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1A

SIMULTMEOUS LINEAR EQUATION

UNIT STRUCTURE

1A.1 Introduction

1A.2 Methods for solving simultaneous linear equations

1A.3 Unit E nd E xercise

1A.1. INTRODUCTION

Equation of a line in a plane is of first degree in x & y and

Conversely every eq uation of first degree in x & y represents a line.

Linear equation:

The general equation of the set of equation ‘ x’ & ‘ y’ variable of the

type

ax by c o with a, b, c real number and at least one of a and b is

not zero, is closely associated with the set of lines in a plane is called

linear equation . Linear equation is obtained by equating to zero a linear

expression.

Linear Expression:

Any expression of the type

ax by c a, b, c in R and at least one

of a and b is non zero, is called linear expression.

1A.2. METHODS FOR SOLVING SIMULTANEOUS

LINEAR EQUATIONS:

*Simultaneous L inear Equation in Two variables :

Let

1 1 1 0 a x b y c and

2 2 2 0 a x b y c . A value for each

variable which satisfies simultaneously both the equations will give the

roots of the equation. There are two methods to solve the given

simultaneous equation

1) Elimination method

2) Cross Multiplication Method

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(1) Elimination Method: -

In this method, two given equations are reduced to a linear

equation in one unknown by eliminating one of the unknow ns and then

solving for the other unknown.

Eg. (i) Solve 2x + 5y = 9 and 3x – y = 5

2x + 5y =9...... (1)

3x – y = 5........ ( 2)

Multiply eqn (2) by 5,

We get,

15 5 25......(3)xy

Adding (1) & (3),

2x + 5y = 9

15x – 5y = 25

17x = 34

2x

Substituting x = 2 in eqn (1),

2x + 5y = 9

2(2) + 5 y = 9

1y

Thus x = 2, y = 1 is the required solution.

(2) Cross-Multiplication Method: -

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

We write the coefficients of x, y, and constant terms and two more

columns by repeating the coefficients of x and y as follows:

1 2 3 4

b1 c1 a1 b1

b2

c2

a2

b2

and the result is given by: munotes.in

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1 2 2 1 1 2 2 1 1 2 2 11 xy

b c b c c a c a a b a b

So, the solution is:

1 2 2 1

1 2 2 1b c b cxa b a b

1 2 2 1

1 2 2 1c a c aya b a b

E.g.

1) Solve 3x + 2y + 17 = 0

5x – 6y – 9 = 0

Solution:

By comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 then using the

result

1 2 2 1 1 2 2 1 1 2 2 11 xy

b c b c c a c a a b a b

1

2 9 17 6 17 5 3 9 3 6 5 2xy

1

84 112 28xy

1.3 4 1xyie

34 xy

Thus x = -3, y = -4 is the required solution.

Simultaneous Linear Equations with Three variables Methods :

Methods: (1) Elimination method

(2) Cross Multiplication method

E.g 1 ) Solve for x, y and z.

3x – 2y + 4z = 1

2x – y + z = 3

x + 3y – 2z = 11

(a) Method of Elimination : Any two of three equations can be chosen

for limi tation of one of the variable.

2x – y + z = 3 ........... (i)

x + 3y – 2z = 11 ........... (ii) munotes.in

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3x – 2y + 4z = 1 ........... (iii)

Multiplying eqn (i) by 2

4x – 2y + 2z = 6...... (iv)

Eliminating variable ‘ z’ by adding (ii) and (i v),

5x + y = 17 ...... (v)

Multiplying eqn (ii) by 2

2x + 6y – 4z = 22... ... (vi)

By Adding (iii) and (vi), 5 x + 4y = 23......... (vii)

Subtracting eqn (vii) from eqn (v),

-3y = – 6

2y

Substituting y = 2 in eqn (v)

5x + 2 = 17

3x

Substituting x = 3 and y = 2 in eqn (i)

2(3) – 2 + z = 3

1 z

x = 3, y = 2, z = -1 is the required solution .

(b) Method of Cross Multiplication:

2x – y + z = 3 …. (i)

x + 3 y – 2z = 11 …. (ii)

3 x – 2 y + 4z = 1 …. (iii)

The equations (i) & (ii) can be written as follows: -

2x – y + (z-3) = 0

x + 3y + (-2z-11) = 0

By cross multiplication,

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x y 1==-1 -2z-11 -3 z-3 z-3 -2 -2z-11 2 3 -1 -1

x y 1==20-z 5z+19 7

20-x=7z

5z+19y=7

Substituting above values for x & y in eqn (iii) i.e 3 x – 2y + 4z = 1

20-z 5z+193 -2 +4z =177

60 - 3z - 10z - 38 + 28z = 7

15z = 7- 22 15z = – 15

1 z

Now,

20 ( 1) 21

77x ∴

3x

5 1 19 14

77y ∴

2y

x = 3, y = 2, z = -1 is the required solution .

3) It the numerator of a fraction is increased by 2 and the denominator

by 1, it becomes 1. Again, if the numerator is decreased by 4 and the

denominator by 2, it becomes

12 . Find the fraction.

Let

xy be the required fraction (given)

211

41

22x

y

x

y

x + 2 = y + 1 i.e. x – y = -1 ..... (i)

2x – 8 = y -2 ∴ 2x – y = 6 .... (ii)

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2x – y = 6 (i)

x – y = –1 (ii)

– + +

x = 7

7x and

Substituting x=7 in eqn (i) we get, 2(7)-y=6

14-y=6

y= 14 - 6 = 8

8y

The fraction =

7

8x

y

4) The age of a man is three times the sum of the ages of his two sons and

5 years hence his age will be double the sum of their ages. Find the

present age of man?

Let ‘x’ years be the present age of man and sum of present ages of

the two sons be ‘y’ years.

Given condition x = 3y .... (i)

x + 5 = 2 ( y + 5 + 5) .... (ii)

Substituting eqn (i) in eqn (ii),

3y + 5 = 2( y + 10)

15y

by substituting y = 15 in (i) x = 3y = 3(15) = 45

45x

Hence, the present age of man is 45 years

1A.3. UNIT END EXERCISE:

Multiple Choice Questions

1) The solution of the set of equations 3 x + 4y = 7, 4x – y = 3 is

[A] (1, -1) [B] (1, 1) [C] (2, 1) [D] (1,-2)

2) The values of x and y satisfying the equations

2, 2 83xyxyy

are given by the pair

[A] (3, 2) [B] ( -2, -3) [C] (2, 3) [D] None of these

3) Solve for x and y: x - 3y = 0, x + 2y = 20 munotes.in

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[A] x = 4, y = 12 [B] x = 12, y = 4 [C] x = 5, y = 4 [D] None of these

4) Solve:

7 8 5 624 3 2x y zx y z

[A] (4, 3, 2) [B] (2, 3, 4) [C] (3, 4, 2) [D] (4, 2, 3)

5) Solve: 3x – 4y + 70z = 0, 2x + 3y – l0z = 0, x + 2y + 3z = 13

[A] (1 , 3, 7) [B] (1, 7, 3) [C] (2, 4, 3) [D] ( -10, 10, 1)

6) Monthly incomes of two persons are in the ratio 4:5 and their monthly

expenses are in the ratio 7:9. If each saves Rs 50 per month find their

monthly in comes.

[A] (500,400) [B] (400,500) [C] (300,600) [D] (350,550)

7) The age of a person is twice the sum of the ages of his two sons and

five years ago his age was thrice the sum of their ages. Find his

present age.

[A] 60 years [B] 52 years [C] 51 years [D] 50 years

8) The sum of the digits in a three digit no. is 12 . If the digits are

reversed, the no. is increased by 495 but reversing only of tens &

units digits increases the number by 36. The number is

[A] 327 [B] 372 [C] 237 [D] 273

Exercise s

1) Solve x + y + z = 5, 2 x – 3y – 4z = -11 and 3 x + 2y – z = -6

2) Find the values of x and y for equations x + 5y = 36 and

5

3xy

xy

3) The wages of 8 men and 6 boys amount to Rs.33. If 4 men earn

Rs.4.50 more than 5 boys. Determine the wages each man and boy.

4) The demand and supply equations for a certain commodity are

4q + 7p = 17 and

7

34qp respectively.

Where p market price; q: quantity . Find the equilibrium price and

quantity .

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1B

QUADRATIC EQUATIONS

UNIT STRUCTURE

1B.1 Introduction and Definitions

1B.2 Unit E nd Exercise

1B.1. INTRODUCTION AND DEFINITIONS

*Polynomial :

An expression of the form a0xn + a 1xn-1 ……… +……. + an-1 x +an,

where

00 a

a1, a2 ....... a n are constants

n : Positive integer

x : Variable (un known)

is called a polynomial in ‘ x’ of degree ‘ n’ and it is denoted by f( x).

Eg. x2 – 3x + 1, 2 x3 – 3x2 + 5,

42 2 3xx polynomials in ‘ x’.

Quadratic Expression :

An expression of the form ax2 + bx + c,

0a and a, b, c

R, is

called a quadratic expression, where R is the set of real numbers.

Eg. (i) 2x2 + 3x + 4

(ii) x2 – 4x + 5 are examples of quadratic expressions .

Quadratic Equation :

The equation of the form ax2 + bx + c = 0,

0a , a, b, c, ∈ R

is called a quadratic equation where R is the set of real numbers. I t is an

equation of degree 2. Generally, coefficients of quadratic equations

encountered are rational numbers .

Eg (i) 3x2 + 7x + 12 = 0

(ii) x2 + 4 = 0 are examples of quadratic equations.

Root of a quadratic equation :

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The value of x which satisfies the quadratic equation ax2 + bx + c = 0

is called the root of given quadratic equation.

Thus, if

x , is the root of quadratic equation

20, ax bx c then

20 a b c

Roots of a Quadratic equation :

The roots of the quadratic equation

ax2 + bx + c = 0 are given by

24

2b b acxa

i.e.

2244,22b b ac b b ac

aa

Sum and Product of the Roots of a Quadratic Equation:

Let

, be the roots of quadratic eqn ax2 + bx + c = 0. If S is the

sum of roots and P the product of the roots of quadratic equation, then the

quadratic equation is

x2 – Sx + P = 0

i.e. x2 – (Sum of roots) x + (Product of roots) = 0

Where,

bSa and

cPa

There are two types of quadratic equations

(1) Pure

(2) Affected

(1) Pure Quadratic Equation:

A quadratic equation is said to be Pure if coefficient of x is zero.

Thus, a pure quadratic equation is of the type ax2 + b = 0;

0a

(2) Affected Quadratic Equation:

A quadratic equation which is not pure is called an affected quadratic

equation. The m ost general form of an affected quadratic equation is

ax2 + bx + c = 0 ;

0 & 0ab

Methods of Solving Pure Quadratic Equations :

Let ax2 – b = 0 be a pure quadratic equation. This implies munotes.in

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22 bbax b x xaa

The roots of ax2 - b are real if ‘ a’ and ‘ b’ are of opposite sign.

Eg. Solve

2 316 9 04xx

Methods of Solving Affected Quadratic Equations :

(1) Method of factorization

(2) Method of perfect square

(1) Method of factorization :

If the expression ax2 + bx + c can be factorized into linear factors

then each of the factors, put to zero, we get two roots

& for the given

quadratic equation

Thus if

2, ax bx c a x x then the roots of

20 ax bx c

are

&

Eg (i) Solve : x2 - 8x + 12 = 0 12

(x – 6) (x – 2) =0

x – 6 = 0 or x – 2 = 0 -6 -2

x = 6 or x = 2

Hence, the r oots of given equation are 6 or 2

(ii) Solve :

28 48 0xx 48

212 4 48 0 0x x x

12 4 12 0 x x x -12 4

4 12 0 xx

4 0 12 0x or x

4 12 x or x

Hence, the roots of given equation are -4 or 12

(iii)Solve :

23 7 2 0xx

23 6 2 0x x x

3 2 1 2 0x x x

3 1 2 0xx

3 1 0 2 0x or x

1 23x or x

Hence, the roots of given equation are -1/3 or -2 munotes.in

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(2) Method of Perfect Square :

When ax2 + bx + c cannot be factorized easily into its linear factors.

If can be solved by the method of perfect square.

Step 1: Let

20 ax bx c be the given equation. Divide both sides of

equation by a , we get

20bcxxaa

0a

Step 2: Transpose the constant term (i.e . the term independent of x) on

R.H.S.

2bcxxaa

Step 3: Add

2

24b

a on both sides to make R .H.S. a perfect square

Thus,

22

2

2244b b c bxxa a a a

2 2

24

24b b acxaa

Thus, a pure equation in the variable

2bxa is

24

22b b acxaa

24

22b b acxaa

24

2b b acxa

Solve:

(i)

22 5 1 0xx cannot be easily factored into linear factors

Comparing with

20 ax bx c

a = 2, b = -5, c = -1

Roots are given by

24

2b b acxa

i.e. 𝑥=−(−5)±√ (−5)2−4(2)(−1)

2(2)

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5 33

44

Thus, the roots are

33 544 and

33 544

Nature of the Roots of a Quadratic Equation :

The roots of the quadratic equation

20 ax bx c

are given by

24

2b b acxa

The quantity, b2 – 4ac is called Discriminant of the quadratic equation

and is generally denoted by

'' .

Thus,

24b ac

Nature of the roots of a quadratic equatio n depends upon its discriminant

for a, b, c

R

(1) If

24 0,b ac roots are real and different .

(2) If

240b ac and is a perfect square, then the roots are real .

different and rational .

(3) If

240b ac and is not a perfect square, then the roots are

real different and irrational .

(4) If

240b ac , then the roots are real & equal .

(5) If

240b ac then the roots are complex and different .

Note :

(i) If a, b, c are rational, then the irrational roots occur in pairs If

one of the root is

, pq the other must be p -iq

(ii) If a, b, c are real, then the complex roots, if any occur in the

other must be p – iq

(iii)If

, are roots of quadratic eqn

20, ax bx c then

0 a x x

Important Results :

(1) If in the quadratic equation

20 ax bx c

0 abc , then one

root of quadratic eqn is unity (1) and the other root is

ca .

(2) If the quadratic eqn

20 ax bx c is satisfied by more than two

values of x i.e it has more than two roots, then it must be an

identity for which a = b = c = 0

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(3) If the roots of quadratic equation

20 ax bx c are real and

distinct, then one root must be greater than

2ba and the other

less than

2ba

Symmetric Function :

Any expression involving

and

as its roots is called a

symmetric function of

and

, if it remains unchanged when

and

are interchanged

Eg.

2 2 2 2,, are all symmetric functions of

and

,

whereas

3 is not a symmetric function since in general

3 need

not be equal to

3

A symmetric functions of

and

can be sol ved with help of

Sum of roots bSa

Product of roots cPa

Examples :

1) Examine the nature of roots of following equation s:

(i)

28 16 0xx

Here, a = 1, b = -8, c = 16

2 24 8 4 16 1b ac

=64-64=0

Discriminate

24 0,b ac the roots are real and equal

(ii)

23 8 4 0xx

Here, a = 3, b = -8, c = 4

2 24 8 4 3 4b ac

64 48

16

Discriminat e

240b ac and a perfect square, the roots are

real, rational and unequal .

2) Solve :

21 1 1274xxxx munotes.in

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21 1 2924xxxx

21 1 29424xxxx [

224 a b a b ab

]

Let

1, xtx

2 29424tt

24 8 45 0tt

24 18 10 45 0t t t

2 2 9 5 2 9 0t t t

2 5 0 2 9 0t or t

5922t or t

Either

1 5 192 2x or xxx

222 5 2 0 2 5 2 0x x or x x

Either

( 5) 25 16 9 81 16

44 x or x

5 3 9 65

44 x or x

∴𝒙=𝟖

𝟒=𝟐 𝒐𝒓 𝒙=𝟐

𝟒=𝟏

𝟐 𝒐𝒓 𝒙= −𝟗+√𝟔𝟓

𝟒,𝒙=−𝟗−√𝟔𝟓

𝟒

1 9 65 9 652, , ,2 4 4 x

Hence, the roots of given equation are 2, ½ , −𝟗+√𝟔𝟓

𝟒,−𝟗−√𝟔𝟓

𝟒

3) Solve the equation :

226 9 4 6 6x x x x

Let

266x x t

34tt

26 9 16t t t

210 9 0 tt

1 9 0tt

t = 1 or t = 9

226 6 1 6 6 9x x or x x

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6 36 4 1 3

1 5 02x x or x

6 4 31,52x or x

1,5 3 2 3x or x

Hence, the roots of given equation are 1, 5, 3+2√3,3−2√3

4) Find value of

6 6 6 ....

Let

6 6 6 .... x

𝒙

6x

26 xx [∵ taking square on both the sides]

260 xx

(x – 3) (x + 2) = 0

32x or x

The value of

6 6 6 .... is 3 or -2

5) It the roots of e q.

20 x px q

'' &

'' and form the equation

whose roots are

&

,

are roots of e qn

20 x px q .

Comparing with ax2 + bx + c = 0. a = 1, b = –p, c = q

pp

and

q

,

bc

aa

Now,

2222

22pq

q

1

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2

2 2Requiredeq 1 0n pqis x xq

[

2Pr ) 0

x Sumof roots x oduct of root ]

2220 qx p q x q

6) Solve :

435 15 9 0x x x

425 3 9 0x x x

4 2 2 26 9 5 3 6 0x x x x x

(𝑥2−3)2−5𝑥(𝑥2−3)+6𝑥2=0

Substitute

23xt

225 6 0t xt x

Roots of above eqn t = 2x or t = 3x

𝑥2−3=2𝑥 𝑜𝑟 𝑥2−3=3𝑥

𝑥2−2𝑥−3=0 𝑜𝑟 𝑥2−3𝑥−3=0

(𝑥+1)(𝑥−3)=0 𝑜𝑟 𝑥=3±√9−4(−3)(1)

2

𝑥=−1,3 0𝑟 𝑥=3±√21

2

Hence, the roots of given equation are -1, 3, 3+√21

2, 3−√21

2

7)

25: 4 3.2 2 0xxSolve

254 3.2 2 0xx

222 3.2 .2 32 0xx

22 12.2 32 0xx

Let 2x = y

212 32 0yy

28 4 32 0y y y

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48y or y

2 4 2 8xxor

222x or

322x

23x or x

8) If

, be the roots of

22 4 1 0xx the value of

22

1422 [∵α+β= −𝑏

𝑎 𝑎𝑛𝑑 𝛼𝛽=𝑐

𝑎]

and

12

3

2 2 3 3 3

3 1 (2) 3 (2)2

12

83

1/ 2

11

1/ 2

22

The value of

22

= -22

9) Divide 25 into two parts so that sum of their reciprocals is

16 .

Let the parts be x and 25 – x

Given condition :

1 1 1

25 6xx

25 1

25 6xx

xx

2150 25 xx

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15 10 15 0 x x x

15 10 0xx

x = 10, 15

Part of 25 are 10 and 15.

10) A piece of iron rod casts ` 69. If the rod was 2 meter shorte r and each

meter unchanged, w hat is the length of rod ?

Let the length of rod be ‘x’ meters

The rate per meter

60

x

New length = ( x – 2)

As cost is same, n ew rate per metre

60

2x

Given condition:

60 6012xx

60 6012xx

12012 xx

120 2 xx

22 120 0xx

12 10 0xx

Either x = 12 or x = -10

The length c annot be negativ e. Therefore, the length of the rod is 12m .

Simultaneous Equations in two unknowns

(1) Linear simultaneous equations

(2) Non-linear simultaneous equation s

Eg. (1) Solve :

24 3 1, 12 13 25x y xy x munotes.in

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From

414 3 1,3xx y y

Substituting the above value of y in

212 13 25xy x

2 4112 13 253xxx

2216 4 13 25x x x

229 4 25 0xx

29 25 1 0xx

25129x or x

In terms of y

31

4yx

3 1 3 1 2514 4 29yyor

43129y or y

25 43Requiredsolution is 1, 1 ,29 29x y or x y

(2) Solve :

2 3 5, 1x y xy

1 2 3 6xy x y xy = 6(1) = 6

(2x) (3y) = 6

2 3 5xy

As

222 3 2 3 4 2 3x y x y x y

25 24 [

2x + 3y = 5 & (2 x)(3y) = 6]

1

2 3 1xy

∴ 2𝑥−3𝑦=1 𝑜𝑟 2𝑥−3𝑦=−1

Now 2𝑥+3𝑦=5,2𝑥−3𝑦=1 𝑜𝑟 2𝑥+3𝑦=5,2𝑥−3𝑦=−1

∴ 𝑥=3

2,𝑦= 23 𝑜𝑟 𝑥=1,𝑦=1 ⁄⁄

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Hence, the required solution is

1, 1xy

or

3 22, 3xy

(3) Solve

227, 133 x y xy x y xy

2 22x y xy x y xy

x y xy x y xy

133 7 x y xy

7

x y xy

19 x y xy …. (1)

7 x y xy …. (2)

Adding (1) & (2)

2 26xy

∴

13 xy …. (3)

Subtracting (2) from (1)

2 12xy

6 36 xy xy …. (4)

Substituting

n 36 in eq yx (3)

3613 xx

213 36 0xx

29 4 36 0x x x

9 4 9 0 x x x

94x or x

49y or y

Required solution is

9, 4xy

4, 9 or x y

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Simultaneous Equations in three unknowns

Eg: (1 )

5 4 0x y z …… (i)

2 5 4 0xyz …… (ii)

2 2 220x y z …… (iii)

From (i) & (ii)

By cross multiplication ,

16 5 2 20 25 8x y z

11 22 33x y z

23yzxk

, 2 , 3 x k y k z k

Substituting the above values in eqn (iii)

2 2 28 9 0k k k

220k

0k

0, 0, 0x y z

(2)

5,x y z …… (i)

8 y z x …… (ii)

9z x y …… (iii)

Adding (i) & (ii ),

13 xy xz yz ………. (iv)

Subtra cting ( iii) from (iv ),

24xy

2 xy ………………………… (v )

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Similarly adding (2) & (3) & subtracting eqn (1) from the sum,

2 17yz xz xy

5 xz xy

2 12yz

6 yz …………………………….. (vi )

Similarly adding (1) & (3), and subtracting eqn (2) from sum

2 14xz xy yz

8 xy yz

26xz

3xz ………………………………… (vii )

12233x y y

z x z [

From (v), (vi) and (vii )]

23yzxk

, 2 , 3 x k y k z k

Substituting the above values in eqn (i)

5 x y z

2 3 5k k k

222 3 5kk

255k

21k

1 k

1 2, 3 x y z

Required solution :

1, 2, 3x y z

OR

1, 2, 3 x y z

(3)

245 x xy xz

275 yx y yz munotes.in

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2105 zx zy z

45 x x y z ……………………………….. (1)

75 y x y z ………………………………. (2)

105 z x y z ………………………………. (3)

Adding we get,

2225 x y z

15 x y z

3, 5, 7 x y z [

From (1), (2) and (3)]

Required solution :

3, 5, 7x y z

3, 5, 7 or x y z

1B.2. UNIT END EXERCISE

*Multiple Choice Questions :

(1) If

2 3 22 3 .2 1 0,xx then values of x are

[A] 0, 1 [B] 1, 2 [C] 0, 3 [D] 0, -3

(2) If

be the roots of eqn

22 4 3 0,xx the values of

22 is

[A] 5 [B] 7 [C] 3 [D] -4

(3) The equation

24 2 5 0 x p x p has equal roots, the value of

p will be

[A]

1 [B] 2 [C]

2 [D] – 2

(4) If the roots of eqn

222 1 0 x p x p are real, then

11 A 1 B 4 C D 44p p p p

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(5) If the roots of the eqn exceeds the other by 4, then the value of m is

A 10 B 11 C m = 9 D m = 12 mm

(6) The area of a rectangular field is 2000 m2 and its perimeter is 180 m.

Form a quadratic eqn by taking the length of field as x and solve it to

find length of breadth of field. The length and breadth are

A (205 ,80 ) B (50 ,40 ) C (60m,50m) D None m m m m

Exercise:

1)

2

2114 0 Solvefor x xxxx

2) Solve

22

22113

11xx

xx

3) Solve

213320 xx

4) Solve

223 18 3 4 6 4x x x x

5) If

and

be the roots of

27 12 0.xx Find the equation

whose roots are

2 and

2

6) If are roots of

22 3 7 0.xx the valves of

and

33

7) Solve

2229, 3 x y x y

8) Solve

14 x y xy

2284 x y xy munotes.in

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9) Solve

3 2 0, 4 3 0x y z x y z

3 3 3467 x y z

10) Solve

23 xy x y

41 xz x z

27 yz y z

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2A

Determinants

UNIT STRUCTURE

2A.1 Objectives

2A.2 Introduction

2A.3 Evaluation of Determinant

2A.4 Properties of Determinant

2A.5 Minors and co -Facto rs

2A.6 Cramer’s Rule for solving Linear equations

2A.7 Unit End Exercise

2A.1. OBJECTIVES

In this chapter a student has to learn the

Concept of Determinant.

Minors and co -Facto rs

Applications of Determinant in solving Linear equations

2A.2. INTRODUCTION

* Determinants:

The determinant has definite value. In determinant number of rows and columns are always

equal .

e.g.

11 12

1 11 12 21 22

21 22 22

11 12 13

2 21 22 23

31 32 33 33

11 12 13 14

21 22 23 24

3

31 32 33 34

41 42 43 44 4, , , .

. 2

2 ; x

xd

xn

nddeterminant of order and

daaD is a a a a are called its elementsaa

a a a

D a a a is and

a a a

a a a a

a a a aDa a a a

aeterminant of o

ae

ard

ar

4 4 .thdeterminant ofs order i

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2A.3. EVALUATION OF DETER MINANT

(a)Second order determinant:

The value of determinant of order 2 ,

abAcd is

det( )abA A ad bccd

e.g.

The value of

23

13A is

23det( ) (2)(3) ( 3)( 1) 6 3 313AA

The value of determinant is 3.

(b)Third order determinant:

The value of determinant of order 3

is given by

1 2 3

2 3 1 3 12

1 2 3 1 2 3

2 3 1 3 12

1 2 3det( )a a ab b b b bbA A b b b a a ac c c c ccc c c

E.g.

The value of A= |321

−510

3−14| is given by

3 2 1

det( ) 5 1 0

3 1 4

1 0 5 0 5 13211 4 3 4 3 1AA

det (A)=3(4 -0)-2(-20-0)+1(5 -3)=3(4) -2(-20)+1(2)=12+40+2=54

1 2 2

1 2 2

1 2 2a a a

A b b b

c c cmunotes.in

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The value of determinant is 54.

2A.4. PROPERTIES OF DETER MINANT

(1) The value of determinant is not attend by changing the rows into the corresponding

columns and the columns into the corresponding rows.

1 2 3 1 1 1

1 2 3 2 2 2

1 2 3 3 3 3a a a a b c

b b b a b c

c c c a b c

(2) If two rows or two columns of a determinant are identical, the determinant has the

value zero.

1 2 3

1 2 3

1 2 30a a a

b b b

b b b (Since R 2=R3) OR

1 2 2

1 2 2

1 2 20a a a

b b b

c c c (Since C 2=C3)

(3) If two adjacent rows or columns of the determinant are interchanged, the value of the

determinant so obtained is the negative of the value of the original determinant.

1 2 3 1 2 3

1 2 3 1 2 3

1 2 3 1 2 3a a a b b b

b b b a a a

c c c c c c

(4) If the elements of any row or column are multiplied by the same factor, the value of the

determinant so obtained is equal to the value of the original determinant multiplied by

that factor.

1 2 3 1 1 1

1 2 3 2 2 2

1 2 3 3 3 3ma ma ma a b c

b b b m a b c

c c c a b c

(5) Sum of determinants:

If any element in any row (or columns) consists of the sum of two terms, the

determinant can be expressed as the sum of two other determinants whose other

rows ( or columns) remain the same, while the remaining row (or column) consists of

these terms respectively.

Thus,

1 1 2 3 1 2 3 1 2 3

1 1 2 3 1 2 3 1 2 3

1 1 2 3 1 2 3 1 2 3a a a a a a a a

b b b b b b b b

c c c c c c c c

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*PRODUCT OF DETERMINANT :

The product of two determinants is possible only if both the determinants are of

same order.

Let A and B two determinants of the order 3.

1 2 3

1 2 3

1 2 3a a a

A b b b

c c c and

1 2 3

1 2 3

1 2 3l l l

B m m m

n n n

Then,

1 2 3 1 2 3

1 2 3 1 2 3

1 2 3 1 2 3

1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3

1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3

1 1 2 2 3 3 1 1 2 2 3 3 1 1 2 2 3 3a a a l l l

AXB b b b X m m m

c c c n n n

a l a l a l a m a m a m a n a n a n

b l b l b l b m b m b m b n b n b n

c l c l c l c m c m c m c n c n c n

=

' ' '

1 1 1 2 1 3

' ' '

2 1 2 2 2 3

' ' '

3 1 3 2 3 3R R R R R R

R R R R R R

R R R R R R ;

where ,

1R,

2R,

3R are rows of A and

'

1R ,

'

2R,

'

3R are rows of B .

2A.5. MINORS AND CO -FACTO RS

(a) Minor of an element:

Consider the determinant A of order n written as

11 12 13 1n

21 22 23 2n

n1 n2 n3 nn nna a a a

a a a a

A =

a a a a

Then

ijM is called the minor of the element

ija of determinant A, where

ijM is obtained by

deleting

thi row and

thj column of A of the order (n -1)X(n -1).

E.g. Consider the determinant of order 3. munotes.in

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11 12 13

21 22 23

31 32 33 3x3a a a

A = a a a

a a a

M 11 = Minor of an element a 11

22 23

32 33aa = aa

Similarly,

M 12 = Minor of an element a 12

21 23

12

31 33aaM = aa

E.g. Let,

258

A = 1 3 2

0 4 6

11 12 133 2 1 2 1 3M = , M = , M = 4 6 0 6 0 4

21 22 235 8 2 8 2 5M = , M = , M = 4 6 0 6 0 4

Similarly we can find M 31, M 32, M33.

(b) Co-factor of an ele ment:

If A is the determinant of order n and

ijC denotes C o-factor of the element

ija and is obtained

by multiplying the minor

ijM multiplies by by (−1)𝑖+𝑗.

ij

ij ijC = 1 M

Where

ijM is minor of

ija .

If

1 1 1

2 2 2

3 3 3a b c

A = a b c

a b c

11C =

The co -factor of

11

1a =(-1)

22

33bc

bc munotes.in

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12C = The co -factor of

12

1b =(-1)

22

33ac

ac

13C = The co -factor of

13

1c =(-1)

22

33ab

ab

E.g. Consider,

A= |134

021

376|

12

12 12

12

12 C = 1 M

01 C = 136

31 0 3

1 3 3

11

11 11C 1 M

11

1121C = 1 . 76

= 1 12 7 =5

2A.6. CRAMER’S RULE FOR S OLUTION OF LINEAR EQ UATIONS:

(a) Cramer’s Rule for solution of Linear equation in two variables :

The solution system:

1a x+

1by =

1c ……. (1)

2ax+

2b y=

2c ……. (2)

is given by

,y xD DxyDD ;

Where

11

22abDab ,

11

22xcbDcb ,

11

22yacDac

1 1 1 1

2 2 2 2

1 1 1 1

2 2 2 2,c b a c

c b a cxya b a b

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E.g. Solve:

3x+2y=5 …….(1)

4x+ y =3 …….(2)

Solution:

By using Cramer’s Rule,

,y xD DxyDD

Where D=

32

41 = 3 - 8 = -5,

256 5 113xD ,

359 20 1143yD

Thus we get, D=-5 ,

xD = 1,

yD = -11

Substituting in

,y xD DxyDD we get

1 1 11 11,5 5 5 5xy

Therefore, solution is x =

1

5 , y =

11

5

(b)Cramer’s Rule for solution of Linear equation in three variables :

The solution system:

1ax+

1by+

1cz =

1d ……. (1)

2ax+

2by+

2c z =

2d ……. (2)

3ax+

3by+

3c z =

3d ……. (3)

is given by

, , ,y x zD D Dx y zD D D ,

Where

111

222

3 3 3a b c

a b c D

a b c ,

111

222

3 3 3xd b c

d b c D

d b c ,

1 1 1

2 2 2

3 3 3ya d c

a d c D

a d c ,

1 1 1

2 2 2

3 3 3za b d

a b d D

a b d

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1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3

1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2

3 3 3 3 3 3 3 3 3,,d b c a d c a b d

d b c a d c a b d

d b c a d c a b dx y za b c a b c a b c

a b c a b c a b c

a b c a b c a b c

e.g. Solve the following:

3x+ 2y+ z =10

5x+ 3y+2z=17

7x+ 8y+ z = 26

is given by

,,y x zD D Dx y zD D D

Where

3 2 1

5 3 2 3(3 16) 2(5 14) 1(40 21)

781D

=3(-13)-2(-9)+1(19)

=-39+18+19 =-2

10 2 1

17 3 2 10(3 16) 2(17 52) 1(136 78)

26 8 1xD

=10( -13)-2(-35) + 1(58)

= -130+70+58 = -2

3 10 1

5 17 2 3(17 52) 10(5 14) 1(130 119)

7 26 1yD

= 3(-35)-10(-9) + 1(11)

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=-105+90+11 = - 4

3 2 10

5 3 17 3(78 136) 2(130 119) 10(40 21)

7 8 26zD

=3(-58) -2(11) + 10(19)

= -174-22+190

[ 35]

= - 196+190

= -6

Thus we get,

D=-2 ,

xD= -2,

yD = -4,

zD = -6

Substituting in

,,y x zD D Dx y zD D D we get

2 4 61, 2, 32 2 2x y z

Therefore, solution is x=1, y=2 and z=3.

2A.7. UNIT END EXERCISE

i) Find the value of determinant

2 1 1

A = 4 3 1

2 5 1

ii) Write the minors and co -factors of the elements of the determinant

2 3 2

A = 1 4 1

5 6 8

iii) Solve the following system of equations by using Cramer’s rule:

3x+y= 19

3x-y = 23

iv) Solve the following system of equations by using Cramer’s rule:

x+ 2y+ 3z=6

2x+ 4y+z=17

3x+ 2y+9z=2 munotes.in

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v) Find product of the following two determinants:

3 2 1

5 3 2

781 and

258

B = 1 3 2

0 4 6

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2B

MATRICES

UNIT STRUCTURE

2B.1 Objectives

2B.2 Introduction

2B.3 Definitions

2B.4 Illustrative examples 2B.5 Rank of matrix

2B.6 Canonical form or Normal form

2B.7 Normal form PAQ

2B.8 Let Us S um Up

2B.9 Unit E nd Exercise

2B.1. OBJECTIVES

In this chapter a student has to learn the

Concept of adjoint of a matrix .

Inverse of a matrix .

Rank of a matrix and methods finding these.

2B.2. INTRODUCTION

At higher secondary level, we have studied the definition of a

matrix, operations on the matrices, types of matrices inverse of a matrix

etc.

In this chapter, we are studying adjoint method of finding the

inverse of a square matrix and also the rank of a matrix.

2B.3. DEFINITIONS

*Matrix:

A matrix is a rectangular grid of numbers, symbols or expressi ons

that is arranged in a row or column format enclosed in Square or curved

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A system of

m n numbers arranged in the form of an ordered set of m

horizontal lines called rows & n vertical lines c alled columns is called a

matrix of order

m n .

A matrix with m rows and n columns is a matrix of order m x n is

written as

11 12 1

21 22 2

12..............

................

: : ................ :

................n

n

m m mn mxna a a

a a a

a a a

OR

11 12 1

21 22 2

12..............

................

: : ................ :

................n

n

m m mn mxna a a

a a a

a a a

* Note:

i) Matrices are generally denoted by capital letters.

ii) The elements are generally denoted by corresponding small letters.

iii) A matrix has no numerical value.

Types of Matrices:

1) Rectangular matrix :-

Any m x n Matrix where

mn is called rectangular matrix.

For e.g .

A=

232 3 4

1 2 3

2) Column Matrix :

It is a matrix in which there is only one column.

311

2

4A

3) Row Matrix:

It is a matrix in which there is only one row.

135 7 9A

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4) Square Matrix :

It is a matrix in which number of ro ws equals the number of

columns. i.e. it is n x n matrix of order n .

e.g.

2223 A46

Matrix A is a square matrix of order 2.

5) Diagonal Matrix:

It is a square matrix in which all non -diagonal elements are zero.

e.g.

33200

A 0 1 0

0 0 0

6) Scalar Matrix:

It is a square diagonal matrix in which all diagonal elements are equal.

e.g.

33500

A 0 5 0

005

7) Unit Matrix:

It is a scalar matrix with diagonal elements as unity.

e.g.

331 0 0

A 0 1 0

0 0 1

8) Null Matrix:

A matrix whose all elements are zero is said to be Null matrix.

e.g. munotes.in

## Page 39

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33000

A 0 0 0

000

9) Upper Triangular Matrix:

It is a square matrix in which all the elements below the principle diagonal

are zero.

e.g.

331 3 0

A 0 0 1

005

10) Lower Triangular Matrix:

It is a square matrix in which all the elements above the principle diagonal

are zero.

e.g.

330 0 0

A 3 4 0

1 3 2

11) Symmetric Matrix:

If for a square matrix A,

A TAthen A is symmetric

e.g.

1 3 5

A 3 4 1

5 1 9

is symmetric matrix.

12) Skew -Symmetric Matrix :

If for a square matrix A,

–A TA then it is skew -symmetric matrix.

e.g.

0 5 7

5 0 3

7 3 0A

is skew -symmetric matrix.

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Note : For a skew -Symmetric matrix, diagonal elements are always zero.

*Some more types of Matrices:

(a)Transpose of Matrix:

It is a matrix obtained by interchanging rows into columns or columns

into rows.

e.g.

231 3 5 A 3 7 9

321 3

A 3 7

5 9TA Transpose of

(b)Determinant of a Square Matrix:

Let A be a square matrix then

A = determinant of A i.e det A= A

If (i) then

0A matrix A is called as non -singular and

If (i) then

0,A matrix A is singular.

Note : For non -singular matrix A-1 exists.

(c) Co-factor Matrix :-

A matrix C =

ijC where

ijC denotes co -factor of the element

ija .

of a matrix A of order n x n, is called a co -factor matrix.

In above matrix

1 3 4

A = 0 2 1

3 7 6

Co-factor matrix is munotes.in

## Page 41

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5 3 6

C = 10 6 9

3 1 2

111

222

333A B C

C = A B C

A B C

Similarly for a matrix, A =

12

39

the co -factor matrix is

C=

93

21

(d) Adjoint of Matrix :-

If A is any square matrix then transpose of its co -factor matrix is called

Adjoint of A.

Thus in the notations used,

Adjoint of

TAC = transpose of its co -factor matrix

⟹𝐴𝑑𝑗 𝐴= [𝐴1𝐴2 𝐴3

𝐵1𝐵2 𝐵3

𝐶1 𝐶2 𝐶3]

Adjoint of a matrix A is denoted as Adj. A

Thus if,

1 3 4

A = 0 2 1

3 7 6

then Co-factor of matrix

5 3 –6

A = 10 6 9

3 1 2

Adj.

5 10 –3

A = 3 6 1

6 9 2

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Note :

If

2×2abA = cd

then Co-factor of matrix

d-cA = -b a

Adj.

d-bA = -c a

(e) Inverse of a square Matrix: -

Two non -singular square matrices of order n , A and B are said to

be inverse of each other if,AB=BA=I, where I is an identity matrix of

order n. Inverse of A is denoted as A-1 and read as A inverse.

Thus , AA-1=A-1A=I

Inverse of a matrix can also be calculated by the Formula.

A-1 =

1

A Adj. A where

A denotes determinant of A.

Note:

(i)From this relation it is clear that A-1 exist if and only if

A0

i.e. A-1 exist if and only if A is non -singular matrix.

(ii)An easy method to find the inverse of the second order matrix:

Let

2×2abA = cd

then,

1

2×2d1

a A-bA = -c

i.e. interchange the position of a and d and change signs of b and c and

divide by

A .

*Properties of Matrix:

[1] Addition of matrices:

Addition of two matrices is possible if the number of row and

columns of two matrices are equal .

e.g.

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3 2 1

5 1 0

3 1 4A

and

120

5 1 3

3 2 4B

Then

4 4 1

0 2 3

6 1 8AB

[2] Multiplication of matrices:

The Multiplication of two matrices A and B is defined as AB. AB

exists if number of columns of matrix A is equal to number of rows of

matrix B. It is not necessary that AB=BA.

Sometimes BA may not exists.

eg.

3 2 1

5 1 0

3 1 4A

and

120

5 1 3

3 2 4B

Then

3 10 3 6 2 2 0 6 4 6 10 10

. 5 5 0 10 1 0 0 3 0 0 9 3

3 5 12 6 1 8 0 3 16 10 13 13AB

2B.4. ILLUSTRATIVE EXAMPL ES

Example 1: Find the inverse of the matrix by finding its adjoint .

2 1 3

3 1 2

1 2 3A

Solution: We have,

2 3 4 1 9 2 3 6 1A

2 7 15

6A

Here,

0A

1A Exists

Transpose of matrix A = AT

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T2 3 1

A 1 1 2

3 2 3

We find co -factors of the elements of A T (Row -wise)

. . 2 1, . . 3 3, . . 1 1

. . 1 7, . . 1 3, . . 2 5

. . 3 5, . . 2 3, . . 3 1C F C F C F

C F C F C F

C F C F C F

∴ Adj A= [−1 3 −1

−7 3 −5

5 −3−1]

∴ A-1 = 1

|𝐴|,𝐴𝑑𝑗 𝐴= 1

6[−1 3 −1

−7 3 −5

5 −3−1]

Example 2: Find the inverse of matrix A by Adjoint method, if

0 1 2

A = 1 2 3

3 1 1

Solution: Consider

0 1 2

A = 1 2 3

3 1 1

= 0 1 1 8 2 5

= 0 8 10

= 2

Co- factor of the elements of A are as follows

11

1123C = 1 . 111

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12

1213C = 1 . 831

13

1312C = 1 . 531

21

2112C = 1 . 111

22

2202C = 1 . 631

23

2301C = 1 . 331

31

3112C = 1 . 123

32

3202C = 1 . 213

33

3301C = 1 . 112

Thus,

Co-factor of matrix

1 8 5

C = 1 6 3

1 2 1

And Adjoint of A= CT

11 1 1 1 1 1

1= 8 6 2 A 8 6 225 3 1 5 3 1

Note: - A Rectangular matrix does not possess inverse.

Properties of Inverse of Matrix: -

i) The i nverse of a matrix is unique

ii) The inverse of the transpose of a matrix is the transpose of inverse

i.e.

T 1 1 T(A ) (A )

iii) If A & B are two non -singular matrices of the same order

1 1 1(AB) B A

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This property is called reversal law.

Definition: -Orthogonal matrix: -

If a square matrix satisfies the relation

TAA I , then the matrix A

is called an orthogonal matrix and also

T1AA

Example 3:

Show that

Cosθ Cosθ A = Sinθ Cosθ

is orthogonal matrix.

Solution:

To show that A is orthogonal i.e . To show that

TAA I

Cos SinA = Sin Cos

T Cos SinA = Sin Cos

T Cos Sin Cos SinAA = Sin Cos Sin Cos

22

22Cos Sin Cos Sin Sin Cos

Sin Cos Cos Sin Sin Cos

10= I01

A is an orthogonal matrix.

Check Your Progress:

Q. 1) Find the inverse of the following matrices using Adjoint

method, if they exist. munotes.in

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i)

12,22

ii)

23,41

iii)

cos sin,sin cos

iv)

1 3 2

3 0 5 ,

2 5 0

v)

cos sin 0

sin cos 0 ,

0 0 1

vi)

1 2 3

2 3 1

3 1 2

vii)

1 1 1

1 2 3

2 1 3

Q.2) If A =

cos sin,sin cos

B =

1 tan2,

tan 12

C =

1 tan2,

tan 12

prove that A= B.C-1

Q. 3) If

4 3 3

A = 1 0 1

4 4 3

, prove that Adj. A= A

Q. 4) If

1 2 1

A = 0 1 1

1 1 2

, verify if

(AdjA)' (AdjA')

Q.5) Find the inverse of

1 2 1

A = 0 1 1

2 2 3

, hence find inverse of

3 6 3

A = 0 3 3

6 6 9

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2B.5. RANK OF A MATRIX

a) Minor of a matrix

Let A be any given matrix of order mxn. The determinant of any

sub-matrix of a square order is called minor of the matrix A.

We observe that, if ‘r’ denotes the order of a mino r of a matrix of

order m x n then

1 r m , if m1 r n , if n

e.g. Let

1 3 1 4

A = 4 0 1 7

8 5 4 3

From matrix A we get four 3rd Order determinants .

1 3 1 3 1 4 1 1 4 1 3 4

4 0 1 , 0 1 7 , 4 1 7 , 4 0 7

8 5 4 5 4 3 8 4 3 8 5 3

1 3 0 1 3 4, , , 1 , 0 , 3 ,4 0 5 4 0 7

Are some examples of minors of A.

b) Definition – Rank of a matrix :

A number ‘r’ is called rank of a matrix of order mxn if there is

almost one minor of the matrix which is of order r whose value is non -zero

and all the minors of order greater than ‘r’ will be zero.

e.g.(i) Let

102

A = 2 4 1

357

121 0 0 2A = 4, A = 82 4 4 1

etc.

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102

A = 2 4 1 1 23 2 2 19 0

357

Rank of A= 3

(ii)

1 1 2

A = 1 2 3

0 1 1

Here,

1 1 2

A = 1 2 3 1 1 1 1 2 1 0

0 1 1

Third order determinant is zero .

rank of A < 2

111A = 1 012

Thus minor of order 3 is zero and at least one minor of order 2 is non -zero

Rank of A = 2.

Some results:

(i)Rank of null matrix is always zero.

(ii)Rank of any non -zero matrix is always greater than or equal to 1.

(iii)If A is a non -singular n x n matrix then Rank of A is equal to n and if

A is n x n unit matrix then rank of A is equal to n .

(iv)Rank of transpose of matrix A is always equal to rank of A.

(v)Rank of product of two matrices cannot exceed the rank of both of the

matrices.

(vi)Rank of a matrix remains unaffected by elementary transformations .

Elementary Transformations:

Following changes made in the elements of any matrix are called

elementary transactions.

(i) Intercha nging any two rows (or columns) . munotes.in

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(ii) Multiplying all the elements of any row (or column) by a non -zero

real number.

(iii) Adding non -zero scalar multi ples of all the elements of any row (or

columns) into the corresponding elements of any another row (or

column).

Definition: - Equivalent Matrix:

Two matrices A and B are said to be equivalent if one can be

obtained from the other by a sequence of elementary t ransformations. Two

equivalent matrices have the same order & the same rank. It can be

denoted by A ~ B

[It can be read as A equivalent to B]

Example 4: Determine the rank of the matrix.

1 2 3

A = 1 4 2

2 6 5

Solution:

Given

1 2 3

= 1 4 2

2 6 5A

2 2 1 3 3 1R R - R & R R - 2R

We get,

1 2 3

0 2 1

0 2 1

Here two columns are identical . Hence 3rd order minor of A vanished

A 3

Here, 2nd order minor

131001

(A) 2

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Thus , 3rd order minor is zero and at least one minor of order 2 is non -

zero.

Hence the rank of the given matrix is 2.

2B.6. CANONICAL FORM OR NO RMAL FORM

If a matrix A of order m x n is reduced to the form

rIo

oo

using a

sequence of elementary transformations then it called canonical or normal

form. Ir denot es identity matrix of order ‘r’ .

Note: - If any given matrix of order m x n can be reduced to the canonical

form which includes an identity matrix of order ‘ r’ then the matrix is o f

rank ‘r’.

Example 5: Determine rank of the matrix A if

2 1 3 6

A = 3 3 1 2

1 1 1 2

Solution:

2 1 3 6

A = 3 3 1 2

1 1 1 2

13R R

1 1 1 2

3 3 1 2

2 1 3 6

2 1 3 1R 3R , R 2R

1 1 1 2

3 6 2 4

0 1 5 10

23R 7R

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1 1 1 2

0 1 33 66

0 1 5 10

1 2 3 2R R , R R

1 0 32 64

0 1 33 66

0 0 28 56

31R28

1 0 32 64

0 1 33 66

0 0 1 2

1 3 2 3R 32 R , R 33 R

1 0 0 0

0 1 0 0

0 0 1 0

3Io

Rank of A=3

Example 6: Determine the rank of matrix

1 2 7

A = 2 4 7

3 6 10

Solution:

1 2 3

A = 2 4 7

3 6 10

2 1 3 1R 2R , R 3R munotes.in

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1 2 3

2 4 7

3 6 10

32RR

1 2 3

0 0 1

000

12R 3R

1 2 0

0 0 1

000

21C 2C

1 0 0

0 0 1

000

23CC

1 0 0

0 1 0

000

2I0

Rank of A= 2

Example 7: Determine the rank of matrix A if

1 1 2 4

2 3 1 1A =3 1 3 2

6 3 0 7

Solution: munotes.in

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1 1 2 4

2 3 1 1A =3 1 3 2

6 3 0 7

2 1 3 1 4 1R 2R , R 3R , R 6R ,

1 1 2 4

0 5 3 7

0 4 9 10

0 9 12 17

23RR

1 1 2 4

0 1 6 3

0 4 9 10

0 9 12 17

1 2 3 2 4 2R + R , R 4R , R 9R

1 0 8 7

0 1 6 3

0 0 33 22

0 0 66 44

43R 2R

1 0 8 7

0 1 6 3

0 0 33 22

0 0 0 0

31R 11

1 0 8 7

0 1 6 2

0 0 3 2

0 0 0 0

34C - C

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1 0 1 7

0 1 3 3

0 0 1 2

0 0 0 0

1 3 2 3R + R , R 3R

1 0 0 5

0 1 0 3

0 0 1 2

0 0 0 0

4 1 2 2C - 5C 3C 2C

1 0 0 0

0 1 0 0

0 0 1 0

0000

3I0

00

Rank of A= 3

Check Your Progress :-

Reduce the following to normal form and hence find the ranks of the

matrices.

i)

1 2 3

3 1 2

ii)

2 3 4

4 3 1

1 2 4

iii)

3 4 6

5 5 7

3 1 4

iv)

1 2 3 0

2432

3 2 1 3

6 8 7 5

v)

2 1 3 6

3 3 1 1

1 1 1 2

vi)

1 2 1 0

3 2 1 2

2 1 2 5

5 6 3 2

1 3 1 3

2B.7. NORMAL FORM PAQ

If A is any m x n matrix of rank ‘r’ then there exist non-singular

matrices P and Q such that, PAQ is in normal form.

rI0PAQ00

, where I r is the unit matrix of order r, hence

A = r munotes.in

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We observe that, the matrix A can be expressed as

A = I m In ………… (i)

Where I m and In are the identity matrices of order m and n

respectively. Applying the elementary transformations on this equation. A

in L.H.S. can be reduced to normal form. The equation can be

transformable into the equations.

rI0PAQ00

………… (ii)

Note that, the row operations can be performed simultaneously on

L.H.S. and pre -factor in R.H .S. [i.e. Im in equation (i) ] and column

operations can be performed simultaneously on L.H.S. and post factor in

R.H.S. [i.e. In in equation (i)]

Examples 8: Find the non -singular matrices P and Q such that PAQ is in

normal and hence find the rank of A.

i)

2 1 3

A 3 4 1

1 5 4

Solution: Consider

A= I 3 AI3

2 1 3 1 0 0 1 0 0

3 4 1 = 0 1 0 A 0 1 0

1 5 4 0 0 1 0 0 1

13R R

1 5 4 0 0 1 1 0 0

3 4 1 = 0 1 0 A 0 1 0

2 1 3 1 0 0 0 0 1

2 1 3 1C 5C , C 4C

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1 0 0 0 0 1 1 5 4

3 11 11 = 0 1 0 A 0 1 0

2 11 11 1 0 0 0 0 1

23R R

1 0 0 0 0 1 1 5 4

1 0 0 = 1 1 0 A 0 1 0

2 11 11 1 0 0 0 0 1

2 1 3 1 R R , R 2R

1 0 0 0 0 1 1 5 4

0 0 0 = 1 1 1 A 0 1 0

0 11 11 1 0 2 0 0 1

32C C

1 0 0 0 0 1 1 5 1

0 0 0 = 1 1 1 A 0 1 1

0 11 0 1 0 2 0 0 1

31 R 11

1 0 0 0 0 1 1 5 1

0 0 0 = 1 1 1 A 0 1 1

0 1 0 1 2 0 0 1 011 11

23R R

0 0 1 1 0 0 1 5 1

12 0 1 0 = 0 A 0 1 111 11

0 0 0 0 0 1 1 1 1

Thus ,

0 0 1 0 0 1

11 2 1 2 P = 0 and P = 0 = 11 11 11 11 11

1 1 1 1 1 1

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1 5 1 1 5 1

Q = 0 1 1 and Q = 0 1 1 = 1

0 0 1 0 0 1

P and Q are non -singular matrices . Also Rank of A = 2

ii)

2 1 3 6

A 3 3 1 2

1 1 1 2

Solutions:

Consider :

1 0 0 01 0 00 1 0 0A= 0 1 0 A 0 0 1 00 0 10 0 0 1

1 0 0 02 1 3 6 1 0 00 1 0 03 3 1 2 0 1 0 A 0 0 1 01 1 1 2 0 0 10 0 0 1

13RR

1 0 0 01 1 1 2 0 0 10 1 0 03 6 2 4 0 1 0 A0 0 1 02 1 5 10 1 0 00 0 0 1

2 1 3 1 4 1C C , C C , C 2C

1 1 1 21 0 0 0 0 0 10 1 0 03 6 2 4 0 1 0 A0 0 1 02 1 5 10 1 0 00 0 0 1

2 1 3 1R 3R , R 2R

1 1 1 21 0 0 0 0 0 10 1 0 00 6 2 4 0 1 3 A0 0 1 00 1 5 10 1 0 20 0 0 1

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23R 6R

1 1 1 21 0 0 0 0 0 10 1 0 00 0 28 56 6 1 9 A0 0 1 00 1 5 10 1 0 20 0 0 1

43C 2C

1 1 1 01 0 0 0 0 0 10 1 0 00 0 28 0 6 1 9 A 0 0 1 20 1 5 0 1 0 20 0 0 1

32C 5C

1 1 4 01 0 0 0 0 0 10 1 5 00 0 28 0 6 1 9 A 0 0 1 20 1 0 0 1 0 20 0 0 1

231R , R 128

1 1 4 0 0 0 11 0 0 00 1 5 0 3 1 90 0 1 0 A 0 0 1 2 14 28 280 1 0 01 0 2 0 0 0 1

23RR

1 1 4 01 0 0 0 0 0 10 1 5 00 1 0 0 1 0 2 A 0 0 1 20 0 1 0 3 1 90 0 0 114 28 28

31 1 4 00 0 10 1 5 0I 0 = 1 0 2 A 0 0 1 23 1 90 0 0 114 28 28

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0 0 1

1 P = 1 0 2 , P283 1 9

14 28 28

1 1 4 0

0 1 5 0Q = , Q 10 0 1 2

0 0 0 1

P & Q are non - singular. Also, Rank of A = 3.

Check Your Progress :

A) Find the non -singular matrices P and Q such that PAQ is in normal

form and hence find rank of matrix A.

i)

1 0 2

2 3 4

3 3 6

ii)

1 2 3 2

2 3 5 1

1 3 4 5

iii)

3 1 1

1 1 1

1 1 1

iv)

2 3 4 7

3 4 7 9

5 4 6 5

v)

1 3 5 7

4 6 8 10

15 27 39 51

6 12 18 24

2B.8. LET US S UM UP

Definition of matrix & its types of matrices .

Using Adjoint method to find the

1A by

using formula

11A adjAA

Rank of the matrix using row & column transformation

Using canonical & normal form to find Rank of matrix.

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2B.9. UNIT END EXERCISE

i)Find the inverse of matrix

1 2 3

456

7 8 9A

if exists.

ii) Find Adjoint of Matrix

1 1 1

0 2 1

2 1 1A

iii)Find the inverse of A by adjoint method if

1 0 2 1

1 1 0 1

1 0 1 2

2 3 1 0A

iv) Find Rank of matrix

1 2 3

456

7 8 9A

v) Prove that the matrix

0

0

0 0 1Cos Sin

A Sin Cos

is orthogonal .

Also find

1.A

vi) Reduce the matrix

0 1 3 1

1 0 1 1

3 1 0 2

1 1 2 0A

to the normal form

and

find its rank.

vii) Find the non - singular matrix

ρ and α such that

ρ A α is the normal

form when

1 1 1

1 1 1

3 1 1 A =

.Also find the rank of matrix A.

viii) Under what condition t he rank of the matrix will be 3? munotes.in

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2 4 2

2 1 2

10 A =

ix)If

1 1 1 1 2 1

2 3 4 & Y 6 12 6

3 2 3 5 10 5X =

Then show that

xy yx where

denotes Rank.

x) Find the rank of matrix

8 3 6 1

1 6 4 2

7 9 10 3

15 12 16 4 A =

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3A

SIMPLE INTEREST

&

COMPOUND INTEREST

UNIT STRUCTURE

3A.1 Introduction

3A.2 Unit End Exercise

3A.1. INTRODUCTION

Suppose we invest money in the bank for a specific period. At the

end of the period, the bank not only returns our money, invested, but in

addition will give some ‘extra money ’ for using our money which was

kept with them. That extra money we earn is calle d interest.

Many times we borrow some money from our friends or relatives

for personal needs. A large amount may be needed which can be taken

from banks, financial institutions etc. as a loan. In this case we have to pay

extra money w hile repaying the loa n. That extra money paid for makin g

use of their money is called I nterest.

Some useful terms:

(i) Principal (P): The amount or the sum of money which is borrowed,

invested i n a bank or l anded is called the principal (P).

(ii) Interest (I): The ‘extra money’ paid in addition to the principal, is

called Interest (I).

(iii) Amount (A): The total money, including Principal and interest is

called Amount (A).

(iv) Period (‘n’ or ‘t’): The specified pre -decided period is called the

period of inv estment, it is expressed in ‘years’ and it is denoted by

‘n’ or ‘t’.

(v) Rate of Interest: The intere st to be paid by the concerned p arty is

calculated as a ‘Percentage’ of the Principal for specific time period, munotes.in

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at the pre -decided yearly rate. This rate is called the Rate of Interest

per year. It is usually expressed as a percentage of the principal.

The Interes t is calculated in two ways as S imple Interest and

Compound I nterest.

*Simple Interest: (I)

If the interest is charged or calculated on the prin cipal, then it is called

simple interest.

The simple interest I on the principal P at the rate of r %, for the period n

(years) is given by

𝐼= 𝑃×𝑛×𝑟

100 OR 𝐼=𝑃×𝑛×𝑖 Where 𝑖= 𝑟

100

*Amount (A) at the end of n (years) is given by

A P I

100

1100

1

1pnrP

nrP

P in

A P i n

Examples :

(1) If Rs.5, 000 is invested at 5% per annum. Find the amount after (1)

One year (ii) Five years (iii) 6 months (half yearly) (iv) 4 months.

Soln:

If A = Amount, P = Principal, r = rate of interest,

100ri ,

n = time period in years. Then

1 A P in

Here P = 5000, r = 5%

i = 0.05

(i) After one year, n = 1

A = P (1 + in) = 5000(1 + (0.05) × 1)

= 5000(1 + 0.05)

= 5000(1.05)

= 5250

(ii) After five years, n = 5

A = P (1 + in) = 5000(1 + (0.05) × 5) munotes.in

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= 5000(1 + 0.25)

= 5000(1.25)

= 6250

(iii) After six months,

61

12 2n years

A = P (1 + in) = 5000(1 + (0.05) (1/2))

= 5000(1 + 0.025)

= 5125

(iv) After 4 months,

41

12 3n years

A = P (1 + i n) = 5000(1 + (0.05) (1/3))

= 5000

0.0513

= 5000(1.0167)

= 5835

(2) Find the simple interest of Rs.2000 for 5 years at 6% per annum.

Also find amount after 5 years.

Soln:

Given : P = Rs.2000, n = 5 years, r = 6% p.a. i.e. i = 0.06

Simple Interest

I P n i

2000 5 0.06

.600Rs

Amount after 5 years = A = P + I

= 2000 + 600

= Rs.2600

Hence the simple Interest is Rs.600 and the amount after 5 years is

Rs.2600.

(3) At what simple interest rate will Rs.6, 000 get Rs.1, 080 as simple

interest in 3 years.

Soln:

Given P = Rs.6000, n = 3 years, I = Rs.1080

I = P × n × i

1080 = 6000 × 3 × i

1080 = 180 00 i

1080 60.0618000 100i

The simple interest rate is 6% per annum. munotes.in

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(4) What sum of money will amount to Rs.6050 in 3 years at 7% p.a.

simple interest ?

Soln:

Given A = Rs.6050, n = 3, r = 7%

70.07100i , P = (?)

A = P (1 + n i)

6050 = P (1 + 3 × 0.07)

6050 = P (1 + 0.21)

6050 = P (1.21)

∴

605050001.21P

Rs.5000 is the required sum of money.

(5) In how many years will Rs.3, 500 amount to Rs.4, 200 at 5% p.a.

simple Interest ?

Soln:

Given: P = 3,500, A = Rs.4, 200, r = 5%p.a.

50.05100i , n = (?)

A = P (1 + ni)

4200 = 3500( 1+ 0.05 × n)

42001 0.053500n

1.2 1 0.05 n

0.05n 1.2 1 0.2

0.05 0.2 n

0.240.05n

4n

The time required is 4 years .

(6) A sum of money amounts to Rs.6, 600 in 2 years and Rs.7 , 200 in 4

years. Find the sum and the rate of simple interest.

Soln:

Let P = Principal, r = Rate of S.I. Per annum

We have

A P I

100pnrAP munotes.in

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6600 =

2

100prP

26600100prP …. (1)

7200 =

4

100prP

47200100prP …. (2)

Subtracting (1) from (2)

427200 6600100 100 pr prpp

2600100pr

Substituting

2

100pr in eqn. (1)

6600 600

.6000P

P Rs

Also

2600100pr

2 60000

Pr 30000pr

Substitute P = 6000,

We get 6000 r = 30,000

3000056000r

The rate of simple interest is 5% p.a.

(7) Mr. Amit lent Rs.17, 000 for 3 years and Rs.12, 000 for 4 ye ars at

the same rate of simple interest. Find the rate if the total interest

received was Rs.9900.

Soln:

Let r be the common rate of simple interest percent per annum.

For the first loan, P = Rs.17000, n = 3 year

17000 3

51000I p n i

i

i

For the second loan, P = Rs.12000, n = 4 years

12000 4

48000I p n i

i

i

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Total interest = 51000 i + 48000 i = 99000i

But this is given to be Rs.9900

9900 0 i = 9900

99000.199000i

r = 10%

The common rate of S.I. is 10% p.a.

(8) Hina and Mita borrowed Rs.8000 and Rs.15, 000 respectively at the

same rate Simple Interest. After 3 years Hina repayed the loan by

giving Rs.10160. How much amount should Mita pay after

142

years, to pay off the loan, including simple interest ?

Soln:

Let r be the rate of interest p.a.

For Hina:

1..100

8000 3

100

240p n rSI

r

r

Now the S .I., she paid = 10160 – 8000 = 2160

2160 = 240r

r = 9

For Mita :

2..100

15000 4.5 9

100

6075p n rSI

The total amount = 15 , 000 + 6075

= Rs.21, 075

Mita should pay Rs.21 , 075 after

142 years to pay off the loan,

including Simple Interest.

*Compound Interest (C. I.) :

If periodically the interest due is added to the principal and the

interest for the next period is calculated on this addition, then it is called as munotes.in

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compound interest. Since the compound interest is the interest on interest

over a period of time, it depend s on the frequency of the interest

redeemed.

e.g. If Rs.50,000 you deposit in a bank for 2 years a t 7% p.a.

compounded annually. The interest will be calculated in the following

way.

Interest for the First Year:

I = P × i × n = 50,000 × 0.07 × 1 = Rs.3, 500

Interest for the Second year.

Here for calculating interest for the second year principal would

not be the initial deposit, but Principal for calculating second year will be

initial deposit plus interest for the first year. Therefore, principal for

calculating second year interest would be

Rs.50000 + Rs.3500 = Rs. 53 , 500

Interest for the second year = 53500 × 0.07 × 1

= Rs. 3 , 745

Total interest = Interest for first year + Interest for 2nd year.

= Rs.3500 + Rs.3745

= Rs.7 , 245

At the same time the simple interest for 2 years of R s.5000 at 7% p.a. is

S.I. = P × n × i

= 50000 × 0.07 × 2

= Rs.7 , 000

So compound interest for a principal is more than simple interest on

the same amount for the same period.

So we can summarize the main difference between simple interest and

compound interest is that in simple interest the principal remains constant

throughout whereas in the case of compound intere st principal goes on

changing at the end of specified period.

The Formula for amount A is

11100 n

n rA P P i

, Where

100ri , A = Amount. munotes.in

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P = Principal, r = rate of interest per period, n = period of time

The formula for comp ound interest is

1

11n

nCI A P

P i P

Pi

Note: In case of compound interest calculations, it is easier to first

calculate the amount A.

The compound interest for kth period is calculated as follows :

Interest for kth period =

11kP i i

The interest can be compounded year ly, half -yearly quarterly or, monthly,

then amount A at the end of n years is given by

1mniAm

Where

100ri rate of interest p.a.

n = number of years

m = number of times the interest is compounded per year.

e.g. m = 2 if interest is compound half yearly

m = 4 if interest is compound quarterly

m = 12 if interest is compound monthly.

Examples:

(1) Find the compound interest and the amount after 3 years on a

Principal of Rs.15000 at 10% p.a.

Soln: Given : P=Rs.15,000, n=3 years, r = 10%

100.10100i

A= P (1 + i)n C.I. = A – P

= 15000 (1 + 0.10)3 = 19965 – 15000

= 15000 (1.10)3 = Rs.4965

= 15000 (1.331)

= 19965

Thus the compound interest is Rs.4965 and the amount after 3 years is

Rs.19, 965.

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(2) What sum of money will amount to Rs.40, 31, 078.40 in 3 years at 8%

p.a. comp ound interest?

Soln:

Given: n = 3 years, r = 8% = i = 0.08, A = 40, 31, 078.40, P = (?)

A = P (1 + i)n

40, 31,078.40 = P (1+0.08)3

40, 31, 078.40 = P (1.08) 3

40, 31, 078.40 = P (1.259712)

40,31,078.4032,00,0001.259712P

The required sum is Rs.32, 00, 000

(3) At what rate of compound interest would an amount double itself in 4

years ? (Given:

1

32 =1.2611,

1

32 =1.1892)

Soln:

Given that A = 2P & n = 4 years

4

41

21

2 (1 )nA P i

p p i

i

1

421

1.892 1

0.1892i

i

i

18.92%r

The required rate of compound interest is 18.92%

(4) The bank offer fixed deposits for 4 years. Under the following

schemes.

(i) At 12%, if the interest compounded annually.

(ii) At 11% if the interest compounded half -yearly?

State which scheme is more beneficial to the public?

n = 4 years for both schemes.

Soln: munotes.in

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Suppose P = 100 then

(i) For the interest compounded annually,

P = 100, n = 4, r = 12% i.e. i = 0.12

A = P (1 +i)n

= 100 (1 + 0.12)4

= 100(1.12)4

= 100 (1.573519)

= 157.3519

Interest on first scheme = A – P

= 157.3519 – 100 = 57.3519

(ii) For the interest compounded half yearly i.e. m = 2

i = 0.11, n = 4 years

2

12niAP

=

240.11100 12

=

8100 1 0.55

=

8100 1.055

=

100 1.5346865

=

153.46865

Interest on second scheme = A – P

= 153.46865 – 100

= 53.46865

Since interest on first scheme is more than the interest on second

scheme, therefore the first scheme is more beneficial to the public.

(5) A man borrowed a certain amount for 2 years from his friend at 3%

and had to pay a simple interest of Rs.120. He o nce again took a loan of

the same amount for 4 years from a bank at 15% interest, compounded

quarterly. Find the interest he will have to pay to the bank.

Soln: munotes.in

## Page 73

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The problem is in two parts.

(i) First find the principal P by using simple interest for mula for gi ven

S.I = 120, n = 2 & r = 3% i.e. i = 0.03

S.I. = P× n × i, Where S.I. = 120, n = 2 & i = 0.03

120 = P × 2 × 0.03

120 = 0.06 P

120 1200020000.06 6

2000P

P

(ii)The loan of same amount P =2000 (P we found in (i) ),

n=4years, r=15%

i = 0.15 and compounded quarterly.

By using formula of A & C.I. find the final answer.

P = 2000, i = 0.15, n = 4, m = 4

441

0.152000 14mniApm

= 2000(1 + 0.0375)16

= 2000(1.0375)16

= 2000(1.8022278)

= 3604.4556

& C. I = A – P

= 3604.4556 – 2000

= 1604.4556

Nominal and Effective rate of Interest :

Suppose the interest is compounded in times a year

Let i = nominal rate (stated rate) of interest per R e.1 per year

ie = effective rate of interest per Re.1 per year

P = Re. 1

N = 1 year

Then Accumulated value A after 1 year

(i) Using the nominal rate of interest

1mpAm munotes.in

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(ii) Using the effective rate of interest

111ee A i i

Equating (i) & (ii)

11m

eiim

i.e.

11m

eiim

Ex.: Find the effective rate equivalent to the nominal rate 16% p.a. when

compounded (i) half yearly (ii) quarterly

Soln:

160.16100i

(i) Interest compounded half years, m = 2

11m

eiim

2

20.16112

1 0.08 1

1.1664 1

0.1664

Effective rate of interest per Re.1 per year is 0.1664

Effective rate of interest percent per year

= 100 × 0.1664 = 16.64%

(ii) Interest is compounded quarterly, m = 4

4

411

0.16114

1 0.04 1m

eiim

41.04 1

1.16985856 1

0.16985856

Effective rate of interest percent per year is

= 100 × 0.16 985856 = 16. 98585690 munotes.in

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Ex: Which rate yields more interest?

5.8% compounded half -yearly or 6% compounded quarterly .

Soln:

(i) The effective rate of 5.8% comp ound half -yearly

i = 0.058 and m = 2

2

211

0.0.058112

1 0.029 1m

eiim

= 1.0588 41 – 1 = 0.058841

Effective rate 5.8841% p.a.

(ii) The effective rate of 6% compounded quarterly i = 0.06 and m = 4

4

411

0.06114

1 0.015 1

1.06136355 1m

eiim

ei = 0.06136355

Effective rate 6.13 6355% p.a.

6.136355 > 5.8841

Hence 6% compounded quarterly yields more interest than 5.8%

compounded half -yearly.

* Future Value

An amount (Accumulated Amount) of a sum of money including the

interest amount after specified period at a given rate of interest is

called Future Value.

If principal P is kept in a fixed deposit for n years 1% rate of

interest, compounded annuall y then Future Value is calculated by

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.1nF V A P i

i.e. sum due = Principal

1ni

Ex: Find the future value of Rs.24, 500 kept as a fixed deposit, after

years at 7% p.a. compounded annually.

Soln: P = Rs.24, 500, n = 7 years , r = 7% i = 0.07, A = (I)

A = P (1+i)n

= 24500(1+0.07)7

= 24500(1.07)7

= 24,500(1.60578 )

= Rs.39341.65

Ex: Mr. Mehta was approached by a person with two schemes, as he

wanted to invest Rs.1, 20, 000. In Schemes A, the period was 8 years

with 9% rate p.a . compounded annually. In schemes, the period was

10 years with 8% compounded inter est p. a. Advise him about the

choice of scheme w.r.t. the amount to be received.

Soln:

Scheme A:

P = 1, 20, 000, n = 8 years, r = 9%

i = 0.09, A =?

A = P (1 + i)n

= (1, 20, 000) (1+0.09)8

= 1, 20, 000(1.09)8

= 1, 20, 000(1.9925626.4168)

= 239107.52

Scheme B:

P = 1, 20, 000,n = 10 years, r = 8%

i = 0.08

A = P (1 + i)n

=1, 20, 000(1 + 0.08)10

=1, 20, 000 (1.08)10

=1, 20, 000(2.15892499725 )

=259070.9 = 259071

In scheme B Mr. Mehta received more amount, so choice is scheme B.

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* Present Value :

The present value concept is useful when we wish to target for an

amount A after n years and wish to know what amount p should be

invested presently to achieve the target. Then A, the amount is called sum

due and P is called its present value (Present worth) or discounted value.

* Discounting:

The process of finding the present value of a sum due is called

discounting.

Present Value =

Sum due

1ni

i.e.

..

1nAPV

i

Ex: Find the Present Value of Rs.14, 641 at 10% rate of inte rest, payable 4

years from now.

Solun:

Here A = Rs.14641, r = 10%

i = 0.10, n = 4 year

4

4..

1

14641

1 0.10

14641

1.10

14641

1.4641nAPV

i

= 10,000

Present value is Rs.10, 000

Ex: Sohail pr omised to pay Amir Rs.15, 000 after 3 years with compound

rate of interest 8% p.a. He also promised to pay Aakash Rs.20, 000

after 4 years with compound rate of interest 9% p.a. Find the present

worths of these payments. Also find the total present worth of the

money Sohail has to pay.

Solun: munotes.in

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P.V. of Payment of Amir:

15000, 3, 8% 0.08 A n r i

3

3..

1

15000

1 0.08

15000

1.08

15000

1.259712nAPV

i

P.V. =

.10,907.484Rs

P.V. of Payment of Aakash:

20000, 4, 9% 0.09 A n r i

4..

1

20000

1 0.09

20000

1.41158161nAPV

i

P.V. =

.14,168.504Rs

Total present worth of the money Sohil has to pay

= 11,907.484 + 14,168.504=Rs.26, 075.988

Ex: Mr. XYZ has to pay an institution Rs.16, 800 after 3 years. He offers

to pay the institution now at the present value at interest compounded

8%p.a . What amount should he pay now ?

Soln:

Here A = 16800, i = 0.08, n = 3

3..

11

16800

1.08

16800

1.259712

13336.3816nAPV

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Mr. XYZ should pay Rs. 13336.3816 now .

Ex: Mr. PQR has to pay an institution Rs.13336.38 now.

If he agrees to pay a lump sum after 3 years with interest compounded at

8% p.a. What is the amount that he will have to pay ?

Soln:

P = 13336.38, n = 3, i = 0.08

F.V. = P (1 +i)n

= 13336.38(1+0.08)3

= 13336.38(1.08)3

= 13336.38(1.259712)

= 16799.9979 = Rs.16, 800

Mr. PQR will have to pay Rs.16, 800.

Ex.: Mr. Das has to pay an institution Rs.10, 000 at the end of 2 years and

Rs.6, 000 at the end of 3 years from now. If he opts for paying a lump

sum at the end of 3 years, what will be the future value at that time at

interest compounded 8% p.a. ?

Soln:

The amount of the payment of Rs.6000 paid at the end of 3 years is

Rs.6000 itself.

The amount of the payment of Rs.10, 000 at the end of 2 years. Its

amount at the end of 3 years is

A = P (1 + i)n

A = 10000(1.08)1

= 10000 × 1.08

= 10,80 0

The total future value of all payments at the end of 3 years

= 6000 + 10,800

= Rs.16, 800

Mr. Das has to pay Rs.16, 800 at the end of 3 years.

Ex.: Mr. Patel has to pay an institution Rs.10, 000 at the end of 2 years

and Rs.6000 at the end of 3 years from now. If he decides to settle the munotes.in

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payments now, what is the present value at interest compounded

8% p.a. ?

Soln:

1st Payment :

Here A = 1 0000, n = 2 & i = 0.08

..

1nAPV

i

210000

1.08

210000

1.08

10000

1.1664

= 8573.3882

2nd Payment:

A=6000, n=3, i=0.08

3

3..

11

6000

1.08

6000

1.08

6000

1.259712nAPV

= 4762.99344

Total present value of all payments taken together

= 8573.3882 + 4762.99344

= 13336.38164

Present value is Rs.13336.38164 at interest compounded 8% p.a.

3A.2. UNIT END EXERCISE:

(1) At what rate will the simple interest on Rs. 15,000 for 4 years are

equal to the simple interest on Rs. 16,000 for 3 years at 10% p.a.?

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(2) A principal amounts to Rs.9, 680 after 3 years and to Rs.10, 800 after

5 years. Find the principal and rate of interest.

(3) Anita and Amisha borrowed Rs.8, 000 and Rs.15, 000 respectively at

the same rate of simple interest. After 3 years Anita repaid the loan by

giving Rs.10, 160. How much amount should Amisha pay after four

and half years, to pay off the loan, including simple interest?

(4) The simple interest at 20% p.a. on a certain sum of money for 4 years

is Rs. 25,600. Find compound interest on the sum at the same rate for

the same period?

(5) Mr. XYZ wants to purchase smart phone after 4 years which will cost

him Rs. 25,000. How much money he should invest in bank at present

so as to receive Rs. 25,000? If the bank is giving 12% per year rate of

compound interest.

(6) A particular sum of money amounts to Rs.7, 69,824 in 2 years and Rs.

8, 31,409.92 in 3 years. Find the sum and compound interest rate.

(7) On what sum of money will be the difference between the simple

interest and compound interest for 2 years at 4% p.a. be Rs. 56?

(8)The simple interest and the compound interest on a sum of money at a

certain rate for 2 years is Rs. 1260 and Rs.132 3 respectively. Find the

sum and the rate.

(9)A sum of Rs. 6,55,000 is invested in a fixed deposit giving 10% p.a.

compound interest. Find the interest in 4th year.

(10) Find the maturity amount of a 2 year fixed deposit of Rs. 10,000 at

10% p.a. if the interest is compounded semi -annually.

(11)Find the effective rate equivalent to the nominal rate 16%p.a. when

Compounded (i)half yearly(ii)quarterly

(12)Which rate yields more interest :5.8% compounded half -yearly or 6%

compounded quarterly?

(13)Find the future value of Rs. 20,000 after 4 years if the compound

interest rate is 10%.

(14)Find the present value of Rs. 35, 730.48 to be paid three years from munotes.in

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now w ith the rate of compounding at 6% p.a.

(15) Mr. ABC estimates that after 3 years he would requires 50, 00,000

for his new business. He wishes to put aside money now, invested in

an instrument giving interest 7% p.a. compounded half yearly to meet

his requirement then. How much money should he invest presently?

(16)A person is supposed to pay a bank Rs. 5000, Rs. 6000 and Rs. 7000

at the end of 1,2and 3 years r espectively. He offers to settle the

payment now itself. How much will he have to pay now, with rate of

compounding at 12% p.a.?

(17) A person is supposed to pay a company Rs. 5000, Rs. 6000 and Rs.

7000 at the end of 1,2and 3 years respectively. He asks the company

if he can settle the payment by directly by paying a lump sum at the

end of 3 years. The company puts a condition that he should pay a

compound interest at 12% p.a. What amount will he have to pay at

the end of 3 years?

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3B

Annuit y

UNIT STRUCTURE

3B.1. Introduction

3B.2. Type of Annuity

3B.3. Examples

3B.4. Unit E nd Exercise

3B.1. THE CONCEPT OF ANNUITY:

In real life, we all must have gone through with different situations

where we do not have enough amount of money but still we want to buy

things for our use. At that time we borrow the amount and to repay back we

fix some time in which we retu rn fixed amount at regular intervals.

These equal amounts being returned at specific periods and introduces the

concept of annuity.

Pension honoured by the employer to the retired personnel, instalment

to the loans, insurance premiums, salary of the employees distributed by the

company, etc. are few examples of annuity.

* Characteristics of Annuity:

Annuity is a series of payments.

Annuity paid is of equal amounts and is fixed.

Annuity is paid at equal interval of time. It could be either annually,

semi-annually, quarterly, monthly etc.

Annuity can be paid either at the beginning or at the end of each

payment period.

Annui ty payments are periodic in nature.

*Annuity : The series of payments made at successive interval of time is called

an Annuity.

3B.2. TYPES OF ANNUITY:

*Uniform Annuity:

If the payments of an annuity are of equal amounts for equal interval of

time and continue for entire term period, it is called uniform annuity. E.g.

Loan instalments.

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*Varying Annuity:

If the amount of payments of an annuity are unequal and no n-uniform.

Though the payments are made at regular intervals but their amounts are not

same is also known as non -uniform or varying annuity. E.g. Dividends on

bonds, mutual funds.

*Annuity Certain :

If the payments are made or received after a fixed period of time, it is known

as Annuity Certain. Any form of loan instalments or advancements, bank

recur ring are examples of annuity certain.

*Contingent Annuity :

An annuity with number of payments depending on happening an event is

Contingency Annuity i.e. Annuity that does not begin making payments to the

annuitant or the beneficia ry until a certain stated event occurs. E.g.

Annuities that do not begin payments until an individual's retirement or

death.

*Perpetual Annuity (perpetuity): An annuity supposed to go on perpetually

or endlessly is a Perpetual Annuity or a Perpetuity. i.e. The payments

continues forever. E.g. Pension, Cap rate in real estate and dividend stream on

shares.

*Immediate Annuity ( Ordinary annuity) : If the annuities are paid at the end

of each period, it is known as an Immediate Annuity. It is also known as an

Ordinary Annuity. E.g. repayment of loan instalments.

*Annuity Due : If Payments are paid at the beginning of each period, it is

called an Annuity Due. E.g. payment of life insurance policies.

*Deferred Annuity :

If the periodic payments are not made in the beginning of some time periods

and thereafter continues periodically is called Deferred Annuity. The period

during which payments are not made are known as period of deferment. E.g.

In general repayment of home loan instalments begins after the loan has been

disbursed.

*Accumulated Values (or Future value) (Immediate Annuity):

Let each annuity be of Rs. C, rate of in terest per unit per annum be i.

Payments are made at the end of each period, then amount or accumulated

value A

Accumulated Value

. . (1 ) 1n CF V ii

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*Present Value (Immediate Annuity):

Let C -an annuity, n -no. of time periods, i -rate of interest per unit p.a.

. . 1 (1 )n CPV ii

*Accumulated Values (or Future value) (Annuity due):

Let each annuity be of Rs. C, rate of interest per unit per annum be i.

Payments are made at the beginning of each period, then amount or

accumulated value A

Accumulated Value

(1 )(1 ) 1n Ciii

*Present Value (Annuity due):

Let C -an annuity, n -no. of time periods, i -rate of interest per unit p.a.

(1 ). . 1 (1 )n CiPV ii

*Present value of Deferred Annuity:

A deferred annuity is characterized by a payment which is made at some later

date, rather than the beginning or end of the time period.

Let us consider, ‘m’ to b e the deferment period and for ‘n’ periods payments

are due to be made at the en d of each periods after deferment period. The

following formula calculates the present value of deferred annuity.

(1 ). . 1 (1 )m

n CiPV ii

*Amortization of the Loan :

The process of gradual elimina tion of a loan through regular i nstalments

that are sufficient to cover both the principal and the interest, is called as

amortization of the loan.

* Each EMI consists of two parts, one representing interest on the outstanding

balance loan and the other representing part of the principal to be repaid.

*Amortization Table:

The calculations of an amortized loan may be displayed in an

amortization table. The table lists relevant balances and dollar amounts for

each period. Each period is a row in the table, while the columns are typically

current loan balance, total monthly payment, interest portion of payment,

principal portion of payment and ending outstanding balance. The ending

outstanding loan balance of one period becomes the current loan balance for

the next.

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*Equated Monthly Instalments (EMI):

The loan taken from ban ks or any financial institutions repai d as an immediate

annuity in equal instalments with the unit time per iod of loan. This instalment

is known as EMI.

*Interest on reducing balance method :

The method using the present value of annuity using co mpound interest to

calculate the EMI is called the method of reducing balance.

is called the method of reducing balance.

3B.3. EXAMPLES:

[1] Himanshi opened a recurring deposit in a bank for 3 years with

payments of Rs.

4000 paid at the end of each year. Find the money obtained at the end

of period with

6% p.a.

Solution:

Annuity Immediate -Payment at the end of the year

Here C=Rs.4000, i =0.06, n=3 years

3

3. . (1 ) 1

4000(1 0.06) 10.06

66666.6667 (1.06) 1

66666.6667 1.191016 1

66666.6667 (0.191016)

.12734.40n CF V ii

Rs

Hence, Himanshi obtained Rs. 12734.40 at the end of 3 years at 6% p.a.

[2] Rashmi deposits Rs.6000 at the end of every month for 4 years with

9% compound

interest p.a. What is the total amount she will receive at the end of the

period?

Solution: munotes.in

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Annuity Immediate -Payment at th e end of the month

Here C=Rs.6000, i=0.09/12=0.0075, n=4x12=48month

48

48. . (1 ) 1

6000(1 0.0075) 10.0075

8,00,000 (1.0075) 1

8,00,000 (1.43140533 1

8,00,000 (0.43140533

345124.264n CF V ii

Thus , Rashmi will receive the final amount Rs. 345124.264 after 4

years.

[3] Find the present value of immediate annuity of Rs.16, 000 per year

for 3 years at 10%

p.a.

Solution:

Here C =Rs.16, 000, n=3 years, i=0.10, P=?

. . 1 (1 )n CPV ii

3

316000. . 1 (1 0.10)0.10

160000[1 (1.10) ]PV

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31160000 1(1.10)

1160000 11.331

160000[1 07513]

160000[0.2487]

39792

P.V. is Rs.39792

[4] A man purchases a house and takes a mortgage on it for Rs. 10, 00,000 to

be paid off in 4 years by equal annual payments payable at the end of each

year. If the interest rate is 6% p.a., find the sum of money that he will pay each

year.

Solution:

Here P=Rs. 10, 00, 000, n=4 years, i=0.06, C=?

10,00,000 1 (1 )n Cii

410,00,000 1 (1 0.06)0.06C

410,00,000 0.06 [1 (1.06) ] XC

4160,000 1(1.06)C

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160,000 11.262477

60,000 [1 0.79209]

60,000 [0.20791]

60,000 0.20791

60,000288586.400.20791C

C

C

C

C

Hence, each instalment would be of Rs. 288586.40

[5] Miss. MNO purchased a home -theatre on instalment basis such that Rs.

6000 and the remaining amount to be paid in 4 equal quarterly instalments of

Rs.3000 each payable at the end of each quarter. Find the cash price of the

system if the rate of compound interest is 7% p.a.

Solution:

Here C= Rs.6, 000, n=1x4=4 years, i=0.07/4 =, P=?

4

4

4. . 1 (1 )

30001 (1 0.0175)0.0175

171428.57[1 (1.0175) ]

1171428.57 1(1.0175)

1171428.57 1(1.07185903n CPV ii

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171428.57[1 0.9329585]

171428.57[0.0670415]

.11492.82847Rs

Hence, Cash price of the system=initial payment + P.V.

=Rs.6000+Rs.11492.82847

=Rs.17492.83

[6] Find the accumulated value of an annuity due of Rs.1000 per annum for 3

years at 10% p.a.

Solution:

Annuity due -Payment at the end of the year

Here C=Rs.1000, i=0.10, n=3 years

3

3(1 ). . (1 ) 1

1000(1 0.10)(1 0.10) 10.10

1000(1.10)(1.10) 10.10

11001.331 10.10

11,000[0.331]

.3641n CiF V ii

Rs

The accumulated value of an annuity due is Rs.3641

[7] A person plans to put Rs.200 at the beginning of each year in a deposit

giving 2% p.a. compounded annually. What will be the accumulated amount

after 2 years?

Solution: Annuity due -Payment at the end of the year

Here ,C=Rs.200, i=0.02, n=2 years munotes.in

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2

2(1 ). . (1 ) 1

200(1 0.02)(1 0.02) 10.02

200(1.02)(1.02) 10.02n CiF V ii

2041.0404 10.02

10200[0.0404]

.412.08Rs

[8] Find the present value of annuity due of Rs. 100 p.a. for a period of 4 years

if interest is charged at 8% p.a. effective rates.

Solution: Annuity due: yearly payment, C=100, n=4 years, i=0.08,P=?

(1 ). . 1 (1 )n CiPV ii

4

4100(1 0.08)1 (1 0.08)0.08

100(1.08) 110.08 (1.08)

108 110.08 1.36049

1350(1 0.735029)

1350(0.2650)

357.75

Therefore, P.V. of annuity due is Rs.357.75 munotes.in

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[9] A deferred annuity is purchased that will pay Rs. 10,000 for 15 yea rs

after being deferred for 5 years. If money is worth 6% compounded quarterly,

what is the present value of this annuity?

Solution:

Deferred Annuity: Quarterly payment

C=10,000, n=15(4) =60, m=5(4) = 20,i=0.06/4=0.015

20

60(1 ). . 1 (1 )

10,000(1 0.015)1 (1 0.015)0.015m

n CiPV ii

20

60

20 6010,000(1.015)1 (1.015)0.015

10,000 110.015(1.015) (1.015)

10,000 110.015(1.346855) 2.4432198

10,000 110.015(1.346855) 2.4432198

10,000 110.015(1.346855) 2.4432198

10

,000 110.0202028 2.4432198

494980.8937 1 0.409296

494980.893(0.590704)

.292387.19Rs

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Present value of this annuity is

.292387.19Rs

[10] A deferred annuity is purchased that will pay Rs. 500 for 10 years after

being deferred for 6 years. If money is worth 3% compounded annually, what

is the present value of this annuity?

Solution:

Deferred Annuity: C=500, m=6, n=4,i=0.03

6

10

6

10

6 10(1 ). . 1 (1 )

500(1 0.03)1 (1 0.03)0.03

500(0.03)1 (1.03)0.03

500 110.03(1.03) (1.03)

500 110.03(1.19405) 1.343916

5001 0.7440940.03(1.1945)

5000(0.035821)m

n CiPV ii

.255591

13958.2926(0.255591)

.3567.6084Rs

Present value of this annuity is

.3567.6084Rs

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[11] A deferred annuity is purchased that will pay Rs. 5000 for 4 yea rs after

being deferred for 2 years. If money is worth 4% compounded annually, what

is the present value of this annuity?

Solution:

Deferred Annuity: C=5000, m=2, n=4,i=0.04

(1 ). . 1 (1 )m

n CiPV ii

2

4 5000(1 0.04)1 (1 0.04)0.04

2

4

245000(1.04)1 (1.04)0.04

5000 110.04(1.04) (1.04)

5000 110.04(1.0816) 1.16986

50000.8548030.043264

115569.527(0.854803)

.98789.19Rs

Present value of this annuity is

.98789.19Rs

3B.4. UNIT END EXERCISE:

(1) Find the accumulated value after 4 years of an immediate annuity of Rs.

20,000 p.a. with interest Compounded at 6% p.a.

(2) Divya deposited Rs.2000 at the end of each year, for 2 years in a

compa ny and received Rs.4200 as the accumulated value. Find rate of

compound interest.

(3) Manini deposits Rs. 500 with 12%compound interest for 3 years. Find

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(4) Sarita invested Rs.1000 at the end of every month for 4 years at 12 %

p.a. compound interest. Find the amount she will receive at the end of

the period.

(5) Ms. Rukmini plans to save for her daughter’s higher studies. She wants

to accumulate an amount of Rs. 1,00,000 at the end of 4 years. How

much should she invest at the end of eac h year from now, if she can get

interest compounded at 7% p.a.?

(6) Ms.Sima deposited Rs.20,000 at the end of every year for two years. The

rate of interest is 10% p.a., compounded half -yearly. What is the amount

accumulated at the end of 2 years?

(7) Find the accumulated value at the end of 4 years and the present value of

an immediate annuity of Rs.50,000 p.a. for 4 years at 4% p.a.

(8) Kartik purchased a TV set and paid Rs. 5,000 immediately, anther Rs.

5,000 after a year and Rs.5,000 after 2 years and thus became debt free.

Find the price of TV set if compound interest charged was 3.5%p.a.

(9) How much money should a person invest at 7% p.a. compound interest

so that he would get an annuity of Rs. 1, 00,000 at the end of each year

for the next four years after which his principal money will be over?

(10) A TV is purchased for Rs.5,000 cash down and Rs. 10,000 at end of each

month, for 4 months. Find the cash price of the TV if the payments

include interest payment at 12% p.a. compounded monthly.

(11) A man purchases a house and takes a mortgage on it for Rs. 10, 00,000

to be paid of in 4 years by equal annual payments payable at the end of

each year. If the interest rate is 6% p.a., find the sum of money that he

pays each year.

(12) Miss. MNO purchased a refrigerator with a down payment of Rs. 2500

and the remaining amount to be paid in 6 equal monthly instalments of

Rs.1000 each. Find the price of the fridge if the company wants to earn

12% p.a.

(13) A company decide to set aside a certain sum at the end of each year to

create a sinking fund, which should amount to Rs.5 lakhs in 4 years at

12% p.a. Find the amount to be kept aside each year.

(14) Find the present value of an immediate annuity of Rs. 30,000 p.a. for 3

years with interest compounded at 8% p.a.

(15) Raju took a loan of Rs.1,20,000 from a friend for a period of 9 months.

Compute the EMI at 10%p.a.using Reducing balance method.

(16) Mr. Bhatt wants to take a loan of Rs.4 lakhs, which he intends to return

after 4 years, with interest. Ban k A offers him the loan of 4 lakhs at 6%

p.a., flat interest rate and bank B offers him at 8% p.a., on monthly

reducing balance. Comparing the EMI’s decide about the choice of bank

he should make.

(17) A loan of Rs. 1,00,000 is to be repaid in 4 years in 4 equal instal ments ,

with the first instalment at the end of the first year. The rate of interest is

10% p.a. (a) Find the yearly instalment using interest on reducing munotes.in

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balance. (b)Find the interest and principal repayment for each mont h.

[Make the amortization table. ]

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4

LIMITS AND FUNCTIONS

UNIT STRUCTURE

4.1 Introduction

4.2 Types of Function

4.3 Concept of Limit of a Function

4.4 Solved Examples

4.5 Unit and Exercises

4.1 INTRODUCTION

In this chapter we learn the concepts of functions and their limit

Concept of a functions

Dependent and Independent Variables

Consider an equation y = 3x + 4

As we assign different value to x, we obtain the corresponding

values of y in such a relation, x is called as the independent variable and y

is called the dependent variable . For each value of x we get a unique value

of y and we say that y is a function of x and denote it as y = f(x)

Let f be a function from set A to Set B . This is denoted as

: A Bf

.The set A is called the domain of the function and set B i s

called the co-domain of the function . Further set

( )| R f x x A is

called the range of the function.

Example s:

1. Write the domain and range of the function y = f(x) = x2 + 5.

Where

0 5, x Ix

y = f(x) = x2 + 5, Where

0 5, x Ix

Domain = {0, 1, 2, 3, 4, 5,}

Here, f(0)=5, f(1)=6,f(2)=9,f(3)=14,f(4)=21 and f(5)=30

Range = {5, 6, 9, 14, 21, 30}

2. If f(x) = x2 + 3x -1. Find f (0), f (1), f (x+1)

f (0) = (0)2 + 3(0) - 1= -1 munotes.in

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f (1) = 1+3 -1 = 3

f(x+1) = (x+1)2 + 3(x+1) -1

= x2 + 2x + 1+3 x +3–1

= x 2+5 x +3

Examples for Practice :

(1) Find the domain and range for the following functions

(i)

6( ) 3 51xf x xx

(ii)

( ) | | 2 2f x x x

(2) For the function f (x) = 3 x2 +2x -1 find f (2), f (-3) f(0) and f (x + 4)

4.2 TYPES OF FUNCTIONS

I Constant Function :-

For every value of x

f(x) takes the same value . e.g. y = f (x) = 8

II Linear Function :-

Consider the function y = f(x) = ax + b where a and b are real

numbers and x is a variable . Such a function in called as a linear function

the graph of this function will be a straight line and the power of x is 1.

III Quadratic F unction :-

Consider the function y = f (x) = ax2 + bx + c where a, b, c are real

numbers

0a and x is a variable . The highest degree of x is 2 Hence

such a function is called a quadratic function the graph of this function is a

parabola .

IV Polynomial Function :-

A function of the type

y = f (x) = a0 + a1x + a 2 x2 + ......... + anxn is called a polynomial

function where a0, a1, a2,. .......an are real numbers and x is a variable.

The constant function, linear function and quadratic function are

special cases of polynomial function.

V Exponential function :-

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A function of the type f( x) = ax where a is a positive real number

0a

and x is a rational number, is called exponential function.

If a = e where e is the natural logarithmic base whose value is

approximat ely 2.71828183, then we get the exponential function as y = ex

VI Logarithmic Function :-

A function of the type f(x) =

logax where a is positive real number

1a

is called Logarithmic function

Function s in Economics: -

In business activity we use terms like price, demand, supply, revenue,

cost, profit price, demand and supply are related to each other and one can

be expressed as a function of the other. Similarly revenue and cost are

related to the number of units produced and sold. The various function in

Economics are as follows

I) Demand Function :-

Let p denote the price of a commodity whose demand is D. Then the

two variable p and D are related to each other and we can write the

relation as p = f (D) & D = g(p) . It is a convention to write the demand

function s p = f (D). If we plot Demand D on X axis and price P on the Y

axis, the dema nd cur ve appears as shown below:

DP

op=f(D)

The curve indicates that if price decreases the demand increases and

as price increases demand decreases .

II) Supply Functions :-

Let p denote the price of a commodity whose supply is S. Then the

two variables are related to each other and we can write the relation as

p = f (s) or s = g(p). It is a convention to write the supply function a s p =

f(s). If we plot supply S on X axis and price p on Y axis, then the supply

curve appears as shown below.

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SP p=f(S)

* Point of Equilibrium price (Demand = Supply) :-

If both the graphs are plotted on the same co -ordinate axes, the point

of intersection of the two curves is the point of equilibrium price.

III) Cost Function :- Let x denote the quantity of goods produced at price

p. The total cost of producing the goods consists of the fixed cost and

variable cost .

Thus Total cast = Fixed cost + Variable Cost

i.e. Total cost = Fixed Coast + x. P

Fixed cost will be the cost when no goods are produced i .e. when x =0

IV) Revenue Function :- Let x denote the quantity of goods sold at price

p then total revenue generated can be written as R= p. x.

V) Profit Function :- Let C denote the total cost of x units of goods

produced and sold and let R denote the total revenue generated.

Then Profit = R – C.

4.3 CONCEPT OF LIMIT OF A FUNCTION

Let x be a variable and a be a constant, then

xa means x tends to

a i.e. x approaches a, but

xa

xa means x approaches a from the right i.e.

xa , x > a

_ xa means x approaches a from the left i.e

xa , x < a

* Limit of a function :

Consider a function y = f(x) then as

xa if

()f x a then we say

that

lim

xaf x b

, where a and b are constant s.

* Left hand limit and right hand Limit :

12 lim ( ) lim ( )

x a x af x b f x b

Further if b 1 = b 2 = b we say that

lim ( )

xaf x b

Example: y = f(x) = x +1

X 1 1.5 1.9 1.99 2.0001 2.01 2.1 2.5

f(x) 2 2.5 2.9 2.99 3.001 3.01 3.1 3.5

In the above example we observe that as

2 ( ) 3x f x

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* Rules for finding Limits :

If f(x) and g( x) are two functions then

(i)

lim ( ) lim lim ( )

x a x a x af x g x f x g x

(ii)

lim ( ) lim lim ( )

x a x a x af x g x f x g x

(iii)

lim . ( ) limf(x).lim ( )

x a x a x af x g x g x

(iv)

lim ( )()lim( ) lim ( )xa

xa

xafxfx

g x g x Where g(x) = 0

* Methods for finding Limits :-

1. Substitution Method :- In this method, we substitute the limiting

value of x, in the given function to obtain the limit.

Example :

2

2

251lim2

2 5 2 12, 2 2 022

4 10 1

4

15

4xxx

x

x x x

2. Factorization Method :- In this method, we factories the numerator

and denominator, cancel the common factor and then substitute the

limiting value of x to obtain the limit of the function

Ex

3

327lim3xx

x

In this example if we substitute x = 3, we will obtain the value of the

function as

0

0 form which is called as indeterminate form the other

indeterminate forms are

00,1 ,0 , ,0 ,

In the above example (x - 3) is a common factor , so we proceed as

follows .

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3

3

2

327lim3

3 3 9

lim 3, 3 3 03x

xx

x

x x x

x x xx

Now substituting the l imiting value of x we obtain the limit of the

function as

=

22

3lim 3 9 3 3 3 9 9 9 9 27

xxx

3. Simplification Method :- In this method we simplify the function

using the common denomin ator cancel the common factor and then

substitute the limiting value of x to find the limit of the function

Ex.

32214lim22x x x x

2214lim2 ( 2)x x x x

2

224lim( 2)xx

xx

22

lim

xx

2( 2)

2x

xx

[ 2, 2 2 0]x x x

22 2 414 2

4. Rationali zation method :- In this method we rationalize the

numerator or denominator by multiplying and dividing by the

rationalizing factor, then simplifying and canceling the common

factor and substituting the limiting value of x to obtain the limit of the

function.

Ex. =

2

31 10lim3xx

x

22

2 3( 1 10)( 1 10)lim

( 3)( 1 10)xxx

xx munotes.in

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22

2 3

2

2 3( 1 10)( 1 10)lim

( 3)( 1 10)

( 1 10)lim

( 3)( 1 10)x

xxx

xx

x

xx

22

2 3

2

2 3( 1 10)( 1 10)lim

( 3)( 1 10)

9lim

( 3)( 1 10)x

xxx

xx

x

xx

3(3lim

xx

)( 3)

(3x

x 2)( 1 10)x

[ 3, 3, 3 0]x x x

3 3 6 3 3 10

10 10 10 2 10 10

* Limit to infinity :-

So far we have discusse d problems where the value of a limit of a

function was finite . However sometimes the variable x may take values

which go on increasing indefinitely and we say that

x and then find

the limit of the function

Ex.

2

3( 1)( 4)lim3 5 1xxx

xx

32

344lim3 5 1xx x x

xx

Dividing the numerator and denominator by x3 we get

=

23

234 1 41

lim513xx x x

xx

11[ , 0, 0]xxx

=

1 0 0 0 1

3 0 0 3 munotes.in

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* Limit of exponential series :-

1.

1lim 1n

nen

Using Binomial theorem

231 1 2 1 1 1 11 1 . .......2! 3!nn n n n nnn n n n

1 1 21 1 1

1 1 ........2! 3!n n n

1 1 1lim 1 1 1 ............ e2! 3!n

n n

12, , ....... 0 As nnn

Thus

1lim 1n

nen and

1

0lim 1n

nne

2.

lim 1 lim 1xnnxx

nnxxenn

3.

221 log log ..........2!x xa x a a is called Exponential series.

32

2 log 1log log ........2! 3!xxa axaax

2

01lim log 0 , ........ 0x

xaa As x x xx

4.

23

1 ...............2! 3!x xxex

211 ................2! 3!xe x x

x

01lim 1

x

xe

x

23As 0 , .......... 0x x x munotes.in

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4.4 SOLVED EXAMPLES

(i)

2

251lim21xxx

x

4 10 1 13

4 1 5

(ii)

2

237 12lim56xxx

xx

33

lim

xx

4

3x

x2x

3, 3, 3 0x x x

34132

(iii)

2

224lim11 18xx

xx

22

lim

xx

2

92x

xx

2, 2, 2 0x x x

44

77

(iv)

2111lim1 3 2xx x x

111lim( 1) 1 2x x x x

121lim12xx

xx

11

lim

xx

1x2x

1, 1, 1 0x x x

1112

(v)

22411lim3 4 13 36xx x x x munotes.in

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411lim4 1 4 9x x x x x

491lim4 1 9xxx

x x x

428lim4 1 9xx

x x x

424

lim

xx

4x 19xx

4, 4, 4 0x x x

22

5 5 25

(vi)

2

011lim

xxx

x

22

0 21 1 1 1

lim

11xx x x x

x x x

2

0 211lim

11xxx

x x x

2

0 2lim

11xxx

x x x=

0 21lim

11xxx

x x x

0 21lim 0, 0

11xxxx

xx

1 0 1

2 1 0 0 1

(vii)

226 10lim4xxx

x

2 26 10 6 10

lim

4 6 10xx x x x

x x x

2 26 10lim

4 6 10xxx

x x x

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2 224lim

4 6 10xx

x x x

222

lim

xx

22xx 6 10 2x

2, 2, 2 0x x x

2 2 1 1

8 8 4 8 8 2 4 8 8

(viii)

1

04 5 2limxxx

x x

1

04 1 5 1 2 2limxxx

x x

04 1 5 1 2 2 1

limx x x

x x

02 2 1 4 1 5 1limxxx

x x x x

0 0 04 1 5 1 2 1lim lim 2limx x x

x x x x x x

log 4 log5 2log 2

01lim logx

xaax

log 4 log5 log 4

22log 2 log 2 log 4

log 5

(ix)

2

036lim2xx

x x

09 1 6 1lim2xx

x x

001 9 1 6 1lim lim2xx

xx xx

1log9 log62

01lim logx

xaax munotes.in

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13log22

log log logmmnn

(x)

2

25 8 3lim3 7 21xxx

xx

Numerator and denominator both divided by x2

2

2835

lim7 213xxx

xx

211As 0, 0xxx

5

3

(xi)

21 2 3 ........ nlim

n n

21lim2nnn

n

1

2nnn

1 1 1lim 122n n

1As 0nn

(xii)

31.3 3.5 5.7 ...........lim

n n

32 1 2 1lim

nnn

n

2

341

lim

nn

n

2

3341lim lim

nnn

nn

334 1 2 1lim lim6nnn n n n

nn

21 2 1

6

1n n nn

n

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24 1 1 1lim 1 1 2 lim6nn n n n

41 2 06

211As 0, 0nnn

84

63

4.5 UNIT AND EXERCISES

i)

2

2235lim6xxx

xx

ii)

3

244 48lim7 12xxx

xx

iii)

2

13 2 7lim1xxx

x

iv)

2414lim26xx x x

v)

033lim

xxx

x

vi)

2

520lim

3 1 4xxx

x

vii)

41.4 2.5 3.6 ............lim

n n

viii)

3 3 3

41 2 3 .........lim

n n

ix)

32

325lim41nn n n

n

x)

25lim12nnn

n n n

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xi)

206 3 2 1limx x x

x x

xii)

1

04 5 6 3limx x x x

x x

xiii)

2

0lim 1 2x

xx

xiv)

5lim 1n

n n

xv)

41lim 1n

n n

xvi)

3

03lim1n

nn

n

xvii)

2log log2lim2xx

x

2. For what values of x will the expression

23

21x

xx

tends to

infinity?

3. If f(x) = x2, show that

02 (2)lim 12

hf h f

h

* Continuity :

A function f(x) is said to be continuous at x = a if

lim ( ) ( );lim ( ) lim ( ) ( )

x a x a x af x f a f x f x f a

Example

(i) Consider

( ) 1 0f x if x

00if x

Examine whether the function is continuous at x = 0

00lim ( ) lim1 1 0 (0)

xxf x f

Function is not continuous at x = 0

(ii) Consider f( x) = x +1 -1 < x < 0

1, 0 1 x

, 1 2xx

Examine whether f(x) is continuous at x = 0 and x = 1 munotes.in

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00lim ( ) lim 1 1

xxf x x

00lim ( ) lim1 1

xxfx

And f (0) = 1

Function is continuous at x = 0

Try the above example at x = 1

Conclusion :

At the en d of this chapter we have under stood the following concepts.

(1) Functions, their types

(2) Functions in Business and Economics

(3) Limit of a function

(4) Left hand and right hand Li mits

(5) Limit to infinity

(6) Continuity of a function

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5A

DERIVATIVE

UNIT STRUCTURE

5A.1 Objectives

5A.2 Introduction

5A.3 Summary

5A.4. Unit End Exercise

5A.1. OBJECTIVES

After going through this chapter you will able to know :

Find rate of change of a function with respect to one variable

Successive differentiation

Physical and Geometrical meaning of Derivative

Computation of Derivative with the rules of differentiation

5A.2. INTRODUCTION

Calculus is the mathematics of change in motion. It is used to

calculate change in displacement velocity with respect to time also used to

change in supply, with respect to price . It helps to calculate maximize

profit and minimize cost. We find slope of any cur ve using it.

We have already seen different type of function and there limit in

previous unit let us now try to find exact rat es of change at a point.

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Consider a function y = f(x) of a variable x. Suppose x change to

small value x o and set x 1 i.e. x1 = x + x o then the increment of x is given by

∆ x = x 1 – x o

As x change, y change from f(x o) to f(x o + ∆ x) i.e.

00fx

xxx fx y

The function is called difference quotient.

* Derivative :

Let y = f(x) be the function of x. To measure rate at which f (x)

change with respect to x. i.e. rate of change of y with respect to “ x”, is

called Derivative and it is denoted by

f (x) or

dy

dx .

Let x change from x

x + h then f(x)

f (x + h), where the

value of h, becomes smaller and smaller, change in y is given by tending h

0

0'( ) lim

hf x h f x dyfxdx h exists is called the derivative of

function f(x) and denoted by f

'(x) or

dy

dx.

Note: If

0lim

hf x h f x

h exists, it is called first principle of

derivative.

Ex.1 . Find

dy

dx , if y = 3x2 + 2 by using first principle of derivative.

Soln: Give n y = f(x) = 3x2 + 2

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f (x+h) = 3(x+h)2 + 2

= 3x2 + 6xh + 3h2 + 2

By definition of first principle of derivative,

0'( ) lim

hf x h f xfxh

=

2 2 2

03 6 3 2 3 2lim

hx xh h x

h

=

2

063lim

hxh h

h

=

063lim

hh x h

h

= 6x

If

232yx then

6dyxdx

Ex.2 . Find

'( )fx from first principles of the derivative.

If

f x x x

Soln :

1/2 f x x x x x =

3/2x

3/2( ) (x h)f x h

Now

0( ) ( )' lim

hf x h f xfxh

3/2 3/2

0lim

hx h x dyfxdx h

Put

x h y

h y x

0 h x h x y x

3/2 3/2

lim

yxyxfxyx

3/2 1 3

2f x x

1limnn

n

xaxanaxa

1/23'2f x x

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3

2f x x

3/2 3then '2dyIf f x x f x xdx

* Physical meaning of Derivative :

We have study velocity acceleration and magnification in our

school section, these all are related with derivative.

A car moves from point P to point Q, if displacement is given by

f(x) in time x then

fx is the velocity of a car. i.e.

df kgfxdx hr km/hrs

Similarly, when change in velocity take place and it is f(x) in time

x. Acceleration is given by

dffxdx

* Geometric meaning of D erivative :

Y

O XB

Ay=f(x)

Tangent

Let y = f(x) be a curve and let A a nd B be two point on curve y = f(x), AB

be chord on the curve y = f(x), draw tangent at point A is define by

Slope of the Chord

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Taking

lim h 0 ,We get

0lim

hf a h f a

h

Slope of tangent at point A =

0lim

hf a h f a

h

=

tandyfxdx

Derivative represent the slope of the curve y = f(x)

Note : i) If the tangent at point A is parallel to x-axis

then

0 tan 0 0dy

dx

ii) If the tangent at point A is parallel to y – axis

then

90º tan infinitedy

dx

Basic formula for Derivatives

i) If

nyx

1n dynxdx

ii) If y = k where k = constan t

0dy

dx

iii) If

xye

x dyedx

iv) If

xya

logx dyaadx

v) If

y log x

1 dy

dx x

*Algebra of Derivatives :

I) Sum Rule of Derivative :

If u and v are differentiable function of x and y = u + v then

dy du dv

dx dx dx

Similarly, if y = u – v then

dy du dv

dx dx dx

II) Derivatives of Scalar multiplication :

If y = k .u where k is constant and u is function of x, then munotes.in

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dy dukdx dx

III) Product Rule of Derivative :

If u and v are differentiable function of x and y = u

v then

dy dv duuvdx dx dx

IV) Derivative and Division of two function :

If u and v are differentiable function of x and y =

u

v

0v then

2du dvvudy dx dx

dx v

when

0v

Ex.3. Find,

dy

dx if

5 23 2 log 63xy x e x

Soln: Given function ,

5 23 2 log 63xy x e x

Differentiatin g with respect to x

5 23 2 log 63x dy dx e xdx dx

5 23 2 log 63 x d d d dx e xdx dx dx dx

4 213 5 2 03 xxex

4 215 23x dyxedx x

Ex.4. Find,

dy

dx if

21 log y x x

Soln: Given function ,

21 log y x x

Diff. w.r.t. x

221 (log ) log 1dy d dx x x xdx dx dx munotes.in

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=

2 11 log 2x x xx

dy

dx

=

212 logxxxx

Ex.5. Find,

dy

dx if

21

23xyx

Soln: Given function

21

23xyx

Let

21 ux and

23vx

Diff. w. r. t. x

2du dvvudy dx dx

dx v

dy

dx

22

22 3 1 1 2 3

23

ddx x x xdx dx

x

2

22 3 (2 ) 1 (2)

23

x x x

x

22

24 6 2 2

23x x x

x

2

22 6 2

23dy x x

dx x

Ex.6. Find,

dy

dx if

1

12xy

x

Soln: Given function

1

12xy

x

1 Let u x

and

12vx

Diff. w. r. t. x. munotes.in

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2du dvvudy dx dx

dx v

2(1 2 ) 1 1 1 2

12

ddx x x xdy dx dx

dxx

=

211(1 2 ) 1 .2.

22

12xx

xx

x

21111

2

12

xx

x

=

23

2

12x

x

=

23

2 1 2xx

Ex.7. If f(x) = 2 x 3 – 21 x 2 + 72 x + 17

Find the values of x, such that

'0fx

Soln: Given function

322 21 72 17 f x x x x

Diff. w.r. t. x

32

322 21 72 17

2 21 72 17df x x x xdx

d d d df x x x xdx dx dx dx

=

22 3 21 2 72 1 0 xx munotes.in

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2( ) 6 42 72f x x x

'fx

6 3 4xx

but given that

'0fx

6 3 4 0xx

30x OR

40x

3x OR

4x

Check your progress :

Q.1. Differentiate the following function w.r.t. x

a)

3

22 6 33xy x x

2: 4 6dyAns x xdx

b)

423 y x x x

3: 4 6 6 1 dyAns x x xdx

c)

3 2 43 4 1 y x x x

6 5 3 3: 7 18 16 3 6dyAns x x x x xdx

d)

221 y x x x

32: 4 6 2 2dyAns x x xdx

e)

2

22

7xxyx

3

222 14 14:

7dy x xAnsdx x

f)

1y x x

13:

2dy xAnsdx x

g)

1xyx

21:(1 )dyAnsdx x

Q.2. If

1xfxx , show that

' 1 0 f

Q.3. If

3/2y x x , show that

42

xdy

dx

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* Chain Rule :

(Differentiation of a function of a function)

This is also called function of a function which is very useful. We can

say that y is function of u and u is function of x, so that y is of the type of

function of a function, it is denoted by,

y f g x

This is also called as composite function of x. In composite function y

is not directly function of x, but it is a function of x c onnected through

sum other variable.

If y is function of u and u is function of x then derivative of y with

respect to x is equal to the product of the derivative of y with respect to u

and derivative of u with respect to x.

i.e.

dy dy du

dx du dx

this result is called chain rule of differentiation.

Ex.8 . Find

dy

dx, if

357yx

Soln.

357yx

Diff. w. r. t. x

dy

dx =

357dxdx

=

23 5 7 5 7dxxdx

=

23 5 7 5 1x

dy

dx =

215 5 7 x

Ex.9. Find

dy

dx, if

25 4 1 y x x

Soln.

25 4 1 y x x munotes.in

## Page 122

122

Diff. w.r. t . x

dy

dx =

25 4 1dxxdx

=

2

215 4 1

2 5 4 1

dxxdx xx

=

215 2 4 1

2 5 4 1x

xx

=

210 4

2 5 4 1x

xx

=

252

5 4 1x

xx

=

252

5 4 1x

xx

∴ 𝑑𝑦

𝑑𝑥= 5𝑥−2

√5𝑥2−4𝑥=1

Ex.10. If

1yx

x

Prove that

12dyxxdx x

Soln.

1yx

x

Diff. w.r. t . x

dy

dx =

1 dxdx x

=

3/211

2 2 x x

dy

dx =

11

22x x x

2x

dy

dx =

1x

x

Ex.11. If =

xx

xxeeyee

find

dy

dx munotes.in

## Page 123

123

Soln.

xx

xxeeyee

Diff. w.r. t . x

dy

dx

=

2

x x x x x x x x

xxdde e e e e e e edx dx

ee

=

2x x x x x x x x

xxe e e e e e e e

ee

=

22

2x x x x

xxe e e e

ee

dy

dx =

24xx

xxee

ee

dy

dx =

24

xxee

Ex.12. Find

dy

dx, if

23log3xyx

Solun.

23log3xyx

log log logmmnn

log 2 3 log 3y x x

Diff. w. r. t. x

dy

dx

=

log 2 3 log 3dxxdx

=

11. 2 3 . 32 3 3ddxxx dx x dx

=

11312 3 3xx

dy

dx =

31

2 3 3xx

munotes.in

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124

Ex.13. If

2

32 3 5

log

1

xx

y

x find

dy

dx

Soln.

2

32 3 5

log

1

xx

y

x

2 3log 2 3 5 log 1 y x x x

2 11log 2 log(3 5) log( 1)23y x x x

Diff. w.r. t . x.

dy

dx

=

2 11log 2 log 3 5 log 123d d dx x xdx dx dx

dy

dx

=

2

21 1 1 1 1. 2 3 5 12 2 3 5 3 1d d dx x xx dx x dx x dx

22 3 1

2 2 3 5 3 1dy x

dx x x x

Check your progress :

Q.1.Find

dy

dx

a) If

3/42log3xxyex

3 1 1.14 2 3dyAnsdx x x

b) If

22logy x x a

221.dyAnsdx xa

c) If

22

22logx x ay

x x a

222.dyAnsdx xa

munotes.in

## Page 125

125

d) If

logmx mxy e e

.mx mx

mx mxm e e dyAnsdx ee

e) If

5logyx

51. logdyAns edx x

f) If

log3xxy

log. 3 log3 1 log xx dyAns xdx

g) If

422 5 3y x x

32. 4 4 5 2 5 3dyAns x x xdx

h) If y=

2

21x

xe

2

21

222.

1x

xdy xAns edx x

i) If

223xxy

22. 3 log3 2 2xx dyAns xdx

j) If

y a a x

1.

4dyAnsdxa x a a x

k) If

2251xy x e

2

25

52

2. 10 1

1x

x dy xeAns x e xdx x

l) If

1

1x

xeye

3/2 1/2.

11x

xxdy eAnsdx ee

Q.2. If

log 1 1y x x munotes.in

## Page 126

126

Show that

210

1dy

dx x

Q.3. If

1log1xyx

Prove that :

21

1dy

dx x

*Implicit Functions :

An equation in the form y = f(x) defines y as an explicit function of x

and an equation in form of x = g (y) defines x as an explicit function of y.

A function or relation in which the dependent variable is not

isolated on one side of the equation i.e. an equation is in the form

f(x, y) = 0, defines y as an implicit function of x.

Ex.1 4. If

x y a find

dy

dx

Soln.

x y a

Diff. w. r. t. x on both sides,

ddx y adx dx

110

22dy

dx xy

11

22dy

dxyx

y dy

dx x

dy y

dx x munotes.in

## Page 127

127

Ex.1 5. Find

dy

dx

If

222 2 2 0 ax hxy by gx fy c

Solun.

Given :

222 2 2 0 ax hxy by gx fy c

Diff. w.r.t . x

222 2 2 0dax hxy by gx fy cdx

2 2 2 2 2 0 0 d d dax h xy by y g f ydx dx dx

2 2 1 2 2 2 0dy dy dyax h x y by g fdx dx dx

0dy dy dyax hx hy by g fdx dx dx

dyhx by f ax hy gdx

ax hy g dy

dx hx by f

* Higher order derivative OR successive differentiation:

Let y = f( x) be a function of x, defined over an interval, then 1st order

derivative is given by

'fx while differentiate with respect to x.

i.e.

y f x

dyfxdx

Now

fx

again diff. w.r.t. x

We get

2

2''d y dfxdx dx

It is called second order derivative of f(x) munotes.in

## Page 128

128

In this manner by successive differentiation of the function is given by

n

n

ndyfxdx

Ex.1 6. Find

2

2dy

dx

If

7 3 25 4 7 20y x x x

Soln.

Given :

7 3 25 4 7 20y x x x

Diff. w.r. t. x

7 3 25 4 7 20dy dx x xdx dx

=

7 3 25 4 7 20 d d d dx x xdx dx dx dx

=

625 7 4 3 7 2 0x x x

6235 12 14 dyx x xdx

2

62

2(35 12 14 ) d y dx x xdx dx

=

6235 12 14 d d dx x xdx dx dx

=

535 6 12 2 14 1xx

2

5

2210 24 14 dyxxdx

Note :

1)

2

2dy

dx is the derivative of

dy

dx w. r. t. x

2)

2 2

2d y dy

dx dx

Ex.1 7. If

21p

y x x

Prove that

2

22

210d y dyx x p ydx dx munotes.in

## Page 129

129

Soln.

21p

y x x

Diff. w.r. t. x

1

2211

pdy dp x x x xdx dx

=

1

2

2211

21 pxp x x

x

=

1

2

211

1 pxp x x

x

=

2 1

2

211

1

pxxp x x

x

=

1

22

211

1p

p x x x x

x

2

21

1

p

p x xdy

dx x

2211pdyx p x xdx

Again D iff. w.r. t. x

2 1

2 2 2

2221 1 1

11px dy d y xx p x xdx dx xx

22 1

2 2 2

222111

11 px dy d y x xx p x xdx dx xx

222

2

2221

1

11

p

p x xx dy d yxdx dx xx

22

2

2221

11x dy d y p yxdx dx xx

munotes.in

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130

Multiplying by

21x and re -arranging

2

22

210d y dyx x p ydx dx

Ex.1 8. If

3 1log y x thenx

Prove that

2

2230d y dyxdx x dx

Soln.

3 1log yxx

3 1log log log y x x xx

Diff. w.r. t. x

3( log )dy dxxdx dx

233 log log dy d dx x x xdx dx dx

=

321log 3x x xx

=

223 logx x x

223 logx x x

=

2 3 33log log x x x y x xx

2

33

3dyxydx x

dyx y xdx

Again differentiating both sides with respect to x

2

2

233d y dy dyxxdx dx dx

OR

2

2

22 3 0 d y dyxxdx dx

Now dividing throughout by x, we get the required equation, munotes.in

## Page 131

131

2

2230d y dyxdx x dx

Ex.1 9. If

y ax b x (a, b are constants)

Prove that

2

2

220d y dyx x ydx dx

Soln. Given :

y ax b x

2

2Substituting valueof , ,

dy d yydx dx

dy

dx =

2ba

x

1/2

2bxa

...... (i)

Differentiating again

2

3/2

2 3/21

2 2 4 d y b bxdx x

...... (ii)

2

2

22d y dyx x ydx dx

2

3/224 2 bbx x a ax b xx x

2

2Substituting valueof , ,

dy d yydx dx

=

2

3/222bx b xax ax b xx

=

2bxax2bxax bx

=

2

2bxbx

=

b x b x

= 0, Hence proved.

munotes.in

## Page 132

132

5A.3. SUMMARY

In this chapter we have learned.

First principal of derivative

Physical and geometrical meaning of derivative

Rules of Differentiation

Chain rules of differentiation

Differentiation of Implicit function

Successive differentiation

5A.4 . UNIT END EXERCISE:

(a) Find

dy

dx for the following :

1)

6 3 25 7 9 11 y x x x

52.30 21 18 Ans x x x

2)

11y x xx x

21.1Ansx

3)

2/36 7log 6 21x

e y a x x

5/3 7.6 log 4xAns a a xx

4)

2axy

ax

2.aAns

x a x

5)

22logy x x a

221.Ans

xa

6)

2log 1xy e x

2

22. e log 1

1x xAns x

x

7)

2251.xy x e

2

25

52

2.10 1

1x

x xeAns xe x

x

b)

5 4 26 4 2 5 9y x x x x . Find

dy

dx at x = -1 [Ans : 55]

c) Find the slope of the curve

232 y x x Hint: (find

dy

dx at x = 1)] munotes.in

## Page 133

133

d) Find the rat e at which the function f(x) =

5 3 23 7 9 x x x changes

with respect to x.[Ans.

425 9 14x x x ]

e) If

1logyx

x , prove that

1

21dy x

dx x x

Hint: y = log 1 1/ 2log xx

f) If

2xyx prove that

1dyx y ydx

g) If

xxy e e prove that

24dyydx

h) If

22y a x prove that

0dyyxdx

i) Find

dy

dx , when,

1)

5 5 2 250 x y ax y

24

422.

2axy xAns

y ax y

2)

2y xy x

44

.

2y x x

Ans

x y x

3)

22logxy x y

2

221

.

12yx

Ans

xy

4)

logxy xey

j) If

1 1 0x y y x prove that

21 1 0dyxdx

k) If

21 xy prove that

220dyydx

l) If

21m

y x x show that

21x

2

2

20d y dyx m ydx dx

munotes.in

## Page 134

134

5B

APPLICATION S OF

DERIVATIVE

UNIT STRUCTURE

5B.1 Objectives

5B.2 Introduction

5B.3 Increasing and Decreasing Function

5B.4 Maxima and Mini ma

5B.5 Summary

5B.6 Unit End Exercise

5B.1. OBJECTIVES

After going through this chapter you will able to know :

The relation between average revenue and marginal revenue

Relationship between average cost and marginal cost

Relation between marginal product and marginal cost.

Relation between production, cost and revenue functions.

Relation between marginal propensity to consume and marginal

propensity to save.

Calculate concavity, con vexity and point of Infle ction of function. munotes.in

## Page 135

135

Calculate maximum and minimum value of function at a particular point.

5B.2 INTRODUCTION

In the study of all economic problems it is essentially a problem of

finding out the rate of change. It may be the rate of change in the

dependent variable (demand) with respect to the change in the explanatory

variable (price) or it may be the rate of change in the endogenou s variable

(national income) with respect to change in the exogenous variable

(government expenditure) or a particular parameter (marginal propensity

to consume). The mathematical tool which is used to find out the

magnitude and direction of change in a pa rticular variable due to change in

the value of other variables of parameters is broadly known as the

technique of derivative or differentiation. The concept of derivative is

used to deal with a variety of economic problems.

*Demand Function :

It is relation between demand and price of commodity. Let P be

the price and D be demand of commodity. Then we can write :

D = f (P)

Also write a s P = f (D) munotes.in

## Page 136

136

Price

Demand

Note: Demand function is always decreasing function.

*Supply Function :

It is relationship between supply and price of commodity. Let P be

the price and S be supply commodity. Then we can write

S = f (p)

Price

Supply

Also write as P = f (s)

Note : Supply function is always increasing function.

*Equilibrium Price :

When demand and supply of goods are equal that condition at

point is called Equilibrium point. For Demand D an d Supply S for

particular goods, c urves are intersect each other, then the point of

intersect ion is called Equilibrium point on price.

munotes.in

## Page 137

137

Price

D = SEpD = S

*Cost Function :

The amount required to produce x units of goods, is called cost. It

C denotes the production cost at x units of goods, then it can be

written as

C(x) = Fixed cost + Variable Cost

i.e. C(x) = a + b x when x > 0

*Revenue Function :

The amount received from selling the product. It is denoted by R

let x units are sold at price P, then Revenue is R = p x = demand × price

R(x) = P x

*Profit Function :

Let R be revenue and C be cot function for x unit of goods.

Profit = Revenue – Cost

RC

x R x C x

munotes.in

## Page 138

138

*Average Cost :

The cost per unit is called Average Cost. Let C be cost function of

x units then,

Total Cost CAC Average Cost = = No. of Units x

*Marginal Cost :

The rate of change of cost (C) with respect to quantity : ‘ x’ is

called Marginal Cost it is denoted by MC.

d dcMC C xdx dx

*Marginal Average Cost :

The rate of change at Average cost (AC) with respect to qua ntity x

is called Marginal Average Cost.

d AC dMAC ACdx dx

*Marginal Revenue :

The rate of total Revenue with respect to the quantity x i.e.

d dRMR R xdx dx

In marginal analysis at firm operating under.

i) Perfect competition and

ii) Monopoly

I) Perfect Competition :

i) In this case the price is constant then

munotes.in

## Page 139

139

Average Revenue =

R PxAR Pxx

= Constant

ii) Marginal Revenue=

dR dMR P x Pdx dx

= Constant

i.e. In Perfect Competition AR = MR = P = Constant

II) Monopoly –

In this case a monopoly is a sale suppliers of goods produced be

fixed the price ‘P’ of good accordingly to the demand in the market. Thus

price is not constant.

i)

PxAR Px, When P is function of x

ii)

d d dPMR R Px P xdx dx dx

Thus

AR MR

* Marginal Revenue of Product (MRP) :

Let R be the revenue of x units with price P when P is function x

and

dR

dt is Rate of change of Revenue with respect to employee t’

is called Marginal Revenue.

Ex-1. For a certain product, cost function is given by

C = 3x4 – 5x2 + 50x + 20 . Find Average Cost, Marginal Cost and

Marginal Average Cost when x = 5.

Solun: Given Cost Function ,

423 5 70 20C x x x

42Total Cost 3 5 70 20AC = x x x

xx

=

2 203 5 70 xxx munotes.in

## Page 140

140

3

x=520AC = 5 5 5 705

3 125 25 70 4

375 25 74

= 424

423 5 70 20dc dMC x x xdx dx

312 10 70 MC x x

3

5 ( ) 12 5 10 5 70 1520x MC

dMAC ACdx

=

3 203 5 70dxxdx x

2

2

2

2209 5 0

2095xx

xx

2

5 220( ) 9 5 5 219.2

5x MAC

Ex-2. If the demand function is given by

23 5 25P x x When x is the demand

Find : i) Revenue function

ii) Average revenue function

iii) Marginal Revenue when x = 10

Solun:

Curve deman d function

23 5 25P x x

Revenue function

R pD

=

23 5 25x x x

=

323 5 25x x x

munotes.in

## Page 141

141

Average Revenue function

AR =

R PDPDD

=

23 5 25xx

Marginal R evenue =

dR dRdx dx

=

323 5 25dx x xdx

=

29 10 25xx

2

10 ( ) 9 10 10 10 15xMR

=

900 100 25 825

Ex-3. The manufacture r x units of articles at a cost (12 x + 95) per units

and the demand function if P = 47 x – 45, when P is price and x is demand.

Find x for which the total p rofit is increasing.

Solun :

Let no. of units = x

Cost per units = 12 x + 95

Total Cost = C = (12 x +95) x = 12x2 + 95x

Demand function P = 47x – 45

Revenue Function = R = PD

= (47 x – 45) × x

= 47x 2 - 45x

Profit Function f(x) =

RC

=

22(47 45 ) 12 95x x x x

=

235 140xx

235 140xx munotes.in

## Page 142

142

Diff. w. r. t. x

' 70 –140df x xdx

For increasing Function

'0 fx

. . 0diedx

70 140 0x

70 140x

2x

The total profit is increasing for

325 45 120 30 R x x x

5B.3. INCREASING AND DECR EAS ING FUNCTION

*Increasing Function :

If Y = f(x) is a function of x in the interval (a, b) and if Y increases

as x increases in (a, b) then Y is called the increasing function o f x

in the interval (a, b).

Let f(x) be increasing function in the interval (a, b) if

12,,x x a b

such that

1 2 1 2 x x f x f x

*Decreasing function :

If Y = f(x) is a function of x in the interval (a, b) and if Y,

decreases as x is increases and vice versa in (a, b) then Y is called the

decreasing function of x in the interval (a, b)

Let f (x) be decreasing function in the interval (a, b) if

12,,x x a b

such that

1 2 1 2x x f x f x

5B.4. MAXIMA AND MINIA

*Maximum Points :

A function f( x) is said to have a maximum value in an interval I

around x = a, if a

I, and if f (a)

f(x), x

I then f (a) is called the munotes.in

## Page 143

143

maximum value of f(x)

I and a is called the p oint of maximum at

f(x) in I.

*Minimum Point :

A function f(x) is said to have a minimum value in interval I

around x = a, if a

I and if f (a) < f (x), x

I, then f (x) is called the

minimum value of f (x) in I and a is called the point of minimum of

f (x) in I.

Note :

If Y = f (x) where the slope of f (x) is neither positive nor negative

but it is zero i.e. at this point target is parallel to x -axis.

i.e.

0dy

dx or

'( ) 0fx at x = 0

* Second Derivative Test for Maxima and Minima :

If Y = f (x) is continuous differential function at neighbourhood

of a point ‘a’

i) f(x), is Maximum at x = a

If

' 0and '' 0 f a f a

ii) f(x) is minimum at x = a,

'0 fa and

'' 0 fa

Following steps are to be followed

Step – I First find under derivative or f(x) i.e.

fx

Step – II taking f

'(x) = 0 find x = a, b, c root

Step – II find second order derivative i. e. f

'

'(x)

Step - IV

(a) f(x) is maximum at x = a if

f

'

' (x = a) < 0

(b) f (x) is minimum at x = b if f

'

' (x = b) > 0

Ex. 4. Find values of x for which function munotes.in

## Page 144

144

f(x) =2x3 – 3x2 – 72 x + 100 is

(i) Increasing (ii) Decreasing.

Solun:

Given: f (x) =2 x 3 – 3 x 2 – 72 x + 100

Diff. w. r. t. x

'fx 6 x 2 – 6 x – 72

i) f(x) is increasing if

'f (x) > 0

i.e. 6

x2 – 6

x – 72 > 0

6(

x2 –

x – 12) > 0

6(

x – 4) (

x + 3) > 0

i.e.

x > 4 an

x > –3

f(

x) is increasing when

x > 4 or

x > -3

ii) f(x) is decreasing if

'f (

x) < 0

6

x2 – 6

x – 72 < 0

6(

x2 –

x – 12) < 0

6 4 3 0xx

x < 4 as x < –3

f(

x) is decreasing when

x < 4 or

x < –3

Ex.5 . Find the local maxima and local Minima for the function

326 9 7 f x x x x

Solun:

Let given

326 9 7 f x x x x …. I

Diff. w.r.t. x

2' 3 12 9f x x x

"( ) 6 12f x x munotes.in

## Page 145

145

Taking

'0fx we get

23 12 9 0xx

23 4 3 0xx

3 1 0xx

3x or

1x

1 [ " ] 6 1 12 6 0x fx

fx is maximum at

1x

321 1 6 1 9 1 7 11f

fx is maximum at x = 1 & Maximum value = f(1)

321 1 6 1 9 1 7 11f

fx is maxima at

1x & Maximum value is 11

3 [ " ] 6 3 12 18 12 6 0 x fx

f (x) is minimum at x = 3 and minimum value = f(3)

f (3) = (3)3 – 6(3)2 + 9(3) + 7 = 7

f (x) is minimum at x = 3 and minimum value is 7

Ex.6. For the certain product total cost

329 24 17 c x x x Total Revenue

325 45 120 30 R x x x

. Find x for which the profit is maximum and

minimum.

Solun:

Here

329 24 17 C x x x

325 45 120 30R x x x

Profit function:

x R C

=

325 45 120 30x x x –

329 24 17x x x

=

325 45 120 30 x x x –

329 24 17 x x x

=

324 36 96 13x x x …. I

munotes.in

## Page 146

146

2' 12 72 96x x x …. II

" 24 72xx

Taking

'0x we get

2212 72 96 0 12 6 8 0 x x x x

4 2 0xx

4x OR

2x

For x = 2

" 2 24 2 72 48 72 24 0x

x is maximum at

2x and maximum value =

(x = 2)

322 2 36 2 96 2 13 93x

x is maximum at

2x and maximum profit is 93

" 4 24 4 72 24 0x

x is minimum at x = 4 and minimum value =

4x

324 4 4 36 4 96 4 13 77x

x is minimum at x = 4 & minimum value is 77

Ex.7. A manufacturer determines that t he employees will produce a total

of x units of a product per day, where x = 5t, if the demand equation for

the product is P = -2 x + 100 to find the M arginal Revenue Product when

t = 3. Interpret your result ….

Solun:

Here ,

2 100 Px &

5xt

2 100 R Px x x

22 100 R x x

22 5 100 5 R t t munotes.in

## Page 147

147

250 500 R t t

250 500 R t t

Diff. w. r. t. t

100 500 dRtdt

3100 3 500 200

tdR

dt

If 4th employee is hired the extra revenue generated is approximate 200.

Ex.8. Given a consumption function.

C

6000200010xx

1) Find out marginal propensity to consum e (MPC) and marginal

propensity to save when x = 90

2) Also show that MPC & MPS move in the opposite direction when

income (x) changes

Solun:

I. Given

6000200010Cx

2000

10dCMPCdx x

2 906000 30.65 10 90

xMPC

S is saving function defined as

S x C

=

6000200010xx

6000200010 ddMPS S xdx dx x munotes.in

## Page 148

148

=

260001 0 1

10 x

90xMPS = 1 – 0.6

= 0.4

II.

2

33 26000 1200020

10 10d d CMPCdx dx xx

2

3 26000 120002010 10 d d SMPSdx dx x x

Since

22

220 0.d C d Sand MPC and MPSdx dx move in the opposite

direction as x change s.

Ex.9. Given the total Cost Function

3

21000 100 103xC x x

Find :

i) Marginal Cost Function

ii) The slope of marginal cost function

iii) Output at which marginal cost is equal to average variable cost.

Solun :

i) Given the total cost function

3

21000 100 103xC x x

3

21000 100 103 d d xMC C x xdx dx

=

2100 20 xx

ii) Slope of MC =

' , 1f b f af x xba munotes.in

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Slope of MC =

20 2 x

iii)

3

2 xC 1000 100 x 10 x3

Where

3

2V 100 103xx x x and F.C. = 1000

23100 10 / 3 Vx x x xAVCxx

=

2

100 103xx

As per given condition we want to find our output at which

MC = AVC

2

2100 20 100 103xx x x

2210 03xx

210 03xx

20 10 03x OR x

0x OR

30152x

0x OR

15x

Concavity and convex pointed production function.

The concavity and conve xity are used widely in economic theory

and are also central to optimization, thereby A function of a single

variable is concave if every line segment joining, two points on its graph

does not lie above the graph at any point. Similarly, a function of a single

variable is conve x if every line segment joining two point on it graph does

not live below the graph at any point.

The concavity and conve xity are important in optimization theory

because as we see, a simple condition is sufficient and necessary for

maximize of a differen tiable concave function and for a minimize of a

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differentiable function is zero is a maximize of the function and every

point at which the derivative of con vex differentiable function i s zero is

minimized the function.

f(b)

af(a)

O bSlope = f (a)1

M=f(z)

From the graph we say that:

The differentiable function of a si ngle variable defined as an open

interval I is con vex on I if and onl y if

''f b f afxba for

xI

The differentiable function of a single variable defined as an open interval I

is convex if and only if

𝑓(𝑏)−𝑓(𝑎)

𝑏−𝑎 ≥𝑓′(𝑥),𝑥∈1

*Inflexion Point :

The point C is an inflexion point of successive differentiable

function f of a simple variable if f

'

' (c) = 0, for some value of a and b with

a < c < b

Test of concavity and convexity of a curve by second order

derivative.

i) If

"0 fx then it is con vex

ii) If

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iii) If

"0 fx then it is point of inflexion.

Ex.10. Suppose that the total consumption expenditure (C) in (thousand

Rs.) of a family is given by the function

2330 60 2 C x x x when x

denotes the family monthly disposable income in (thousand Rs) Does the

curvature of the consumption function change and at what level of income

it change?

Solun : Given consumption function

Let

2330 60 2 Y C x x x

diff w.r.t. x

𝑑𝑌

𝑑𝑥=30+120 𝑥−6𝑥2

again diff w.r.t . x

𝑑2𝑌

𝑑𝑥2=120 −12𝑥

Taken at S econd order derivative zero i.e.

2

20dy

dx

120 12 0 x

10x

Now for all 0 < x < 10 the second order derivative is positive i.e. The

consumption function is strictly conve x until x = 10

For all x > 10 the second order derivative is negative i.e. the consumption

function is strictly concave after x = 10.

Which implies that at x = 10 the inflexion point occurs or the curvature of

the consumption function changes at x = 10.

*Elasticity of demand: munotes.in

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We are well known with demand law in economics that “If the price

increases the quality demanded will decrease”. Thus the ratio of the

proportionate change in the quantity demand to the proportionate change

in the price is called elasticity of demand.

Let the demand change from D to D+ BD, where price changes from P to

P + BP, there elasticity of demand is given by

Taking limit

0 BP we get

Elasticity of demand =

0lim

BPP BD

D BP

=

P dD

D dP

But according to deman d law if price increases then demand falls i.e.

Elasticity of demand =

p dD

D dp

Elasticity of deman d is denoted by

p dD

D dp

Note:

i) If

| | 0d then it is perfectly inelastic

ii) If

0 | | 1d then it is inelastic

iii) If

0 | | 1d then it is unit elastic

iv) If 𝑛𝑑>1 then it is elastic

v) If

| | 1d then it is perfectly elastic

/

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Ex.11. Find elastic of demand, if Demand function

D = 25 – 11p + p2 when p = 4 and comment on it.

Solun : Given demand function

245 11D P p

Diff w.r. to p

0 11 2 2 11dDPPdp

22 1145 11P dD PPD dp p p

=

2

22 11

45 11PP

PP

When p = 4 then elastic at demand is

2

22 4 11 4 124 0.705817 45 11 4 4P

Here

1O Hence it is inelastic demand.

Relation between marginal revenue, average revenue and

elasticity of demand is,

Let

= elasticity of de mand

MR = Marginal Revenue

AR = Average Revenue then,

By definition of elasticity of deman d,

P dD

D dp

&

TRAR TR AR DD

ddMR TR D ARdD dD

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=

()ddD AR AR DdD dD

=

.dAR D A RdD

[∵

TRTR D P DP P ARD

AR P]

𝑀𝑅 =𝐴𝑅+𝐷𝑑

𝑑𝐷(𝐴𝑅)

= 𝑃+𝐷𝑑𝑃

𝑑𝐷

= 𝑃+𝐷

𝑃×𝑃×𝑑𝑃

𝑑𝐷

= 𝑃[1+𝐷

𝑃×𝑑𝑃

𝑑𝐷]

[∵𝑛=−𝑃

𝐷×𝑑𝐷

𝑑𝑃]

=𝑃(1−1

𝑛)

11 MR AR

Impact of Excise Tax:

An excise tax is a tax changed on each unit of a good or service that

is sold. This is not the same as sales tax in that it is received per unit of the

good rather than as a percentage of the sales. The excise tax can be

received on either the buyer or the seller of the commodity.

Increase the tax rate can either increase or decrease total tax Revenue

depending upon,

i) The elasticity of demand

ii) The elasticity of supply

iii) The size of tax base

A subsidy shift either the demand or supply cause to the right

depending upon whether the buyer or seller receives the subsidy. If it munotes.in

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is the buyer receiving the subsidy the demand cur ve shift right,

lending to an increase in the quantity of demand increasing while

equilibrium price decrease.

Impact of Subsidy:

Marginal listening on production will shift the supply curve to

the right until the vertical distance between the two supply cure is

equal to the per unit subsidy when other things remain equal this will

decrease price paid by the consumer and increase the price received

by the producers. Similarly, a marginal subsidy on consumption will

shift the demand curve to the right when other things remain equal.

This will increase the price received by producers by the same

amounts, as if subsidy had been granted to producer. However in this

case new m arket price will be the price received by producers.

Sales Tax effect on supply and demand:

Most states impose sales tax on some goods and services as a

means of generating revenue. However, sales taxes also influence

consumer behaviour.

As sales tax causes the supply curve to shift inward it has a

secondary effect on the equilibrium price for a product equilibrium

price is the price at which the producers supply matches. Consumer

demand at a stable price since sales tax increases the price of goods,

it causes the equilibrium price to fall. This may mean that it becomes

more difficult consumers change their buying behaviour to purchase

less of the more expensive goods.

While sales tax affects supply directly it only has an indicate

effect on consumer d emand. Sales tax also impact consumers buying

power.

Inventory Control: munotes.in

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It is the quantity Q, which when purchased in each under

minimizes the total cost T, incurred in obtaining and storing material

of a certain time period to fulfil given date of dema nd for the

material during the time period. The cost of the stock is called

Inventory cost and the management of inventory is called inventory

control.

Ex.1 2. The material demanded 1000 units per year the cost price of

material ` 2 per unit; and it cost ` 50 to make the factory ready for

production run of the items regardless of the number of units x

producers in a run; and the cost of staring material is 25% yearly on

the rupee value of the average inventory x on hand.

i) Find the economic under quantity an d total cost corresponding

to that

ii) Find the total cost when each under places for 10000 units.

Solun:

No of production per year =

1000

x

Cost for production =

1000 2500025xx

Variable cost = 2 per units : 20,000

Inventory cost on 100 units = 25

Inventory cost of x units

125100 4xx

Total cost =

2500010004xCx

Diff w.r. t . x

225000 104 dC

dx x

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Taking 𝑑𝐶

𝑑𝑥=0 we get

225000 104 x

X = 6250

𝑑2𝐶

𝑑𝑥2=50,000

𝑥3

2

23

625050,000

(6250) xdc

dx >0

Which is positive

C is minimum when x = 6250

Minimum cost

25000 625010006250 4C

= Rs. 62566.5

When x = 10000 units then,

25000 10000100010000 4C = Rs. 3502.5

5B.5 . SUMMARY

In this chapter we have learned

Marginal Cost, Marginal Revenue

Marginal Average Cost, Marginal Average Revenue

Marginal Analysis of a film operating

Marginal Revenue of products

Maxima, Minima, increasing and decreasing function

Marginal propensity to consume, marginal propensity to save,

concavity and convexity and point of inflection for profit function

Optimal trade in time, effect of taxation and subsidies, effect of

excise tax, imposition of sales tax munotes.in

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Inventory control

5B.6. UNIT END EXERC ISE

1) Show that the scope of average cost curve is equal to

1MC ACx

for the total cost function

32C ax bx cx d .

2) The total revenue received from the sale of x units if a products is

given by. Find (a) the Average Revenue (b) Marginal Revenue at x

= 5 (c) the actual Revenue from selling 51st Unit.

Ans. : (a)

22 12 6 / R x x x x (b) 212 (c) 214

3) The manufacturing cost of an item consists of ` 6000 as over heads,

material cost ` 5 per unit and labour cost `

2/ 60x units produced.

Find how many units must be produced so that the average cost is

minimum.

Ans: x = 600

4) A company charges `700 for a mobile hand set on an order of 60 or

less sets. The charges Reduced by `10per set for each set ordered

in excess of 60. Find the largest size order company should allow

so as to received a maximum revenue.

Ans: x = 65 Sets.

5) Give the price equation P = 100 – 2Q wher e Q is quantity

demanded, find (a) the marginal Revenue (b) point elasticity of

demand when a = 10 (c) nature of the commodity.

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6) A monopolist’s demand function is P = 300 – 5x. Find (a) the

marginal Revenue function (b) the relation between the slope of

MR and AR and (c) at what price is the marginal Revenue zero.

7) A manufacturer determine t employees will produce a total x units

of a product per day, where x – (200 t – t2). If the demand equation

for the produce is P = -0.1 x + x + 70, determine the marginal

revenue product where t = 40.

8) If c = 7 + 0.6I – 0.25 is a consumption function, find marginal

propensity to consume and marginal propensity to save. When I =

16.

9) Given the production function Q = SL1/2 and the price equation P

= 200 – 2Q, obtain the marginal revenue product of labour (L)

when L = 25

10) Prove that marginal cost is 1/x (MC – AC) for the total cost

function c(x) =

223 2 4 7xxx

11) The total function is given by

2325000 1000 5003C x x x x

12) A company has examined its cost structure and revenue structure

and have determined that C the total cost, R total revenue and x the

number of units produced are related as,

2100 0.015Cx &

3Rx

Find the production rate x that will maximise profits of the

company. Find that profit also find profit when x = 120.

13) The cost of producing x units is given by

320.001 0.3 30 42 C x x x x munotes.in

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Determine where this cost function if concave up and where it is concave

down also find the inflexion point.

14) If the price per unit (p) is given by p = 5(2 – x) and the total cost

by c(x) = 10 + 3x2 – 2x3, where x is the number of units produced .

Determine the optimum level of price and output f or profit

maximization.

15) If the total cost funct ion is

323 4 2C q q q . Find at what level

of output average cost will be mi nimum and what level will it be?

16) The total profit y(in rupees) of a company from the manufacture

and sale of x bottles is given by if

2

2 80400xx

17) The cost of function c(x) of a firm is given by

231300 102C x x x x

(1) Outpur at which marginal cost (MC) is minimum

(2) Output at which average cost (AC) is minimum

(3) Output at which AC=MC

18) For a firm under perfect comp etition, it is given that

2

25 28 273xC x x

, where p is price per unit, x is the units of

output, c is the total cost of x units

(1) Find the quantity produced at which profit will be maximum

amount of maximum profit.

(2) What happens to equilibrium output and maximum profit if

P = 12. munotes.in

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19) A manufacturer determines that his total cost function is

2

2 3003xCx

, where x is the no. of units produced. At which

level of output will a verage cost per unit be minimum?

20) A firm produces 36,000 items per year. It costs Rs.250 to make the

machine min, regardless of the number (x) of items produced in a

nw. The cost of storage is 50 paisa per year an average inventory

x/2 on hand. The cost of material per item is Rs.5. Find economic

lot size.

21) For a particular process, the average cost is gi ven by

280 12C x x

, where C is the average cost (per unit) x the

number of units produced. Find the minimum value of the average

cost the no. of units to be produced.

22) If the demand la w is

25

1xP find the elasticity of demand with

respect to price at the point p = 3.

23) A demand function is given by

225 4x p p , where x is the

demand of the goods at price p. Calculate the elasticity of demand

at price p = 5 and p = 8.

24) The demand y for a commodity when its price x is given by

2

1xyx

Find the elasticity of demand when the price is 3 units.

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5C

PARTIAL DERIVATIVE

UNIT STRUCTURE

5C.1 Objectives

5C.2 Introduction

5C.3 Definitions

5C.4 Summary

5C.5 Unit End Exercise

5C.1 . OBJETIVES

After going through this chapter you will be able to.

Partial derivative with total differentiation, second order partial

derivative.

Elasticity of demand with application of partial derivative

Production function: -

Marginal productivity of labour and capital, average product of labour and

capital

Some economics application

5C.2. INTRODUCTION

In previous chapter we have learnt differential calculus and its

application with one variable of the form y = f(x) but in real life, there are

so many c ase of multivariable . For example production may be treated as

a function of labour & capital and price may be a function of demand and munotes.in

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supply. In such a case we use partial derivative two or more variables are

these but we will study here two and three variable function.

5C.3 DEFINITIONS

Partial derivative: -

Z = f( x, y) be a function of two independent variables x and y. The

derivative of f( x, y) with respect to x, keeping y constant is called partial

derivative of z with respect to x and is den oted by

.xzfor or fxx

Simi larly when the derivative of f(x, y) with respect to y, keeping x

constant is called partial derivative of z with respect to y and is den oted by

.yzfor or fyy

Thus in terms of limit we can write,

0,,lim

xf x x y f x y f

xx

limit exists

0,,lim

yf x y y f x y f

yy

Successive partial derivative:

Let

, z f x y be two variable function and

&zz

xy

the first order

partial derivative & both are function of x & y respec tively further be

differentiate partially with respect to x & y, we get second order partial

derivative

i.e.

2

2............. Izzfxxx z x

2

2............. IIzzfyyy y y munotes.in

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2

............. IIIzzfyyx y x y

2

............. IVzzfyyy x x y

III & IV are called mixed partial differentiation.

NOTE: If f, fx & fy are continuous then fxy = fyx

Ex. 1) Find all second order partial derivative for

22, 30 4 7f x y x y xy x

Soln: Given function

22, 30 4 7f x y x y xy x

2230 4 7xff x y xy xxx

0 2 0 4 7 xy

2 4 7xf x y

2230 4 7yff x y xy xyy

0 0 2 4 0 yx

24yf y x

2 4 7xxff x yx x x

2 0 0 2

2xxf

24yyff y xy y y

2 0 2

2yyf

2 4 7yxff x yy x y

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4

4yxf

24xyff y xx y x

=0-4

4

4xyf

Ex 2) If

21

12z

xy y

Prove that

23.zzx y y zxy

Soln:

Given function

21

12z

xy y

12 2. 1 2i e z xy y

322 2 11 2 1 22zxy y xy yxx

32 2 11 2 22xy y y

32 2. 1 2 (I)zx xy xy yx

Now,

3

22 211 2 1 22zxy y xy yyy

3

2 211 2 2 22xy y x y munotes.in

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3

2 211 2 2)(2xy y x y

322 2. 1 2 (II)zy xy y xy yy

Subtracting (II) from (I) we get,

332 2 2 22.

1 2 1 2zzxyxy

xy xy y xy y xy y

3 3 32 2 2 2 2 2 21 2 1 2 1 2xy xy y xy xy y y xy y

322 212y xy y

23yz

Hence,

23.zzx y y zxy proved.

*Homogenous function :

A function

,f x y of two variables is said to be homogeneous functio n in

which the power of each ter m is same. i.e . x & y of same degree.

If f (x, y) is homogeneous function of two variables x & y with degree n ,

If

, , for 0nf x y f x y

211:,eg f x yx xy

Here

,f x y is a homogeneous function of degree – 2.

*Euler’s t heorem on homogeneous functions:

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Let

,f x y be a homogeneous function of degree n, with

1 2,, ..........,n x x x

variables then

1 2 1 2

12. . ..... . , ,.......nn

nf f fx x x n f x x xx x x

If z is homogeneous function of x, y, of degree n and

( ), z f u then

. ( )..()z z n f uxyx y f u

Ex. 1) If

2 2 21 1 log log,xyf x yx xy x y

then prove that

. . 2 0ffx y fxy

Soln:

2 2 21 1 log log,xyf x yx xy x y

2 2 2 2log1 1 1 1.

1y

x

y x x x yx x

,f x y

is a homogeneous function of degree – 2

By Euler’s theorem.

..ffx y nfxy

. . 2ffx x fxy

. . 2 0ffx f fxy

Ex. 2) Verify Euler’s theorem for the function

1144

1155xyz

xy

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Soln: Here

1144

1155xyz

xy

The numer ator is a homogeneous function of degree

14 and denominator

is homogeneous function of degree

15

By Euler’s theorem,

1. . .20zzx y zxy

Now,

1144

1155z x y

xx xy

1 1 1 1 1 1 1 15 5 5 5 4 4 4 4

21155.ddx y x y x y x ydx dx

xy

1 1 3 4 115 5 5 4 4 4

2115511.45x y x x y x

xy

1 1 1 1 1 15 5 5 4 4 4

2115511..45.x x y x x yzxxxy

Similarly,

1 1 1 1 1 15 5 5 4 4 4

2115511..45.y x y y x yzyyxy

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..zzxyxy

1 1 1 1 1 1 1 15 5 5 5 4 4 4 4

2115511.45x y x y x y x y

xy

1144

115511

45xy

xy

1.20z

Here, Euler’s theorem ,

1..20zzx y zxy is verified .

Ex. 3)

If

33

log .xyzxy Show that

. . 2zzxyxy

Soln:

33

logxyzxy

Here z is not homogeneous function but if

3

3

33

21

1zyxxxy yu e x fx xy yxx

Here z is homogenous function of degree 2 .

1.. . 2. 2z

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Here

. . 2zzxyxy

Check your progress:

(1) If

33

33xyzxy show that

. . 3zzx y zxy

(2) If

22xyf

xy , prove that

3..2zzx y zxy

(3) If

,yufx prove that

. . 0uuxyxy

(4) Verify Euler’s theorem for

(1)

2224 ax xy by

(2)

44xy

xy

(5) Verify Euler’s theorem for function

3344

3355xyz

xy

Maximum and minimum values of two variable function

, z f x y

Let

, z f x y be two variables function for which continuously partial

derivative exists in interval (a, b) .

Consider

2

2A ( , )xxff a bx

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2

B ( , ).xyff a bxy

2

2( , )yyfC f a by

IInd derivative test conditi ons for maxima and minima.

(1) If

20 & 0 AC B A then z is relative m inimum.

(2) If

2AC - B < 0 relative maximum.

(3) If

2AC - B < 0 it is neither minimum nor maximum. It is called saddle

point.

(4) If

2AC - B < 0 then it is maximum, minimum or saddle poin t.

Ex. 1) Find the points of maximum & minimums for the function

3 2 2 23 3 3 4 z x xy x y

223 3 6zxyxx

66zxy yy

Now,

0 & 0zz

xy

223 3 6 0xyx

.... (I)

6 6 0xy y .....(II)

Solving (I) & (II) we get,

0 1 2x x x

01yy

0,0 2,0 , 1,1 , 1, 1

are the stationary points.

A 6 6xxfx

0,0 6 2,0 6xx xxff

1,1 0 1, 1 0xx xxff

6xy B f y

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1,1 6 1, 1 6xy xyff

C 6 6yyfx

0,0 6 2,0 6yy yyff

1,1 0 1, 1 0yy yyff

For (0, 0)

2 2AC-B 6) ( 6 0 36 0& A<0

z is relatively maximum at (0,0)

0,0 4z

For (2, 0)

2 2AC-B 6 6 0 36 0& A 0

z is relatively minimum at (2,0)

2,0 0z

For (1,1)

2 2AC B 0 0 6 0 36 36 0

Which is saddle points .

For (1, -1)

2 2AC - B 0 0 6 0 36 36 0

Which is saddle points.

Examine maxima & minima for the following functions:

(1)

223 2 4 7 12z x y xy x y z is minimum at (1,2) z = 3

(2)

222 2 2z x y x y z is maximum at (1,1) z = 4

(3)

333 z x y xy z is maximum at (1,1) & z = -1

(4)

336 63 12 z x y x y xy

z is maximum (3,3) z = 540 munotes.in

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z is maximu m (-7,-7) z = 7 84

*Production function :

The C obb - Dougles production function of the economy as a whole is

given by

P AL K , 1

Where P is total production, L is the quantity of labour , k is the quantity of

capital and

A, , are constants .

Marginal pro ductivity of labour & capital

Assume Q = f (L, K) is the production function where the amount

produced is given as a function of the labour and capital used.

Where Q (Total output) = f (L, K).

The partial derivative

&QQ

LK

gives marginal product of labour &

capital respectively for the C obb-Dougles production function.

1MPK .Aab Q bQb L KKk

and

1MPL Aab Q aQa L KLk

Thus for the C obb-Dougles production function, the marginal product of

capital respect to labour is a constant times the average product capital

with respect to labor .

Q

KQ K

Q QK

K

Which is called the Capital elast icity of pr oduct, it is equal to the ratio of

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Ex 1) The fo llowing is a linear homogenous production function

222 x aL hLK bk

where X, L, K represent output, labour and capital

respectively. Show that

..XXL K XLL

Soln: Given homogeneous & production function

122 22 X aL hKL bK

Marginal production labour

122 22XaL hKL bKLL

1

2 2 2 2 212 . 22aL hKL bK aL hKL bKL

2 2 2 21 2 2

2 22aL hK aL hK

aL hKL bK aL hKL bK

Marginal product capital

122 22XaL hKL bKKK

22

221122 2aL hKL bKK aL hKL bK

2 2 2 222

2 2 2bK hL bK hL

aL hKL bK aL hKL bK

2 2 2 2.

22L aL hK K bK hL XXLKLL aL hKL bK aL hKL bK

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22

222

2aL hKL bK

aL hKL bK

122 22 aL hKL bK

=

X

..XXL K XLK . Hence proved

*Principle of marginal rate of technical substitution (MRTS)

The principle of marginal rate of technical substitution is based on the

production function wher e two functions can be substituted in variable

proportions in such a way as to produce a constant level of output .

Definition : “The marginal rate of technical substitution is the amount of

an output that a firm can give up by increasing the amount of the other

input by one unit and still remain on the came isoquant”.

The marginal rate of technical substitution between two functions

capital can be substituted for capital in the production of goal x without

changing the quantity of output . As we move along an isoquant

downwards to the right each point on it represents the substitution of

labour for capital.

The marginal rate of technical substitution of labour for capital is the

slope or gradient of the isoquan t at a point a ccordingly ,

Slope of MRTS k

KK

LL

The marginal rate of technical substitution can also be expressed as the

ratio of the marginal physical product labour to the marginal physical

product of capital. munotes.in

## Page 177

176

i.e. MRTS LK

MPL

MPK

Though the output remains constant the process of substitution being

change.

Isoquants: -Isoquant is the curve respectively all possible efficient

combination of inputs needed to produce a certain quantity of specific

output or output combination.

The main assumptions of iso-quant curves are as follows:

(1) Only two functions are used to product into small points.

(2) Factors of production can be divided into small parts.

(3) Technique of production is constant.

(4) The substitution between the two factors is technically possible.

(5) Under t he given technique, factors of production can be used with

maximum efficiency.

Properties of Isoqua nt curve:

(1) Isoquant curve slope downwards from left to right because MRTS of

labour for capital diminishes.

(2) Isouants are coming to the origin because of the MRTS diminishes

along the isoquant.

(3) Two isoquant curves represent higher level of output.

(4) Isoquants need not be parallel to each other.

(5) No Isoquants can touch either a xis.

(6) Each Isoquant is oval -shaped.

Utility function: munotes.in

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177

Let

( , , ) u f x y z be a function of three variables x, y & z then partial

derivatives of u, i.e

,&u u u

x y z

are the marginal functions of x, y, z

respectively.

Let

( , ) u f x y be the total utility function of a consumer, whe re x & y are

the quantities of two commodities (or goo ds) q 1, q2 which be consumes.

Then

u

x

is the (partial) marginal utility of x and

u

y

is the (partial)

marginal utility of y .

Marginal Utility:

In the theory of economics behaviou rs utility functions relates to total

utility (u) obtained from the consumptions of a given quantity (u) . Thus

given the utility function u = u(Q), the additional derivative from an

additional infinitesimal cons umption of Q is given by the derivati ve

'( )uuQQ

Which is called marginal utility. Furthers, the change in marginal utility

due to infinitesimal change in Q is given by the second order derivative

2

2''( )uuuQQ Q Q

If marginal utility (mu) declines as Q increases, t hus

''( ) 0uQ indicating

the operation of the law of diminishing marginal utility.

Derivative of shape of indifference curve:

The concept of partial derivative and total derivative can be used to

find out the shape of an indifference curve in connection with consumer’s

behaviou rs as isoquant in connection with a production function. Both the

indifference curve and the isoquant a re always com plex to the origin in

order that utility on total output should remain constant.

munotes.in

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178

For setting convexity of a curve, the first derivative should be negative

and the second order derivative should be positive .

Consider a utility function when the consumer consume s two

commodities x & y such that

( , ) u u x y ..... (I)

The convexity of indifference curve given by

2

20 & 0yy

xx

By taking total differential of utility function we get,

0uudu dx dyxy

x

yuu dy x

y dx u

y

.... (II)

When u x & u y is marginal utility of x & y respectively

( , ) & ( , )x x y yu u x y u u x y

again differentiate eqn II with respect to x

2

2x

yu d y dy

dx x dx x u

2

322

()x y xy y x

yx x yy u u u u u u u

u

The sign of above equation indicate whe ther the difference curve is

conv ex or concave. We can show that the value of the expression in the

numerator is positive u y > 0 and u y3 > 0.Hence

2

20dy

dx

munotes.in

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179

Ex Chick whether the indifference curve of convex or concave.

21

2u x y is convex or concave.

Soln: Given utility function :

21

2u x y

122xu x y x y

2

21 1 1

242yxux

yy

1

2xx xyxu y u

y

232 2

311.248yyxu x y

y

22 2

232x y xy y xx x yy

yu u u u u u u dy

dx u

22 2 22

3

3

21 1 124 8 4 2

1

4x x x xx y x y y

y y y y

x

y

4 4 4

64 8 16

64x x x

y y y

x

yy

munotes.in

## Page 181

180

220y

x

Since both x & y are positive

Therefore, the indifference curve of the utility function is convex to the

origin .

*Marginal Rate Substitution :

If the utility function is given level of utility or satisfaction from various

combinations a1 & a2 respectively. Havens , the locus of all such

combination is an indifference curve.

A

OB

Commodity xCommodity y

The total differential of the utility function

U ( , ) isu x y

.uudu dx dyxy

But U along the indifference curve is zero.

du o

0du udx dyxy

x

yu dy

dx u munotes.in

## Page 182

181

The negative value of

dy

dx is called the marginal rate of Substitution of

commodity a1 for a2.

The Hessian matrix and determining optimization.

The Hessian matrix of

( , ) z f x y is defined to be

( , )xx xy

f

yx yyff

H x yff

at any point at which all the second partial derivative of f exists.

Determinant of

( , )xx xy

f

yx yyff

H x yff

( . ) ( . )xx yy xy yxf f f f

Test for optimization: -

1st find boundary point by using

( , ) 0f x y . To identity, if a point ( x, y,)

with zero gradie nt is local maximum and minimum, check the Hassian

determinant.

(a) If H ( x1, y1) is positive then (x1, y1) is a local minimum.

(b) If H (x1, y1) is negative definite then (x1, y1) is a local maximum.

Ex Find the local extreme of

33( , ) 3f x y x y xy

2

20 33,0 33xyf x y

yx

23 3 0xy ....(I)

23 3 0yx .....(II )

Solving (I) & (II) we get,

, 0,0 1,1x y or

63,36xH x yy munotes.in

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182

030,0 930Det H

1st minor = det (H1) = 36 & 2nd minor = det |H2| = -9

631,136H

1st minor = det (H1) = 36 & 2nd minor = det |H2| = 36 - 9 = 25 > 0

(1,1)H is positive definite which implies that (1,1) is local

maxim um f(x, y)

Budget line: -

A graphical depiction of the various combinations of two selected

products that a consumer can afford at specified prices for the products

given their particular income level. When a typical business is analysing a

two product budget line, the amount of th e first product are plotted on the

horizontal x axis and the amounts of the second product are plotted on the

vertical y axis.

The problem is about how much goods a person can buy with limited

income. Assume: on saving, with income I, only spend money on goods x

& y with the price Px & Py.

Thus, budget constraint is

.xyP x P y I

Suppose

2 1, 8xyP P I then

28xy

The slope of budget line is

dy Px

dx Py

Bundles below the line are affordable . munotes.in

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183

Constrained Optimization with lagrange ’s multipliers:

Suppose f(x,y) is a given two variable function and g(x, y) = 0 is

constraints on the variable x & y.

Using the Lagrange’s multipliers method to find the constrained maxima

or minima of f ( x,y) by using Lagrange’s function

, y, , ,L x f x y g x y

The following steps are to be followed:

Step (1) Taking lagrange’s multiplier

( , , ) ( , ) ,L x y f x y g x y

Step (2) Find

,xyL L and L

Step (3) Taking

0 & 0 & 0xyL L L we get stationary points and

value of

Step (4) Find

,L ,L & ,xx xy yy x yL g g

Step (5) Taking determinant

0 9 9

9

9xx xy

xy yyxy

D x L L

y L L

Step (6)

(i) D > 0, (a, b) then given function f has a maximum at the

stationary point.

(ii) If D < 0 then the given function f has minimum at that

stationary point.

Ex 1) The cost function of a product is made from two row material x

and y. The profit function is given by 115 x + 117 y – x2 – 2y2. If we want to

manufacture 98 units of both products , taken together per day, f ind the

number of units of each type to be manufactured by th e company to get

maximum profit.

Soln:

Given cost function

22, 115 117 2f x y x y x y and

constraints is

, 98g x y x y munotes.in

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184

Lagrange’s function

22, , 115 117 2 98L x y x y x y x y

taking 1st order partial derivative

115 2xLx

117 4yLy

98 L x y

taking

0, 0& 0xyL L L we get,

115 2 0 117 4 0 98 0 x y x y

115

2x

..... (I)

117

4y .....(II)

98 xy

..... (II I)

Substituting I & II in III we get

115 1179844

230 2 117 392

227 3

45 3

15

Substituting

15 in I & II we get

x = 65 & y = 33

65,33

is the stationary point.

2 0 4 1, 1xx yx yy x yL L L g g

0 9 9 0 1 1

9 1 2 0

9 1 0 4xx xy

xy yyxy

D x L L

y L L

2 0 1 0 1 20 1 10 4 1 4 1 0

0 1 4 1 0 2 munotes.in

## Page 186

185

4 2 6 0

The function f(x, y) is maximum at (65, 33) &

maximum value of f is

f

2265,33 115(65) 177(33) (65) 2(33)

= 4933

Ex 2) Find the maximum and minimum distance of the point (3, 4, 12)

from the sphere

2 2 21 x y z

Soln: Let the co -ordinates of the given point be ( x, y, z ) then its distance

D for (3, 4, 12)

2 2 23 4 12 D x y z

2 2 2, , 3 4 12f x y z x y z

2 2 21 x y z

2 2 2g( ) x 1x y z

, , , , , , ,L x y z f x y z g x y z

2 2 2 2 2 23 4 12 1x y z x y z

2 3 2xL x x ..... (I)

2 4 2yL y y ..... (II)

2 12 2zL z z ..... (III )

taking

0, 0, 0x y zL L L

3

1x ..... (IV)

4

1y ..... (V)

12

1z ..... (VI)

2223 4 121111

2229 16 1441

111

21 169 munotes.in

## Page 187

186

1 13

Substituting

1 13 in IV,V &IVI

3 4 12 3 4 12,,13, 13, 13 13 13 13

and

3 4 12,13, 13 13

The optimum distance

2 2 23 4 123 4 1213 13 13D

12

The maximum distance =

2 2 23 4 123 4 1213 13 13

14

5C.4. SUMMARY

In this chapter we have learned:

Partial derivative of Ist & IInd order.

Elasticity of demand with the partial derivative.

Production function, MPL and MPK.

Utility function, Marginal utility

Marginal rate of technical substitution

Isoquant properties of Isoquant

Indiffe rence curve

Optimization test y Hassian matrix

Budget line

To optimization by Lagrange’s multipliers

5C.5. UNIT END EXERCISE:

(1) Calculate

&zz

xy

, function

2

1xzxy

Ans:

22

1x x y zz

x xy ;

2

21zx

y xy munotes.in

## Page 188

187

(2) If

() u f x where

22r x y prove that

22

221"( ) '( )yyffxy

(3) Verify Euler’s theorem for the function

(I)

221zx xy y (II)

44xyzxy

(4) If

22

logxyzxy prove that

1zzzyxy

(5) The production function of a firm is given by

3 1444 , 0, 0.Q L K L K

Find the marginal productivity of labor and

capital. Also show that

.QQL K QLK

(6) If

526log( ) 18z xy x y . Find the value

22

&..zz

x y y x

State the conclusion to be drawn from the result.

(7) If utility function is

1 2 1 2 logu ax bx x x . Find the marginal

utility.

(8) If

yufx . Show that

. . 0uuxyxy

(9) If

3 2 3( , ) 2 11 3f x y x x y prove that

. . 3 ( , )ffx y f x yxy

(10) If

2logyyxx prove that

. . 24yyxyxy

(11) Given the production function

2 3 2 31.08 0.03 .68 0.08 .Q L L l K K

Find the quantities of labor

and capital that maximize output Q. Ans

24, 24LK

(12) Given a total cost function

27 2 64c x xy . Find combination of

input x & y must be produced to meet the requirement of 77 units

that maximizes the cost of fulfillin g the requirement.

(13) Find the all 2nd order partial derivative at given point. munotes.in

## Page 189

188

(1)

22xyz x yyx

(2)

3 2 2 323 z x x y xy y

(14) If the production function is given as

0.4 0.7100Q L K . Find the

marginal productivity of labor & capital.

(15) Find the marginal productivity of labor and capital at (5, 1) of

production .

3369 Q L L K LK K

(16) Examine for maxima & minima for

(1)

22, 6 4f x y x x y y

(2)

33, 12 17f x y x x y y

(17) Optimize the cost function subject

222 26 10x x y ky to the

constraint x + y = 20.

(18) A monopolist charges different prices in the two markets where his

functions are:

1 1 2 2 1 221 0.1 , & 50 0.4 ; ,x P x P P P being prices

and x1, x2 be quantities demanded. His total cost function is TC =

10x + 2000, where x is total output Find the prices that the

monopolist should charge to maximize his profit .

Ans : max profit = 67.5.

munotes.in

## Page 190

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6A

INDEFINITE and DEFINITE

INTEGRATION

UNIT STRUCTURE:

6A.1 Objectives

6A.2 Introduction

6A.3 Unit End Exercise

6A.1. OBJECTIVES

After studying this chapter, you should be able to understand:

To find the indefinite integral of a given function

To state the standard indefinite integrals

To evaluate definite integrals

6A.2. INTRODUCTION

After having learnt what is meant by differentiation, we come to the

reverse process of it, namely integration.

Consider the following examples:

(1) If f(x)=x , then

'( )fx =1,Question: What is the function whose

derivative is 1? Ans: x

(2) If f(x)=x3, then

'( )fx =3x2,Question: What is the function whose

derivative is 3x2? Ans: x3 munotes.in

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190

(3) If f(x)=x3/2,then

'( )fx =

3

2x1/2,Question: What is the function whose

derivative is

3

2 x1/2? Ans: x3/2

The answers which we find (the functions x, x3, x3/2) are called

primitives or anti -derivatives or integrals of the given function .

* Definition of integral of a function:

If f(x) and g(x) are two functions such that

( ) ( )dg x f xdx then we

define integral of f(x), w.r.t x to be the function g(x). This is put in

notation form as

( ) ( )f x dx g x ,

read as integral of f(x) w.r.t. x is g(x).The function f(x) is called the

integrand. Presence of dx indicates that the integration is to be taken

with respect to the variable ‘x’. The process of finding the primitive or

integra l of a function is called integration. Thus integration is the

inverse process of a differentiation.

323dxxdx

233x dx x

,

But

3 2 3 27 3 , 5 3 ,ddx x x xdx dx

32 , 3, In general where c is any realdx c x nudxmber

Hence, in general, we write ,

233x dx x c .

The number c is called the constant of integration.

, ( ) ( )dHence if g x f x thendx

( ) ( )f x dx g x c ,

For different values of c, we get different integrals of g(x).

() . ( ) is called indefin f x d ite integ xg r xc al

munotes.in

## Page 192

191

*Integrals of standard functions:

If a and b are any non -zero real numbers,

11()1. , 1 ( ) , 11 ( 1)nn

nn x ax bx dx c n ax b dx c nn a n

1 1 12. log log()dx x c dx ax b cx ax b a

13.x x ax b ax be dx e c e dx e ca

4.log logx bx k

x bx k aaa dx c a dx ca b a

1

22115. tanxdx cx a a a

22116. log2xadx cx a a x a

22117. log2axdx ca x a a x

22

2218. log dx x x a c

xa

22

2219. log dx x x a c

xa

1

22110. sinxdx ca ax

2

2 2 2 2 111. sin22x a xa x dx a x ca

2

2 2 2 2 2 212. log22xaa x dx a x x a x c

2

2 2 2 2 2 213. log22xax a dx x a x x a c

'( )14. log ( )()fxdx f x cfx

'( )15. 2 ( )

()fxdx f x c

fx

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## Page 193

192

*Rules of Integration:

If f(x) and g(x) are two real valued functions such that

()f x dx and

( ) ,g x dx exist then

1. [ ( ) ( )] ( ) ( )f x g x dx f x dx g x dx

2. ), . ( ) ( wher k f x e k is a real c dx k ons f x dx tant

,

Example s:

Integrate the following w.r.t. where x is given by:

(i)

3+1 4

3 3x + 4 3x + 43x + 4 dx = + C = + C3 3+1 12

(ii)

22 4 23 5 9 30 25 x dx x x dx

=

429 30 25 1x dx x dx dx

=

53

9 30 2553xxxc

=

53910 255x x x c

(iii) If

32' 4 3 2 f x x x x k and

0 1 ; 1 4ff ,

Find

fx .

' 3 24 3 2 f x x x x k

32' 4 3 2 f x f x dx x x x k dx

=

224 3 3 2 1x dx x dx x dx k dx

=

4 3 2

4 3 24 3 2x x xkx c

4 3 2 f x x x x kx c munotes.in

## Page 194

193

Now

4 3 21 0 0 0 0 0

1f k c

c

4 3 24 1 1 1 1 1f k c

111 kc

1 1 ( 1)kc

4 2 4 2

2kk

k

4 3 221 f x x x x x

*Methods of Integration :

In this method, we reduce the given function to standard form by

changing variable x to t, using some suitable substitution

xt .

Result:

If

xt is differentiable function of t

Then,

' f x dx f t t dt

Corollary : 1

1

' , 1 01n

n fxf x f x dx c nn

Corollary : 2

'log | |fxdx f x cfx

Corollary: 3

'2fxdx f x c

fx

*Evaluate the following:

(1)

log7 1 logxxx dx

Soln:

log7 1 logxxLet I x dx

Put x log x = t. Different w.r.t. x munotes.in

## Page 195

194

1log 1dtxxx dx

1 log x dx dt

log7t 77log 7 log 7xx

t

eeI dt c c

2)

1

ndxxx

Soln:

1

1

1

11

1

1

1 1

11

11

1 1n

n

n

n

n

n

nLet I dx

xx

xdxx

nxdxnx

dxdxdxn x

'

1 1log | 1| [ log | | ]1n fxx c f x cn f x

munotes.in

## Page 196

195

3)

5

51x

xedx

e

Soln:

Let

5

51x

xeI dx

e

Put

51xet . Differentiate w.r.t. x,

5555xx dte dx dt e dx

1/2 1

1/2 1 1 1 1 1 1

1 5 5 5 512tI dt dt t dt c

tt

1/2

1/2 5 1 2 2 215 1/ 2 5 5 5x tc t c t c e c

(4)

11xe

xeexdxex

.

Soln:

Let

11xe

xeexI dxex

11

1

1xe

xe

x

e

xe

xe

xee e xI dxex

exedxex

e exdx

e e x

()1xe

xedexdxdxe ex

'1log [ log | | ]xe fxe x c f x ce f x

munotes.in

## Page 197

196

Some Special Integrals :

1)

1

2211tanxdx cx a a a

2)

2211log2xadx cx a a x a

3)

2211log2axdx ca x a a x

4)

1

221sinxdx ca ax

5)

22

221log dx x x a c

ax

6)

22

221log dx x x a c

xa

7)

2

2 2 2 2 1sin22x a xa x dx a x ca

8)

2

2 2 2 2 2 2log22xaa x dx a x x a x c

9)

2

2 2 2 2 4 2log22 xax a dx x a x x a c

Ex.: Find the following integrals :

1)

2 2

2 21 1 1 1 1

49 9 49 9 9 7

9 3I dx dx dxxx x

11

221 1 1 1tan tan77933xxc dx cx a a a

11 1 3 3 1 3tan tan9 7 7 21 7xxcc munotes.in

## Page 198

197

2)

2 2 2 21 1 1 1 1

1 4 1 4 4 14 2I dx dx dxx x x

22111 2log1 1 422 2

11log2x

c

x

xadx cx a a x a

1 2 1log4 2 1xcx

3)

2 2 211

25 5I dx dx

x x

22

221log 25 log 25x x c dx x x c

ax

4)

22

623 1 1xxI dx dx

x x

Put

3 2 233dtx t x dx dt x dx

1

211sin33 1dtI t c

t

13 1sin3xc

1

221sinxdx ca ax

5)

21

48I dxxx

2 2 211

4 4 4 8 22dx dxxx x

1 12tan22xc

1

2211tan xdx cx a a a munotes.in

## Page 199

198

6)

249 x

Let

249 I x dx

2223 x dx

2

21 3 9 34 9 sin2 2 2x x xxc

2

2 2 2 2 1sin22x a xa x dx a x ca

27) 9 4 x dx

Let

2232 Ix

2

22 399 4 log 3 9 422xxx x x c

2

2 2 2 2 2 2log22xax a dx x a x x a c

28) 4 5 x x dx

Let

245 I x x dx

24 4 1 x x dx

2221x dx

22 214 5 log 2 4 522xx x x x x c

29) 2 ax x dx

2

2 2 22

2Let I ax x dx

a a ax x dx

2 2 2

2 22 a a ax x dx

a a x dx

munotes.in

## Page 200

199

2

212 sin22a x a a xax x ca

10) 3 5 x x dx

28 15 x x dx

21 8 15x x dx

21 { 8 16 16 1 } x x dx

221 4 1 x dx

2214 x dx

2

2 21 1 144 1 4 sin2 2 1xx x c

1 413 5 sin 422xx x x c

.

* Integration by parts:

The method of integration by parts is used when the integrand is expressed

as a product of two functions, one of which can be differentiated and the

other can be integrated conveniently.

If u and v are both functions of x, then

. du vdx u vdx u vdx dxdx

Note:

(1) When integrand is a product of two functions, out of which the

second has to be integrated ( who’s integral is known), hence we

should make the proper choice of first and second function.

(2) We can also choose the first function as the function which comes

first in the word ‘ LIATE’ where

L-logarithmic function

I-the inverse function

A-the algebraic function munotes.in

## Page 201

200

T-the trigonometry function

E-the exponential function

Examples .: Find the following integrals.

1)

22 xx e dx

Let

22 xI x e dx

22 xu x v e

& by using

.duu v dx u v dx v dx dxdx

I

2 2 2 2 xx dx e dx x e dx dxdx

22

2222 xxeex x dx

22

2

2x

x xexe dx

Again by using integration by parts method for

2xx e dx

2 2 2 21

2x x x dx e x e dx x e dx dxdx

2

2 2 21122x

xx ex e x e dx dx

2 2 2 21 1 1

2 2 2x x xx e x e e dx

2

2 2 21 1 1

2 2 2 2x

xx ex e x e c

2 2 2 21 1 1

2 2 4x x xx e x e e c

I

2 2 2 21 1 1

2 2 4x x xx e x e e c

2)

logx dx munotes.in

## Page 202

201

Let

log I x dx

1.log x dx

Let

log , 1u x v &

duuv dx u v dx v dx dxdx

log 1 log 1dI x dx x dx dxdx

1logx x x dxx

log 1x x dx

logx x x c

3)

32xx dx

Let

32xI x dx

Taking

3&2xu x v by using integration by parts

3322xx dx dx x dx dxdx

332213 log 2 3 log 2xx

x dx

3

3 2123 log 2 3log 2x

x xI dx

3

3 1 1 223log 2 3log 2 3 log 2x

xxc

3

3

21223log 2 9 log 2

x

xxc

4)

lognx x dx

Let

lognI x x dx munotes.in

## Page 203

202

Taking

log &nu x v x integrating by parts

log lognn dI x x dx x x dx dxdx

111log11 nnxxx dxn x n

11log11n

n xx x dxnn

111log1 1 1nnxxxcn n n

11log11nxI x cnn

5)

2log 1 x dx

Let

21 log 1 I x dx

Taking

2log 1 1u x v integrating by parts

22log 1 1 log 1 1dI x dx x dx dxdx

2

22log 11xx x x dxx

2

2

2log 1 21 xx x dxx

2

2

211log 1 21xx x dxx

2

21log 1 2 1 21x x dx dxx

21log 1 2 2 tanx x x x c

6)

log log xdxx munotes.in

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203

log log 1log logxI dx x dxxx

Taking

1log log &u x vx , Integrating by parts

11log log log logdI x dx x dx dxx dx x

1log log log loglogx x x dxxx

1log log log x x dxx

log log log log x x x c

log log log 1x x c

*Integration by Partial Function:

If f(x) and g(x) are two polynomials, then

()

()fx

gx is a

rational function, where g(x)

0.

If deg of f(x) < deg of g(x), then

()

()fx

gx is a proper rational function.

It can be expressed by partial fractions using following table.

Where A, B, C and D used in the table are real numbers.

Rational Form Partial Form

1)

2Px qx c

x a x b x c

A B c

x a x b x c

2)

2

2Px qx r

x a x b

2A B c

x a x b xa

3)

2

3Px qx r

x a x b

23A B c D

x a x b x a x a

4)

2

2Px qx r

x a x bx c

2A Bx c

x a x bx c munotes.in

## Page 205

204

Where

2x bx c cannot be

factorized further

Examples :

i) Evaluate

1 2 1xdxxx

Let

1 2 1 1 2 1x A B

x x x x

Multiplying both sides by

1 2 1xx , we have

2 1 1 x A x B x …. (i)

Putting

1x we get

1 2 1 1 1 1AB

13A

1

3A

Putting

1

2x in eqn (i) we get

1 1 12 1 12 2 2AB

13

22B

12 13 32B

13B

11

1 2 1 3 1 3 2 1xdx dxx x x x

1 1 1 1

3 1 3 2 1dx dxxx

1 1 1log 1 log 2 13 3 2xx

11log 1 log 2 136 x x c

ii) Evaluate

3dx

xx munotes.in

## Page 206

205

3 211 1dx dx dx

x x x x x xx

Let

1

1 1 1 1A B c

x x x x x x

Multiplying both sides by

11x x x , we have

1 1 1 1 1A x x Bx x c x x

Putting

0, 1x and -1 we have

11AA

1122BB

1122cc

31 1 1

2 1 2 1dxdxx x x x x

log 111log log 12 1 2 xxx

11log log 1 log 122x x x c

12log log 1 log 12x x x c

2 1log log 1 log 12 x x x c

22 1log log 12 x x c

2

21log21xcx

iii) Evaluate

222

12xdx

xx

Putting

2xt

2x dx dt

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## Page 207

206

12dtItt

Now

1

1 2 1 2AB

t t t t

1 2 1A t B t ………. (1)

Putting

1 t in eqn (i) we get

1A

Putting

2 t in eqn (i) we get

1 B

Substituting the values A & B in eqn (1) we get

1

1 2 1 2 dt dtI dxt t t t

log 1 log 2t t c

22log 1 log 2x x c

iv) Evaluate

25

12xdx

xx

Let

225

12 1 2 2x A B c

xx x x x …. (1)

25 2 1 2 1x A x B x x C x

Putting

2 x we get

2 5 0 0 2 1 ABc

3 c

3 c

Putting

1 x we get

21 5 1 2 0 0 A B c

241A

4A

Comparing coeff of

2x on both sides of equation

225 4 4 3 2 1x A x x B x x c x

24 3 4 2 x A B x A B C A B C

0 AB

AB

BA munotes.in

## Page 208

207

4 B

Substituting value of A, B & C in eqn (1)

225 4 4 3

12 1 2 2xdx dx dx dxxx x x x

21 1 14 4 312 2dx dx dxxx x

14log 1 4log 2 32x x cx

34log 1 4log 22x x cx

v) Evaluate

2

221

4 25xdx

xx

Consider the integrated & replace

2x by

t only.

(Here we are not substituting

t for

x )

Then the integrated becomes

1

4 25 4 25t A B

t t t t

………. ( x)

1 25 4t A t B t

………. (i)

Putting

4 t in eqn (i), we get

13 217AA

Putting

25t in eqn (i), we get

25 1 0 21 AB

2424 2121BB .

Substituting the values A & B and replacing t by x in eqn (x);

We get ,

munotes.in

## Page 209

208

2

22

221

4 25

1 24

7 21

4 25xdx

xx

dx dxxx

221 1 24 1

7 4 21 25 dx dxxx

11 1 24tan tan14 2 105 5 xxc

11 1 1 24 1tan tan7 2 2 21 5 5xxc

vi) Evaluate

28

24dx

xx

Let

2 28

24 24A Bx cdxxx xx ………. ( x)

Then

28 4 2A x Bx c x ………. (i)

Putting

2 x in eqn (i) we get

8 8 1 AA

Comparing coeff of

2x ,

x and constant number on both sides

of (i)

228 4 2 2Ax A Bx cx Bx c

22 4 2 x A B x C B A c

Comparing coeff of

201 x A B A B B

Comparing coeff of

2 0 2 1 0 2 x B c c C

Compare constant form

4 2 8 4 1 2 2 8AC

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209

Substituting value of A, B & C in equation ( x), we get

2 28 1 2

24 24xdx dx dxxx xx

221122 4 4xdx dx dxx x x

221 1 2 122 2 4 4xdx dx dxx x x

21 11log 2 log 4 2. tan2 2 2xx x c

21 1log 2 log 4 tan22xx x c

2' 2For using log4fx xdx dx f x cx f x

and

1

2 2 21 1 1[ using tan ]4xFor dx dx cx x a a a

*Definite Integration :

In geometrical and other applications of integral calculus, it

becomes necessary to find the difference in the values of the integral of a

function f (x) between two assigned values of an independent variable x,

say a, b. The difference is called the defin ite integral f (x),

where a & be are finite number [a, b], over the interval [ a, b] and is

denoted by

b

af x dx . Thus

b

af x dx b a where

x is an

integral of

fx .

The number a is called lower limit and b is called upper limit of

definite integral. munotes.in

## Page 211

210

b

af x dx is called definite integral because the indefinite constant of

integration does not appear in it.

Since

b

b

a

af x dx x c b c a c b a

Here arbitrary constant c disappears in the process.

*Fundamental theorem of Integral Calculus:

Let f be the continuous function defined on [a ,b].

If

( ) ( ) ,f x dx x c then

( ) ( ) ( )b

af x dx b a

There is no need of taking the constant of integration c, because c gets

eliminated.

*EXAMPLES:

Evaluate the following integrals :

3

2

2

32

2

32

2

64

42(1)

2

1

2

1

2

112x

x

xe dx

e

e

ee

ee

munotes.in

## Page 212

211

1

2

1

1

2

1(2) 3 2

9 12 4x dx

x x dx

132

19 12 432xxx

132

1

3 2 3 23 6 4

3 1 6 1 4 1 3 1 6 1 1x x x

3 1 6 4 3 1 6 1 4 1

3 6 4 3 6 4

1 13

1 13 14

3)

124

1x dx

124

11

1 12x

124

112x

11/2 22 4 1

2 2 1

2 1 2

munotes.in

## Page 213

212

23

2

3

2

2

31 2

32(4)

12

2

2x

xe dx

e

ee

ee

(5)

1

01

1dx

xx

1

0

1

01

11

1

1xx

dx

x x x x

xxdx

1

0

111 12 2

001

1x x dx

x dx x dx

113 32 2

00

1 13 32 2

0 01

3322

22133x x

xx

33 3322 22221 1 1 0 1 033

3 32 2

32

32222 1 133

22 1 13

2223

munotes.in

## Page 214

213

1222 2 13

4213

(6) If

2

038a

x dx , find the value of a

3

0

3

0383

8a

ax

x

(7)

2

1logx dx

2

1log log 1 log x 1dx x dx dx dxdx

(using integration by parts)

1logx x x dxx

22

2

1

111log 1 logx dx x x x dxx

22

11logx x x

2log 2 1 log1 2 1

2 log 2 0 1

2log 2 1

log 4 1 .

(8)

3

2

11

1dx

xx munotes.in

## Page 215

214

Let

2 21

1 1A Bx c

xx xx

211A x Bx C x ………. (i)

Putting

01xA

Comparing coeff of

2ofxx on both sides

0 & 0 , 1 A B C B

3

2

1

33

2

111

1

1

1dx

xx

xdx dxxx

332

1

111log 12 dx xx

3 32

1 11log log 12 xx

1log 3 log 1 log 10 log 22

1 10 1log 3 log log 3 log 52 2 2 .

UNIT END EXERCISE:

[A] Evaluate the following integral:

[1] .x xdx

21[2] x dxx

22[3]xxe e dx

2

[4]1xdxx

2[5]4x

xedxe

21[6]

1dx

xx

2

22(3 2 5)[7]( 1) ( 5)xxdxxx

2(3 2)[8]( 2) ( 3)xdxxx

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## Page 216

215

(3 2)[9]( 2)( 3)xdxxx

2[10] .axx e dx

2[11] (log ) x dx

21[12]

23dx

xx

21[13]8 20dxxx

42[14]1xdxxx

2

222[15]( 4)( 9)xdxxx

2[16] 9 5 x dx

2[17] 4 5 x x dx

2[18] 4 3 2 x x dx

1[19]1x

xedxe

1[20]

1dx

xx

2[21] ( 1)xx e dx

log( )[22]logxdxx

32[23]( 1)xdxx

2[24]65x

xxedxee

1[25]logdxxx

[B] Evaluate the following :

1

01[1]25dxx

1

201

2 2 12 dxxx

4

2 01

233 dx

xx

2

21log4xdxx

3

22 15xdxx

9

461dx

x

2

1log 7 x xdx

4

0( 1)( 4)8xxdx

x

123

0(2 ) 9 x x dx

1

01[( 2)( 1)10] dxxx

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## Page 217

216

6A

Application of

Integration

UNIT STRUCTURE

6B.1 Introduction

6B.2 Unit End Exercise

6B.1. INTRODUCTION

Applications of Integration :

*To find the cost function when Marginal cost is given :

If C repres ent the total cost of producing an output x, then marginal

cost is given by

dcMC C MC dx kdx

The constant of integration k can be evaluated if the fixed cost (i.e.

the cost when x = 0) is given further, average cost AC ca n be obtained

from the relation :

CACx

Ex: The marginal cost function of a product is given by

2100 10 0.1dcqqdq

, where q is the output obtain the total and the

average cost function of the firm under the assumption that is fixed cost is

Rs. 500 .

Soln. munotes.in

## Page 218

217

2100 10 0.1dcMC q qdq

Integrating both sides w. r. t. q we have

2100 10 0.1 C q q dq

23

100 10 0.123qqqk

Now the fixed cost is 500 i.e. when q = 0, C = 500

500k

Hence , total cost function is

3

2100 5 50030qC q q

Average cost is

2500100 5930CqACqq

Ex.: The marginal cost function of manufacturing x shares is

26 10 6 xx

. The total cost of producing a part of shares is 12. Find the

total and average cost function.

Soln.

26 10 6dMC c x xdx

26 10 6 C x x dx

23

6 10 623xxxk

Where k is the constant of integration

Now

12C when

2x

232212 6 2 10 623k

12 12 20 16 k

12 12 20 16 4k munotes.in

## Page 219

218

The total cost fu nction is

236 5 2 4c x x x .

To find the total revenue function and demand function when

marginal revenue function is given.

If R is the total revenue when the output is x, then the marginal

revenue MR is given by

dRMRdx .

Hence, if the MR is given, then the total revenue R is the indefinite

integra l of MR w .r.t. x

i.e.

R MR dx k , where k is the constant of

integration, which can be evaluated for the fact that the total revenue R is

zero when output x is zero.

Since

R px , the demand function can be easily obtained as

Rpx .

Ex: If the marginal revenue function for output is given by

265

2MR

q

, find the total revenue functions by integration. Also

deduce the demand function.

Soln :

265

2MR

q

∴ 𝑹=∫(𝑴𝑹) 𝒅𝒒

∴ 𝑹= ∫(𝟔

(𝒒+𝟐)𝟐−𝟓)𝒅𝒒=−𝟔

𝒒+𝟐−𝟓𝒒+𝒌

Since total revenue is zero at

0q , we get

60 5 02k

03 k

3k munotes.in

## Page 220

219

6532Rqq

6352Rqq

Also we know,

R P q

63536 252qR qPq q q q q

3 2 6 3 6 65522q q

q q q q

352qPqq is the required demand function.

Ex: If the marginal revenue function is

121

2dRMR qdq where R

stands for total revenue .what is the demand function?

Soln :

121

2dRMR qdq

1 121211

1 22 12qR q dq k

12

1/21

122qR

R q k

Total revenue R = 0 at q = 0

120 0 0R k k

12Rq is the Total Revenue function .

121122 RqP q p qqq

is the required demand function.

*Maximum Profit :

Suppose we want to find out the maximum profit of a firm when

only the marginal cost and the marginal revenue functions are given munotes.in

## Page 221

220

equating marginal c ost to marginal revenue we can find the output that

maximizes total profits.

Profit

P R C , where P = total profit, R = total revenue, C = total cost.

dP dR dC

dx dx dx ; x = output …. (i)

Integrating eqn (i),

dR dcP dx dx k R C kdx dx

when k = constant of integration, can be found from the additional

information given.

Remarks :

(i) If may be noted that profit is maximized when marginal revenue

equals marginal cost given the assumption of pure competition

total pro fit is the integral of marginal revenue minus marginal cost

from zero quantity for which profit is maximized.

(ii) To determine profit maximizing output, first find second derivative

of

MR MC i.e. second derivat ive of total profit i.e.

''px

If

'' 0px then maximum profit at x.

(iii) Total profit zero indicates no profit and total profit negative signify

a loss.

Ex:

The ABC Co. Ltd. has approximated the marginal revenue

function for one of it s product by

220 2 MR x x . The marginal c ost

function is approximated by

281 16 MC x x . Determine the profit

maximize output and the total profit at the optimal output.

Soln:

Solving for profit maximizing output, set

MR MC

2 2 2 220 2 81 16 20 2 81 16 9 MR MC MR MC x x x x x x x x munotes.in

## Page 222

221

281 36 3 0 3 9 0 3 9 x x x x x or x

2( 81 36 3 ) 36 6 ''ddMR MC x x x p xdx dx

'' 3 36 6 3 18 0p

3 At x Profit is minimum.

'' 9 36 6 9 18 0p

9 At x Profit is maximum .

Total Profit

9

2

081 36 3 x x dx

923

081. 36. 3.23xxx

923

081 18x x x

2381 9 18 9 9

0 . ;Which indicates no profit

To find the consumption function when the marginal propensity to

consume (MPC) is given :

If p is the consumption when the disposable income of a person is

x, the marginal propensity to consume (MPC) is given by

dpMPCdx

Hence if MPC is given, the consumption p is given the indefinite

integral of M.P.C. w.r.t. x. i.e. munotes.in

## Page 223

222

p MPC dx k

The constant of integration k, can be evaluated if the value of p is

known for some x.

Ex. If the marginal propensity to save (MPS) is 1.5 + 0.2 x-2, when x is the

income. Find the consumption function, given that the consumption is 4.8

when income is ten.

Soln. Now “derivative of consumption function w.r.t. output represents

marginal propensity to consume”.

21.5 0.2dpMPS xdx

2 1 1

21.5 0.2 1.5 0.2 1.5 0.22 1 1xxp x dx x k k

0.21.5p x kx

Now p = 4.8 when x = 10

0.24.8 1.5 10 kx

10.18 k

Hence the consumption function is

0.21.5 10.18pxx

In order to find CS under monopoly, i.e. to maximise profit we must have

MR = MC

2144 96 12 56 4 x x x

212 100 88 0xx

23 25 22 0xx

002213x x OR x x munotes.in

## Page 224

223

When

2

0 0 01, 12 2 100x D x p

1

2

0144 48 4 1 100 CS x x dx

123

0

04 64144 48. 4. 100 144 24 100 units2 3 3 3 xxx

Again when

2

0 0 022 44 64; 123 3 9x p D x

22/3

2

022 64144 48 439 CS x x dx

=

22/33

2

048 1408144 42 3 27 xxx

=

2322 22 4 22 1408144 243 3 3 3 27

=

3168 11616 4 10648 1408

3 9 3 27 27

=

85536 104544 42592 4224

81

=

19360

81units

Consumer’s Surplus :

Consumer’s Surplus

0

00

0x

CS D x dx x p

Ex.: The demand law for a commodity is

220p D D . Find the

consumer’s surplus when the demand is 3.

220 p f D D D

When demand

03 D the price

2

020 3 3 8 p

Consumer’s Surplus =

1

00

0D

f D dD p D munotes.in

## Page 225

224

=

3

2

020 8 3 D D dD

=

9 2720 3 2423D

960 9 242

=

120 9 18 48

2

45

2

Ex: Demand and supply functions are

222 D x x and

56 4 S x x

respectively. Determine CS under monopoly (so as to

maximise the profit) and the supply function is identified with the

marginal cost function.

Soln.

() TR x D x

=

212 2xx

=

2144 48 4x x x

=

23144 48 4 TR x x x

2144 96 12dMR TR x xdx

Since the supply price is identified with MC, we have

. . 56 4M C x

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## Page 226

225

Producer’s Surplus :

Producer’s Surplus

0

00

0 x

PS x p S x dx

Ex. Find the producer surplus under the pure competition for demand

function

821Px and supply function

132Px where p is price

and x is quantity.

Soln.: Under pure competition, market equilibrium conditions can be

obtained by equating the demand and supply.

812312xx

16 4 1 3 1 x x x

216 4 4 4 3 x x x

28 9 0xx

9 1 0xx

1x or

9 x

9x

is inadmissible as quantity cannot be negative

1x

When

1x

882 2 4 2 21 1 1px

Producer surplus :

0

00

0x

p x S x dx

=

1

03122xdx munotes.in

## Page 227

226

=

12

012322xx

112322

17222

724

71244

Ex. The demand and supply function under perfect competition are

216yx

and

224yx respectively. Find the market price

consumer’s surplus and producer’s surplus.

Soln:

Demand function:

216yx …. (1) Substracting (1) from 2

Supply function :

224yx …. (2)

20 12 3 x

0 2xx

when

2

0 16 2 12yy

Thus when the quantity demanded or supplied is 2 units the price is 12

units.

2

2

016 2 12 CS x dx

=

23

016 243xx

=

832 243

=

165333

munotes.in

## Page 228

227

Producer’s surplus :

=

2

2

02 12 2 4 x dx

=

23

024 2 43xx

=

224 8 4 23

=

1624 83

=

3210.673

*The Learning Curve:

In any environment if a person is assigned to do the same task, then after a

period of time, there is an improvement in his performance. If data points

are collected over a period of time, the curve constructed on the graph will

show a decrease in effort per unit for repetitive operations. This curve is

very important in cost analysis, cost estimation and efficiency studies.

This curve is called the learning curve. The learning curve shows that if a

task is performed over and over than less time will be required at each

iteration .

The rate of reduction in direct labour requiremen ts is

described by a curve called Learning curve. The general form of the

function is usually taken as:

f(x)=A.x

Where f(x) is the number of hours direct labour required to produce the xth

unit,-1<0 and A>0.

The total number of labour hours required to produce units numbered ‘a’

through ‘b’ is

( ) .bb

aaN f x dx A x dx

munotes.in

## Page 229

228

y

xfxAxa

Ex.:

After producing 35 units the production manager of x company

determines that its production facility is following a learning curve of the

form

0.5100 f x x where

fx is the rate of labor hours required to

assemble the xth unit. How many total labor hours should they estimated

are required to produce an additional 25 units.

Soln.:

60

0.5

351000 N x dx

6060 0.5 6010.5 2

3535 3510001000 10000.5 0.5xx dx x

11222000 60 35 2000 7.746 5.916 3660 hours

*Rate of Sales:

When the rate of sales of a product is a known function of x, say

fx

where x is a time measure, the total sales of this product over a time

period T is

0T

f x dx .

munotes.in

## Page 230

229

Ex.:

Suppose the rate of sales of a new product is given by

200 90xf x e

where x is the number of days the product is on the

market. Find the total sales during the first 4 days. Given :

40.018 e .

Soln.:

The total sales

44

0090200 90 2001x

x ef x dx e dx x

440

0200 90 200 4 90 0 90xx e e e

4800 90 90 710 90 0.018 711.62 units e

.

6B.2. UNIT END EXERCISE:

[1] The marginal cost function of a firm is given by MC=3000.e0.3x+50,

when x is quantity produced. If fixed cost is Rs. 80,000, find the total

cost function of the firm. [TC=10000e0.3x+50x+70,000]

[2] Find the total cost function and demand function if marginal revenue

is MR=7 -4x-x2.

[3] A company determines that the marginal cost of producing x units of

a particular commodity during a one -day operation is MC=16x -

1591, where the commodity is fixed at Rs. 9 per unit and the fixed

cost is Rs. 1800 per day.(a)Find the cost function.(b)Find the

revenue function.(c)find the profit function. (d)What is the

maximum profit that can be obtained in a one -day operation?

[Hint:(a)C (X)=

2( ) (16 1591) 8 1591MC dx x dx x x k munotes.in

## Page 231

230

(a)Now the fixed cost is 1800. i.e. when x= 0, c=1800. Therefore,

k=1800.

Hence the total cost function is C(x)=

28 1591 1800xx .

(b)R( x)=9 x

(c) P( x)=R( x)-C(X)=

28 1600 1800xx

(d)P

' (x)=-16 x +1600=0

x =100.

Also p

'

' ( x) = -16<0

The maximum profit can be obtained in

one day is P(100)=Rs.78,200]

[4] The marginal cost of a production of a firm is given as

C’(x) =5 +0.13x. Further, the marginal revenue is R

' (x)=18. Also it

is given that C(0)=Rs.120. Compute the total profit.

[Sinc e profit is maximum, where MC=MR.

i.e. 5+0.13x=18

x=100

1 ( ) '( ) 18 18R x R x dx dx x k , where put k 1=0,asunder pure

competition,

TR = Output x Price

R(x)=18 x.

2

2 ( ) '( ) (5 0.13 ) 5 0.132xC x C x dx x dx x k .But given that C(0)=120

k2=120.

C(x)=5 x+0.065 x 2+120.and P( x)=R( x)-C(x)

=13x -0.065 x 2-120.

Total profit when x =100,P(100)=Rs. 530]

[5] Find the consumer surplus and producer surplus under pure

competition for demand function

821px and supply function

3

2xp , where p is price and x is

quantity.

[Under pure competition, market equilibrium conditions can be

obtained by equating the demand and supply. CS=8log2 -4 and

PS=1/4.]

munotes.in

## Page 232

231

[6] Find the consumer surplus and producer surplus defined by demand

curve D(x)=20 -5x

And supply curve S(x)=4x+8.

4/3

040 4[ (20 5 ) ,33CS x dx X and

4/3

0160(4 8)9PS x dx ]

[7] Under a monopoly, the quantity sold and market price are

determined by the demand function. If the demand function for a

profit maximizing monopolist is P=274 -x 2 and MC=4+3 x, find

consumer’s surplus.[TR=P.x=274x -x3,MR=274 -3x2,the monopolist

Maximizes profit at MR=MC. X 0=9and p 0=193

And

9

2

0(274 ) 193 9 486] CS x dx X

[8] The production manager of an electronic company obtained the

following function

0.5( ) 1356.4f x x , where f(x) is the rate of labour

hours required to assemble the unit of a product. The function is

based on the experience of assembling the first 50 units of the

product. The company was asked to bid on a new order of 100

additional units.

Find the total la bour hours required to assemble 100 units

150

0.5

50[ ( ) . 1356.4 ]bb

aaN f x dx A x dx x

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