Page 1
1
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
munotes.in
Page 2
2
(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
Page 3
3
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
Page 4
4
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,
munotes.in
Page 5
5
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)
Subtracting (i) from (ii) munotes.in
Page 6
6
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
Page 7
7
[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 .
munotes.in
Page 8
8
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 :
munotes.in
Page 9
9
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
Page 10
10
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
Page 11
11
(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)
munotes.in
Page 12
12
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
munotes.in
Page 13
13
(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
Page 14
14
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
226 5 1 6 3 0x x or x x munotes.in
Page 15
15
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
munotes.in
Page 16
16
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
8 4 8 0 y y y munotes.in
Page 17
17
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
225 150 0xx munotes.in
Page 18
18
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
Page 19
19
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 ⁄⁄
munotes.in
Page 20
20
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
munotes.in
Page 21
21
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 )
munotes.in
Page 22
22
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
Page 23
23
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
munotes.in
Page 24
24
(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
Page 25
25
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
munotes.in
Page 26
26
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
munotes.in
Page 27
27
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
Page 28
28
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
munotes.in
Page 29
29
*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
Page 30
30
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
Page 31
31
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
a b a b munotes.in
Page 32
32
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
munotes.in
Page 33
33
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)
munotes.in
Page 34
34
=-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
Page 35
35
v) Find product of the following two determinants:
3 2 1
5 3 2
781 and
258
B = 1 3 2
0 4 6
munotes.in
Page 36
36
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
brackets. munotes.in
Page 37
37
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
munotes.in
Page 38
38
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
39
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.
munotes.in
Page 40
40
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
41
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
munotes.in
Page 42
42
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.
munotes.in
Page 43
43
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
munotes.in
Page 44
44
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
munotes.in
Page 45
45
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
munotes.in
Page 46
46
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
Page 47
47
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
munotes.in
Page 48
48
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.
munotes.in
Page 49
49
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
Page 50
50
(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
munotes.in
Page 51
51
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
munotes.in
Page 52
52
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
Page 53
53
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
Page 54
54
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
munotes.in
Page 55
55
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
Page 56
56
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
munotes.in
Page 57
57
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
munotes.in
Page 58
58
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
munotes.in
Page 59
59
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
munotes.in
Page 60
60
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.
munotes.in
Page 61
61
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
Page 62
62
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 =
munotes.in
Page 63
63
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
Page 64
64
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
Page 65
65
= 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
Page 66
66
(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
Page 67
67
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
munotes.in
Page 68
68
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
Page 69
69
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
Page 70
70
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.
munotes.in
Page 71
71
(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
Page 72
72
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
73
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
Page 74
74
(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
Page 75
75
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
using the formula. munotes.in
Page 76
76
.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.
munotes.in
Page 77
77
* 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
Page 78
78
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
munotes.in
Page 79
79
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
Page 80
80
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.?
munotes.in
Page 81
81
(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
Page 82
82
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?
munotes.in
Page 83
83
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.
munotes.in
Page 84
84
*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
munotes.in
Page 85
85
*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.
munotes.in
Page 86
86
*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
Page 87
87
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
munotes.in
Page 88
88
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
munotes.in
Page 89
89
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
munotes.in
Page 90
90
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
Page 91
91
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
Page 92
92
[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
munotes.in
Page 93
93
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
munotes.in
Page 94
94
[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
the final amount if payment is at the end of each quarter. munotes.in
Page 95
95
(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
Page 96
96
balance. (b)Find the interest and principal repayment for each mont h.
[Make the amortization table. ]
munotes.in
Page 97
97
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
Page 98
98
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 :-
munotes.in
Page 99
99
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.
Generally supply and price increases or decreases together . munotes.in
Page 100
100
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
munotes.in
Page 101
101
* 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 .
munotes.in
Page 102
102
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
Page 103
103
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
Page 104
104
* 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
Page 105
105
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
Page 106
106
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
munotes.in
Page 107
107
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
Page 108
108
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
munotes.in
Page 109
109
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
munotes.in
Page 110
110
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
Page 111
111
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
munotes.in
Page 112
112
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.
munotes.in
Page 113
113
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
munotes.in
Page 114
114
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
munotes.in
Page 115
115
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
f a h f aABh munotes.in
Page 116
116
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
Page 117
117
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
Page 118
118
=
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
Page 119
119
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
Page 120
120
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
munotes.in
Page 121
121
* 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
Page 124
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
Page 130
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
Page 149
149
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
differe ntiable function i.e. every point at which the derivative at a concave munotes.in
Page 150
150
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
"0 fx then it is con vex munotes.in
Page 151
151
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
Page 152
152
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
/
/BD D P BD
BP P D BPmunotes.in
Page 153
153
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
munotes.in
Page 154
154
=
()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
Page 155
155
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
Page 156
156
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
munotes.in
Page 157
157
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
Page 158
158
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.
munotes.in
Page 159
159
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
Page 160
160
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
Page 161
161
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.
munotes.in
Page 163
162
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
Page 164
163
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
Page 165
164
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
040 munotes.in
Page 166
165
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
Page 167
166
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:
Statement: munotes.in
Page 168
167
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
munotes.in
Page 169
168
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
munotes.in
Page 170
169
..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
zn f z z z exyx y f z e munotes.in
Page 171
170
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
munotes.in
Page 172
171
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
0,0 0 2,0 0xy xyff munotes.in
Page 173
172
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
Page 174
173
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
the marginal product of capital to the average product of capital . munotes.in
Page 175
174
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
munotes.in
Page 176
175
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
Page 178
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
Page 179
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
Page 180
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
Page 183
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
Page 184
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
Page 185
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
189
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
Page 191
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
munotes.in
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
Page 204
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
munotes.in
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
munotes.in
Page 210
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
munotes.in
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
munotes.in
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
munotes.in
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
munotes.in