusiasm. We sincerely thank our publishers for their untiring efforts in bringing out this book excellently.

-Authors

(vii)-

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Contents Preface ...................................................................................................... (vii)

Part - A Remedial Mathematics CHAPTER

1

Equations Reducible to Quadratics Equations ................... ..................... 1 CHAPTER

2

Simultaneous Equations ............................... .......................................... 35 CHAPTER

3

Determinants ........... ................................................................................ 51 CHAPTER

4

Matrices ........... ........................................................................................ 89 CHAPTER

5

Mensuration and its Pharmaceutical Applications ................................ 129 CHAPTER

6

Objective and Prerequisite of an Ideal Measure .................................. 147

(ix)

(x)

Contents CHAPTER

7

Trigonometry ......................................................................................... 189 CHAPTER

8

Logarithm .............................................................................................. 226 CHAPTER

9

•

I

Analytical Plane Geometry .................................................................... 251 CHAPTER

10

Calculus ................................................................................................ 287 CHAPTER

11

Differentiation ........................................................................................ 309 CHAPTER

12

Integral Calculus ................................................................................... 372

Part - B Advanced Mathematics CHAPTER

13

Differential Equations of the First Order and First Degree .................. 437 CHAPTER

14

Linear Differential Equations of Higher Order with Constant Coefficients ........................................................................... 501 CHAPTER

15

Simultaneous Linear Differential Equations with Constant Coefficient ............................................................................. 554

(xi)

Contents

CHAPTER

16

Laplace Transforms and its Applications .............................................. 575 CHAPTER

17

Biometrices Errors and Approximations ............................................... 657 CHAPTER

18

Data Organization ................................................................................. 662 CHAPTER

19

Measures of Dispersion ........................................................................ 680 CHAPTER

20

Theory of Probability ............................................................................ 705 CHAPTER

21

Skewness and Kurtosis ........................................................................ 766 CHAPTER

22

Correlation and Regression .................................................................. 779 CHAPTER

23

Method of Least Square ........................................................................ 814 CHAPTER

24

Statistical Inference ............................................................................... 827 CHAPTER

25

Analysis of Variance (Anova) ................................................................ 872 Question Bank .........................................................

L ••• . . . . . . . . . . . . . . . . . . . . . . . . . . . .

885

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Part - A

Remedial Mathematics

"This page is Intentionally Left Blank"

CHAPTER

1 EQUATIONS REDUCIBLE TO QUADRATICS EQUATIONS

Numbers The.numbers 1,2,3 ...

00

are called natural numbers or positive integers.

~ is called afraction where a and b are any two positive integers. b

The number system consisting of positive and negative fractions is called rational

number. The rational number is of the form ~ where a and b are integers and b:;; 0. That b is -4, -2, -1,0, 1,2,3,4 ... are called integers. The number

.fi, .fj,

.J5 ...

which cannot be expressed as a (i.e. ratio of two b integers), are called irrational numbers. The rational numbers, positive and negative, irrational numbers and zero constitute the real number.

Types of Numbers Natural number or N 1,2,3,4, .....

2

Comprehensive Remedial Mathematics for Pharmacy

Real number or R The set of rational and irrational numbers i.e., R = Q numbers.

U

In where Ir denotes the Irrational

Rational number or Q 3

1

1

1

4

2

2

3

..... --,--,0, -, - .....

Primes or P 2,3,5, 7, 11, 13 .....

Composite numbers 4•. 6,8,9,10,12 ..... Even numbers

2, 4, 6, 8, 10, 12 ..... Odd numbers

1,3,5, 7, 9, 11 .....

Complex Numbers The numbers of the form x + iy where i = ~ and a and b are real numbers are called complex numbers.

Useful Formulae

·';(~~·,br = a2iit b~_ . ~ 2~b :' ,~.(a -:>p)2==~t;:~ b2 '~ .2~;:\?':1\J~

';.;1-£a ,i bf + (a'- bi

=' 2a , +'2b , 2

- ' (a + biiT'% (a-'bi = 4ab i , .

=l

81l- 18y + 1 = 0 9y .(9y - 1) - 1 (9y - 1) = 0 (9y - 1) (9y - 1) = 0 9y - 1 = 0 or 9y - 1 = 0

1

1

9

9

y= - ory=-

3~ =.!. 9

3 " Hence·

x

or 3x

= .!. 9

x

= T2 or 3 = T2

x=-2 orx =-2 x';-2

Base same then powerequaf

Equations Reduci~le to Quadratics EquCltions

y- 2 + T~ = 10

Problem 2:

Solve

Solution:

The given equation can be written as 2

Y. Let

21

X

3 + T = 10

by exponent rule

..... (1)

x

3 = y, put in equation (1) y. 3 2 + y-l = 10

or

1

9y + - = 10 Y 9/ - lOy + 1 = 0 9/-9y-y+ 1 =0 9y (y - 1) -1 (y - 1) = 0 (9y-l)(y-l) = 0 9y - 1 = 0 and y - I = 0

1

Y = - and y =- 1 9 3~

= -1 9

and 3x = 1

x=-2 and x= 0 Hence

Problem 3:

x = 0,-2

Solve for x (x+ 1)(x+2)(x+3)(x+4)+ 1 =0

... Type - V

[(x + l)(x + 2)] [(x +.3)(x + 4)] + 1 = 0

Solution:

[x 2 + 2x + x + 2] [x 2 + 4x + 3x + 12] + 1 = 0 (x2 + 3x + 12)(x2 + 7x + 12) + 1 = 0 Which is not similar, so the given equation can be written as [(x + l)(x + 4)] [(x + 2)(x + 3)] + 1 = 0

,

Let

.

,

(x- + 5x + 4) (x- + 5x + 6) + I = 0 x2 + 5x = y, put in (1) ~

./

(y +4)(y+ 6)+ 1 =0

y2 + 4y + 6y + 24 + 1 = 0

y2 + lOy + 25

= 0

..... (1)

22

Comprehensive Remedial Mathematics for Pharmacy

l

+ 5y + 5y + 25 = 0

y (y + 5) + 5 (y + 5) = 0 (y + 5)(y + 5) = 0 (y+ 5/ = 0

.: x2 + 5x = y

Y= - 5 Now

x 2 + 5x = - 5 x2 + 5x + 5 = 0

-5±~25-4(1)(5)

-5±.Js

2 (1)

2

x=--~---

Problem 4:

Solve

Solution:

Suppose

..... (1)

~3x2_4x-6 =y so that 3x2 - 4x - 6 = y2 and so 3x2 - 4x = y2 + 6, putting these values in (1) y+l+6=18 l+y-12=O y2 + 4y - 3y - 12 = 0

Y (y + 4) - 3 (y + 4) = 0 (y - 3)(y + 4) = 0

Y - 3 = 0 and y + 4 = 0 Y = 3 andy=-4

~3X2 -4x-6 =3

and

squaring the both sides 3x2 -4x-6 = 9

and

3x2 - 4x - 15 = 0

and

3x2 - 9x + 5x - 15 = 0

and

3x(x - 3) + 5(x - 3) = 0 and

3x2 - 4x - 6 = 16 3x2 - 4x - 22 = 0 x=

4±~(-4)2 -4(3)(-22) 2(3)

x=

4 ± "'16 + 264 6

Equations Reducible to Quadratics Equations

(x - 3)(3x + 5) = 0

and

-5

x=3 andx=3

x=

x=

4 ± -J280

6 4 ± -J4 x 70 2x3

2±J70 3

. set IS . { 3, -5, 2+J70 , 2-J70} . :. The solutIOn 333

Problem 5:

Solve J(x 2 + 5x + 1) -1 = 2x

Solution:

We have J(x 2 + 5x + 1) = 2x + 1, squaring both the sides.

... Type-VI

x2 + 5x + 1 = 4x2 + 1 + 4x - 3x2 + X = 0 3x2 -x = 0 x(3x-l)=0 x = 0, 3x-1 = 0

1

x= 0 orx=3

Hence the solution set{ 0,

Problem 6:

-J 4 -

Solution:

The given equation can be written as

x +

-J9 + x

k}

-J9 + x

= 5

= 5-

-J 4 -

x , squaring the both sides

9 + x = 25 + 4 - x - 2 (5) -J 4 - x 2x - 20 = -1O-J4- x 2 (x-1O)=-1O-J4-x x - 10 = - 5 2

-J 4 -

x , squaring the both sides

x + 100 - 20x = 25 (4 - x)

23

2-4

Comprehensive Rehiedial Mathernatic:s for Phannacy x2 + 100 -:- 20x = 100 - 25x

x2 + 5x = 0 x (x + 5) = 0

x= 0, x=-5 Hence th:e-solution set {a, - 5}

Problem 7:

x (2x + l)(x - 2)(2x - 3) = 63

Solution:

The given equation can be written as (x+O)

(x+!J(X-2)(x-~1 2 -: 2)

=63 4

Therefore taking together first and fourth and second and third factors, we get {x( x-%J} {(

x+~J(X-2)} =61

(X2 -%x J (X2 -%X-l) Putting

=

61

..... (A)

2 x - %x = y, put in (A) y(y-l)=

463

4y2 - 4y - 63 = 0

41 -18y + 14y~-63 ='0' 2y(2y~,9)+7

(2y-9)=fr .

(2y - 9) Oy + 7)'= 0

9 2

-7 2

y= - ory=-

9 Now, when y = 2 .._ r 2 '3x'~ 9"-' Then x ----

2

2

"

-,

,

' ",

-;~,. ~ ,

{:

,Equations Reducible to Quadratics Equations

25-,

2x1 - 3x- 9 = 0 2Xl - 6x + 3x - 9 = 0 2x (x - 3) + 3 (x - 3) = 0 (x - 3)(2x + 3) = 0 .., x= 3 orx=-J

2

Again, When

-7

y=2 2 3 x - -x

Then

-7

=-

2

2

2x" - 3x + 7 =,0

..... (B)

Its discriminant = b2 - 4ac

=(-3i-4

x

2

x

7

=-47 < 0 :. Equation (B) has no teal roots. Hence real roots the given equation are 3; - 3 2 3 :. solution set {3, -2 }

~5X2

- 6x + 8 - .JSx 2 - 6x -7 = 1

Problem 6:

Solve

Solution:

LetA= .J5x l -6x+8 ,B = .J5x 2 -6x-7 then

A-B=l

.... Oi)

A2 - B2 = 15

..... (iii)

Dividing (iii) by (ii) A + B = 15 then A

... Type VI

=8 . .J5x 2 -6x+.&

=

8

26

Comprehensive Remedial Mathematics for Pharmacy Squaring the both sides

5x2 - 6x + 8 = 64 5x2 - 6x - 56 = 0 5x2 - 20x + 14x - 56 = 0 5x(x - 4) + 14(x - 4) = 0 (x-4)(5x+ 14)=0 -14 x=4 or x=-5 -14) :. Solution set = ( 4, -5-

Problem 1.4 1.

x+2 x-2

x-2 x+2

5 6

-----=-

ADS:

(

2 -5 , 10)

2. 2X4 + 9x3 + 8x2 + 9x +2 = 0 ADS:

(-2+F3,-2-J3)

ADS:

(-.J2,.J2) imaginary

ADS:

(-1, 1)

A DS.. [2 " _! 5+J41 ,5-J41j --2 4 4

Equations Reducible to Quadratics Equations 7.

27

(x+ l)(x+2)(x-5)(x-6)= 144 ADs: (2. 7, -3)

9 16

8. x( x + 1)( x + 2)( x + 3) = -

Ans: 9.

(_~, -3 +

JiO, -JiOj

-3 222

x 2 +6X+2~X2 +6x =24 ADs: (2. -8, -3 + 3.J5 . -3 - 3.J5)

10.

~3X2 +7x-l +~3X2 +7x-l0 =9 ADs: (2,- 1:)

11.

~2y+8 = 20-y ADs: (14)

12.

~4x2+x-2=12x-l ADs:

(~)

13. (2x-l) = ~3X2_2

ADs: (1,3) 14.

.J3x+l0 +.J6-x =6 ADs: (2. 5)

15.

.Jx+5 +.Jx+21 =.J6x+40 ADs: (4)

Solution of Word Problems involving Quadratic Equations We consider such word problems in the examples given below. Example 1:

The product of Ramu's age (in years) five years ago with age 9 years later is 15. Find Ramu's present age.

Solution:

Let Ramu's present age = x His age 5 years ago = x - 5

28

Comprehensive Remedial Mathematics for Pharmacy His age 9 years later = x + 9 According to the problem Therefore (x - 5) (x + 9) = 15 x 2 + 4x - 45 = 15 i.e.,

x 2 + 4x - 60 = ,0

Solving

x 2 + lOx - 6x - 60 = 0 x( x + 10) - 6( x + 10) = 0 (x - 6) (x + 1'0) = 0

x=-10

x= 6 or

Since x is the present age of Ramu, it cannot be negative. Therefore, we reject the solution x = -10. Thus Ramu's present age is 6 years (check - Ramu's age 5 years ago

= 6 - 5 = 1 year

Ramu's age 9 years later = 15 years Their product = 15; as ~equired in the problem

Example 2:

Find two natural numbers which differ by 3 and whose squares' have the sum 117. ', ;

Solution:

Let the larger of the two numbers be x Then the other number''''' 5c - 3 Sum of their squares = 117 x-, + (x - 3y = ,117

"

2x" - 6x - 108 ~

~

~.

"r"'"

Divlding both sides 'by -2, ' ,

•

'.

,

r

=0 '

x 2 -3x-54=0 x 2 - 9x + 6x - 54 x(x - 9) + 6(x -

=0 9) = 0

(x - 9)(x + 6) = 0

The solution of this quadratic are x = 9 or x =--6

Equations Reducible to Quadratics Equations

29

Since x must be natural number. it cannot be negative and so we reject the solution x = -6 Therefore x = 9 Therefore. the larger number = 9 The other number = x - 3 = 6 ,

,

,

'

Therefore, the numbers required are 6 and 9.

Check: Sum of their squares = 36 + 81 = 117 as required and they differ bY,3 ...

Example 3:

The product of tWo successive mUltiples of 5 multipliers. .

Solution:

Let the two required numbers be 5x and 5(x + I)

IS

300. Determine the

According to the problem Sx

x

S(x + 1) = 300

x(x + I) = 12

x2 + x -12 = 0 (x-3)(x+4)=0 x-3=0

x=3

x+4=0

x = -4 which is rejected as -4

Thus, the numbers are S x 3. 5(3

+1)

i.e., 15,20.

Example 4:

Divide 16 into two parts su\:h that twice the square of the larger part exceeds the square of the smaller part by 164.

Solution:

Let the larger part be x So the smaller part

is

0< x < 16

16 - x

According to the problem

2(xi = (16 -

X)2

+ 164

2X2 = 25~ - 32x + x 2 + 164

x2 + 32x - 420 = 0 x 2 + 42x - lOx - 420 = 0 (x + 42)(x - 10) = 0 x=-42

or

x = 10

30

Comprehensive Remedial Mathematics for Pharmacy But x = -42 does not satisfy 0 < x < 16 so it is rejected . .". X =

10, which is the largest part

.". smaller part = 16 - 10 = 6

Example 5:

A fast ~rain 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 kmlhr less than that of the fast train, find the speed of two trains.

Solution:

Let the speed of fast train = x kmlhr Then the speed of slow train = (x - 10) kmlhr Time taken for journey of 600 km by fast train = -600 h ours

(T· distance) .: lme =-speed

x

And time taken for journey of 600 km by slow train 600

= - - hours

x-10 Since the fast train takes 3 hours less than the slow train 600 x

=

600 -3 x-l0

600(x - 10) = 600x - 3x(x -10) 600x - 6000 3x x

2

2 -

-

(Multiplying by x(x - 3)

2

= 600x - 3x + 30x

30x - 6000 lOx - 2000

=0

=0

(x - 50) (x + 40) = 0 x = 50 or

x = -40

Rejecting x = -40, as the speed of train cannot be negative . .". speed of fast train = x = 50 kmlhr Speed of slow train

= x-I 0 = 40 kmlhr

Example 6:

A motor-boat whose speed is 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of water.

Solution:

Let x km/hr be the speed of water. The speed of motor boat in still water is 15 kmlhr

Equations Reducible to Quadratics Equations

----------------------------~

31

Therefore, its speed downstream is (IS + x) kmlhr and the speed upstream is (IS - x) kmlhr Time taken for going 30 km downstream = Time taken for going 30 km upstream =

~ IS + x

~ IS - x

hours

hours

Since the total time is given to be 4 hours 30 minutes i.e.,

.2. 2

hours

30 30 9 - - + - - = - hours IS+x IS-x 2 30(lS-x)+30(IS+x) 9 = (1S + x)(IS - x) 2

900

9

22S-X2

2

1800 = 9 (22S - x2) 200 = 22S - x2 x2 = 2S x = S or x =-S

[.,' X

:t= -S as x > 0]

x=S Hence the speed of water = S kmlhr. Example 7:

The hypotenuse of a right triangle is 1m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.

Solution:

Let x m be the shortest side of the right triangle

= (2x - l)m and third side = (x + l)m.

then hypotenuse

using pythagorus theorem

(2x - Ii = x2 + (x + 1)2 4x2 - 4x + 1 = x2 + x2 + 2x + 1 2X2 - 6x = 0 2x (x- 3) = 0

,32

CQmprehensiveRemedial Mathematics for Pharmacy x = 0 or x := 3 but x ~ 0 m is in inadmissible as x is the length of a side. Thus x = 3 m is the length of the shortest side of the triangle . ... Length of hypotenuse = 2x - 1 =2x3-1=5riJ. Length of third side

=x+l=4m , ,

ExampleS:

The sides (in cm) of a right triangle containing the right angle are 5x and 3x - 1. If the area of the triangle IS 60 C1112 , find 'the sides' of the triangle.

Solution:

Given that area of right triangle ABC = 60 cm2 A ~

0,

r x

r')

L -_ _ _ _ _ _ _ _ _ _ _ _

~~

B x+a+b+c

b

x+a+b+c

=0

Taking (x + a + b + c) common from CI

1

b

c

=>(x+a+b+c) 1 x+b 1 by

R2~

b

c

=0

x+c

Rz-R I andR3 ~ R3-RI

b c =>(x+a+b+c) 0 x

o

0 =0

0 x

on expanding, (x + a + b + c) x 2 = 0 Z X =

Hence

0

x=0

or

x + a + b + c=0

or

x = -(a + b + c)

Applications of Determinants 1. Area of Triangle: If the vertices of a triangle are (Xl, YI) (xz, yz) and (X3, Y3) respectively. Then area of ~ is 1

=-

2

In determinant form

~ = 1. 2

[Xl (yz -Y3) + X/Y3 - YI) + X3(YI - yz)] Xl X2

YI Yz 1

X3

Y3

1

Detenninants

65

Condition for collinearty: Three points A(x\, YI), B(X2' Y2) and C(X3, Y3) are collinear if area of MBC = 0 XI YI i.e. X2 Y2 X3

I 1=0

Y3

2. Crammer's rule: Consider three linear simultaneous equations are alx + bly + CIZ = d l

..... (1)

a2X+ b2y+ C2Z= d2

..... (2)

a3x+ b3Y+C3Z=d3

..... (3)

al If

L\= a 2

bl

ci

dl

b2 c2

III = d 2

a 3 b3 al L\2 = a 2 a3

c3

dl

CI d 2 c2 d 3 c3

bl

ci

b 2 c2

d3

b3

c3

al

bl

dl

L\3 = a 2 b 2 a 3 b3

d2 d3

Then by Crammer's rule, we have L\ L\ L\ x= _I y= _2 Z= _3 (L\;t:O) L\' L\' ,1'

Case I

If L\ ;t: 0, then, in this case the given system of equations is called consistent and has unique solution.

Case II

If L\ = 0 and any of L\\, L\2, L\3 is non zero, then, the given system of equation is inconsistent i.e., it has no solution.

Caselli

If L\ = 0 and all of L\\, L\2, L\3 are zero. Then given system of equation is consistent and has infinite number of solutions.

Case IV

In case of linear equation of two variables i.e., alx + bly = CI and a2X + b2y = C2. Then

bll

b

2

I

66

Comprehensive Remedial Mathematics for Phannacy

Example 14: Find the area of the triangle with vertices (-2, -3), (3, 2), (-1, -8)

Solution:

Area of

~=

Xl Yl 1 - X2 Y2 2 X3 Y3 -2

1 -2 =

-3 2 1

3 -1

1

-8

2. [-2 (2 + 8) + 3 (3 + 1) +1(-24 + 2)] 2

1 [-20+12-22] 2

=-

1 =2

x

(-30) = -15 = 15

Example 15: Show that the following points are collinear: A(b, c + a), B(c, a + b), C(a,b+c) Solution:

b 1 Area of MBC = - c 2 a By

C2

~

~=

c+a

1

a+b 1 b+c

1

C2 +Cl

b c+a+b 1 1 - c a+b+c 1 2 a a+b+c 1

Taking (a + b + c) common from C2 b 1 1 1 ~=-(a+b+c)c 1 1 2 all ~

=0

(by identical property)

Since area of MBC is zero, then points are collinear.

Detenninants

Example 16: Solve the following equation with the help of detenninants. x-3y+ z= 2 3x+y+z= 6 5x +y+ 3z= 3

-3 Solution:

Let

~=

3

1

5

3

= 1(3 - 1) + 3 (9 - 5) + 1(3 - 5) = 2 + 12 - 2 = 12 ~I

2 -3 1 = 6 1 1 313 = 2(3 - 1) + 3 (18 - 3) + 1(6 - 3) = 4 + 45 + 3 = 52 2

~2

=

3 6 1 533

= 1(18 - 3) -2(9 - 5) + 1(9 - 30) =15-8-21 =-14

1 -3 2 ~3 = 3 6

5

3

= 1(3 - 6) +3(9 - 30) +2(3 - 5)

= -3 -63 -4 = -70 ~

52 13 =- =~ 12 3 ~ -14 -7 y=_2=_=_ ~ 12 6 ~ -70 -35 z= _3 = _ = _ ~ 12 6

By Crammer's rule, x =

_I

67

68

Comprehensive Remedial Mathematics for Pharmacy

Example 17: Solve the system of equations using Cramer's rule 2x + y-z= 3 x + 2y + 2z=4 x+y+z=2

2 Solution:

Let

~=

1 -1 2

2

=2(2-2)-1 (1-2)-1(1-2) =0+1+1=2

3 ~l

-1

= 4 2

2

2 = 3(2 - 2) -1(4 - 4) -1(4 - 4) =0-0-0=0

2 3 -1 ~2

=

1 4

2

1 2

1

=2(4-4)-3(1-2)-1(2-4) =0+3+2=5

2 ~3

=

1 3 2 4

112 =2(4-4)-1(2-4)+3 (1-2) =0+2-3=-1

x=~=O ~

y=

~

_2

~

z = ~3 ~

5 =_ 2

=.::.! 2

Detenninants

69

Example 18: Solve 2x ~ 3y + z = 1, x + Y + Z = 2, 3x - 4z -17 = 0 by using Crammer's rule. Solution:

Given equations are 2x- 3y + Z = 1 x+y+z=2 3x- 4z = 17

2 -3 ~

= 1 3

1

1

0-4

= 2(-4) +3(-4 -3) + 1(-3) =-8-21-3 =-32 1 -3 1 ~l

= 2 17

0

-4

= 1(-4) + 3 (-8 - 17) + 1(-1 7) = -4 -75 -17 = -96 ~2 =

2

1

1

1

2

1

3 17 -4 = 2(-8 -17) -1(-4 -3) +1(17 - 6)

= -50 + 7 + 11 = -32 2 -3 Ll3

= 1 3

1

1 2 0 17

= 2(17) +3(17 - 6) + 1(-3) = 34 + 33 - 3 = 64 x= ~= -96 =3 ~ -32

y=~2=-32=1 ~

-32

~3 64 z=-=-=-2 ~ -32

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Comprehensive Remedial Mathematics for Pharmacy

Example 19: Solve the equation using Crammer's rule' x+y+z=1 2x-y + z= 2 x + 2y+ 2z=3 1

Solution:

d= 2 -1

1

1

2 2

= 1(-2 -2) -1(4 - 1) + 1(4 + 1) =-4-3 + 5 =-2 1

dl

= 2 -1 1 3

2 2

= 1(-2 -2) -1 (4 - 3) + 1(4 +3) =-4-1+7=2 1 1 d2

=

2

2

132 = 1 (4-3)-1 (4-1)+1 (6-2) =1-3+4=2

1 d3

=

1

2 -1 2 123

= 1(-3 -4) -1(6 -2) + 1(4 + 1) =-7-4+5=-6 x=

~=~=-1 d

y= d 2

d

-2

=~=-1 -2

z= d 3 = -6 =3

d

-2

Detenninants

Example 20: Solve the system of equations using Cramer's rule 2y-3z= 0 x + 3y=-4 3x +4y= 3 Solution:

Let

0 2 -3 A= 1 3 0 3 4 0 =0-2(0)-3(4-9)

= 15

o Al

2 -3

= -4 3

0 340

= 0 -2(0) -3(-16 - 9) = 75

o

0-3

A2 = 1 -4

0

3

0

3

= 0-0-3 (3 + 12) =-45 020 A3 = 1 3 -4 3 4 3 = 0 - 2 (3 + 12) + 0 = -30 x= ~= 75 =5 A 15 y=

-45 =-=-3 A 15

A

_2

z= A3 = -30 =-2 A 15

71

72

Comprehensive Remedial Mathematics for Pharmacy

Some Practice Questions

Example 1:

Evaluate

Solution:

By

265 240 219 240 225 198 219 198 181

~=

=

25 21 219 15 27 198 21 17 1.81

4 21 9 -12 27 -72 4

by

R2~

17

11

R2 + 3R1 and R3~ R3-Rl

21 9 0 90 -45 o -4 2 4

=

=4190 -

-4

45

1

2

= 4(90 x 2 - 45 x 4) = 4(180 - 180) = 0

Example 2:

Evaluate

I: ::: - : :

Solution:

Let

u

A

~:I

--Ia - ib - c + idl c+id a+ ib

= (a - ib) (a + ib) - (-c + id) (c + id) = a2 _ i2 b2 _ (i2 d2 _ c2) = a2 + b2 _ (--d2 _ c2) = a2 + b2 + c2 + d 2

(": i2 = -1)

Determinants

73

!I

I~

Example 3:

Find all the co-factors of

Solution:

Co-factor Cll = (_1)1+1,4 = 4 Cl2 = (_1)1+2, 1 =-1 C21 = (_1)2+1,2 =-2

= (-li+ 2, 3 = 3

Cn

Example 4:

60

12

8

Show that 35

21

4 =0

9

2

17

= 60(21

Solution:

x

2-9

x

4)-12(35

x

2-17

x

4)+8 (35

x

9-21

= 60 x (42 - 36) -12 (70 - 68) +8 (315 - 357) = 60

6 - 12

x

2+8

x

x

(-42)

= 360 - 24 - 336 = 360 - 360 = 0

Example 5:

Prove that w

w Solution:

By

CI

w2

= 0, where w is a cube root of unity,

2

w

~

CI + C2 + C3

1+W+W2 d= 1+w+w2 l+w +w2 W w2 = 0 w2 1 0

0 d=O

1

w

W w2 w2 1 1

w

x

17)

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Comprehensive Remedial Mathematics for Pharmacy

Example 6:

Solution:

Evaluate

a-b

b-c

c-a

b- c

c- a

a- b

c-a

a-b

b-c

Apply

000 ~=

b-c

c-a

a-b

c-a a-b

b-c

~=O

Example 7:

Solution:

"b+c

a

a

b

c+a

b

c

c

a+b

Prove that

Applying Rl

~

Rl - R2 - R3, we get

o ~=

=4abc

-2c

-2b

b c+a

b

c

c a+b

b I-2bIb c+al = 2c b c c Ic a+b

On expanding

= 2c[b(a + b) -bc] -2b [bc - c(c + a)] = 2c[ab + b2- bc] -2b [bc - c2 -ac] = 2abc + 2b2c - 2bc2 -2b2c + 2bc2 + 2abc =4abc

Example 8:

x+A. Show that x x

Solution:

Applying Cl

~

x x+A.

x x

x

x+A.

Cl + C2 + C3

3x+A. ~=

= A.2(3x + A.)

x

3x+A. x+A. 3x+A.

x

x x x+A.

Determinants Taking (3x + A) common from c], we get x

x

= (3x + A) 1 x + A

x

x

1

X+A

x· x

= (3x + A) 0 A 0

o

0 A

11~

~I

On expanding, we get d = (3x + A).

= (3x + A) A2 4

4

Example 9:

Solve the equation 1 -2

=0

x2

1 2x

Solution: 1

o

=>1

-6

o -3 =0 2 x -4

2x-4

on expanding, we get, -6(x2 - 4) +3(2x - 4) = 0 - 6x2 + 24 + 6x - 12 = 0 6x2 - 6x -12 = 0 6(x2 - x -2) = 0 x2-x-2 = 0 x2 -2x + x -2

=0

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

or

x+l=O

x=+2

or

x=-1

75

76

Comprehensive Remedial Mathematics for Pharmacy

1 bc a(b + c) Example 10: Prove the following results 1 ca b( c + a) = 0 1 ab c(a+b)

Solution:

Applying

C2

~ C2

+ C3, we get

1 bc + ab + ac a(b + c) ~ = 1 ca + bc + ba b( c + a) 1 ab+ca+cb c(a+ b) Taking (ab + bc + ca) common from C2, we get 1 1 a(b + c) = (ab + bc + ca) 1 1 b(c+a) 1 1 c(a + b) ( ... by property of identical column) = (ab + bc + ca) x 0 =0

Example 11: Solve the following system of equations x + 3y + 3z = 1,

Solution:

x + 4y + 3z = 0,

From the given equations 1 3 3 ~= 1 4 3 1 3 4 = 1(16 - 9) -3(4 - 3) +3 (3 - 4) =7-3-3=1 ~l

133 = 0 4 3 234 = 1(16 - 9) -3(0 - 6) +3(0 - 8) = 7 + 18 - 24= 1 1 1 3

~2

=

1 0 3 1 2 4

= 1(0 - 6) -1(4 - 3) +3(2 - 0) =-6-1+6=-1

x+ 3y +4z=2

Determinants 1 3 ~3

=

77

1

1 4 0

3 2 = 1(8-0)-3(2-0)+1(3-4) =8-6-1=1 using Crammer's rule ~l

1 1

x=-=~

y=

~ _2 ~

x=1

-1 =_

y=-1

1

~3 1 z=-=~ 1

z=1

Example 12: The total sale (s) in lakhs of rupees of a firm selling two drugs p and q is given by the relationship. s = ap + bq + c Data for the sales of first three months is given:

Months

Total sale

P

q

2

12

2

3

2

13

6

2

3

IS

S

3

U sing determinant method, determine the sales in next months when it sells 4 units of p and S units of q.

Solution:

Given the equation s = ap + bq + c For the one month, s = 12, p= 2, q = 3 ..

2a+3b+c=12

..... (1)

For the second month, s = 13, P = 6, q = 2 6a + 2b + c = 13

..... (2)

For the third month, s = IS, p = S, q = 3 Sa+3b+c=IS

..... (3)

78

Comprehensive Remedial Mathematics for Pharmacy From the given equations, ~=

2 3 6 2 1

5 3 = 2(2 - 3) -3(6 - 5) + 1(18 - 10) =-2-3 + 8 = 3 12 3 1 ~l

=

13 2 1 15 3 1

= 12(2 - 3) -3(13 - 15) + 1(39 - 30) =-12+6+9=3 2 12 1 ~2

= 6 13 1 5 15 1 = 2(13 - 15) -12(6 - 5) + 1(90 - 65)

= -4 -12 2 3 ~3 = 6 2 5 3

+ 25 = 9 12 13 15

= 2(30 - 39) -3(90 - 65) +12(18 - 10) =-18-75 +96=3 Using Cramer's rule

a=

~=l=1

b=

~=2=3

c=

~=l=1

~

~

~

3

3

3

Now, for the next month, if p = 4, q = 5

Determinants Then total sale, s = ap + bq +

79

C

=lx4+3x5+1 =4+15+1 S = 20 lakhs

Example 13: Show that the following system of equations is inconsistent 2x +y = 3, Solution:

4x+2y= 5.

From the given equations d= dl =

I! ~I

=

4 - 4 =0

I~ ~I = 6 - 5 = 1

Here d = 0 but dl *- 0 :. The given system is inconsistent Hence, it has no solutions

Example 14: Solve the system of equations x + 2y = 3 and 4x + 8y

12 using

determinants. Solution:

From given system of equations

d=

I~

!I

=8 - 8 =0

3 dl = 1 12

21 = 24 - 24 = 0 8

d2= 11

31=12-12=0

4 12

Here

d = d 1 = d2 = 0

Therefore, the system is consistent and has infinite number of solutions.

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Comprehensive Remedial Mathematics for Pharmacy

Example 15: Show that the points (8 -1), (4, 7) and (6,3) are collinear Xl YI I Solution:

1 we have, Area of ~ = - x 2 2

Y2

1

X3

Y3

1

8

-1

4

7

1

2 6

3

1

1

==

~

[8(7-3)+1 (4-6)+1(12-42)] 2 1 = - [8 x 4-2-30] 2

1

= - [32 - 32] = 0

2

.

Since, area of triangle is zero, hence the given points are collinear

Example based on the Pharmaceutical Applications of Determinants Example 1:

A mixture is to be made of milk, water and sugar. The three materials contains quantities P, Q, R as shown in table. How to form a mixture which will have 8 gms of P, 5 gms of Q and 7 gms of sugar.

Material

P

Q

R

Milk

1

3

4

Water

2

Sugar

5

2 1

Solve by using Crammer's rule.

Solution:

Let milk is x, water is Y and sugar is z in a mixture. Then we have the equation x + 3y + 4z= 8 2x+y + 2z= 5 5x+y+z= 7

Determinants By using Cramer's rule

134 ~=

2

1 2

5

1 1

= 1(1 =

2) -3(2 - 10) +4(2 - 5)

-1 + 24 -12 = 11 834

~l

== 5 1 2 711

= 8(1 - 2) -3(5 - 14) +4(5 -7) = -8

+ 27 - 8 = 11

184 ~2

= 2 5 2 571

= 1(5 - 14) -8(2 - 10) + 4(14 -25) = -9 + 64 - 44 = 11

1 3 8 ~3

= 2 1 5 5 1 7

= 1(7 - 5) -3(14 - 25) +8(2 - 5)

= 2 + 33 -

24 = 11

x=

~=.!..!.=1

y=

~ _2 ~

~

11

11 =-=1 11

~3 11 z= -=-=1 ~ 11

81

82

Comprehensive Remedial Mathematics for Pharmacy

Example 2:

The monthly expenditure in an Industry for 3 months is given below according to the staff employed ..

Month

Employees

Total monthly Salary

Clerks

Typists

Peons

April

4

2

3

4900

March

3

3

2

4500

Feb

4

3

4

5800

Calculate the salary for each type of staff by using determinant Solution: ,

Let the salary of clerks, typists and peons are a, b, c respectively. Then

4a + 2b + 3c = 4900 3a + 3b + 2c = 4500 4a + 3b + 4c = 5800

By using Cramer's rule 423 il= 3 3

2

434 = 4(12 - 6) -2(12 - 8) +3(9 - 12) = 7 4900 2 3 ill = 4500

3 2

5800 3 4 = 4900(12 - 6) -2(18000 - 11600) +3 (13500 - 17400) =4900 4

4900 3

il2 = 3

4500 2

4

5800 4

= 4(18000 - 11600) -4900(12 - 8)+3 (17400 - 18000) =4200

Determinants 4 ~3

= 3 4

2

4900

3 4500 3

5800

= 4(17400 -13500) -2(17400 - 18000) + 4900(9 - 12) = 2100 a= ~= 4900 =700 ~ 7 b= ~= 4200 =600 ~ 7 c = ~= 2100 =300 ~ 7 :. Salary of clerks, typists and peons are Rs 700, Rs 600, and Rs 300

Exercise 3.1 1

5

1

1. Evaluate 5 2 0 3 0 6

2. Evaluate each of the following determinants. (i)

x+l

x2

2

x-I

..

(n)

Icosecx cotx I (iii) cotx cosecx

31 37 92

3. Evaluate 31

58

71

31

55

74

22 32 4. Evaluate 22 3 2 4 2 3 2 4 2 52 12

I~

_wwl

83

84

Comprehensive Remedial Mathematics for Pharmacy w3 5.

w2

Show that w 3

w x

w

2

-6

= 3, w is the cube root of unity

w 1

6. If = 2x

- 3 3

o

2

= 0, find the value of x. seca

0

7. Show that tan a

-:-seca

tan a

= sec 2a

0

8. If 4

-3

2

-1

2

5 2

3

find the cofactor of elements 1, -3,2 and also find the value of determinant. 9. Find the ratio of the cofactor and minor of 4 in determinant

-8

0 -1

0

4

2

-1

0 ab 2

0

ac 2

10. Show that a 2 b 0 a 2c cb 2 bc

ll. Show that ca ab

12. Prove that

a 1 b 1 c

3 bc 2 = 2a b 3c"

0

,

a2

a3

b b2 = 1 b 2 c c2 c2

b3

a

a-

a

2

bc

b

2

ca =0

c

2

ab

c3

Detenninants b 2 +c 2

ac

ab

bc

13. Prove that

ac

bc

=

a2 + b2

1 a

bc

14. Without expanding, prove that 1 b

ca

1 c

ab

[Hint: multiply RJ, R2, R3 by a, b, c and x

3

7

15. Solve 2

x

2

7

6

x

4a2b2c2

11 a

a

2

b b2

=1

1 c divi~e

c2

by abc]

=0

[Hint: R\ ~ R\ + R2 + R3]

-6

x

-1

16. Show that x = 8 is a root of the equation 2 -3x x-3 and solve it. -3 2x x+2 17.

Solve the following for x. (i)

x+7

3x

3x

x+7

3x

3x

3x 3x =0 x+7

x (ii) 1 b 1 c

x3 b 3 =0 c3

18. Find the value ofx so that the points A(3, -2), B(x, 2) and C(8, 8) lie on a line.

19. Find k so that the points are collinear (3, -2) (k, 2) and (8, 8). 20.

If points (a, 0), (0, b) and (p, q) are collinear, prove that £. + 1 = 1 . a b

21. Solve the following system of linear equations using Cramer's rule: x + 2y + 3z = 6, 2x + 4y + Z = 7 and 3x + 2y + 9z = 14

85

86

Comprehensive Remedial Mathematics for Pharmacy

22. Solve the following system of equations using Cramer's rule: 5x -7y + z = 11, 6x - 8y - z = 15 and 3x + 2y - 6z = 7. 23. Does the following system of equations have a unique solution? x-y+z=-1 3x-y+ 2z= 1 2x + z= 3 24. Find the value of k for which the following system is consistent 2x - y + 3 = 0, kx - y + 1 = 0, 5x - y - 3 =

°

[Hint: solving equations (i) and (ii), we get x = 2, Y = 7 by Cramer's rule and substitute in equation (iii) we get k = 3] 25. (i) Can we add two determinants of order (2 x 3) and (3 x 2) (ii) what about the value of determinant when row and columns are interchanged. 26. By Cramer's rule solve the following system of equations. (i) 3x + y = 19, 3x - y = 23 (ii) x-2y=4,

-3x+5y=-7

27. By Cramer's rule solve the system of equations x + Y = 5, y + z = 3, z + x = 4 28. Solve the following equations using Cramer's rule 4 5 6 8 - - + - - = 3 and - - + - - = 5 x+2 y-l x+2 y-l [Hint Let _1_=a, _1_=b then4a+5b=3,6a+8b=5] x+2 y-l 29. Solve the equations using Cramer's rule 1 3 3 4 -+-=5 and ---=2 x y x y

Determinants a2

(b + C)2

a-J

b2 c-0

(c+a)2

a2

bc

, c- +ac

b2

ac

30. Prove that

= 2abc(a + b + C)3 b2 (a+b)2

c-J

31. Prove that a 2 +ab

b 2 + bc

ab

C

= 4a2b 2c2

J

b+c

c+a

a+b

a

b

c

32.' Prove that q+r

r+p

p+q =2p

q

r

y+z

z+x

x+y

x

y

z

p

33. If a

'#

p, b

'#

b c

q, C '# r, and a

q

c

a

b

r

=0

p q r evaluate - - + - - + - p-a q-b r-c [Hint: by Rl ~ Rl - R2 and R2 ~ R2 - R3 and expanding after divide by (p -q) (q - b) (r- c) we get _p_+_q_+_r_ = 2] p-a q-b r-c

34. If x, y and z are distinct and X

x2

~ = Y y2

1+x3 1 + y3 = 0, then prove that xyz =-1.

abc 35.

Prove that b

c

a

cab

= 3abc -

a 3 - b 3 - c3

87

Comprehensive Remedial Mathematics for Pharmacy

88

Answers 1.

24

2.

(i) -(1 +X2)

3.

0

(ii) 1

(iii) 1

4. -8 6.

0

8.

Cll

9.

1 :-1

= -12, Cl2 = -2,

C\3

= 23, L1 = 40

15.

x = 2, x = 7, x =-9

16.

5/3, 1,2

17.

(i)-1

18.

x=5

19.

5

21.

x=y=z=1

22.

x= l,y=-l,z=-1

23.

No

25.

(i) No

26.

(i) x = 7, Y = -2 (ii) x = -6, y =-5

27.

x = 3, y = 2, z = 1

28.

x=-4, y=2

29.

1 x="2,y=l

2. 2.

, 2' 2

(ii) 0, -1, -12

(ii) same or equal

CHAPTER

4 MATRICES

The theory of matrices came into existence in the nineteenth century as a result of the 'works of mathematician Arthen Cayley. To defme the matrix, we define the following terms. (i) Row:

A horizontal line is called a row. for example a, b, c,

)

(ii) Column: A vertical line is called a column.

Ia,

-l-~ (iii) Array:

A collection of numbers arranged in rows and columns forms an array.

Matrix A collection of mn numbers arranged in the form of an ordered set of m rows and n columns in a rectangular array is called an m x n matrix. Generally we denote the matrix by rectangular [ ] or square brackets ( ), but another way to represent it by IIII

Example: (i) Matrix of order m

A=

x

all a 2l

n. a l2 a 22

a l3 a 23

a ln a 2n

90

Comprehensive Remedial Mathematics for Pharmacy (ii)

Matrix of order 2 B

(iii)

=

x

3

[~ ~ ~l

Matrix of order 3 x 2

It is clear that order of matrix can be written by m number of rows and n is number of columns.

x

n, where m is

Types of Matrices 1.

Row Matrix: It contains one row Example:

2.

A = [1 23]1'3

B = [1, 2, 3, - - - n]lxn

Column Matrix: It contains one column

Example:

A=

r~]

l3

2 B= 3 3xl

n

3.

Zero Matrix or Null Matrix: A matrix whose all elements are zero is called zero matrix. The order of zero matrix can be made as we can. Example:

4.

nxl

A=

r~l lolxI

Square Matrix: A matrix is called square matrix if it has number of rows are equal to number of columns. Example:

A=

fl

L2

1

2 4

J 1

_x~

B= la, bl c

i

a2

3

b2

ab 3 ]

c2

c3

3x3

Matrices Generally we write square matrix of order n, at the place of n

x

91

n square matrix.

Example: square matrix of order 4 is

I~

A~ l~

2 3 4 3

4

5

6

7

8

0

1 2

Note: In every square matrix there are two diagonals. The diagonal which starts from the left hand top comer to end at the right hand bottom comer is called the principal diagonal or 'leading diagonal' i.e. A=

all a 2l

a l2 a 22

a3 a 23

a 3l

In short elements where i =j are formed principal diagonal of square matrix. 5.

Diagonal Matrix: If all the elements of a square matrix are zero except those in the principal diagonal is called a diagonal matrix.

Example

[~ ~J.[~ r ~j

are diagonal matrices. They can also denoted as diagonal (1,2), diagonal (2, 1,5) 6.

Scalar Matrix: A diagonal matrix whose diagonal elements are all equal, is called a scalar matrix.

Example:

7.

2 0 0OJ [o 0 2

A= 0 2

3x3

Unit Matrix: A diagonal matrix is called unit matrix if its principal diagonal elements are 1. It is denoted by In, where n is order of square matrix.

(a)

(b)

12 =

[o1 0] 1

2x2

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Comprehensive Remedial Mathematics for Pharmacy

(c)

[, =

l~

0

1 0

1 0 0 0

~L

(d)

14

1 0 0 0 0 1 0 0

=

0

0

0

1

4x4

Note: For unit matrix in general we can define A = [alJ] = 1 if i = j

o if 8.

i;t: j

Triangular Matrix: If in a square matrix all the elements are above or bellow the principal diagonal are zero, is called a triangular matrix. There are two types of triangular matrix; (b) Lower triangular

( a) upper triangular

(a) Upper triangular: A square matrix having all elements below principal diagonal are zero is called upper triangular matrix.

A=

all 0

a l2

a 13

a 22

a 23

aln a 2n

0

0

a 33

a 3n

0

0

0

ann

it is clear that A = [alJ] = 0 for i > j then A is upper triangular (b) Lower triangular: A square matrix having all the elements above the principal diagonal is zero is called lower triangular matrix.

A=

all

0

0

0

a 2l

a 22

0

0

a 3l

a 32

a 33

0

ann an! a n2 A = [alJ] = 0 for i < j is lower triangular Example of upper and lower triangular matrices. I 2 3 4 1 0 A=

0

3 4

5

0

0

2

3

0

0

0

5

Upper triangular

B=

0

0

2

3 0

0

4

5 6

0

7

8 9 10

Lower triangular

Matrices

9.

93

Equal Matrix: Two matrices A and B are said to be equal matrix if (i)

Their order are exactly same.

(ii) Their corresponding elements are equal.

Example:

A

=

[~

!]

B=[:

:]

A=Bif a=l, b=2, c=3,d=4

10.

Singular Matrix: A square matrix is said to be singular matrix if its corresponding determinant is zero, that is IAI = 0 Example:

[~

!]

(i)

A=

(ii)

A~ r~ :J 2

5

8

11.

Non-singular or Regular or Investigable Matrix: A square matrix is said to be non-singular if its corresponding determinant is not zero i.e. IAI::j:. 0

Example:

A

=

[~

!]

B~r; : :J

Addition and Subtraction of Matrices Two matrices A and B can be added or subtracted if their order are same. If order of matrix A is m x n, order of matrix B is m x n, then order of A ± B is also m x n.

Example:

94

Comprehensive Remedial Mathematics for Pharmacy

Properties of Addition of Matrices If, three matrices A, B, C are given of same order then (i)

A + B = B + A (commutative law)

(ii)

A + (B + C) = (A + B) + C (Associative law)

(iii)

Cancellation law

(iv)

(a) A + B = A + C

~

B = C (left cancellation)

(b) B + A = C + A

~

B = C (Right cancellation)

Additive inverse: For every matrix A, there exists a matrix - A such that A+(-A)=O where A = [a'J]mxn, and -A = [-a'J]mxn then -A is called additive inverse of A.

(v)

Distributive law: IfK is any number then K (A + B) = KA + KB

Multiplication of Two Matrices Two matrices A and B can be multiplied if number of columns in A are equal to number ofrows in B. i.e. [A]myn [B]n

X2 .... Xn respectively. Then the weighted arithmetic mean are W.A.M.= wlxl+w2x2+·····+wnxo =1:wx WI +W2 + ..... +wo 1:w

Objective and Prerequisite of an Ideal Measure

157

Example 7:

The arithmetic mean of 3 sets are 25, 10 and 15 whose corresponding number of observations are 200, 250 and 300. Find the combined mean.

Solution:

Let

XI

= 25,

fl

= 200, f2 = 250, f3 = 300

x2

=10,

X3

=15

· d ant .h · mean = fix i +f2x1- +f3 x 3 metIc T hen com b me fl + f2 + f3 200x 25 + 250x 10+ 300x 15 =

200+ 250+300 5000 + 2500 + 4500

=-------

750

= 12000 =16 750

Example 8:

There are 45 students in a class, of which 15 are girls. The average weight of 15 girls is 45 kg, 30 boys is 52 kg. Find the mean weight in kg, of the entire class.

Solution:

Here

= 15, X2 = 30 WI = 45, W2 = 52

XI

. Mean weight

= w IXI +w 2X2 XI +X2 15 x 45 +30x 52 = 15 +30 = 2235 = 49.67 kg

45

Example 9:

In an industry there are 10 officers, 60 clerks and 400 workers. Their average salaries are Rs. 627, Rs. 230 and Rs.120 respectively. Find the salary of all employees of the industry taken together.

Solution:

Here

= 10, W2 = 60, W3 = 400 XI = 627, X2 = 230, X3 = 120

WI -

W

WIX I +W 2X2 +W3 X3 = ----'---.:.....--=--=--=---=--

=

WI +W 2 +W 3 10x627+60x230+400xI20 10+60+400

= 6270 + 13800 + 48000 = 68070 =144.8 470

470

Rs.

158

Comprehensive Remedial Mathematics for Pharmacy

Example 10: Find the average marks of students from the following table. Marks

No. of students

Marks

No. of students

Above 0

80

Above 60

28

Above 10

77

Above 70

16

Above 20

72

Above 80

10

Above 30

65

Above 90

8

Above 40

55

Above 100

0

Above 50

43

Solution: The given data is in the form of more than type cumulative frequency distribution. Let us arrange the marks in groups which may be 0-10, 10-20 .... etc. Now, number of students getting marks above 0 is 80 and above 10 is 77. Hence, number of students getting marks between 0 - 10 is 80 - 77 = 3 and so on. Marks

Mid-value x

f

x-a u=--

x-a

h

fu

0-10

5

80-77 = 3

-50

-5

-15

10-20

15

77-72=5

--40

--4

-20

20 -30

25

72-65 =7

-30

-3

-21

30-40

35

65 - 55 = 10

-20

-2

-20

40-50

45

55-43=12

-10

-1

-12

50-60

55

43 -28 = 15

0

0

0

60-70

65

28 - 16 = 12

10

1

12

70-80

75

16-10=6

20

2

12

80-90

85

10-8=2

30

3

6

90-100

95

8-0=8

40

4

32

Total

N=80

Lfu

Mean=a+h. Here

N

a = 55, h = 10 Mean = 55+ 10x(-26) 80 = 55 -3.25

= 51.75

Eru =-26

Objective and Prerequisite of an Ideal Measure

159

Merits, Demerits and uses of Arithmetic Mean Merits 1. 2. 3. 4. 5. 6. 7.

The arithmetic mean is rigidly defined. It is based on all the observations made. It is easily calculated from the given data. It is determinate i.e., it is not indefinite. It is least affected by fluctuations of sampling. It lends itselfto algebraic treatment. Its general nature is readily comprehensive.

Demerits I. The arithmetic average may not be represented in the actual data. 2. It gives greater importance to bigger items of a given series. 3. It can hardly be located by inspection, mode and median can be. 4. It can ignore any single item only at the risk of losing its accuracy. 5. It may sometimes give fallacious conclusions. 6. It cannot be calculated if the extreme class is open e.g. below 5 or above 70. Uses 1. Arithmetic mean is preferred when the frequency distribution symmetrical. 2. The use of arithmetic mean is recommended when we want a measure of central tendency that have greatest stability. ~ 3. When other constants such as standard deviation, coefficient of correlation are to be computed later on, we use arithmetic mean. 4. It used in case of average output, average imports or exports, average cost of production etc. arithmetic mean is the suitable average.

Geometric Mean (G.M.): Ifx" X2 ...... Xn be the n values of the variate X, none of them being zero, then the geometric mean, G is defined by I

G = (XI , x 2 .... ··x n )-;;log G

•

1

= - (log XI + log X2 + ..... + log xn) n

1 n log G =- Llogx 1 n

G

1=1

= Antilog r..!.. IIOgX Ln

1=1

I ]

160

Comprehensive Remedial Mathematics for Pharmacy

If X\, X2 ..... Xn occurs f\, f2 ........ fn times respectively and N is the total frequency then G = Antilog

[~N t

fl log XI]

1=1

Example 11:

Find geometric mean of 4,8, 16 I

Solution:

G = (4x 8x 16)3 I

= (22 X 2 3 X 24 )3 I

= (29)3 =23 = 8

Example 12:

Find geometric mean for the following data 85,70,15,75,500,8,45,250,40,36.

Solution: log x

x

1.9294

85 70

1.8451

15

1.1761

75

1.8751

500

2.6990

8

0.9031

45

1.6532

250

2.3979

40

1.6021 1.5563

36

:Elogx = 17.6373 I

Geometric mean G = (XI x 2....xn)~ 1 log G = -Llogx n 17.6373 Iog G = 10 G

= Antilog (1.76373)

= 58.03

Objective and Prerequisite of an Ideal Measure

Example 13:

161

Find the geometric mean of the following distribution: Marks

0-10

10-20

20- 30

30-40

5

8

3

4

No. of students

Solution: Class

Mid value x

f

log x

0-10

5

5

0.6990

flog x 3.4950

10-20

15

8

1.1761

9.4088

20-30

25

3

1.3979

4.1937

30-40

35

4

1.5541

6.2164

N=20

Total

kflogx = 23.3139

Geometric mean log G = ~Lflogx N 23.3139

=

20

log G = 1.1657 G = Anti log (1.1657)

= 14.64

Merits, Demerits and uses of Geometric Mean Merits 1. 2. 3. 4.

The geometric mean is based on all the observations of a series. It is not much affected by fluctuations of sampling. It lends itself to algebraical treatment. It is the most appropriate average when dealing with ratio.

Demerits 1. 2. 3. 4.

It is not simple to understand. It is comparatively difficult to calculate. It does not give equal weight to every item. It can not be calculated if the number of negative value is odd.

Uses 1. Geometric mean is used to find the relative changes, such as the average rate of population growth, average rate of depreciation of machines etc. 2. It is used in the construction of index numbers.

162

Comprehensive Remedial Mathematics for Pharmacy

Harmonic Mean (H.M) : The harmonic mean is the reciprocal of the arithmetic mean of their reciprocals. (i)

For Individual series: Let XI, X2 .... Xn be the n values of variate x. The Harmonic mean denoted by H is given by

n

n

H= 1 1 1 --1 - + - + .... + - LXI x 2 xn X (ii)

For Discrete Series: If XI, X2 ..... Xn have the frequencies flo f2 ..... fn respectively, then Harmonic mean is given by N H=-y, whereN=H

LX

(iii) For Grouped Series: The mid-value are taken as XI, X2 .... Xn with the corresponding frequencies flo f2 .... fn. Then harmonic mean is given by

H=~

Li X

Example 14:

Calculate the harmonic mean of the following: 4,8, 16

Solution:

H.M.

n =-----

1 1 1 -+-+XI x 2 X3

=

3 = 3 =3xI6=48=6.855 1 1 1 4+2+1 7 7 -+-+-

4

Example 15:

8

16

16

Find the harmonic mean of the following frequency distribution: Class Frequency

2-4 20

4-6 40

6-8 30

8-10 10

Objective and Prerequisite of an Ideal Measure

163

Solution: Class

x

f x

1

f

-

-

X

2-4

3

20

0.3333

6.666

4-6

5

40

0.2000

8.000

6-8

7

30

0.1428

4.284

8 -10

9

10

0.1111

1.111

Total

N= 100

L.!:. = 20.061 X

Harmonic mean H =

N f

L-X

=

100 20.061

= 4.985

Merits, Demerits and uses of Harmonic Mean Merits 1. It is based on all the value of observations. 2. It is capable of further algebraic treatment. 3. It is the most appropriate average when calculating the average speed of train, the speed is expressed in kilometers per hour etc. Demerits 1. It is difficult to calculate. 2. It is difficult to understand. 3. It gives a very high weightage to small values. Uses 1. When the frequency distribution highly positively skewed. 2. When we have to find average rates and ratios under certain condition e.g. average price, average speed etc.

Median When the observations are arranged in ascending or descending order of magnitude, then the middle value is called median of these observations. In other words, median is that value of the variable which divides a series into two equal parts so that orie-half or more of the items are equal to or less than it. It is denoted by Md.

164

Comprehensive Remedial Mathematics for Pharmacy

Computation of Median 1. Median in Individual Series: Let Xl,

X2 .••. Xn

be the n values of a variate

X

written

in ascending order of magnitude. Then median is given by Median = Value of middle term (a) When n is odd, then median

= n + I th value. 2

(b) When n is even, then median

= ~ {;

th value + (; + 1}h value}

Example 16:

Find the median of the following data 15,35, 18, 19,26,25,20,29,27.

Solution:

Arranging the data in ascending order, we get 15,18,19,20,25,26,27,29,35 Here n = 9, which is an odd number n+ · =me d Ian -1th va Iue 2 9+ 1 = - - th value 2 th = 5 value

Example 17: According to the census of 1981, following are the population figures in thousand, of 10 cities: 2000, 1180, 1785, 1500, 560, 782, 1200,385, 1123,222 Find the median.

Solution:

Arranging the data in ascending order, we get 222,385,560,782, 1123,1180,1200,1500, 1785,2000 Here n = 10, the even number ..

Median =

~[~ th value + (; + 1}h value]

~~[ I~ th VaIUe+e~ +I)" value]

Objective and Prerequisite of an Ideal Measure

165

1

= -. [5 th value + 6th value] 2

=!

[1123 + 1180]

2

2303

.

= - - = 1151.5 thousand

2

2. Median in Discrete Series: Let x\, X2 ...... Xn be the observations in ascending order with corresponding frequencies f\, f2 .... fn respectively. We first find cumulative frequencies. Median is given by the value of n + 1 th term.

2

Example 18: The marks obtained by students of a class are following: Marks

25

24

23

22

21

20

No. of Students

3

5

4

7

3

1- -

Find the median

Solution: Marks

f

Cumulative frequency (c)

25

3

3

24

5

8

23

4

12

22

7

19

21

3

22

20

1

23

Lf=23 ml'd

vaIue =n-+-1 2 23+1

--2

= 12

166

Comprehensive Remedial Mathematics for Pharmacy Median == The value of item against cumulative frequency 12 = 23 Md =23

3. Median in Grouped Series: Median

=

value of N th item, where N is total 2 N frequency. The lowest class for which the cumulative frequency exceeds is 2 called the median class.

Where I = lowest limit of median class f = The frequency of the median class i = width of the median class F = cumulative frequency preceeding the median class N = Total frequency

Example 19:

Find the median for the following distribution:

Wages in Rs

0- \0

10-20

20-30

30-40

40- 50

22

38

46

35

20

No. of workers

Solution: Wages in Rs.

No. of workers f

0- \0

22

Cumulative Frequency (F)

22

10-20

38

60

20- 30

46

\06

30-40

35

141

40-50

20

161

N = 161

Here

N = 161 Median = size of (N; .

1)

item

th th

161+1. . Item = 81 th Items

= size of ( --2- )

Objective and Prerequisite of an Ideal Measure

167

The 81 th item lies in 20 - 30 group. Hence 20 - 30 is the middle class. Then 1= 20, N = 161, F = 60, f = 46, i = 10 N --F Md=/+ _2_-xi f

~-60 =20+-=2_- x10 46 20+ 205 48 = 20 + 4.46 = 24.46

=

Merits, Demerits and uses of Median Merits 1. It is easily understood. 2.

It is very readily calculated and can exactly be located.

3. It is not affected by abnormally large or small values of the variable. 4. It is rigidly defined. 5. It can be determined by mere inspection in a frequency graph. 6. It can be calculated for distribution with open end classes.

Demerits 1. The median does not lend itself to algebraic treatment. 2. It may not be represented by the actual data. 3. In a discrete series it may become indefinite. 4. In case of even number of observations it can not be determined exactly. 5. It does not depend on all the items. Uses Median is suitable in the following situations: 1. When there is an open class at one or both ends of the frequency distribution, provided that the median does not fall in one of those open classes. 2.

When the characteristics like intelligence, honesty etc., cannot be measured numerically but the individual can be ranked in order.

3. When there is suspected heterogeneity. 4.

When the data is concerned with wealth etc.

168

Comprehensive Remedial Mathematics for Pharmacy

Mode

The mode is defined to be the size at the variable which occurs most frequently or the point of maximum frequency. In other words, the mode of the distribution is that value of the variable for which frequency is maximum. It is represented by Mo. Method for computing Mode (i)

For Individual series: Mode is found out by observations of the value which occurs mo.st.

Example 20:

Find the mode of the following data. 0, 1,6, 7,2,3, 7, 6, 6,2,6,0,5,6,

Solution: (ii)

°

Since 6 occurs 5 times and no other item occurs 5 or more than 5 times, hence mode is 6.

For Discrete series: Value with the highest frequency is called mode.

Example 21:

Calculate the mode from the following distribution Marks No. of students

Solution:

Here, highest frequency is 40 and therefore mode is 60.

(iii) For grouped series: If there is a single class with maximum frequency, we call this class as mode class and within this class mode is obtained by the formula f-C 1 Xl• M 0-/ - + 2f -C1 -fl Where f = frequency of the mode-class I = lower limit of the mode-class

LI = frequency of pre mode-class fl = frequency of post mode-class i = width of the class interval

Example 22:

Find the mode of the following distribution:

ubjective and Prerequisite of an Ideal Measure

Solution:

169

Here, the greatest frequency 72 lies in the class 21 - 28. Hence, this is the mode-class. Then 1= 21, f= 72, L[ = 36, f[ = 51, i = 7. mode

Mo = I +

f - C[ 2f -L[ -f[

=21+

X

i

72-36 144-36-51

x7

= 21 + 84 19 =

21 + 4.4

= 25.4

Merits, Demerits and Uses of Mode Merits 1. It is easily understood. 2. It can be determined with considerable accuracy from the well selected sample data. 3. It can be very easily determined from the graph. 4. It can be easily located by mere inspection in certain cases. 5. The extreme items have no effect provided they are not in the mode-class. Demerits 1. It is ill-defined. 2. A clearly defined mode does not always exist. 3. It is not based on all the observations of a series. 4. It is not capable of further mathematical treatment. Uses The use of mode is recommended:1. When a quick and approximate measure of central tendency is desired. 2. When frequency distribution is skew. 3. When a typical average size is required in case of readymade garments, shoes etc. Relationship between Mean, Median and Mode 1. In case of a symmetrical distribution, the mean, median and mode coincide with each other. 2. In case of moderately asymmetrical distribution, we use the formula (an empirical relationship) (ii) Mode = 3 Median - 2 Mean (i) Mean - mode = 3 (Mean - Median)

170

Comprehensive Remedial Mathematics for Pharmacy

Note

(i)

When the frequencies are higher for the lower classes and smaller for the upper classes. Then relationship is Mean> Median> Mode

(ii)

When the frequencies are smaller for the lower classes and higher for the upper classes then relationship is Mean < Median < Mode

Mean

Median

Mode

Example 23:

For a moderately asymmetrical distribution, the median and mean are 30.2 and 32.4 respectively. Calculate mode.

Solution:

Using empirical formula, Mode = 3 Median - 2 Mean

Example 24:

=

3 x 30.2 - 2 x 32.4

=

90.6-64.8

=

25.8

Find out the Mean, Median and Mode from the following:

I Rs. Less than I No.ofReceivers

10 15

20 35

30 60

40 84

50 96

60 127

70 198

80 250

Objective and Prerequisite of an Ideal Measure

Solution: Class

0-10 10-20 20- 30 30-40 40- 50 50- 60 60-70 70 -80

Mid-value x

f

5 15

15 20

25 35 45

25 24

55 65 75

Total

Cumulative f

12 31 71 52 N=250

15 35 60 84 96 127 198 250

d=x-a

x-a u=-h

--40

--4

-60

-30 -20 -10 0 10 20 30

-3 -2

-{ill

-1 0 1 2 3

Let assumed mean a = 45, h = 10 Lfu Mean =a+ - x h N = 45 + 135 x 10 250 = 45 + 5.4 = 50 .4 N

--c

Median Here N

= I + _2__ x h f

= 250 =125, hence median class is 50 -

2 2 Thus 1= 50, f= 31, c = 96 Median

= 50+ 125-96 x10 31 = 50+ 290

31 = 50 + 9.35 = 59.35 Now, by empirical fonnula, Mode

= 3 Median - 2 Mean = 3 x 59.35 - 2 x 50.4 = 178.05 - 100.8

= 77.25

fu

60.

-50 -24 0 31 142 156 Lfu = 135

171

172

Comprehensive Remedial Mathematics for Pharmacy

Example 25:

Find the mean wages of labourers in a pharmaceutical factory for the following distribution. No.of Labourers

Wages in (Rs) 20-30

3

30-40

5

40-50

20

50-60

10

60-70

5

Solution: Wages in Rs

Number of Labourers

Mid point x

fx

20-30

3

25

75

30-40

5

35

175

40-50

20

45

900

50-60

10

55

550

60-70

5

65

325

Total

N=43

Lfx = 2025

~fx

Mean x = N

= 2025 = Rs. 47.09 43

Example 26:

Find out the arithmetic mean for the following Table: Class

Frequency

0-10

8

10-20

16

20-30

25

30-40

30

40-50

15

50-60

6

Objective and Prerequisite of an Ideal Measure Solution:

Class

Mid value (x)

f

d = x -25

x - 25 u=--

0-10

5

8

-20

-2

-16

10 -20

15

16

-10

-I

-16

20 - 30

25 - a

25

0

0

0

30-40

35

30

10

I

30

40-50

45

15

20

2

30

50- 60

55

6

30

3

18

Here

fu

Hu=46

N= 100

Total

i

a = 25, i = 10 Lfu mean x = a + i N 25 + 10 x 46 100 == 25 + 4.6

=

=

Example 27:

29.6

Find the median from the following data of manufacturing company. Mid value

Frequency

115

5

125

26

135

48

145

70

155

118

165

60

175

38

185

22

195

3

173

174

Comprehensive Remedial Mathematics for Pharmacy

Solution:

The cumulative frequency Table is Mid value

Frequency

Cumulative Frequency

115 125 135 145 155 165 175 185 195

5 26 48 70 118 60 38 22 3 N=390

5 31 79 149 267 327 365 387 390

Mid value = N + 1 = 390+ 1 = 391 =195.5 2 2 2 Which comes against the cumulative frequency 267 Median (Md) = 155

Example 28:

50 persons were examined through X-rays and observations are noted as under. Diameter of heart (in mm)

Number of persons

120 121 122 123 124

5 9 14 8 5 9

125

Find the median

Solution:

Making the cumulative frequency Table Diameter of heart (in mm)

Number of persons

Cumulative frequency

120 121 122 123 124

5

5 14

125

9 14 8 5 9 N=50

28

36 41 50

Objective and Prerequisite of an Ideal Measure . Here mid value ,

N +1

50 + 1

175

51

= -2- = -2- = -2 = 25.5

Which comes against the cumulative frequency 28 Median = 122

Example 29:

Solution:

Calculate mean, median and mode of the following data relating to weight of 120 articles. Weight (tn kg)

Number of articles

0-10

14

10-20

17

20-30

22

30-40

26

40-50

23

50-60

18

Mean Weight in kg

Number of articles (1)

Mid point x

fx

0-10

14

5

70

10-20

17

15

255

20-30

22

25

550

30-40

26

35

910

40-50

23

45

1035

18

55

50-60

990

N = 120

Mean x

2:fx N

Hx=3810

3810

= -=-=

120

31.75

Median Weight (in kg) 0-10

Number of articles (1)

Cumulative frequency

14

14

10-20

17

31

20-30

22

53

30-40

26

79

40-50

23

102

50-60

18

120

N= 120

176

Comprehensive Remedial Mathematics for Pharmacy . N 120 MId value = -=-=60

2

2

Which comes against cumulative frequency 79. The median class will be 30 - 40 Here! = 30, i = 10, f= 26, F = 53

!'£ - F Median (Md) = 1+ _2__ x i f -'"'0 - -' + 60-53 x 10 26 = 30 + 70 26 = 30 + 2.69 = 32.69 Mode In the given table the interval 30 - 40 has the maximum frequency :. The mode class is 30 - 40 Here 1= 30, i = 10, f= 26, Ll = 22, fl = 23 f-f Mode (Mo) = I + -1 X i 2f -C 1 -fl 26-22 =30+ x 10 52-22-23 =30+ 40 7 = 30 + 5.71 = 35.71

Example 30:

Find the mean, median and mode of the following data. Students (x)

1 4 7 10 13

16 19 22 25 28

Frequency (0 7

46 165 195 189 89 28 19 9 3

Objective and Prerequisite of an Ideal Measure

Solution: Mean ~ = Lfx = 8436 750 N = 1l.24 Students (x)

1 4 7 10 13 16 19 22 25 28

Fn~guency

fx

Cumulative frequen~

7 184 1155 1950 2457 1424 532 418 225 84 Hx =8436

7 53 218 413 602 691 719 738 747 750

(I)

7 46 165 195 189 89 28 19 9 3 N=750

Median . N 750 MId value = -=-=375

2

2

Which comes against the cumulative frequency 413 :. Median = 10 Mode Mode

Frequency

Students (x)

First two

Leave first

First three

First leave

First two leave

(iv)

(v)

(vi)

(vii)

211

218

(i)

(ii)

(iii)

1 4 7

7 46 165 195 189 89 28 19

53

10

13 16 19 22 25 28

9 3

406

360 384 278

549 473

117 47

56 28

12

306 136 31

177

178

Comprehensive Remedial Mathematics for Pharmacy

Analysis Table

(i)

10

(ii)

7, 10

(iii)

10, l3

(iv)

10, l3, 16

(v)

4, 7, 10

(vi)

7, 10, l3

(vii) 7, 10, l3, 16 It is clear that the class 10 occur maximum number of times. Hence mode is 10. Example 31:

Calculate Mean, Median and Mode of the following distribution Marks

Frequency

10-25

6

25-40

20

40-55

44

55-70

26

70-85

3

85-100

1

Solution: Marks

Frequency (I)

Mid point (x)

fx

Cumulative frequency

10-25

6

17.5

105

6

25-40

20

32.5

650

26

40-55

44

47.5

2090

70

55-70

26

62.5

1625

96

70-85

3

77.5

232.5

99

1

92.5

92.5

100

85-100

N= 100

Lfx = 4795

Mean

~ = rfx = 4795 = 47.95 N

100

Median Mid value

= 100 = 50 2

Objective and Prerequisite of an Ideal Measure Which comes against cumulative frequency 70 :. The median class is 40 - 55

1= 40, i = 15, F = 26, f= 44

Here

N -F Median (Md) = 1+ _2__ x i f

=40+ 50-26 x15 44 =40+8.18=48.18 Mode In the Table, the interval 40 - 55 has maximum frequency.

I = 40, f = 44, LI = 20, fl = 26, i = 15

Here

Mode = I +

f-f

-I

2f -C1 -fl

=40+

X

i

44-20. x15 88-20-26

= 40 + 8.57 = 48.57 Example 32:

Calculate-the mode of the following distribution Class

Frequency

4-8

10

8-12

12

12-16

16

16-20

14

20-24

10

24-28

8

28-32

17

32-36

5

36-40

4

Solution: The interval 28 - 32 has the maximum frequency 17. Here

1= 28, i = 4, f= 17,f-1 = 8, fl = 5

179

180

Comprehensive Remedial Mathematics for Pharmacy Mode = 1+

f - LI 2f -C1 -fl

= 28+

X

i

17 -8 x4 34-8-5

36 =28+·21

= 28 + 1.71 = 29.71 Example 33:

Solution:

Find the mode of the following data Class

FreQuencv

0-5

25

5 -10

15

10 - 15

8

15 -20

3

20-25

2

maximum frequency 25 in class interval 0-5. Here

1= 0, f = 25, LI = 0, fl = 15, i = 5 Mode =

f - LI 2f-L1 -fl

=0+

X1

25-0 x5 50-0-15

= 25x5 = 3.57 35

Example 34: Find the mean of the following distribution Marks obtained

Number of Students

0-10

7

10-20

9

20-30

12

30-40

8

40-50

4

Objective and Prerequisite of an Ideal Measure Solution: Marks obtained

Mid value (x)

Number of students

fx

(I)

0-10 10-20 20-30 30-40 40-50

5 15 25 35 45

7 9 12 8 4 N=40

35 -135 300 280 180 Lfx = 930

Mean = Lfx = 930 = 2325 N 40 Example 35:

Find the arithmetic mean for the following table: Class 0-20 20-40 40-60 60-80 80 -100

Frequency 2 7 10 3 3

Solution: Class 0-20 20-40 40-60 60-80 80-100

Mid value (x) 10 30 50 70 90

-

Lfx

Frequency (I) 2 7 10 3 3 N=25

fx 20 210 500 210 270 Lfx = 1210

1210

Mean X = - = - - =48.4 N 25 Example 36:

Calculate the median from the table given below. Value 7 8 10 9 11 12 13

Frequency 2 1 4 5 6 1 3

181

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Comprehensive Remedial Mathematics for Pharmacy

Solution: Value

Frequency

Cumulative frequency

7

2

2

8

1

3

10

4

7

9

5

12

11

6

18

12

1

19

3

22

13

N=22

Mid value = N = 22 =11 2 2 which comes against cumulative frequency 12 Median =9

Example 37:

Calculate the median for the following data. Wages (Rs.)

Number of workers

10-15

4

15-20

6

20-25

8

25-30

5

30-35

3

35-40

2

Solution: Wages (Rs)

Number of Workers

Cumulative frequency

10-15

4

4 10

15-20

6

20-25

8

18

25-30

5

23

30-35

3

26

2

28

35-40

N=28

Objective and Prerequisite of an Ideal Measure Mid Value = N=28=14 . 2 2 Which comes against cumulative frequency 18 :. median class is 20 - 25 Here

i=20,i=5,f=8,F=10 N -F Median = i + _2__ x i f

=20+ 14-10 x5 8 20 =20+- = 20 + 2.5 = 22.5 8

Example 38:

Solution:

Find the mode from the following Table. Class interval

Frequency

0-15

2

15-30

3

30-45

5

45-60

10

60-75

7

The interval 45 - 60 has maximum frequency 10 Here

i=45,i=15,f=10,f_I=5,fl=7 Mode = i +

f-f -I 2f -C1 -fl

=45+

X

i

10-5 x15 20-5 -7

= 45 + 75 8 =45+9.37 = 54.37

183

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Comprehensive Remedial Mathematics for Pharmacy

Exercises 1. Marks obtained by 9 students in Maths are given below. 52,

75,

40,

70,

43,

40,

65,

35,

48

Calculate the arithmetic mean. 2. Calculate the mean of the following. Height in em

65

66

67

68

69

70

71

72

73

Number of Plants

1

4

5

7

11

10

6

4

2

3. Find the mean for the following distribution. Class

0-7

7 -14

14 - 21

21-28

28-35

35-42

42-49

Frequency

19

25

36

72

51

43

28

4. Find the mean from the following data. Marks

No.of students

Marks

No. ofstudents

Below 10

5

Below 60

60

Below 20

9

Below 70

70

Below 30

17

Below 80

78

Below 40

29

Below 90

83

Below 50

45

Below 100

85

5. If the arithmetic mean of the following frequency distribution is 39.25. Find the missing term : Daily wages (Rs)

25

'30

35

50

60

75

Number of Labourers

10

?

13

8

. 5

4

6. The mean wage of 500 workers in a factory running two shifts of 360 and 140 workers respectively is Rs. 70. The mean wage of 360 workers working in day shift is Rs. 75. Find the mean wage of 140 workers working in the night shift.

Objective and Prerequisite of an Ideal Measure 7. Calculate the mean from the following data: Class Intervals

Frequency

Class intervals

Frequency

35-40

7

60-65

42

40-45

8

65-70

42

45-50

12

70-75

15

50-55

26

75-80

17

55-60

32

80-85

9

8. Find the median of the following 20,

18,

22,

25,

27,

15

12,

9. Find the median from the following Table: Marks

No. of Students

Marks

No. of Students

0-10

2

50-60

20

10-20

18

60-70

6

20-30

30

70-80

3

30-40

45

40-50

35

10. Find the median for the following frequency distribution. Class Frequency

0-6

6-12

12-18

18-22

22-24

24-30

30-36

36-42

5

11

25

20

15

18

12

6

11. Determine the mode from the following figures. 25,

15,

23,

40,

27,

25,

25,

23,

12. Compute the mode of the following:

13. Find the mode from the following Table: Marks

No. of Students

Marks

No. of Students

0-10

2

50-60

20

10-20

18

60-70

6

20-30

30

70-80

3

30-40

45

40-50

25

20

185

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Comprehensive Remedial Mathematics for Pharmacy

14. Find the mode of the following. Marks

1-5

6-10

11-15

16-20

21-25

26-30

31-35

36-40

41-45

No. of Candidates

7

10

16

32

24

18

10

5

1

15. For the following frequency distribution the mode is 47.5. Find the missing frequency. Class

0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

Frequency

7

8

10

-

40

35

10

7

16. Find out the mean, median and mode of the following distribution. Marks

Frequency

10-25

6

25-40

20

40-45

44

55 -70

26

70- 85

3

85 -100

1

17. Compute mean, median and mode for the following data. Marks less than

5

10

15

20

25

30

35

40

Frequency

2

4

14

27

48

64

72

75

18. Calculate mean, median and mode for the data given below. Age less than

25

30

35

40

45

50

55

601

No. of Persons

5

12

22

40

55

67

74

79 1

19. Find the Geometric mean of6, 12,24. 20. The marks obtained by seven students are 5, 10, 15,20,25,30 and 35. Find the Geometric mean and harmonic mean. 21. Calculate Geometric mean and harmonic mean for the data given below.

I ; I 2~

7 11 5

11~

0 118

I~

11:

I

Objective and Prerequisite of an Ideal Measure

187

22. Find geometric mean for the following frequency distribution. Class

Frequency

0-10

10-20

20-30

30-40

40-50

1

2

6

6

5

23. Find the harmonic mean for the following frequency distribution. Class

Frequency

40-50

50-60

60-70

70-80

80-90

90-100

19

25

36

72

51

43

24. If the mean of the following distribution is 6. Find the value of A.

25. The mean of 15 observations is 10. If2 is subtracted from each observation, Find the new mean. . - :Ex l 120 [Hmt: :Ex = 15 x 10 = 150, New sum :Ex l = 150 - 15 x 2 = 120 :. x = == 8] n 15 26. If the mean of6, 8, 5, 7, x and 4 is 7. Find x. 27. The mean of a set of 67 values is 35. If each of the value is multiplied by 4. What will be mean of the set of new values. 28. The mean of 40 observations was 160. It was detected on rechecking that the value of 165 was wrongly copied as 125 for computation of mean. Find the correct mean. 29. The arithmetic mean of a set of 40 values is 65. If each of the 40 values is increased by 5. What will be the mean of set of new values. 30. The sum of deviation of a set of n values xl. X2 •••.• Xn measured from 50 is -10 and the sum of deviations of the values from 46 is 70. Find the value ofn and the mean. 10 70 [Hint: x=50- and x =46+n n 10 70 :. 50-- = 46+- => n = 20 and x

n

n

= 49.5]

31. A car owner buys petrol at Rs. 7.50, Rs. 8.00 and Rs. 8.50 per litre for the three successive years. Compute the average cost per litre of petrol when he spends Rs. 4000 each year. 32. Number 50, 42, 41, 2x + 10, 2x -8, 12, 11, 8,6 and if their median is 25. Find x.

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Comprehensive Remedial Mathematics for Pharmacy

33. Find the median of the following data. 41,43,127,99,71,92,71,58,57 If 58 is replaced by 85. What will be the new median. 34. Out of 10 observations arranged in descending order the fifth and sixth observations are 13 and 11 respectively. What is the median value of all the ten observation? 35. Explain mode with example. Explain where mode is better suited in comparision to other measures of central values.

Answers

57~

1. 52

2. 69.19

3. 26.5

4. 48.41

5. 6

6.

7. 61.84

8. 20

9. 36.56

10. 21

11. 25

12. 6

13. 36

14. 18.66

15. 25

16. 47.95,48.18,48.57

7

17.21.9,22.6,23.1

18. 37.59,39.86,38.64

19. 12

20. G.M = 16.9, H.M = 13.5

21. F.M=7.604,H.M=7.401

22.28.08

23. 71.46

24.7

26. 12

27. 140

31. Rs.7.98

32. 12

33. 61, 71

34. 12

28. 161

29.70

CHAPTER

7 TRIGONOMETRY

Introduction Trigonometry is the science of measuring triangles. It is a branch of mathematics which deals with the measurement of angles and the problems applied with angles. System of Measurement ofAngles: There are three system of measurement of angles.

1. Sexagesimal system (Degree System): In this system, an angle is measured in degree, minutes and seconds. A complete rotation describes 360°. We represent the degree, minutes and seconds by the symbols 0,'," respectively. :. 1 right angle = 90° 1 degree (10) = 60' (60 minutes) 1 minute (1')

= 60"

(60 seconds)

2. Centesimal System (Grade System): In this system, right angle is divided into 100 equal parts, each part is called a grade. Each grade is subdivided into 100 equal parts, called minutes and each minutes are divided into 100 equal parts, called seconds. 1 right angle = 100g 1 grade (1 g)= 100' 1 minute

(I')

=

100"

3. Circular system (Radian System): In this system an angle is measured in radians. A radian is an angle subtended at the centre of a circle by an arc whose length is equal to the radius ofthe circle. 1 radian = 1c.

190

Comprehensive Remedial Mathematics for Pharmacy

Relation between degree and radians: Let r be the radius of circle and I be the length of arc (I) length of arc. Then angle (8) = . radians.

(bp

radIUS (r)

1t radian = 180°

here

180° 1 radian =- - , 1°= ~ radian 180 ' 1t 22

where 1t =

-

1t radian = 200g = 180°

U

7

Example 1: Express the following angles in radian measure: (i) 45° (ii) 18°20' Solution:

we know that 180° = 1t radians (i)

45° = ~ 180

(ii)

18020'

x

45

= .!: radians 4

= 18 20 degree = (55)°

60 .: 180° = 1t radians 55)° = ~ ( 3 180

Example 2:

~ = 1I1t 3

radians

108

Find the degree measure corresponding to the following radian measure: (i)

Solution:

x

3 •

(:

r

(ii)

we know that 1tc = 180° (i)

(ii)

(-2t= 180 x{-2) degree 1t - -

180x2x 7 d 22 egree

= - 114.54° =-144° 32'24"

0 .. (°.54 = 54 x 60' = 32.40'] 0.40' = 0.40 x 60 = 24"

Trigonometry

191

Example 3: Change 80g to degree Solution:

we know that 1g

=( -90 )0 100

80g = 90 x 80 = 72° 100 Example 4:

Find the radius of a circle in which a central angle of 45° intercepts an arc of 187 cm.

Solution:

Let r be the radius of circle. Let I = 187

~x45 =~ radian 180

Now

4

I r= - => r= 187x4 => r= 187x4x7 =238cm a 1t 22

Example 5:

A train is traveling on a curve 700 m radius at 14 kmIhr. Through what angle will it return in one minute.

Solution:

Given, radius = 700 m Speed of train = 14 km/h · · :. d Istance covered·In 1mInute = 14 x 1000 m

60

1= 700 m 3

I Now, we know that a = -

r

. =-700 x - 1 radians 3 700

a = .!.. 3

radian

Trigonometrical ratio or Trigonometrical Functions In a MBC, BC = Perpendicular (p) AB = base (B), AC = Hypotenuse (H)

. a = -P SIn H' B cos = H' P

a

tana= -

B'

cosec a =H P H sec = B B cot a = P

a

LJP

1\

B

B

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Comprehensive Remedial Mathematics for Pharmacy

Fundamental Identities 1. sin8

=

1 cosec9 1 2. cos 8 = - sec8 1 3. tan 8 = - cot8

4. sin2 8 + cos2 8 = 1

5. sin2 8 = 1 ~OS2 8 6. cos2 8 = 1- sin2 8

7. 1 + tan2 8 = sec2 8 8. sec2 8 -tan2 8 = 1 9. 1 + cor 8 = cosec2 8 2

10. cosec 8 - cor 8

=1

Example 6: (i) if sin 8

= ~, fmd the value of cos 8 and tan 8.

(ii) If tan A =

5

~,find the value of3 sin A + 4 cos A 4

Solution: (i)

3 cos281 = 5' 2 -8 cos 8 = ± .J'-1--si-n.8 = sm

-

=±FG) =±~1- :5 =±~=±~ 3

tan 8 = sin 8 cos8

= "5 = ~ 4

5

4

.28 sm

Trigonometry 3 tan A =-,

(ii)

tJ3 C

4

AC 2 = AB2 + BC2

= (4)2 + (3i

= 16 + 9 = 25 AC=5

Pi.

. A =3 cos A =4 S10

5'

5

3 sin A + 4 cos A

=3 x ~ + 4 x .i '5

5

=~+~=5

Example 7:

Solution:

5 5 1- cosA sin A =--Prove that sinA l+cosA L.H.S. = l-cosA = l-cosA x 1+ cosA sin A sin A 1+ cosA 2 l-cos A =----sinA(1 + cosA)

sin 2 A - sin A(1 + cos A) sin A

= l+cosA Example 8:

Solution:

Prove that

=R.H.S

1+cosA - - = cosec A + cot A l-cosA

L.H.S

=

193

---=

=

l+cosA l+cosA x--1 - cos A 1 + cos A

(I +COSA)2 l-cos 2 A

(l+cosAf sin 2 A l+cosA =--sin A 1 cosA =--+-sinA sinA

=

= cosec A + cot A = R.H.S

4

B

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Comprehensive Remedial Mathematics for Pharmacy

Example 9: If tan 8 + sin 8 = m and tan 8 - sin 8 = n, then prove that Solution:

m2 _n2 = 4.jrnn. L.H.S = m2- n2 = (tan 8 + sin

8i - (tan 8 - sin 8i

= tan2 8 + sin2 8 + 2 tan 8 sin 8 - (tan2 8 + sin2 8 -2 tan 8 sin 8) = 4 tan 8 sin 8 R.H.S = 4& =4~(tan8+sin8)(tan8-sin8) = 4~tan2 8 -sin 2 8

--4

. 28 cos 28

Sill - - S i l. l 28

2 =4 sin 8(-1--1) cos 2 8

=4 sin b ~sec28-1 = 4sin8~tan2 8 = 4 sin 8 tan 8

L.B.S = R.B.S

Example 10: If cos 8 + sin 8 = .fi cos 8, Prove that cos 8 - sin8 = .fi sin 8 Solution:

Given cos 8 + sin 8 = .fi cos 8 Squaring on both the sides (cos 8 + sin 8i = (.fi cos 8i cos2 8 + sin2 8 + 2 sin 8 cos 8 = 2 cos2 8 2 sin 8 cos 8 = 2 cos2 8 - cos2 8 - sin2 8 2 sin 8 cos 8 = cos2 8 - sin2 8 2 sin 8 cos 8 = (cos 8 - sin 8) (cos 8 + sin 8) 2 sin 8 cos 8 = (cos 8 - sin 8) .fi cos 8

(.: cos8 + sin8 = .fi cos9) cos 8 -

. 8

Sill

2sin8cos8 = ----c=---

.ficos8 cos 8 - sin8 = .fi sin 8

195

Trigonometry

. tan8+see8-1 1+sin8 Exampl e 11 : Sa Ive - - - - tan 8 - see8 + 1 eos8 tan 8 + see8-1 L.H.S =- - - - Solution: tan 8 - see8 + 1

= =

(tan 8 + see8) - (see 2 8 - tan 2 8)

(": see 2 8 - tan 2 8 = 1)

tan 8 - see 8 + 1 (tan 8 + see8) - (see8 - tan 8)(see8 + tan 8)

tan 8 -see8 + 1 (tan 8 + see 8)(1- see 8 + tan 8) =~---~----~ (tan 8 - see8 + 1)

= tan 8 + see 8 sin8 1 =--+-eos8 eos8

= 1 + sinO = R.H.S cosO

Trigonometric Table

~

00

sin 8

0

T-ratio

eos 8

1

0 30 (

%)

1 2

.J2

J3

1

-

tan 8

eat 8

see 8

0

00

1 00

1

60

0 (; )

J3

-

900(~) 1

2

1

-

0

2

.J2

1

1

00

J3 J3

J3

1

1

0

2

.J2

2

00

.J2

2

1

J3 eosee 8

45°(~)

2

2

J3

J3

196

Comprehensive Remedial Mathematics for Pharmacy

Example 12: .Find the value of sin2 45° + cos2 45° + tan2 30°

,in'45' + 00,'45' + tan' 30'

Solution:

~ C~

r Jzr r +(

111 =-+-+2 2 3 3+3+2 =--6

8

4

·6

3

=-=-

Example 13: If

tan2 9 + sec 9 = 5. Find cos 9.

Solution:

tan2 9 + sec 9 = 5

Given

sec2 9 - 1 + sec 9 = 5 sec2 9 + sec 9 - 6 = 0 sec2 9 + 3 sec 9 - 2 sec 9 - 6 = 0 sec 9 (sec 9 + 3) -2(sec 9 + 3) = 0 (sec 9 + 3) (sec 9 - 2) = 0 sec9+3=0 or

sec9-2=0

sec 9 =-3

or

1 cos 9 = - 3

or

sec9=2 1 cos 9 = 2

Example 14: If A = 30°, verify that cos 3A = 4 cos 3 A - 3 cos A. Solution:

Given A = 30° L.H.S = cos 3 A = cos 3 x 30° = cos 90° =0

+(

~

Trigonometry R.H.S.

197

= 4 cos3 A - 3 cos A = 4 cos3 '30° - 3 cos 30°

=4( ~J -3X~ =4x

3J3 _3J3 8

2

3J3 3J3 2

2

=0 L.H.S = R.H.S

Trigonometric Functions of angles II quadrant I quadrant Station All i.e. sin and cosec i.e. all positive are positive III quadrant Ticket i.e. tan and cot are positive

IV quadrant Collector i.e. cos and sec are posItive

Note: To memories the signs ofT-ratios in different quadrant. "All stations Ticket Collector" . 1. In pt quadrant, all the trigonometric functions are positive with angle (90 - 8) and (360 + 8) 2. In lind quadrant, only sin 8 and cosec 8 are positive with angle (90 + 8) and (180 - 8). 3. In IIIrd quadrant, only tan 8 and cot 8 are positive with angle (180 + 8) and (270 -8). 4. In IV th quadrant, only cos 8 and sec 8 are positive with angle (270 + 8) and (360 - 8). (i)

In case of 90° and 270°, the trigonometric functions will be changed i.e., sin 8 ~ cos 8, tan 8 ~ cot 8, cosec 8 ~ sec 8.

(ii)

In case of 180° and 360°, the trigonometric functions will not be changed i.e., sin 8 ~ sin 8, cos 8 ~ cos 8, tan 8 ~ tan 8 etc.

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Comprehensive Remedial Mathematics for Pharmacy

~ -e

sin

cos

tan

cot

cosec

sec

--cot 8 tan 8 -tan 8 --cot 8 cot 8 tan 8 -tan 8 --cot 8 cot 8

--cosec 8 sec 8 sec 8 cosec 8 --cosec 8 -sec 8 -sec 8 --cosec 8 cosec 8

sec 8 cosec 8 --cosec 8 -sec 8 -sec 8 --cosec 8 cosec 8 sec 8 sec 8

Anele

90° -8 90° + 8 180° - 8 180° + 8 270° -8 270° + 8 360° - 8 360° + 8

-sin 8 cos 8 cos 8 sin 8 -sin 8 --cos 8 --cos 8 -sin 8 sin 8

cos 8 -tan 8 sin 8 cot 8 - sin 8 -cot8 -cos 8 -tan 8 --cos 8 . tan 8 -sin 8 cot 8 sin 8 --cot 8 -tan 8 cos 8 cos 8 tan 8

Example 15: Find the following (i) sin (1920°) (ii) tan (765°) (iii) cosec (870°) Solution:

(i)

sin (1920°) = sin(3600

x

5 + 120°)

= sin 120° = sin(900 + 30°)

ofi

=cos30 = 2 tan (765°) = tan (360°

(ii)

x

2 + 45°)

= tan 45° = 1 (iii)

cosec (870°) = cosec (360°

x

2 + 150°)

= cosec 150° = cosec (90° + 60°)

= sec 60° = 2 Example 16: Prove that sin(2700 - 8) sin (90° - 8) - cos(2700 - 8) cos (90° + 8) + 1 = 0 Solution:

L.H.S = sin(2700 - 8) sin (90° - 8) - cos(2700 - 8) cos (90° + 8) +1 = --cos 8. cos 8 - (-sin 8) (-sin 8) + 1 = --cos2 8 - sin2 8 + 1 = -(cos2 8 + sin2 8) + 1 =-1 + 1

=0

Trigonometry

Example 17: Prove that sin 60° cos 30° - cos 150° sin 120° = Solution:

~

2

L.H.S = sin 60° cos 30° - cos 150° sin 120° =

sin 60° cos 30° - cos (90° + 60°) sin (90° + 30°)

=

sin 60° cos 30° - (-sin 60°) cos 30°

fjfjfjfj

=-x-+-x-

2

2

2

=~+~=~ 4

Example 18: Prove that

Solution:

4

2

=R.H.S

2

tan(3600 + 8) + cot(900 - 8) = 2 sin 8 cos 8 l+tan(l800 -8)cot(2700 +8) L.H.S = tan(3600 + 8) + cot(900 - 8) 1 + tan(1800 - 8)cot(2700 + 8) tan 8 + tan 8 1+ (-tan8)(-tan8)

=-------

2tan8

=---=-2

l+tan 8

=

2tan8 sec 2 8

2sin8 _ cos8 --1cos 2 8

= 2 sin 8 cos 8 =

R.H.S

Trigonometric ratios of sum and difference of angles

1. sin(A + B) = sin A cos B + cos A sin B 2. sin (A - B) = sin A cos B - cos A sin B 3. cos (A + B) = cos A cos B - sin A sin B 4. cos (A - B) = cos A cos B + sin A sin B

199

200

Comprehensive Remedial Mathematics for Pharmacy 5. tan (A + B)

= tan A + tanB I-tanAtanB

6. tan(A-B) = tanA-tanB 1+ tan A tanB 7. cot (A + B) = cotAcotB -1 cot A + cotB 8. cot(A-B)= cotAcotB+l cotA-cotB 9. sin(A + B) . sin (A - B) = sin2 A - sin2 B = cos2B - cos2A 10. cos(A + B) . cos(A - B) = cos2A - sin2B = cos2B - sin2A 11. tan(A + B + C) = tan A + tan B + tan C - tan A tan B tan C I-tan A tan B-tan Btan C -tan C tan A

Example 19: Obtain the values of sin 15°, cos 15°, sin 75°, cos 75°. Solution:

(i)

sin 15° = sine45° - 30°)

= sin 45° cos 30° 1

.J3

1

cos 45° sin 30°

1

.J3-1 2J2

=-x---x-=-2 2

J2

(ii)

J2

cosl5° = cos(45° - 30°)

= cos 45° cos 30° + sin 45° sin 30° 1

.J3 2

1

1 2

.J3+1 2J2

=-x-+-x-=--

J2

(iii)

J2

sin 75° = sine45° + 30°)

= sin 45° cos 30° + cos 45° sin 30° 1

.J3

1

.J3+1 2J2

1

=-x-+-x-=-2 2

J2

(iv)

J2

cos 75° = cos( 45° + 30°)

= cos 45° cos 30° 1

.J3

1

sin 45° sin 30° 1

.J3-1

=-x---x-=-2 2

J2

J2

2J2

Trigonometry

201

Example 20: Prove that sin 105° + cos 105° = cos 45° Solution:

L.H.S = sin 105° + cos 105°

= sin (60° + 45°) + cos (60° + 45°) =

sin 60° cos 45° + cos 60° sin 45° + cos 60° cos 45° - sin 60° sin 45°

Jj 1 1 1 1 1 Jj 1 =--x--+-x--+-x-----x--

212212212 212

1 1 2 =--+--=--

212 212 212 1

= J2 = cos 45° =

R.H.S

Example 21: If A + B = 45°, Prove that (1 + tan A) (1 + tan B) = 2 Solution:

Given A + B = 45° ~

tan(A + B) = tan 45° tanA+tanB =1 I-tanAtanB tanA+tanB= I-tanAtanB 1 +tanA+tanB+tanAtanB=2 (1 + tan A) + tan B( 1 + tan A) = 2 (1

+ tan A) (1 + tan B) = 2

Example 22: Prove that sin(A + B) sin (A - B) = sin2 A - sin2B = cos2B Solution:

C~S2 A

L.H.S = sin(A + B) sin(A - B) = (sin A cos B + cos A sin B) (sin A cos B - cos A sin B) =

sin2 A cos 2 B - cos2 A sin2 B

= sin2 A(1- sin2 B) -

(1 - sin2 A) sin2 B

= sin2 A - sin2 A sin2 B - sin2 B + sin2 A sin2 B =

sin2 A - sin2 B

= (1 - cos 2 A) - (1 - cos2 B)

= cos2 B -

cos2 A

= R.H.S

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Comprehensive Remedial Mathematics for Pharmacy

Example 23: Prove that tan 58 - tan 38 - tan 28

tan 58 = tan (38 + 28)

Let

Solution:

= tan 58 tan 38 tan 28.

tan 58

=

tan 38 + tan 28 1- tan 38. tan 28

tan S8 - tan 58 tan 38 tan 28 = tan 38 + tan 28 tan 58 - tan 38 - tan28 = tan 58 tan 38 tan 28

Formulae to transform the product into sum or difference 1. 2 sin A cos B = sin (A + B) + sin (A - B) 2. 2 cos A sin B = sin (A + B) - sin(A - B) 3. 2 cos A cos B = cos(A + B) + cos (A - B) 4. 2 sin A sin B = cos(A - B) - cos(A + B) Transforming of sums and difference into product C+D C-D . C + Sill . D = 2 Sill . - cos - 2 2

1·

Sill

2·

Sill

. C

. D

- Sill

=

2

C+D. C-D cos - - Sill - -

2

2

C+D C-D 3. cos C + cos D = 2 cos - - cos - 2 2 . C+D . D-C 4 · cos C -cos D = 2 Sill - - S I l l - 2 2 Example 24: Solution:

Pr~ve that sin 20° sin 40° sin 60° sin 80° = ~ 16

L.R.S = sin 20° sin 40° sin 60° sin 80°

= sin 20° sin 40° x

J3 2

x

sin 80°

=J3 sin 20° x! (2 sin 40° sin 80°) 2

=J3 sin 20° 4

2

[cos (80° - 40°) - cos(800 + 40°)]

Trigonometry

= J3 sin 20°

[cos 40° - cos 120°]

4

=

~ sin20

= J3

cos 40° +

(

~)

[sin 60° - sin 20° + sin 20°]

8

=

0

J3

sin 600 =

8

J3 x J3 = ~ = R.H.S 8

2

16

Example 25: Prove that sin 51t - cos 41t = J3 sin ~ 18

Solution:

9

9

"51t 41t L.H.S = Stn--cos18 9

[": cos 8 = sin "51t "1t =Stn--Stn18 18

51t

1t

-+-

= 2 cos 18

1t

18 sin 18

2 =

51t

--18

2

1t" 1t 2 cos - Stn-

6

9

= 2 x J3 sin 1t = J3sin 1t = R.H.S 2

9

9

(~ - 8 ) ]

203

204

Comprehensive Remedial Mathematics for Pharmacy

Example 26: Prove that

sin A + sin3A cosA+cos3A

L.H.S.

Solution:

= tan 2A

= sin A + sin3A cosA+cos3A A-3A . A+3A cos--2 SIn = 2 2 A+3A A-3A cos--2 cos

2

2

sin2A =- = tan 2A = R.H.S cos2A

Exampl e 27 : Prove that =

cos38 + 2cosS8 + cos78 cosS8 =--cos8 + 2cos38 + cosS8 cos38

L.H.S

Solution:

= cos38+cos78+2cosS8 cosS8 + cos8 + 2cos38 2cos 78 + 38 .cos 78 - 38 + 2cosS8

= __~2~_-=~2~_ ____ S8+8 S8-8 2cos--cos-- + 2cos38

2

2

2cosS8cos28 + 2cosS8

=--------2cos38cos28 + 2cos38 2cosS8(cos 28 + 1) = -----'------.:.. 2 cos 38(cos 28 + 1)

cos 59 - - - = R.H.S cos 39 To express the trigonometrical ratios of angle 2A in terms of trigonometrical ratios of angle A. 1. sin 2A = 2 sin A cos A 2. cos 2A = cos2A - sin2 A

= 2 cos2 A-I = 1-2 sin2 A 3. tan 2A

=

2tanA 2 I-tan A

4. sin 2A

=

2tanA 1+tan 2 A

Trigonometry

5. cos 2A =

I-tan 2 A l+tan 2 A

1- cos2x

6. sin2x

=- - -

7. cos 2x

= 1 + cos2x

2

2

Proof: 1. We know that sin(A + B) = sin A cos B + cos A sin B

Put B = A, we get sin 2A = 2 sin A cos A 2. Let cos(A + B) = cos A cos B - sin A sin B Put B = A, we get cos2A = cos 2A - sin2 A Now again

cos2A = cos2A - (1- cos 2A) 2

cos2A = 2 cos A - 1 again

cos2A

= cos2A =

sin2 A

(1 - sin2 A) - sin 2 A

cos2A = 1 - 2 sin2 A 3. Let

tan (A + B)

Put B = A, we get tan 2A = = 4. Let

= tan A + tanB 1- tan A tanB 2tanA 2

I-tan A sin 2A = 2 sin A cos A 2sinAcosA

= sin 2 A + cos 2 A

dividing by cos 2A, we get 2sinA sin2A =

5. Let

=

cosA 2 sin A + 1 cos 2 A 2tanA . 2A = sm --l+tan 2 A cos2A = cos2A - sin2A =

cos 2 A -sin 2 A sin 2 A+cos 2 A

205

206

Comprehensive Remedial Mathematics for Pharmacy dividing by cos2A, we have 1-tan2 A cos 2A = - - : 1+tan 2 A

Angle 3A in terms of angle A

1. sin 3A = 3 sin A - 4 sin3 A

2. cos 3A = 4 cos3 A - 3 cos A _ 3tanA-tan 3 A 3 . tan 3A 2 1-3tan A Example 28: Find the value of (i) sin 180(ii) cos 18°, (iii) sin 36° (iv) cos n° (v) sin 54° Solution: (i) sin 18°

Let

A = 18° 5A= 90° 2A + 3A= 90° 2A= 90° -3A sin2A = sin(900 - 3A) sin2A = cos 3A 2 sin A cos A = 4 cos3 A - 3 cos A 2 sin A = 4 cos2 A - 3 2 sin A = 4(1 - sin2 A) - 3 2 sin A = 4 - 4sin2A-3 4 sin2 A + 2 sin A-I = 0 . A

Sill

-2±~4+16

=----8

-2±.J20 8 . A

Sill

-1±J5

=--4

since 18° is acute angle, hence sin 18° is not negative ..

sin180 = = J5 -1 4

Trigonometry (ii)

cos 18° Since

cos e = ~1 - sin 2 e cos 18° = ~1 - sin 2 18°

~~ =~1- 5+1-2.Js 16

=~1- 6-2.Js 16

=~10+2.Js 16

.. (iii)

cos 18° = ~~10 + 2.Js 4

cos 36° Let

cos 36° = cos 2 x 18° = 1 - 2 sin2 18°

~ 1-2[ Fs4-1 J 6-2.Js 8 2+2.Js .Js +1 --= 4 8

=1-

..

sin 36° = ~1- cos 2 36° =

1-[FstJ

=~1- 6+2.Js =~~10-2.Js 16 4 (iv)

cos

n° = cos(900 = sin 18° .Js -1 4

---

18°)

207

208

Comprehensive Remedial Mathematics for Pharmacy sin ~4° = sin(900 - 36°)

(v)

cos 36°

=

J5 +1 4 Example 29: Find the value of (i) sin 22f (ii) tan 22f 2 2 Solution:

(i)

45° cos 45° = 1 -2 sin2 2 1 . 2 45° -=1-2sm -

fi

2

2 sin2 45° = 1__1_

fi

2 Sin2 45° =

2

fi -1 2fi

fi fi

x

. 2 45° 2-fi sm - = - -

2

sin 22f

2

4

= J2-fi 4

45° 2tantan 450 = __--=2'---:2 45° I-tan 2

(ii)

=>

2tan': tan A = 22A A [ I-tan 2

1° 2tan221= 2 1° 2 I-tan 222

10

1 - tan 2 22-

2

= 2 tan

10

22-

2

10 1° tan2 22- + 2 tan 22- - 1 = 0 2 2

1

209

Trigonometry

1° 1° 2 tan 22- + 2 tan 22- + 1 = 2 2

1°

2

(tan 22- + 2

Ii = 2

1°

tan 22 - + 1 = 2

1°

(adding 2 on both sides)

fi

fi

tan 22- = -1 2 Example 30: Prove that 4 sin A sin(60 - A) sin(60 + A) = sin 3A

Solution:

L.H.S = 4 sin A sin (60 - A) sin (60 + A) = 4 sin A (sin2 60 - sin2 A) = 4 sin A

(%- - sin 2 A J

= 3 sin A - 4 sin3 A

= sin 3A = R.H.S Example 31: If 0 < x < 2 1t, find sin ~ and cos ~. When tan x = - 4 . x lies in IT quadrant. - 2 2 3 Solution:

-4

Given tan x = -

3

-3

cosx= Now

. x

sm-=

for II quadrant, cos x is negative

5

~1-COSX

2

(':Sin 2 ;

2

=Jl:~

A

=~5~2 =~= Js again,

x ~1 +cosx cos -=

2

2

=

= 1-~OSX J

-~

~ _5 2

=~ 5~2·=H Js =

:/14

~c

210

Comprehensive Remedial Mathematics for Pharmacy

~. Find the value of tan A

Example 32: If

sin2A =

Solution:

Given sin2A =

5

2tanA

--::--= 2

~ 5

4

1+ tan A 5 10 tan A = 4 + 4 tan2 A 4 tan2 A-I 0 tan A + 4 = 0 4tan2 A-8tanA-2tanA+4=O 4 tan A(tan A - 2) -2(tanA - 2) = 0 tan A-2 = 0 or

4tanA-2=O

tanA=2

or

r

tanA=2

Example 33: Prove that cos 4x -= 1 - 8sin2 x cos2 x Solution:

L.H.S = cos 4x = cos 2(2x) = 1- 2 sin2 2x = 1 - 2 (2 sin x cos

xi

2

= 1 - 2 (4 sin x cos2 x) = 1 - 8 sin2x cos2 x = R.H.S 1 1 Example 34: If2 cos A = X +- then prove that 2 cos 3A = x 3 +-3 X x Solution: cos 3A = 4 cos3 A - 3 cos A

1 1 3 3 1 =4x-(x+-) --(x+-) 8 x 2 X 1[3 1 1 1] =x +-+3(x +-)-3(x +-) 2 x3 x X (.,' (a+b)3=a3 +b3 +3(a+b» cos 3A

1 -J =21 ( x 3 +~

1 2 cos 3A = x 3 +-3 X

Trigonometry

Example 35: Prove that cos2 48° - sin" 12° Solution:

211

8

L.H.S = cose 48° - sine 12°

1 1 ° .1 ° [.:2COS"A=1+COS2A] = - [2cos- 48 - 2 sm- 12 ] '") 2 sin 2 A = 1- cos 2A

= -1

2

°

°

[cos 96 + cos 24 ]

9_6 _-_2_4 ] = ~[2COS 96 + 24 .cos_ 2

2

2

= cos 60° cos 36° = - . - - = - - = R.H.S 2 4 8 Example 36: Prove that

~2 + .J2 + 2cos48

Solution:

L.H.S =

= 2cos8

~2 +.J2 + 2cos48

= ~2 + ~2 + 2(2cos 2 28 -1)

=~2+~4coS228

=.J2 + 2cos28 = ~2 + 2(2cos 2 8 -1) = ~4COS2 8

= 2 cos 8 = R.H.S Example37: Provethat

cosA sin A . + =smA+cosA 1 - tan A 1- cot A

Solution:

L.H.S. =

cosA + sinA 1 ~ tan A 1- cot A

212

Comprehensive Remedial Mathematics for Pharmacy cosA sin A = 1- sin A + 1 _ cos A cosA sin A sin 2 A

cos 2 A

=

+-----

cosA-sinA sinA-cosA sin 2 A cos 2 A =----cosA -sin A cosA -sinA =----cosA -sinA _ (cos A -sin A)(cosA + sin A)

(cosA-sinA)

= sin A + cos A = R.H.S Example 38: If cot S = Solution:

=~, Find the value of sec 8 and cosec 8. 40

9 BC cot 8 = - = 40 AC In LlABC, AB2 = AC 2 + BC 2

Given

= (40)2

+ (9)2

A

LJ40 B

9

C

= 1600 + 81 = 1681 AB=41 AB 41 seeS = - = BC 9 and

Example 39: If tan Solution:

cosec8

AB

41

AC

40

:=-=-

e ::: ~3 , then find the values of sin 8 + cos 8 and sec 8 + cosec 8.

Given tan e = ~ = AC 3 BC In LlABC, AB2 = AC 2 + BC 2 =4+9

AB=

Jlj

tJ2 A

B

3

C

Trigonometry

.Jl3

. 8 =2,

SIn

sec8= = 3

.Jl3

.Jl3

3 cos8= = -

cosec 8 = = 2

.Jl3'

2 3 sin8+cos8 = - + -

.Jl3.Jl3 5

-.Jl3 and

.Jl3.Jl3 +-

sec 8 + cosec 8 = -

=

3

2

2.Jl3 + 3.Jl3 6

=~JU 6

Example 40: Prove that

cos2A = tan(45° - A) 1+ sin2A cos2A L.H.S = - - 1+ sin2A COS

2 A - sin 2 A

=--~-~~--------

sin 2 A+cos 2 A+2sinAcosA (cosA -sinA)(cosA +sinA) (cosA+sinA)2

=~------~~--~--~

cosA-sinA cosA + sin A

=-----

divide by cos A, we have I-tanA 1+ tan A

=----

=

tan 45° - tan A 1 + tan 45° tan A

= tan( 45° - A) = R.H.S

(': tan 45° = 1)

213

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Comprehensive Remedial Mathematics for Pharmacy

Example 41: Prove that = Solution:

sin2A = cot A l-cos2A

L.H.S =

sin2A l-cos2A 2sinAcosA = -------::-1-(l-2sin 2 A) 2sinAcosA 2sin 2 A cosA = - - =cotA=R.H.S sinA

Example 42: Prove that cos 20° cos 40° cos 60° cos 80° = ~ 16 Solution:

L.H.S = cos 20° cos 40° cos 60° cos 80° = cos 20° cos 40° . ..!.. . cos 80° 2

°

°

°

= -1 cos 20 (2 cos 80 cos 40 ) 4

1 = - cos 20° [ cos 120° + cos 40°] 4 [.: 2 cos A cos B = cos (A + B) + cos (A - B)] 1 1 = -cos200[-- + cos400]

4

2

°

°

-1 1 = -cos20 + -cos20 cos40

8

4

°

= -=-!. cos 20° +.!. (2 cos 40° cos 20°) 8 8 = -=-!.cos200 + .!.cos600 + .!..cos200

8 1 = -cos600 8

8

8

111 =-x-=8 2 16

= R.H.S

Trigonometry

IOn 8n 3n 5n Example 43: Prove that cos- + cos- + cos- + cos- = 0 13

13

13

13

IOn 8n 3n 5n L.H.S =cos-+cos-+cos-+cos-

Solution:

13

13

13

13

= (cos IOn + cos 3n) +(cos 8n + cos 5n)

13

13

13

13

13n 7n 13n 3n = 2cos-cos-+ 2cos-cos26 26 26 26 [.: cosc + cos

n 2

7n 26

0 C~ 0 C; 0] = 2cos

n 2

cos

3n 26

= 2cos-cos-+ 2cos-cos-

= 0

Example 44: Prove that Solution:

(.: cos~ = 0

+ 0 = 0 = R.H.S

sec A -tanA ? = 1 - 2 sec A tan A + 2 tan- A sec A + tan A L. H. S = sec A - tan A x sec A - tan A sec A + tan A sec A - tan A (sec A - tan A) 2 = sec 2 A - tan 2 A

=

sec 2 A + tan 2 A - 2secA tan A

= 1+ tan 2A + tan 2 A - 2 secA tanA

= 1 + 2 tan 2 A -

2 sec A tan A = R.H.S

Example 45: If tan 8 = 1, find the value of sec 28 and cosec 2 8. . I-tan 2 8 Solution: SInce cos2 8 = ? 1+tan-8 1 + tan 2 8 sec 28 = - -2I-tan 8 Since

tan 8 = 1

Now,

2tan8 . 28 =---=-SIn l+tan 2 8

1+1

2

sec 28 =-- =- =00 1-1 0

215

216

Comprehensive Remedial Mathematics for Pharmacy l+tan 2 8 2tan8

=- - -

cosec 28 since

tan8 = 1 cosec 2 8

1+ 1 2 = 2xl 2

= -

=

1

Example 46: Find the value of each of the following.

(i)cos210 0

(ii)tan lIn 5

(iii) sin 5n 3

Solution: (i)

cos 210° = cos(1800 + 30°)

= - 10g12 1728 = 3 .. ) 2-5 -( II

-1

32

=> IOg2 - 1 = - 5 32

(iii) 10-3 = 0.001 => 10gIO 0.001 =-3

Example 2:

Write the following in exponential form: (i) log2 128 = 7

Solution:

(ii) log9 729 = 3

(i) log2 128 = 7 => 27 = 128 (ii) log9 729 = 3 => 93 = 729

Properties of Logarithms: (i) the logarithm of 1 on any base a (a :t: 0) is zero i.e. aO

= 1=>

loga 1 = 0

(ii) The logarithm of any number to the base it self is always I i.e. a 1 = a => loga a = I (iii) Base will never be 00 or O. (iv) The logarithm ofzero on any base is negatively infinite i.e. a-a:> = 0 => loga 0 =

-00

Fundamental laws of logarithm First Law:

The logarithm of the product of two numbers is equal to the sum of the logarithm of these numbers to the same base. i.e.

Proof:

Let

=>

and

logan = Y

m = a" and n = aY

Again by definition loga (mn) = x + y loga (mn) = loga m + loga n

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Comprehensive Remedial Mathematics for Pharmacy

Second Law: The logarithm of the quotient of two numbers is equal to the difference of the logarithm of the numerator and the logarithm of the denominator to the same base. i.e.

Proof:

Let

logam = x and loga n = y

:. By definition,

10~(:)= x-y 10~(: ) = lo~m -logan Third Law:

The logarithm of a number raised to a power is equal to the product of the index of the power and the logarithm of the number to the same base. i.e lo~

Proof:

Let

(mll) = n

lo~m

loga m = x => m = aX mll = (aXt

:. By definition, lo~ (mll) = nx loga (mll) = n 10ga m Fourth Law: The base of logarithm can be changed. If the logarithm of numbers are given to the base a, then the logarithm of those numbers can be obtained to another base b. i.e. 10gb m = 10ga m Proof:

Let

x

10gb a

10gb m = x, 10ga m = y, 10gb a = z bX= m, aY = m and bZ = a

bX = aY

Logarithm

229

x=yz 10gb m = loga m x 10gb a lo&m=

Fifth Law:

log m b 10g b a

The product of logarithm of number b to base a with logarithm of number a to base b is equal to 1. i.e.

Proof:

10gb a = 1

10& b

x

Let

loga b = x ~ aX = b

~

10gb (aX) = 10gb b

~ X

10gb a= 1

~

or

Example 3:

loga b x 10gb a = I 1 10gba=-loga b

Prove that loga (1 + 2 + 3) = 10& 1 + 10& 2 + 10& 3.

L.R.S. = loga (1 + 2 + 3) = loga 6

Solution:

R.H.S. = loga 1 + 10& 2 + 10& 3 = loga (1

x

2

x

3)

(': 10& (m x n x p) = 10& m + loga n + loga p)

= 10&1 6 L.H.S = R.H.S n

Example 4:

If log (m + n) = log m + log n, show that m = - n-l

Solution:

Given

log (m + n) = log m + log n

~

log (m + n) = log (mn)

~

m+n=mn

~

n=mn-m

~

n=m(n-l)

~

m=--

0

0-1

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Comprehensive Remedial Mathematics for Pharmacy

Example 5:

144 = log2x. Find the value ofx, when log2 log2 12 L.H.S = log} 144 = log2 (12 x 12) log} 12 log 212 log) (12)2

Solution:

log2 12 _ 2log 2 12 =2 log2 12 Thus, we have log2 x = 2

=> x = 22 => x=4 Example 6: Solution:

x+y logx + logy 2 ' then prove that x = y Iflog-- = 2 log x + Y = log x + log ~ Given

2

2

x+y 1 log-2- = "2 log (x.y) 1

x+y log-- = log (xy)2 2 x+y .!. -=(XY)2 2

Example 7: Solution:

=>

(x;yr =xy

=> =>

2 x + 2 x +

=>

(x-yi = 0

=>

x-y=O

=>

x=y

1 + 2xy = 4xy 1- 2xy = 0

) 2 a+b loga+logb If a- + b = 7ab, prove that log - - = --==-----=3 2 2 2 Given a + b = 7ab a 2 + b2 + 2ab = 7ab + 2ab

=> =>

(a+ bi=9ab

Logarithm

2

a+b

log -3(

231

)

=

log (ab)

b)

a+ 210g ( -3= loga + 10gb

=>

Iog (-a +3-b) _- loga +2 10gb

Example 8:

If

log(mn) = log m -log n, show that n = 1.

Solution:

Given

log(mn) = log m -log n

=>

log m + log n = log m - log n

=>

210g n = 0

=>

log n = 0

=>

log n ± log 1

=>

n

Example 9:

Prove that

Solution:

L.H.S. A =

(.: log 1 = 0)

=1

x log y- log z • ylog z -log x • zlog x - log y

=1

x log y -log z . ylog z -log x • zlog x -log y

log A = log [xlog y -

log z • ylog z - log x . zlog x - log Y]

= log x log y - log z + log ylog z -log x + log zlog x - log y = (log y - log

= log x log y

t:) log x + (log z

-

log x)log y + (log x -log y) logz

-log x log z + log Y log z

-

log x log y + log x log z

-

log y log z

10gA= 0

=> log A

=

log 1

(': log 1 =;= 0)

=> A= 1 :. L.H.S = R.H.S

Types of Logarithm 1. Natural Logarithm: The logarithm of a number to the base e is called natural logarithm. It is also called Napieian logarithm. The value of e = 2.7186 (approximate)

232

Comprehensive Remedial Mathematics for Pharmacy

2. Common Logarithm: The logarithm of a number to the base lOis called common logarithm. In numerical calculations common logarithms are used. The common logarithm of a number consists of two parts: 1. The integral part of logarithm is called characteristic, and 2. The decimal part is called the mantissa.

To find characteristic of a common logarithm: The following rules are: Rule 1: Number greater than 1. The characteristic of a common logarithm is a positive integer and is one less than the number of digits in the integral part of that number.

Number

Number of digits in the inte2ral part of the number

Characteristic

(i) 2.70827

1

1-1=0

(ii) 56.787

2

2-1=1

(iii) 134.82

3

(iv) 1348.2

4

.

3-1=2 4-1=3

Rule 2: Number less than 1. If the positive number begins with a decimal point, the characteristic of its common logarithm is negative and is numerically one greater than the number of zeros immediately after the decimal point. The negative characteristics are written as a symbol (-) 'bar' over it, i.e. -1, -2, -3 etc., are negative characteristics which are written as 1,2,3 etc., and read as one bar, two bar, three bar etc.

Number

Number of zeros after decimal point in this number (- n)

Adding 1 Charactristic

-

(i) 0.9207

0

0+1=1

-lor 1

(ii) 0.00392

2

2+1=3

-30r3

(iii) 0.0342

1

1+1=2

-2 or 2

(iv) 0.00007027

4

4+1=5

-5 or 5

-

-

-

Logarithm

233

To find mantissa of a common logarithm To find the mantissa of common logarithm of a number, special tables have been formed. These tables are called logarithmic table.

Procedure: (i) Ignore the decimal point, use the digits irrespective of the decimal point. Also ignore zero if it is not a significant figure. In case of more than four significant figures, round off the number up to four significant figures while using the four figure log tables. (ii) The first two digits are used for determining the row of logarithm table. If there is only one digit then we add a zero. (iii) Proceed along that row till that column (in the main body) of third digit is reached. Read offthe figure in that row and column. (iv) Proceed along that row till that common (in the main difference) of fourth digit is reached and read off this figure. (v) Add the two figures obtained in step (iii) and (iv). This is the required mantissa of the common logarithm.

Example 10: Obtain the logarithm of90. Solution:

The given is greater than 1 so characteristic is 2 - 1 = 1

For mantissa: First look down the left column to locate 90, look across the 90 row and locate the entry under column 90 labelled o. Read off this figure 9542. Hence mantissa of 10glO 90 = 0.9542 :. 10glO 90 = characteristic + mantissa =

1 + 0.9542

= 1.9542 !£xample 11: Find the logarithm of 0.0004291 Solution:

characteristic: The given is less than 1 and the number of zeros after the decimal point is 3 hence, the characteristic = - (3 + 1) or - 4 or 4.

Mantissa:

Ignore decimal point consider the number 4291 The figure 42 under 9 in main table

= 6325

The figure 42 under I in main difference adding = 6326 Hence mantissa = 0.6326

234

Comprehensive Remedial Mathematics for Pharmacy :. 10gIO 0.0004291 = characteristic + mantissa =-4 + 0.6326

-

= 4.6326 Antilogarithms: The antilog of a number is that number whose logarithm is the given number. Now to find out antilog of a given number, we start with the digits after the decimal point (mantissa) in the antilog table.

Procedure: 1. In case of negative logarithm, the mantissa must be converted into a positive mantissa before reference to antilog table. 2. The first two digits of mantissa are seen in the first column of the antilog table to locate the relevant row. 3. Read off the figure given before it in the column heading of the third digit. 4. Read off the figure before that row in the mean difference under column heading of the fourth digit. 5. Add the figure obtained in step (3) and (4). The number so obtained is called a desire number. 6. Finally, use the characteristic of the given number for placing the decimal point in result, obtained in step (5). These are three conditions: (a) If characteristic is 0 or positive: In this case the number of digits in the whole part ofthe desire number will be one more than the characteristic. e.g. If characteristic is 2 then adding 1, we get 3. Thus the point of decimal will be after 3 digits to right side in the number obtained from antilog. (b) If characteristic is (-1) or 1: In this case adding 1, we get O. Thus the number will start. from decimal point. (c) If characteristic is negative and more that (-1): In this case the number of zeros just after the decimal point will be one less than the characteristics.

Example 12: Find antilog of (i) - 0.2682 (ii) 3.5386 Solution:

(i) Here, the given number is negative. So, we have to convert the mantissa into positive. This is done by adding and subtracting 1 from this number, thus - 0.2682 = - 0.2682 + 1 - 1 = -1 + (1 - 0.2682) =-1 + 0.7318 -

= 1.7318

Logarithm

235

From antilog table, The figure before. 73 under 1 in main table

= 5383

The figure before .73 under 8 in mean difference Adding

10 = 5393

Thus, desired number = 5393 The characteristic is -1. Therefore, the number must have no digit in the integral part. The number of zeros just after decimal point will (-1 + 1) i.e. O. Hence, antilog (-0.2682) = 0.5393

= 3551

(ii) The figure before .53 under 8 in main table The figure before .53 under 6 in mean difference Adding

5 = 3456

Thus, desired number = 3456 Since the characteristic is 3 or (-3). Then number of zeros just after the decimal point will be (-3 + 1) =-2 -

Hence, antilog (3.5386) = 0.003456. Example 13: Using logarithmic table, find the value of each offollowing

Solution:

(i) 2.38x3.901 (ii)(0.8931)--4 (iii) 4.83 2.38 x 3.901 x=---(i) Let 4.83 2.38x3.901 1ogx= 1o g - - - 4.83

log x = log (2.38

x

0.0148xO.274 0.8735

3.901) -log 4.83

= log 2.38 + log 3.901 -log 4.83 = 0.3766 + 0.5912 - 0.6839 log x = 0.2839 x = Antilog (0.2839) (ii) Let

x= 1.923 x = (0.8931)--4 log x = log (0.8931)--4 = -4 log (0.8931)

236

Comprehensive Remedial Mathematics for Pharmacy = -4 x (-1.9509) = -4 x (-1 + 0.9509) = -4 x - 0.0491 log x = 0.1964 x = Antilog (0.1964) x = 1.571 (iii) Let

0.0148x 0.278 x= / - - - - 0.8735 1 X

10 x=lo (0.OI48 O.278)2 g g 0.8735

=~ 10 (0.0148 X 0.278) 2

g

0.8735

=

2"1 [log (0.0148) + log (0.278) -log (0.8735)]

=

~ 2

(2.1703 +0 1 .4378- 1.9412)

=

~ [(-2 + 0.1703) + (-1 + 0.4378) -

=

2"1 [-2 - 0.3331]

2

= -1.1666 =-1-1 + 1-0.1666 =-2 + 0.8334 log x = 2.8334 -

x = Antilog (2 .8334)

x = 0006814 16 25 81 Example 14: Prove that 7 log2 - + 5 log2 - + 3 log2 - = log22 15 24 80 Solution:

L.H.S

16 25. 81 = 7 log - + 5 log - + 3 Ilog15 24 80

(-1 + 9412)]

Logarithm

=

237

710g[~) + 5 IOg[-+-) + 310 g [-+-) 3x5 2 x3 2 x5 7 [log 24 -log (3 x 5)] + 5 [log 52 -log (2 3x 3)] + 3 [log 34 log (2 4x 5)]

_

7 [4 log 2 -log 3 -log 5)] + 5 [2 log 5 - 3 log 2 -log 3)] + 3 [4 log 3 - 4 log 2 -log 5] = 28 log 2 -7log 3 -7 log 5 + 10 log 5 - 15 log 2 - 5 log 3 + 12 log 3 - 12 log 2 - 3 log 5

= IOg22 = R.H.S Example 15: Find the value of x if 7x + 7x+2 = 250 7x + r+ 2 = 250 Given Solution:

7x + 7' I Does limit exists at x = I lim

Left limit =

Solution:

f(x) =

x~l-

lim

=

h~O

(1- h)

lim x~r-

x2

2

(put x = I-h)

=I Right limit =

=

lim . + f(x) =

x~l

lim h~O

lim x~l

+

[(1 +h)2) + I]

(x 2 + 1) (put x = 1 + h)

=1+1=2 here,

lim x~1

_ f(x)

lim *' x~1 + (fx)

Hence limit off(x) does not exists at x = 1.

Example 2:

I-x lim Let f(x) = -1-,0 :s x:s 1, then evaluate x ~.!.. f(x) if it exists. +x 2

Solution:

Let

f(x)

I-x

=-

l+x

Left limit =

lim 1 - f(x) =

x~-

2

=

lim 1 1- ~ x~l+x

. l-(.!..-h) hm 2

h~OI+(k-h)

2

Calculus

301

1 -+h · 1 = 1m _2_ h~O ~-h 2 1

2 =~

=

3 2

Right limit =

3

lim 1+ f(x) =

x~-

2

lim

lim 1 1+ -

X

x~-I+x

2

1-(k+h)

1+(~+h ) 1-!2 _ 1 - 1+! -3

_

2 Since left limit = Right limit .1 lim Hence 1 f(x) exists and is equal to -. x~3

2

Continuity of a function at a point: A function is continuous at a point a if the left limit and right limit exists and both are equal to the value of function at x = a. i.e.,

lim x~a

Example 3:

_ f(x) =

lim x~a

+ f(x) = f(a)

Test the continuity of the function defined as 2

f(x)

=

{

x -4x+3 x-I

x::/.

x= 1 lim lim x 2 - 4 x + 3 Right limit = +f(x) = +- - - x~1 x~1 x-I -2

Solution:

lim (1+h)2 -4(I+h)+3 (1+h)-1

= h~O

302

Comprehensive Remedial Mathematics for Pharmacy lim h 2 - 2h

=h~O lim

= Left limit =

h~O

lim x~l-

h (h-2) =-2

f(x) =

lim x 2 - 4x + 3 x~1 x-I

lim (l_h)2 -4(l-h)+3

= h~O = = Value f(l) = -2 Since i.e.,

(l-h)-I 2 lim h +2h

h~O

lim h~O

-h -(h+2) =-2

(given)

Left limit = Right limit = value lim x~l

_ f(x) =

lim x~1

+

f(x) = f(l)

f(x) is continuous at x = 1 Example 4:

A function f(x) is defined as

f( x) = ..!. - x when 0 < x < ..!. \ 2 ' 2 1 1 - whenx=2' 2 3 1 = - - x when - < x < 1

2'

2

Prove that f(x) is discontinuous at x = Solution:

..!.

2 .. lim _ lim _ I lefthmlt= 1 f(x)= I (--x) x~x~2 2 2 =

=

hI: o[~-(~-h)] lim h~O

h=O

Calculus

Right limit =

lim\ + f(x) =

lim\ +

x~-

x~-

2

2

= lim

h~O

= lim I-f(xj:;t:

Since

x~-

2

Hence

~( l\.X)

(i _

lim h~O

2

x)

[i-(~+h)l 2 2 (l-h)=l

lim I l+f(x):;t:f(-) x~2

2

. d"IscontInUOUS at x =I IS 2

lim

Ixl does not ~ists

Example 5:

prove that

Solution:

f(x) = Ixl = x (when x > 0) = -x (when x < 0) lim lim left limit = _ f(x) = _ (-x)

x~1

x~1

x~1

lim

= h ~ 0 -(1 - h) =-1

Right limit =

lim x~1

= lim

SInce

x~1

+

lim h~O

f(x) =

lim x~1

(l + h)

_ f(x) :;t:

(x)

=I

lim x~1

+

+

f(x)

The given function does not exists Example 6:

At what point is the following function discontinuous? f(x)=

Solution:

Let

x2 -

,x2+1 x- -3x+2 3x + 2 = x 2 - 2x - x + 2 =x(x-2)-I(x-2) = (x - 2)(x-l)

x= 1,2

303

304

Comprehensive Remedial Mathematics for Pharmacy since f(x) is not defined at x = 1 and x = 2 Hence f(x) is discontinuous at x = 1 and x = 2.

Example 7: Solution:

Prove that the sine function is continuous. lim lim we know that sinx = 0 and cos x = 1 x~O

Then

lim x~a

x~O

sin x =

=

lim h~O

sin(a + h)

lim h~O

= sin a

(sin a cos h + cos a sin h) lim

h~O

cos h + cos a

lim h~O

sin h

sin a x 1 + cos a x 0 sin a sin x is a continuous function = =

Example 8: Solution:

The exponential function is continuous i.e., f(x) = eX. lim we know that eX.= 1 x~O

lim

eX =

x~a

lim

ea+h

h~O

=

lim

a

h~O

= ea.

h

e .e

lim h~O

.e

h

= ea xl

= ea

eX is continuous function.

Exercises 1.

Iff(x) = x2 - 4x + 6, find f(1), f (1) ,find f(2 + h) and f(2+h)-f(2) h

2.

Given f(x) = 3x - 1, if x > 3 = x 2 - 2, if -2 :::;; x :::;; 3 = 2x + 3, if x < -2 Evaluate f(2), f(4), f(-I), f(-3), f(O)

"2

Calculus

305

3.

Determine the domain and range of the following functions. X+7 x 2 -1 (i) -(ii) -(iii).Jx - 3 x-5 x-I 4. Iff(x) = log x, prove that feu v w) = feu) + f(v) + few) 5.

Iff(x)= IOg(I-X),provethatf(a)+f(b)= f(a+b) l+x l+ab

6.

Iff(x) = sin x and F(x) = cos x, prove that [f(x)f + [F(x)f = l.

7.

Determine which of the following are even or odd?

8.

(i)

f(x) = (x4 + 3) ~X6 -3

(ii)

f(x) = x3 (3x + X5)3

(iii)

f(x) = 2x

(iv)

f(x) = 3x - T X

If f(x) = aX, show that f(x) . fey) = f(x + y)

9. If f(x) = 1- x then prove that f(tan 8) = tan l+x 10.

4

Evaluate the limits lim 1 4x 2 -1

(i)

x~2

lim

(iii)

x~O

(ii)

2x-1 ~-~ x

(iv)

x 2 -4x+3 x~I x 2 +2x-3 lim

(v) 11.

(~- 8) lim x~-2

x 3 +8 x+2

x 2 -5x+6 x~2 x 2 -4 lim

(vi)

Evaluate 5

X4 -256

(i) lim

x-4

x~4

(ii)

7

lim x~a

7

x _a x 5 _a 5

(iii)

lim

l-cosx

x~O

Xl

lim x~a

12. Evaluate (i)

lim x~O

OO') ( III

sinpx -.smqx

r1m x~O

. . .Jx-J;.

smx-sma

(ii) .

(lV)

lim

cos ax - cos bx

x~O

x~

1

5

(x + 2)3 - (a + 2)3 x-a

306 13.

Comprehensive Remedial Mathematics for Pharmacy Evaluate lim

0) 14.

1t

lim

1+ cos2x

x-»")

(ii)

(1t-2X)2

Evaluate lim e -e (i) x-»o smx '(

15.

Show that

16.

Evaluate

17.

Evaluate (i) (iii)

-'(

( ii)

1t

I-tanx

x-»4

e smx -1 x-»o x lim

lim Ix-21 - - does not exist. x-»2 x-2 lim

x-»oo

x -I x

4x 2 -5x+7 x-»oo 2x-3 lim

lim x-»O

(ii)

eX cos x

(iv)

lim x-»oo

Discuss the continuity of f(x) = x2+ 1 at x = I x +1

19.

Discuss the continuity of f(x) =

{I-C~SX x-

l

~x) =

x

-::f.

0

x=o

2 Given

~CJx+3-~)

lim x(e" -1) x -» 0 1- cos x

18.

20.

1t

x-4

r:x +cosx

X -::f.

0

x=o Show that f(x) is continuous at x = 0 21.

Show that function is discontinuous at x = 2. f(x) =

x2

307

Calculus

22.

f(x)

If

=

{

l-cos4x . ' x-J

Xi:-

4

x=O

.,

0

Find whether function f(x) is continuous at x = O. 23.

At what points is the following function discontinuous? f(x) =

24.

Find where the following function is discontinuous f( x) =

25.

,x + 1 x- -5x+6

x-I --=-J- - -

x--4x+3

Prove that f(x) defined as below is discontinuous at x = ..!.. 2 f(x) =

if 0 < x O h =

. ../X + h - fx L1 m - - - - h-->O h

=

. ../x+h-fx L 1m h-->O h

x

(~+fx) (../x+h+fx)

x+h-x = Lim --===----==h-->O h(../x+h+fx) = Lim h-->O

--;===-----;=

../X + h + fx

1 - 2fx

Example:

Differentiate sin2 x from first principle

Solution:

Given f(x) = sin 2 x, f(x.+ h) = sin 2 (x + h) by definition dy sin 2 (x+h)-sin 2 x - = Lim - - - - - dx h-->O h . sin (x + h + x) sin (x + h - x) -~--~-~--~ h-->O h

= L 1m

(.: sin 2A - sin 2B = sin (A + B) sin (A - B) . sin (2x + h) sin h = L 1m --'---'--h-->O

=

h

sin-h· (. Sill 2x + h) . L'1 m L 1m h-->O h-->O h

= Lim sin (2x + h) = sin 2x h-->O

Example:

Differentiate eX by first principle

Solution:

Given f(x) = eX, f(x + h) = ex~h

Differentiation by definition d

--.r = dx

ex+h _ex

Lim - - h-->O h '(

x

h

=

. e.e-e LI m - - - h-->O h

=

lim --'---":" h-->O h

eX(eh_l)

h2 ... -1 eX ( l+h+T!+ =

Lim

h

h-->O

~ + ...)

eX .h (I + =

Lim

h

h-->O

=

1

Lim eX h-->O

(I + E.. + ... ) 2!

=

Example:

Differentiate logx by first principle

Solution:

Given f(x) = logx, f(x + h) = log (x + h)

eX

By definition dy = Lim log (x + h) -log x dx h-->O h

= Lim =

h

log ( x + h ) x

Lim

~h

IOg(l

Lim

~h (~_~(~)2 + ... ] x 2 x

h-->O

=

~

h-->O

h-->O

= Lim h-->O

1

x

(.; log m -log n =

+~) x

(~-~~+ x 2 x 2

. .)

109: )

313

314

Comprehensive Remedial Mathematics for Pharmacy

Example:

Differentiate sin x by first principle

Solution:

Given f(x)

= sin x, f (x + h) = sin (x + h) = Lim (sin (x + h) - sin x) h-+O h

2 cos x + h + x . sin _x_+_h_-_x = Lim h-+O

2

[

1

2

h

. h

sm= Lim cos x +.!: . _ _ 2 ( h-+O 2) h 2

=

sin.!: Lim cos ( x +.!: ) . Lim - h 2 h-+O 2 h-+O 2

= cos x

Example:

Differentiate (sec x + cosec x) w.r.t. x

Solution:

Let

y = sec x + cosec x dy d d -=-secx + -cosec x dx dx dx = sec x tan x - cosec x cot x

Formula (i) Differential coefficient of the product of two functions

(product rule)

d d d -[fl (x).f2(x)] = fl (X)-f2(X) + f2(X)-fl (x) dx dx dx (ii) Differential coefficient of the quotient of two functions d d f2 (x) x -dx fl (x) - fl (x) -dx f2 (x) - - - = ----==------=-----'::::..:..-dx f 2 (x) [f2 (X)]2 d fl (x)

(quotient rule)

Differentiation

Example 4:

Differentiate x2 10g x w.r.t. x.

Solution:

d 0 2d do -(x-. logx) = x -log x + log x -xdx dx dx 2

1

= x . - + log x. 2x x = x + 2x log x =x(l + 210g x) Example 5:

Differentiate sin x . log x w.r.t. x.

· o utlOn: Sl

. -de' Sin X. I og) x = Sin dx

X-d

dx

Sin log x + Iog x -d . dx

X

. X. -1 + Iog x. cos x = Sin x

Sin X

= - - + logx. cos x x Example 6:

Differentiate x3 eX sin x w.r.t. x.

Solution:

~(X3. eX. sin x) dx

= x3 ~ (ex. sin x) + eX sin dx

X~X3 dx

~sin x + sin x ~ex] + eX sin x. 3x2 dx

= x [e X dx 3

3

2

= x le x cos x + sin x.e x J + 3x eX sin x = x3 eX cos x + x3 eX sin x + 3x2 eX sin x

Example 7:

Solution:

. . eX Dlfferentlate- w.r.t. x.

x

d X X d x-e -e - x dx dx x2

315

316

Comprehensive Remedial Mathematics for Pharmacy 2

Example 8:

'f:>: . 5x +6x+7 Dl lerentlate 2 w.r.t. x.

Solution:

2 d [5X +6X+7] dx 2X2 +3x+4

2x +3x+4

(2X2 -

+3x+4)~-(5x2 +6x+7)-(5x2 +6x+7}-~(2x2 +3x+4) ~

~x2+3x+4)

=

~

(2X2 + 3x + 4)(5.2x + 6.1 + 0)- (5x 2 + 6x + 7 )(2.2x + 3.1 + 0) (2X2 + 3x + 4)

=

(2x2 +3x+4)(10x+6)-(5x 2 +6x+7)(4x+3) (2X2 +3x+4)

20x 3 +12x2 +30x 2 +18x+40x+24-20x3 - 15x 2 -24x2 -18x-28x-21 (2X2 +3x+4) 3x2+12x+3 (2X2+3x+4)

Example 9:

x dy Ify = - - , then prove that x - = y(1-y). x+5 dx

Solution:

x y=-x+5 d = ---.r dx

(x+5) dy x-x~(x+5) dx dx (X+5)2 (x + 5) .1- x (1 + 0)

(x + 5)2

= (x+5)-x (x + 5)2

dy dx

--

5 (x + 5)2

Differentiation dy 5x x-= 0 dx (x+5)-

x

5 x+5 x+5 =Y(l _ _ X ) x+5 dy x - =y(l-y) dx

Differential Coefficient of a Function of Function Ify = fez), where z = F(x), then by eliminating z, we get y = f {F(x)} . x3,e smx" cos 2x are t he fiunctIOn . 0 f fiunctIon. . e.g. Sill

Again, let

y = f(t), where t = g(x)

Then

dy = dy . ~, this is known as 'chain rule'. dx dt dx

Example 10: Find the differential coefficient of sin x2 w.r.t. x Solution:

Let

.

1

y= Sill xdy d . , -=-SIllXdx dx

Put

x2 = t dy d. - = -Sillt dx dx d. dt = - SIllt.dt dx d 2 = COS t. - x dx

= COS t. 2x = 2x COS x2

317

318

Comprehensive Remedial Mathematics for Pharmacy

Example 11: Differentiate esmx w.r.t. x. Solution:

Let Put

sinx=t y=e t

Now,

dy = dy.~ dx dt dx d

t

d

.

= -e. -smx dt dx = esmx • cos x Example 12: Differentiate log (log sin x) w.r.t. x. Solution:

Let

y = log (log sin x)

Put

log sin x = t Y = logt dy = dy dt dx dt . dx id . = -dogt. - Iogsmx dt dx . =-1 -d Iogsmx t dx

Again, put sin x = u dy 1 d du -=-.-logudx t du dx lId . = -.-.-smx t u dx 1 - - - . - - . cos x log sin x sin x

cotx log sin x

Differentiation

Example 13: Ify = log

l-cosmx dy , then find - . 1+cosmx dx

Solution:

y= log

dy dx

=

1- cos mx 1 + cos mx

""Lsm . ~ -~ mx d 2 -log , mx dx 2 cos- 2

d mx = - log tandx 2

Put

mx tan-=t 2

dy d d mx - = - logt. - tandx dt dx 2 1 d mx = - . - tant dx 2 mx Again put =u 2 dy = ___ . _d_ tan u. _d__ m_x dx tan mx du dx 2 2

___ sec 2 u. m mx 2 tan2

,mx ---.sec-2 mx 2 tan-

m

2

mx cos-

m

2

mx sm-

2

2

m

smmx = m cosec mx

, mx cos-2

319

320

Comprehensive Remedial Mathematics for Pharmacy

Differential coefficient of inverse Trigonometric Functions or Trigonometrical Transformation d . -1 1 - sm x = --:=== dx Jl-x 2

Formula

d

-1

-1

- cos x = ---:=== dx JI-x 2 d -1 1 - tan x = - dx l+x2 d

-1

-1

- cot x = - dx 1+ x 2 d -1 1 - sec x =--=== dx x~ d

-1

-1

- cosec x = --=== dx x~

Differentiation by Trigonometric Transformation

J

. sm . -1 ( --2 2x Example 14: D 1·f-&lerentlate w.r.t. x. l+x

Solution:

Let

. -1 ( 2xy=sm 1+ x 2

Put

X

= tan 8

J

=> 8 = tan- I x

. -I ( 2 tan8 y=sm 1+ tan 28

= sin- I (sin 28) y=28 Y = 2 tan- I x dy = 2~ tan- I x dx dx 2

J (

0: sin 28 = 2 tan 8 ) 1+ tan 28

Differentiation

Example 15: Ify = coe l

(~ + 1], then find dy. x dx

Solution:

Y = cot

_I(~

Put

x

=> 8 = tan- I x

x=tan8 Y = cot

+1]

_1(~I+tan28 +.1] tan 8

= cot-I (sec8.+ 1J 8

tan

--+1 1

= coel co~8 1 [ sm8 cos8 = cot-I

(1 +sincos8 8J 2

= coel

2 cos [ 2 sin

~cos~

-I 8 = cot cot-

2

8 y= 2 1 -I y=-tan x 2

dy 1 d -I - = - - tan x dx 2 dx 1

1

~2

1

321

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Comprehensive Remedial Mathematics for Pharmacy

Example 16: Ify = sin- I (2xJl-

Xl), then find :~ .

y=sin- I (2xJl-x 2 )

Solution:

Put

~ 8 = sin- I x

x= sin 8

y = sin- I (2sin8Jl- sin 2e) = sin- I (2 sin 8 cos 8) = sin- I sin2 8 y=28 y=2 sin- I x dy dx

-

=

2 -sm d ·-1 x dx 2

=-===

~

Example 17: Differentiate sin-I ( x.Jl-x -fxJl-x2 )w.r.t. x. Solution:

Let

y= sin-I ( x.Jl-x -fxJI-X 2 )

Put

x = sin A, fx = sin B

~

B = sin- I fx

A = sin- I x,

y=sin- l (sinAJl-sin 2 B-sinBJl-sin 2 A) = sin- I (sin A cos B - cos A sin B) = sin- I sin (A - B) y=A-B y = sin- I x - sin- I fx dy d. -I d. -I r -=-sm x--sm '"I/X dx dx dx

=

1

Jl-x 2

d. -I dt --sm t . dt dx

put fx =t

Differentiation 1

.~£

1

~1- t 1

323

2

dx

1

1

~1-x2 - .JI-x . 2£ 2 ~x - x 2

Differentiation of Implicit Functions There may be a relation between x and y when it is not possible to express it in the form ofy = f(x) 'or' to solve an equation for y in terms ofx. We call such a relation as implicit function. For dy by differentiating the given relation w.r.t. x dx

Example 18: Find dy , when x2 + y2 = a 2 dx

Solution:

Given

x 2 + i = a2

differentiating both the sides w.r.t. x, we get d 2 dx

d 2 dx

d 2 dx

-x + -y =-a 2x + 2y dy = 0 dx dy 2y - =-2x dx dy = -2x dx 2y

-x y

Ans

Example 19: Find dy , when ax2 + 2hxy + bi + 2gx + 2fy + c = 0 dx

Solution:

Given

ax2 + 2hxy + bi + 2gx + 2fy + c = 0

324

Comprehensive Remedial Mathematics for Pharmacy differentiating both the sides w.r.t. x, we get d dJ d d d d a-x2 +2h - xy+b-y-+2g-x+2f-y+-c=O dx dx dx dx dx dx 2ax+2h(xdY +y~x) +2by dy +2g+2f dy dx dx dx dx 2ax + 2hx dy + 2yh + 2by dy + 2g + 2f dy = 0 dx dx dx (2ax + 2hy + 2g) + (2hx + 2by +2t) dy = 0 dx 2 (hx + by + t) dy = - 2 (ax + hy + g) dx dy = - (ax + hy + g) dx (hx+by+f)

Example 20: Find dy, when 4x2 + 9/ = 36. dx Solution:

Given

4x2 + 9y2 = 36

differentiating both the sides w.r.t. x, we get d 2 d 2 d 4.-x +9-y = - 36 dx dx dx 4.2x + 9.2y dy = 0 dx 8x + 18y dy = 0 dx 18y dy = - 8x dx dy dx

= -8x 18y -4x 9y

+~=O

Differentiation Example 21: Find dy, when ~ + ~ = l. dx x y Solution:

Given

1

1

x

Y

- +-

=

1

y+x = 1 xy y+x=xy differentiating both the sides w.r.t. x, we get dy d dy d -+-x=x-+y-x dx dx dx dx dy +1 =x dy +y dx dx dy dy --x-=y-l dx dx dy (l-x)=y-l dx dy y-l -=-dx I-x Example 22: Solution:

Find dy, when x3 + Sxy + y3 = 64. dx Given x3 + Sxy + l = 64 Differentiating both the sides w.r.t x, we get d 3 d d 3 d - x +S-xy+-y =-64 dx dx dx dx 3X2+S(xdY +Y)+3 y2dY =0 dx dx 3x 2 +Sx dy +Sy+3 y 2 dy =0 dx dx (3x 2 + Sy) + (Sx + 3l) dy = 0 dx (Sx + 3l) dy = -(3x2 + Sy) dx dy -(3x 2 +Sy) -dx SX+3y2

325

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Comprehensive Remedial Mathematics for Pharmacy

Logarithmic Differentiation: If the power of a function of x is also a function of x, then it would be convenient for us if we first take logarithms w.r.t the base e and then differentiate both sides w.r.t x. Let us recall the formulae

(i)

log mn = n log m

(ii)

log(mn) = log m + log n

(iii)

log -

m n

=

log m - log n

Example 23:

If Y = xx, then prove that dy = XX (1 + logx) dx

Solution:

Given y = XX Taking log on both the sides, we get log y = log XX log y = x . log x Differentiating both the sides w.r.t x, we get d d d -logy = x-Iogx + logx-x dx dx dx 1 dy 1 - - = x.-+ logx.l Y dx x 1 dy --=1+10gx Y dx

•

dy = y(l + log x) dx

Differentiation of Parametric equation: when variables x and y are expressed in terms of a thir~ variable like t or e, then this third variable is called parameter. e.g. x = get) and y = h(t), these are called parametric equations. Formula

--~

dy = dy I dt dx dx/dt

Example 24:

Ifx = a sec e,

y = b tan e,

Solution:

Given.x~

y=btane

a sec e,

"

find dy. dx

Differentiation dx - = a sec 8 tan 8, d8 dy dy/d8

327

2 dv -" = b sec 8 d8

---=----

asec8tan8 b =-cosec8 a

Example 25:

1 dy Ifx = at-, y = 2at, then find - . dx

Solution:

Given x = at2

y = 2at

dx - =2at dt dy dv,dt

dy =2a dt

-=-"--

dx

dx,dt 2a 2at

=

t

Differentiation of Infinite Series: When the function is given in infinite series. In this case we use the fact that if a term is deleted from an infinite series it remains uneffected. '"

Example: Solution:

dy , find - . dx

here y = xY ~

log Y = Y log x

Differentiating w.r.t x, we have 1 dy d dy - - = y-logx + logxy dx dx dx

=2::. + logx dy x

dx

(~-IOgX] = 2::.x

dy dx y

328

Comprehensive Remedial Mathematics for Pharmacy dy dx dy dx

[1- YyIOgX) =.rx 1

yx(l-ylogx)

= -----''----

y=Jsinx+~sinx+Jsinx+ .......oo

Example:

If

Solution:

Here

, Prove that dy = cosx

dx

y=~sinx+y

l

= sinx +y

2ydy =cosx+ dy dx dx dy (2y -1) = cosx dx dy cosx ---dx 2y-l ~rxj")

Example:

If y = (~ I

Solution:

Here

D

x

d

y=(~r logy=y log ~ logy=

r2

log x

Differentiating w.r.t x, we get 1 dy y 1 log x dy --=-.-+--ydx 2 x 2 dx

dY[~_IOg~)=L

dx y

2

2x

I

dy [2 -ylogx =L dx 2y ) 2x dy dx

2

show that ~ = y dx x(2-ylogx)

y2

-=----'---

x(2-ylogx)

2y-l

329

Differentiation

Example:

If y = e He ,~e'~ ~,show that dy = _y_ dx l-y

Solution:

Here y = eX+Y Taking log on both sides (.: loge = 1)

log y = (x + y) log e log y = x + y Differentiating w.r.t x, we get

~ dy =1 + dy y dx

dx

dY(~_I)=1

dx y

dY(I-y) =1 dx y dy dx

y I-y

----

y=Jx+~x+.Jx+ ......oo

Example:

If

Solution:

Here y=~x+y

then prove that dy =_·_1_ dx 2y-1 ..... (1)

Taking log on both sides log y =

1 2

- log(x + y)

Differentiating w.r.t x, we get

~ dy =~._I_(I+ dY) y dx

[

I

2 x+y

1

y- 2(x+y)

1

dx

dy 1 dx = 2(x+y)

From eg. (I) x + Y = / Putting in eg. (2) we get

( ~Y __2y21_) dydx -__2y21_

..... (2)

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Comprehensive Remedial Mathematics for Pharmacy

[

2Y-1)dY _ 1 dx - 2y2

2T

dy dx

2y-1

Successive Differentiation: Differentiating the first derivative of a function we obtain second derivative, differentiating the second derivative, we get the third derivative and so on. This process of finding derivatives of higher order is known as successive differentiation.

Example 27:

Find second derivative ofax3 + bx 2 + cx + d

Solution:

Let y = ax3 + bx2 + cx + d Differentiating w.r. t x, we get dy = 3ax2 + 2bx + c dx Again differentiating, we get

~ (dY) dx dx

= 3a (2x) + 2b + 0

d2 y - 2 =6ax+2b dx

Example 28:

d2 If Y = x + tanx, Prove that cos 2 x - { + 2x = 2y dx

Solution:

Given y

= x + tanx

Differentiating w.r.t x, we get dy = 1 + sec2 x dx Differentiating again, we get

~ (dY ) dx dx

= 0 + 2 sec x . sec x tan x

d 2y

--2 = 2 sec2 x tan x dx

2tanx - cos 2 x

Differentiation

Some Practice Questions First Principle

Example 1:

Differentiate cos x by frrst principle.

Solution:

Let f(x) = cos x, f(x + h) = cos(x + h) Then by frrst principle dy

-dx

lim f(x + h) - f(x) h~O

=

=

h

lim cos(x + h) -cosx h~O

h

. x+h+x . x-x-h lim 2 sm 2 sm 2 h

h

lim 2 sin( x +

=

h~O

%)sin( -2

)

h

. h . ( h) lim sm 2 =sm x+- . -h~O 2 h~O h 2 lim

=-sin x

331

332

Comprehensive Remedial Mathematics for Phannacy

Example 2:

Differentiate tan x by first principle.

Solution:

Let f(x) = tan x,

f(x + h) = tan (x + h)

then by first principle dy dx

--

=

lim f(x + h) - f(x) h~O h lim tan(x + h) - tan x h

h~O

sin(x + h) lim cos(x + h) = h~O h

=

sin x cosx

lim sin(x + h)cosx - cos(x + h)sin x h~0 hcosxcos(x + h) lim

sin(x + h - x)

=h ~ 0 hcosxcos(x + h) =

lim sin h lim 1 h ~ 0 h 'h ~ 0 cosx.cos(x + h)

=lx---cosx.cosx

Example 3:

Differentiate sec x by first principle.

Solution:

Let f(x)

= sec x, f(x + h) = sec(x + h)

Then by first principle dy =lim f(x+h)-f(x) dx h-+O h . sec(x+h)-secx =11m --'----'----h-+O

h

Differentiation

-------

= lim cos(x + h)

cosx

h

h.....O

cos x - cos(x + h) = II' m ------'-h ..... O hcosxcos(x+h)

'( x+h), h ,sm2 sm 2 2 HO h cos x cos( x + h)

= lim

, h

sm2 = lim sin(x +.!:) , lim - - - - - - ,lim -h2

h .....O

h ..... O

cosxcos(x+h)

h ..... O

2 smx

=----xl

cosx,cosx

= sec x, tan x Example 4:

Differentiate .Jsin x by first principle,

Solution:

Let f(x) = .Jsin x ,

f (x + h) = ~sin (x + h)

Then by first principle dy dx

= lim f(x + h) h ..... O

f(x)

h

, ~sin (x + h) - .Jsinx =hhm --'-------.....O h '

= I1m

~sin (x + h) - .Jsin x h

h ..... O

'

sin (x + h) - sinx

= I1m -r========:--r===h ..... O h(~sin

(x + h) + .Jsin x)

2 cos (x + .!:), sin.!:

~sin (x + h) + .Jsin x

x -'-;=====-----r==

' 2 2 = I1m -r=====---'r===HO h(~sin (x + h) + .Jsin x)

~sin (x + h) + .Jsin x

333

334

Comprehensive Remedial Mathematics for Pharmacy

.

cos( x+%)

= hm-r======;--.j'==' h~O ~sin (x + h) + sin x

l' sin% lm-h hO 2

cosx

=-==;----==

.J sinx + .Jsinx

= ExampleS: Solution:

cosx 2 .Jsinx

Differentiate sin-1x by fIrst principle. Let f(x) = Sin-1 x = y, f(x + h) = sin-1 (x + h) = y + k ~

x = sin y

x + h = sin (y + k)

Then by fIrst principle dy = lim f(x + h) - f(x) dx h~O h . sin-l (x+h)-sin-1x = 11m ---'-----''---h

h~O

. Y+ k - y =11m ---=----"h~O

(x + h) - x

k

(ash~O ~ k~O)

=lim-----k~O sin (y + k) - sin y

r

n~

k

= ,'To 2COs(Y+

Sin

k 1

· = 11m

k~O

cos

(

k) y+"2

. l'I m2- -

1 1 =--xl =--

cos Y cos Y 1 =-;==== ~l- sin 2 y 1

=--===

F-7

k

k~O. SlD"2

Differentiation

Example 6:

Differentiate sinFx by first principle.

Solution:

Let f(x) = sinFx ,

335

f (x + h) = sin.Jx + h

Then by fi~st principle dy

dx

= lim f(x + h) - f(x) h

h->O

.

sin ~(x + h) - sinfx

= 11m - - - ' - - - - - - -

h

h->O

2

cos( ~ +~)Sin( ~ -~:

= lim--~--------'------'------'h

h->O

(rx+h +fx)

2

=lim

cos

2'

h

h->O

= lim

2 c