NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability
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Page 1
NCERT
SOLUTIONS
CLASS - 12th
aglase .co
SOLUTIONS
CLASS - 12th
aglase .co
Page 2
Class : 12th
Subject : Maths
Chapter : 5
Chapter Name : Continuity and Differentiability
Q1 Prove that the function f(x) = 5x − 3 is continuous at x = 0, at x = − 3 and at x = 5
Answer.
The given function is f(x) = 5x − 3
At x = 0, f(0) = 5 × 0 − 3 = 3
lim x → 0f(x) = lim x → 0(5x − 3) = 5 × 0 − 3 = − 3
∴ lim x → 0f(x) = f(0)
Therefore, f is continuous at x = 0
At x = − 3, f( − 3) = 5 × ( − 3) − 3 = − 18
lim x → − 3f(x) = lim x → − 3(5x − 3) = 5 × ( − 3) − 3 = − 18
∴ lim x → − 3f(x) = f( − 3)
Therefore, f is continuous at x = − 3
At x = 5, f(x) = f(5) = 5 × 5 − 3 = 25 − 3 = 22
lim x → 5f(x) = lim x → 5(5x − 3) = 5 × 5 − 3 = 22
∴ lim x → 5f(x) = f(5)
Therefore, f is continuous at x = 5
Page : 159 , Block Name : Exercise 5.1
Q2 Examine the continuity of the function f(x) = 2x 2 − 1 at x = 3
Answer.
The given function is f(x) = 2x 2 − 1
At x = 3, f(x) = f(3) = 2 × 3 2 − 1 = 17
( )
lim x → 3f(x) = lim x → 3 2x 2 − 1 = 2 × 3 2 − 1 = 17
∴ lim x → 3f(x) = f(3)
Thus, f is continuous at x = 3
Page : 159 , Block Name : Exercise 5.1
Q3Processing
Examine the 100%
math: following functions for continuity.
Subject : Maths
Chapter : 5
Chapter Name : Continuity and Differentiability
Q1 Prove that the function f(x) = 5x − 3 is continuous at x = 0, at x = − 3 and at x = 5
Answer.
The given function is f(x) = 5x − 3
At x = 0, f(0) = 5 × 0 − 3 = 3
lim x → 0f(x) = lim x → 0(5x − 3) = 5 × 0 − 3 = − 3
∴ lim x → 0f(x) = f(0)
Therefore, f is continuous at x = 0
At x = − 3, f( − 3) = 5 × ( − 3) − 3 = − 18
lim x → − 3f(x) = lim x → − 3(5x − 3) = 5 × ( − 3) − 3 = − 18
∴ lim x → − 3f(x) = f( − 3)
Therefore, f is continuous at x = − 3
At x = 5, f(x) = f(5) = 5 × 5 − 3 = 25 − 3 = 22
lim x → 5f(x) = lim x → 5(5x − 3) = 5 × 5 − 3 = 22
∴ lim x → 5f(x) = f(5)
Therefore, f is continuous at x = 5
Page : 159 , Block Name : Exercise 5.1
Q2 Examine the continuity of the function f(x) = 2x 2 − 1 at x = 3
Answer.
The given function is f(x) = 2x 2 − 1
At x = 3, f(x) = f(3) = 2 × 3 2 − 1 = 17
( )
lim x → 3f(x) = lim x → 3 2x 2 − 1 = 2 × 3 2 − 1 = 17
∴ lim x → 3f(x) = f(3)
Thus, f is continuous at x = 3
Page : 159 , Block Name : Exercise 5.1
Q3Processing
Examine the 100%
math: following functions for continuity.
Page 3
(a)f(x) = x − 5
1
(b) f(x) = x − 5 , x ≠ 5
x 2 − 25
(c)f(x) = x + 5 , x ≠ − 5
(d)f(x) = | x − 5 |
Answer.
(a) The given function is f(x) = x − 5
It is evident that f is defined at every real number k and its value at k is k − 5 .
It is also observed that, f(x) = lim x → k(x − 5) = k − 5 = f(k)
∴ lim x → kf(x) = f(k)
Hence, f is continuous at every real number and therefore, it is a continuous function.
1
(b)f(x) = x − 5 , x ≠ 5
For any real number k ≠ 5, we obtain
1 1
lim x → kf(x) = lim x → k x − 5 = k − 5
1
Also, f(k) = k − 5 ( As k ≠ 5)
∴ lim x → tf(x) = f(k)
∴ lim x → kf(x) = f(k)
Hence, f is continuous at every point in the domain of f and therefore, it is a continuous
function.
x 2 − 25
(c) The given function is f(x) = x + 5 , x ≠ − 5
For any real number c ≠ − 5, we obtain
x 2 − 25 (x+5) (x−5)
lim x → ∞f(x) = lim x → ∞ x + 5 = lim x → c x+5
= lim x → ∞(x − 5) = (c − 5)
(c+5) (c−5)
Also, f(c) = c+5
= (c − 5)( as c ≠ − 5)
∴ lim x → ∞f(x) = f(c)
Hence, f is continuous at every point in the domain of f and therefore, it is a continuous
function.
(d)The given function f(x) = | x − 5 | =
{ 5 − x, if x < 5
x − 5, if x ≥ 5
This function f is defined at all points of the real line.
Let c be a point on a real line. Then, c < 5 or c = 5 or c > 5
Case 1 : c < 5
Then, f(c) = 5 − c
1
(b) f(x) = x − 5 , x ≠ 5
x 2 − 25
(c)f(x) = x + 5 , x ≠ − 5
(d)f(x) = | x − 5 |
Answer.
(a) The given function is f(x) = x − 5
It is evident that f is defined at every real number k and its value at k is k − 5 .
It is also observed that, f(x) = lim x → k(x − 5) = k − 5 = f(k)
∴ lim x → kf(x) = f(k)
Hence, f is continuous at every real number and therefore, it is a continuous function.
1
(b)f(x) = x − 5 , x ≠ 5
For any real number k ≠ 5, we obtain
1 1
lim x → kf(x) = lim x → k x − 5 = k − 5
1
Also, f(k) = k − 5 ( As k ≠ 5)
∴ lim x → tf(x) = f(k)
∴ lim x → kf(x) = f(k)
Hence, f is continuous at every point in the domain of f and therefore, it is a continuous
function.
x 2 − 25
(c) The given function is f(x) = x + 5 , x ≠ − 5
For any real number c ≠ − 5, we obtain
x 2 − 25 (x+5) (x−5)
lim x → ∞f(x) = lim x → ∞ x + 5 = lim x → c x+5
= lim x → ∞(x − 5) = (c − 5)
(c+5) (c−5)
Also, f(c) = c+5
= (c − 5)( as c ≠ − 5)
∴ lim x → ∞f(x) = f(c)
Hence, f is continuous at every point in the domain of f and therefore, it is a continuous
function.
(d)The given function f(x) = | x − 5 | =
{ 5 − x, if x < 5
x − 5, if x ≥ 5
This function f is defined at all points of the real line.
Let c be a point on a real line. Then, c < 5 or c = 5 or c > 5
Case 1 : c < 5
Then, f(c) = 5 − c
Page 4
lim x → ∞f(x) = lim x → ∞(5 − x) = 5 − c
∴ lim x → ef(x) = f(c)
Therefore, f is continuous at all real numbers less than 5 .
Case II : c = 5
Then, f(c) = f(5) = (5 − 5) = 0
lim x → 5f(x) = lim x → 5(5 − x) = (5 − 5) = 0
lim x → 5f(x) = lim x → 5(x − 5) = 0
∴ lim x → 5f(x) = lim x → c ∗ f(x) = f(c)
Therefore, f is continuous at x = 5
Case III: c > 5
Then, f(c) = f(5) = c − 5
lim x → cf(x) = lim x → ∞(x − 5) = c − 5
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at every real numbers greater than 5 .
Hence, f is continuous at every real number and therefore, it is a continuous function.
Page : 159 , Block Name : Exercise 5.1
Q4 Prove that the function f(x) = x ′′ is continuous at x = n, where n is a positive integer.
Answer.
The given function is f(x) = x n
It is evident that f is defined at all positive integers, n, and its value at n is n n .
( )
Then, lim x → mf(n) = lim x → m x n = n n
∴ lim x → nf(x) = f(n)
Therefore, f is continuous at n, where n is a positive integer.
Page : 159 , Block Name : Exercise 5.1
Q5
Is the function f defined by
f(x) = { x, if x ≤ 1
5, if x > 1
continuous at x = 0 ? At x = 1 ? At x = 2 ?
Answer. The given function f is
f(x) = { x, if x ≤ 1
5, if x > 1
∴ lim x → ef(x) = f(c)
Therefore, f is continuous at all real numbers less than 5 .
Case II : c = 5
Then, f(c) = f(5) = (5 − 5) = 0
lim x → 5f(x) = lim x → 5(5 − x) = (5 − 5) = 0
lim x → 5f(x) = lim x → 5(x − 5) = 0
∴ lim x → 5f(x) = lim x → c ∗ f(x) = f(c)
Therefore, f is continuous at x = 5
Case III: c > 5
Then, f(c) = f(5) = c − 5
lim x → cf(x) = lim x → ∞(x − 5) = c − 5
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at every real numbers greater than 5 .
Hence, f is continuous at every real number and therefore, it is a continuous function.
Page : 159 , Block Name : Exercise 5.1
Q4 Prove that the function f(x) = x ′′ is continuous at x = n, where n is a positive integer.
Answer.
The given function is f(x) = x n
It is evident that f is defined at all positive integers, n, and its value at n is n n .
( )
Then, lim x → mf(n) = lim x → m x n = n n
∴ lim x → nf(x) = f(n)
Therefore, f is continuous at n, where n is a positive integer.
Page : 159 , Block Name : Exercise 5.1
Q5
Is the function f defined by
f(x) = { x, if x ≤ 1
5, if x > 1
continuous at x = 0 ? At x = 1 ? At x = 2 ?
Answer. The given function f is
f(x) = { x, if x ≤ 1
5, if x > 1
Page 5
It is evident that f is defined at 0 and its value at 0 is 0 .
Then, lim x → 0f(x) = lim x → 0x = 0
∴ lim x → 0f(x) = f(0)
Therefore, f is continuous at x = 0
At x = 1
f is defined at 1 and its value at 1 is 1 . lim x → 1 + f(x) = lim x → 1 + (5) = 5
The left hand limit of f at x = 1 is, ∴ lim x → 1f(x) ≠ lim x → 1 + f(x)
lim x → 1f(x) = lim x → 1x = 1
The right hand limit of f at x = 1 is,
Therefore, f is not continuous at x = 1
At x = 2,
f is defined at 2 and its value at 2 is 5 .
Then, lim x → 2f(x) = lim x → 2(5) = 5
∴ lim x → 2f(x) = f(2)
Therefore, f is continuous at x = 2
Page : 159 , Block Name : Exercise 5.1
Q6 Find all points of discontinuity of f, where f is de ned by
f(x) = { 2x + 3, if x ≤ 2
2x − 3, if x > 2
Answer. The given function f is f(x) =
{ 2x + 3, if x ≤ 2
2x − 3, if x > 2
It is evident that the given function f is de ned at all the points of the real line. Let c be a point on
the real line. Then, three cases arise
(i) c < 2
(ii) c > 2
(iii) c = 2
Case (i)c < 2
Then, f(c) = 2c + 3
lim x → cf(x) = lim x → ∞(2x + 3) = 2c + 3
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x < 2
case (ii) c > 2
Then, f(c) = 2c − 3
lim x → ∞f(x) = lim x → ∞(2x − 3) = 2c − 3
∴ lim x → ∞f(x) = f(c)
Then, lim x → 0f(x) = lim x → 0x = 0
∴ lim x → 0f(x) = f(0)
Therefore, f is continuous at x = 0
At x = 1
f is defined at 1 and its value at 1 is 1 . lim x → 1 + f(x) = lim x → 1 + (5) = 5
The left hand limit of f at x = 1 is, ∴ lim x → 1f(x) ≠ lim x → 1 + f(x)
lim x → 1f(x) = lim x → 1x = 1
The right hand limit of f at x = 1 is,
Therefore, f is not continuous at x = 1
At x = 2,
f is defined at 2 and its value at 2 is 5 .
Then, lim x → 2f(x) = lim x → 2(5) = 5
∴ lim x → 2f(x) = f(2)
Therefore, f is continuous at x = 2
Page : 159 , Block Name : Exercise 5.1
Q6 Find all points of discontinuity of f, where f is de ned by
f(x) = { 2x + 3, if x ≤ 2
2x − 3, if x > 2
Answer. The given function f is f(x) =
{ 2x + 3, if x ≤ 2
2x − 3, if x > 2
It is evident that the given function f is de ned at all the points of the real line. Let c be a point on
the real line. Then, three cases arise
(i) c < 2
(ii) c > 2
(iii) c = 2
Case (i)c < 2
Then, f(c) = 2c + 3
lim x → cf(x) = lim x → ∞(2x + 3) = 2c + 3
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x < 2
case (ii) c > 2
Then, f(c) = 2c − 3
lim x → ∞f(x) = lim x → ∞(2x − 3) = 2c − 3
∴ lim x → ∞f(x) = f(c)
Page 6
Therefore, f is continuous at all points x, such that x > 2
case (iii) c = 2
Then, the left hand limit of f at x = 2 is,
lim x → 2f(x) = lim x → 2(2x + 3) = 2 × 2 + 3 = 7
The right hand limit of f at x = 2 is,
lim x → 2 + f(x) = lim x → 2 2(2x − 3) = 2 × 2 − 3 = 1
It is observed that the left and right hand limit of f at x = 2 do not coincide. Therefore, f is not
continuous at x = 2 Hence, x = 2 is the only point of discontinuity of f.
Page : 159 , Block Name : Exercise 5.1
Q7 Find all points of discontinuity of f, where f is de ned by
{
| x | + 3, if x ≤ − 3
f(x) = − 2x, if − 3 < x < 3
6x + 2, if x ≥ 3
Answer. The given function f is
{
| x | + 3 = − x + 3, if x ≤ − 3
f(x) = − 2x, if − 3 < x < 3
6x + 2, if x ≥ 3
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Case I:
If c < − 3, then f(c) = − c + 3
lim x → cf(x) = lim x → ∞( − x + 3) = − c + 3
∴ lim x → cf(x) = f(c)
Case II:
If c = − 3, then f( − 3) = − ( − 3) + 3 = 6
lim x → − 3f(x) = lim x → − 3( − x + 3) = − ( − 3) + 3 = 6
lim x → − 3 + f(x) = lim x → − 3 + ( − 2x) = − 2 × ( − 3) = 6
∴ lim x → − 3f(x) = f( − 3)
Therefore, f is continuous at x = − 3
Case III:
If − 3 < c < 3, then f(c) = − 2c and lim x → cf(x) = lim x → c( − 2x) = − 2c
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous in ( − 3, 3) .
Case IV:
If c = 3, then the left hand limit of f at x = 3 is,
lim x → 3f(x)math:
Processing = lim x → y( − 2x) = − 2 × 3 = − 6
100%
case (iii) c = 2
Then, the left hand limit of f at x = 2 is,
lim x → 2f(x) = lim x → 2(2x + 3) = 2 × 2 + 3 = 7
The right hand limit of f at x = 2 is,
lim x → 2 + f(x) = lim x → 2 2(2x − 3) = 2 × 2 − 3 = 1
It is observed that the left and right hand limit of f at x = 2 do not coincide. Therefore, f is not
continuous at x = 2 Hence, x = 2 is the only point of discontinuity of f.
Page : 159 , Block Name : Exercise 5.1
Q7 Find all points of discontinuity of f, where f is de ned by
{
| x | + 3, if x ≤ − 3
f(x) = − 2x, if − 3 < x < 3
6x + 2, if x ≥ 3
Answer. The given function f is
{
| x | + 3 = − x + 3, if x ≤ − 3
f(x) = − 2x, if − 3 < x < 3
6x + 2, if x ≥ 3
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Case I:
If c < − 3, then f(c) = − c + 3
lim x → cf(x) = lim x → ∞( − x + 3) = − c + 3
∴ lim x → cf(x) = f(c)
Case II:
If c = − 3, then f( − 3) = − ( − 3) + 3 = 6
lim x → − 3f(x) = lim x → − 3( − x + 3) = − ( − 3) + 3 = 6
lim x → − 3 + f(x) = lim x → − 3 + ( − 2x) = − 2 × ( − 3) = 6
∴ lim x → − 3f(x) = f( − 3)
Therefore, f is continuous at x = − 3
Case III:
If − 3 < c < 3, then f(c) = − 2c and lim x → cf(x) = lim x → c( − 2x) = − 2c
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous in ( − 3, 3) .
Case IV:
If c = 3, then the left hand limit of f at x = 3 is,
lim x → 3f(x)math:
Processing = lim x → y( − 2x) = − 2 × 3 = − 6
100%
Page 7
The right hand limit of f at x = 3 is,
lim x → 1f(x) = lim (6x + 2) = 6 × 3 + 2 = 20
It is observed that the left and right hand limit of f at x = 3 do not coincide.
Therefore, f is not continuous at x = 3
Case v:
If c > 3, then f(c) = 6c + 2 and lim x → cf(x) = lim x → ϵ(6x + 2) = 6c + 2
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 3
Hence, x = 3 is the only point of discontinuity of f .
Page : 159 , Block Name : Exercise 5.1
Q8 Find all points of discontinuity of f, where f is de ned by
{
|x|
x if x ≠ 0
f(x) =
0, if x = 0
Answer. The given function f is
{
|x|
x if x ≠ 0
f(x) =
0, if x = 0
{
|x| −x
x
= x = − 1 if x < 0
f(x) = 0, if x = 0
|x| x
x
= x = 1, if x > 0
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Case I:
If c < 0, then f(c) = − 1
lim x → cf(x) = lim x → ∞( − 1) = − 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x < 0
lim x → 1f(x) = lim (6x + 2) = 6 × 3 + 2 = 20
It is observed that the left and right hand limit of f at x = 3 do not coincide.
Therefore, f is not continuous at x = 3
Case v:
If c > 3, then f(c) = 6c + 2 and lim x → cf(x) = lim x → ϵ(6x + 2) = 6c + 2
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 3
Hence, x = 3 is the only point of discontinuity of f .
Page : 159 , Block Name : Exercise 5.1
Q8 Find all points of discontinuity of f, where f is de ned by
{
|x|
x if x ≠ 0
f(x) =
0, if x = 0
Answer. The given function f is
{
|x|
x if x ≠ 0
f(x) =
0, if x = 0
{
|x| −x
x
= x = − 1 if x < 0
f(x) = 0, if x = 0
|x| x
x
= x = 1, if x > 0
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Case I:
If c < 0, then f(c) = − 1
lim x → cf(x) = lim x → ∞( − 1) = − 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x < 0
Page 8
Case II:
If c = 0, then the left hand limit of f at x = 0 is,
lim x → ∞f(x) = lim x → 0( − 1) = − 1
The right hand limit of f at x = 0 is,
lim x → ∞f(x) = lim x → 0 + (1) = 1
It is observed that the left and right hand limit of f at x = 0 do not coincide.
Therefore, f is not continuous at x = 0
Case III:
If c > 0, then f(c) = 1
lim x → ∞f(x) = lim x → e(1) = 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 0
Hence, x = 0 is the only point of discontinuity of f.
Page : 159 , Block Name : Exercise 5.1
Q9 Find all points of discontinuity of f, where f is de ned by
{
x
| x | , if x < 0
f(x) =
− 1, if x ≥ 0
Answer. The given function f is
{
x
| x | , if x < 0
f(x) =
− 1, if x ≥ 0
It is known that,x < 0 ⇒ | x | = − x
Therefore, the given function can be rewritten as
{
x x
| x | = x = − 1, if x < 0
f(x) =
− 1, if x ≥ 0
f(c) = − 1 = lim x → ef(x)
Therefore, the given function is a continuous function.
Hence, the given function has no point of discontinuity.
Page : 159 , Block Name : Exercise 5.1
Q10 Find all points of discontinuity of f, where f is de ned by
If c = 0, then the left hand limit of f at x = 0 is,
lim x → ∞f(x) = lim x → 0( − 1) = − 1
The right hand limit of f at x = 0 is,
lim x → ∞f(x) = lim x → 0 + (1) = 1
It is observed that the left and right hand limit of f at x = 0 do not coincide.
Therefore, f is not continuous at x = 0
Case III:
If c > 0, then f(c) = 1
lim x → ∞f(x) = lim x → e(1) = 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 0
Hence, x = 0 is the only point of discontinuity of f.
Page : 159 , Block Name : Exercise 5.1
Q9 Find all points of discontinuity of f, where f is de ned by
{
x
| x | , if x < 0
f(x) =
− 1, if x ≥ 0
Answer. The given function f is
{
x
| x | , if x < 0
f(x) =
− 1, if x ≥ 0
It is known that,x < 0 ⇒ | x | = − x
Therefore, the given function can be rewritten as
{
x x
| x | = x = − 1, if x < 0
f(x) =
− 1, if x ≥ 0
f(c) = − 1 = lim x → ef(x)
Therefore, the given function is a continuous function.
Hence, the given function has no point of discontinuity.
Page : 159 , Block Name : Exercise 5.1
Q10 Find all points of discontinuity of f, where f is de ned by
Page 9
f(x) =
{ x + 1, if x ≥ 1
x 2 + 1, if x < 1
Answer. The given function f is
f(x) =
{ x + 1, if x ≥ 1
x 2 + 1, if x < 1
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Case I:
( )
If c < 1, then f(c) = c 2 + 1 and lim x → ϵf(x) = lim x → e x 2 + 1 = c 2 + 1
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x < 1
Case II:
If c = 1, then f(c) = f(1) = 1 + 1 = 2
The left hand limit of f at x = 1 is,
( )
lim x → 1f(x) = lim x → 1 x 2 + 1 = 1 2 + 1 = 2
The right hand limit of f at x = 1 is,
lim x → 1 + f(x) = lim x → 1 1(x + 1) = 1 + 1 = 2
∴ lim x → 1f(x) = f(1)
Therefore, f is continuous at x = 1
Case III:
If c > 1, then f(c) = c + 1
lim x → ∞f(x) = lim x → c(x + 1) = c + 1
∴ lim f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1 Hence, the given function f has no point
of discontinuity
Page : 159 , Block Name : Exercise 5.1
Q11 Find all points of discontinuity of f, where f is de ned by
f(x) =
{ x 3 − 3,
x 2 + 1,
if x ≤ 2
if x > 2
Answer. The given function f is
f(x) =
{ x 3 − 3, if x ≤ 2
x 2 + 1, if x > 2
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
{ x + 1, if x ≥ 1
x 2 + 1, if x < 1
Answer. The given function f is
f(x) =
{ x + 1, if x ≥ 1
x 2 + 1, if x < 1
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Case I:
( )
If c < 1, then f(c) = c 2 + 1 and lim x → ϵf(x) = lim x → e x 2 + 1 = c 2 + 1
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x < 1
Case II:
If c = 1, then f(c) = f(1) = 1 + 1 = 2
The left hand limit of f at x = 1 is,
( )
lim x → 1f(x) = lim x → 1 x 2 + 1 = 1 2 + 1 = 2
The right hand limit of f at x = 1 is,
lim x → 1 + f(x) = lim x → 1 1(x + 1) = 1 + 1 = 2
∴ lim x → 1f(x) = f(1)
Therefore, f is continuous at x = 1
Case III:
If c > 1, then f(c) = c + 1
lim x → ∞f(x) = lim x → c(x + 1) = c + 1
∴ lim f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1 Hence, the given function f has no point
of discontinuity
Page : 159 , Block Name : Exercise 5.1
Q11 Find all points of discontinuity of f, where f is de ned by
f(x) =
{ x 3 − 3,
x 2 + 1,
if x ≤ 2
if x > 2
Answer. The given function f is
f(x) =
{ x 3 − 3, if x ≤ 2
x 2 + 1, if x > 2
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Page 10
Case I:
( )
If c < 2, then f(c) = c 3 − 3 and lim x → ∞f(x) = lim x → e x 3 − 3 = c 3 − 3
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x < 2
Case II:
If c = 2, then f(c) = f(2) = 2 3 − 3 = 5
( )
lim x → 2f(x) = lim x → 2 x 3 − 3 = 2 3 − 3 = 5
lim x → 2f(x) = lim x → 2 (x + 1 ) = 2 + 1 = 5
2 2
∴ lim x → 2f(x) = f(2)
Therefore, , is continuous at x = 2
Case II:
If c = 2, then f(c) = f(2) = 2 3 − 3 = 5
( )
lim x → 2f(x) = lim x → 2 + x 3 − 3 = 2 3 − 3 = 5
( )
lim x → 2 + f(x) = lim x → 2 + x 2 + 1 = 2 2 + 1 = 5
∴ lim x → 2f(x) = f(2)
Therefore, f is continuous at x = 2
If c > 2, then f(c) = c 2 + 1
( )
lim x → cf(x) = lim x → c x 2 + 1 = c 2 + 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 2 Thus, the given function f is continuous at
every point on the real line. Hence, f has no point of discontinuity.
Page : 159 , Block Name : Exercise 5.1
Q12 Find all points of discontinuity of f, where f is de ned by
f(x) =
{ x 10 − 1,
x 2,
if x ≤ 1
if x > 1
Answer. The given function f is
f(x) =
{ x 10 − 1,
x 2,
if x ≤ 1
if x > 1
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
( )
If c < 2, then f(c) = c 3 − 3 and lim x → ∞f(x) = lim x → e x 3 − 3 = c 3 − 3
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x < 2
Case II:
If c = 2, then f(c) = f(2) = 2 3 − 3 = 5
( )
lim x → 2f(x) = lim x → 2 x 3 − 3 = 2 3 − 3 = 5
lim x → 2f(x) = lim x → 2 (x + 1 ) = 2 + 1 = 5
2 2
∴ lim x → 2f(x) = f(2)
Therefore, , is continuous at x = 2
Case II:
If c = 2, then f(c) = f(2) = 2 3 − 3 = 5
( )
lim x → 2f(x) = lim x → 2 + x 3 − 3 = 2 3 − 3 = 5
( )
lim x → 2 + f(x) = lim x → 2 + x 2 + 1 = 2 2 + 1 = 5
∴ lim x → 2f(x) = f(2)
Therefore, f is continuous at x = 2
If c > 2, then f(c) = c 2 + 1
( )
lim x → cf(x) = lim x → c x 2 + 1 = c 2 + 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 2 Thus, the given function f is continuous at
every point on the real line. Hence, f has no point of discontinuity.
Page : 159 , Block Name : Exercise 5.1
Q12 Find all points of discontinuity of f, where f is de ned by
f(x) =
{ x 10 − 1,
x 2,
if x ≤ 1
if x > 1
Answer. The given function f is
f(x) =
{ x 10 − 1,
x 2,
if x ≤ 1
if x > 1
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Page 11
Case I:
( )
If c < 1, then f(c) = c 10 − 1 and lim x → ∞f(x) = lim x → ∞ x 10 − 1 = c 10 − 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < 1
If c = 1, then the left hand limit of f at x = 1 is,
( )
lim x → 1f(x) = lim x → 1 x 10 − 1 = 1 10 − 1 = 1 − 1 = 0
The right hand limit of f at x = 1 is,
( )
lim x → 1 + f(x) = lim x → 1 2 x 2 = 1 2 = 1
It is observed that the left and right hand limit of f at x = 1 do not coincide. Therefore, f is not
continuous at x = 1
Case III:
If c > 1, then f(c) = c 2
( )
lim x → cf(x) = lim x → c x 2 = c 2
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1 Thus, from the above observation, it can
be concluded that x = 1 is the only point of discontinuity of f.
Page : 159 , Block Name : Exercise 5.1
Q13 Is the function de ned by
f(x) = { x + 5, if x ≤ 1
x − 5, if x > 1
a continuous function ?
Answer. The given function is
f(x) =
{ x + 5,
x − 5,
if x ≤ 1
if x > 1
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Case I:
If c < 1, then f(c) = c + 5 and lim x → cf(x) = lim x → c(x + 5) = c + 5
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < 1
( )
If c < 1, then f(c) = c 10 − 1 and lim x → ∞f(x) = lim x → ∞ x 10 − 1 = c 10 − 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < 1
If c = 1, then the left hand limit of f at x = 1 is,
( )
lim x → 1f(x) = lim x → 1 x 10 − 1 = 1 10 − 1 = 1 − 1 = 0
The right hand limit of f at x = 1 is,
( )
lim x → 1 + f(x) = lim x → 1 2 x 2 = 1 2 = 1
It is observed that the left and right hand limit of f at x = 1 do not coincide. Therefore, f is not
continuous at x = 1
Case III:
If c > 1, then f(c) = c 2
( )
lim x → cf(x) = lim x → c x 2 = c 2
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1 Thus, from the above observation, it can
be concluded that x = 1 is the only point of discontinuity of f.
Page : 159 , Block Name : Exercise 5.1
Q13 Is the function de ned by
f(x) = { x + 5, if x ≤ 1
x − 5, if x > 1
a continuous function ?
Answer. The given function is
f(x) =
{ x + 5,
x − 5,
if x ≤ 1
if x > 1
The given function f is de ned at all the points of the real line. Let c be a point on the real line.
Case I:
If c < 1, then f(c) = c + 5 and lim x → cf(x) = lim x → c(x + 5) = c + 5
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < 1
Page 12
Case II:
If c = 1, then f(1) = 1 + 5 = 6
The left hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1(x + 5) = 1 + 5 = 6
The right hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1(x − 5) = 1 − 5 = − 4
It is observed that the left and right hand limit of f at x = 1 do not coincide. Therefore, f is not
continuous at x = 1
Case III:
If c > 1, then f(c) = c − 5 and lim x → cf(x) = lim x → c(x − 5) = c − 5
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1 Thus, from the above observation, it can
be concluded that x = 1 is the only point of discontinuity of f.
Page : 159 , Block Name : Exercise 5.1
Q14 Discuss the continuity of the function f, where f is de ned by
{
3, if 0 ≤ x ≤ 1
f(x) = 4, if 1 < x < 3
5, if 3 ≤ x ≤ 10
Answer. The given function is
{
3, if 0 ≤ x ≤ 1
f(x) = 4, if 1 < x < 3
5, if 3 ≤ x ≤ 10
The given function is de ned at all points of the interval [0, 10]. Let c be a point in the interval [0,
10].
Case I
If 0 ≤ c < 1, then f(c) = 3 and lim x → tf(x) = lim x → c → (3) = 3
∴ lim x → cf(x) = f(c)
Therefore, f is continuous in the interval [0, 1).
Case II:
If c = 1, then f(3) = 3
The left hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1(3) = 3
The right hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1 1(4) = 4
It is observed that the left and right hand limits of f at x = 1 do not coincide. Therefore, f is not
Processing math:
continuous at x =100%
1
If c = 1, then f(1) = 1 + 5 = 6
The left hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1(x + 5) = 1 + 5 = 6
The right hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1(x − 5) = 1 − 5 = − 4
It is observed that the left and right hand limit of f at x = 1 do not coincide. Therefore, f is not
continuous at x = 1
Case III:
If c > 1, then f(c) = c − 5 and lim x → cf(x) = lim x → c(x − 5) = c − 5
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1 Thus, from the above observation, it can
be concluded that x = 1 is the only point of discontinuity of f.
Page : 159 , Block Name : Exercise 5.1
Q14 Discuss the continuity of the function f, where f is de ned by
{
3, if 0 ≤ x ≤ 1
f(x) = 4, if 1 < x < 3
5, if 3 ≤ x ≤ 10
Answer. The given function is
{
3, if 0 ≤ x ≤ 1
f(x) = 4, if 1 < x < 3
5, if 3 ≤ x ≤ 10
The given function is de ned at all points of the interval [0, 10]. Let c be a point in the interval [0,
10].
Case I
If 0 ≤ c < 1, then f(c) = 3 and lim x → tf(x) = lim x → c → (3) = 3
∴ lim x → cf(x) = f(c)
Therefore, f is continuous in the interval [0, 1).
Case II:
If c = 1, then f(3) = 3
The left hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1(3) = 3
The right hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1 1(4) = 4
It is observed that the left and right hand limits of f at x = 1 do not coincide. Therefore, f is not
Processing math:
continuous at x =100%
1
Page 13
Case III:
If 1 < c < 3, then f(c) = 4 and lim x → cf(x) = lim x → ϵ(4) = 4
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points of the interval (1, 3) .
Case IV:
If c = 3, then f(c) = 5
The left hand limit of f at x = 3 is,
lim x → 3f(x) = lim x → 3(4) = 4
The right hand limit of f at x = 3 is,
lim x → 3 + f(x) = lim x → 3 + (5) = 5
It is observed that the left and right hand limits of f at x = 3 do not coincide.
Therefore f is not continuous at x = 3
Case V:
If 3 < c ≤ 10, then f(c) = 5 and lim x → cf(x) = lim x → c(5) = 5
lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points of the interval (3, 10)
Hence, f is not continuous at x = 1 and x = 3
Page : 160 , Block Name : Exercise 5.1
Q15 Discuss the continuity of the function f, where f is de ned by
{
2x, if x < 0
f(x) = 0, if 0 ≤ x ≤ 1
4x, if x > 1
Answer. The give function is
{
2x, if x < 0
f(x) = 0, if 0 ≤ x ≤ 1
4x, if x > 1
The given function is de ned at all points of the real line. Let c be a point on the real line.
Case I:
If c < 0, then f(c) = 2c
lim x → cf(x) = lim x → c(2x) = 2c
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < 0
If 1 < c < 3, then f(c) = 4 and lim x → cf(x) = lim x → ϵ(4) = 4
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points of the interval (1, 3) .
Case IV:
If c = 3, then f(c) = 5
The left hand limit of f at x = 3 is,
lim x → 3f(x) = lim x → 3(4) = 4
The right hand limit of f at x = 3 is,
lim x → 3 + f(x) = lim x → 3 + (5) = 5
It is observed that the left and right hand limits of f at x = 3 do not coincide.
Therefore f is not continuous at x = 3
Case V:
If 3 < c ≤ 10, then f(c) = 5 and lim x → cf(x) = lim x → c(5) = 5
lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points of the interval (3, 10)
Hence, f is not continuous at x = 1 and x = 3
Page : 160 , Block Name : Exercise 5.1
Q15 Discuss the continuity of the function f, where f is de ned by
{
2x, if x < 0
f(x) = 0, if 0 ≤ x ≤ 1
4x, if x > 1
Answer. The give function is
{
2x, if x < 0
f(x) = 0, if 0 ≤ x ≤ 1
4x, if x > 1
The given function is de ned at all points of the real line. Let c be a point on the real line.
Case I:
If c < 0, then f(c) = 2c
lim x → cf(x) = lim x → c(2x) = 2c
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < 0
Page 14
Case II:
If c = 0, then f(c) = f(0) = 0
The left hand limit of f at x = 0 is,
lim x → 0f(x) = lim x → 0 + (2x) = 2 × 0 = 0
The right hand limit of f at x = 0 is,
lim x → 0 + f(x) = lim x → 0 + (0) = 0
∴ lim x → 0 + f(x) = f(0)
Therefore, f is continuous at x = 0
Case III:
If 0 < c < 1, then f(x) = 0 and lim x → ∞f(x) = lim (0) = 0
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points of the interval (0, 1).
Case IV:
If c = 1, then f(c) = f(1) = 0
The left hand limit of f at x = 1 is,
lim x → 1f(x) = lim (0) = 0
The right hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1 + (4x) = 4 × 1 = 4
It is observed that the left and right hand limits of f at x = 1 do not coincide. Therefore, f is not
continuous at x = 1
Case V :
If c < 1, then f(c) = 4c and lim x → cf(x) = lim x → c(4x) = 4c
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x , such that x > 1
Hence, f is not continuous only at x = 1
Page : 160 , Block Name : Exercise 5.1
Q16 Discuss the continuity of the function f, where f is de ned by
{
− 2, if x ≤ − 1
f(x) = 2x, if − 1 < x ≤ 1
2, if x > 1
Answer. The given function is
{
− 2, if x ≤ − 1
f(x) = 2x, if − 1 < x ≤ 1
2, if x > 1
The given function
Processing is de ned at all points of the real line. Let c be a point on the real line.
math: 100%
If c = 0, then f(c) = f(0) = 0
The left hand limit of f at x = 0 is,
lim x → 0f(x) = lim x → 0 + (2x) = 2 × 0 = 0
The right hand limit of f at x = 0 is,
lim x → 0 + f(x) = lim x → 0 + (0) = 0
∴ lim x → 0 + f(x) = f(0)
Therefore, f is continuous at x = 0
Case III:
If 0 < c < 1, then f(x) = 0 and lim x → ∞f(x) = lim (0) = 0
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points of the interval (0, 1).
Case IV:
If c = 1, then f(c) = f(1) = 0
The left hand limit of f at x = 1 is,
lim x → 1f(x) = lim (0) = 0
The right hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1 + (4x) = 4 × 1 = 4
It is observed that the left and right hand limits of f at x = 1 do not coincide. Therefore, f is not
continuous at x = 1
Case V :
If c < 1, then f(c) = 4c and lim x → cf(x) = lim x → c(4x) = 4c
∴ lim x → cf(x) = f(c)
Therefore, f is continuous at all points x , such that x > 1
Hence, f is not continuous only at x = 1
Page : 160 , Block Name : Exercise 5.1
Q16 Discuss the continuity of the function f, where f is de ned by
{
− 2, if x ≤ − 1
f(x) = 2x, if − 1 < x ≤ 1
2, if x > 1
Answer. The given function is
{
− 2, if x ≤ − 1
f(x) = 2x, if − 1 < x ≤ 1
2, if x > 1
The given function
Processing is de ned at all points of the real line. Let c be a point on the real line.
math: 100%
Page 15
Case I:
If c < − 1, then f(c) = − 2 and lim x → ϵf(x) = lim x → c( − 2) = − 2
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < − 1
Case II:
If c = − 1, then f(c) = f( − 1) = − 2
The left hand limit of f at x = − 1 is,
lim x → − 1f(x) = lim x → − 1( − 2) = − 2
The right hand limit of f at x = − 1 is,
lim x → − 1 + f(x) = lim x → − 1 + (2x) = 2 × ( − 1) = − 2
∴ lim x → − 1f(x) = f( − 1)
Therefore, f is continuous at x = − 1
Case III:
If − 1 < c < 1, then f(c) = 2c
Therefore, f is continuous at all points of the interval (−1, 1)
lim x → ∞f(x) = lim x → ∞(2x) = 2c
∴ lim f(x) = f(c)
Case IV
If c = 1, then f(c) = f(1) = 2 × 1 = 2
The left hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1(2x) = 2 × 1 = 2
The right hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1 12 = 2
∴ lim x → 1f(x) = f(c)
Therefore, f is continuous at x = 2
Case V :
If c > 1, then f(c) = 2 and lim x → cf(x) = lim x → c(2) = 2
lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1
Thus, from the above observations, it can be concluded that f is continuous at all points
of the real line.
Page : 160 , Block Name : Exercise 5.1
Q17 Find the relationship between a and b so that the function f de ned by
f(x) =
{ ax + 1, if x ≤ 3
bx + 3, if x > 3
is continuous at x = 3
Answer. The given function f is
If c < − 1, then f(c) = − 2 and lim x → ϵf(x) = lim x → c( − 2) = − 2
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < − 1
Case II:
If c = − 1, then f(c) = f( − 1) = − 2
The left hand limit of f at x = − 1 is,
lim x → − 1f(x) = lim x → − 1( − 2) = − 2
The right hand limit of f at x = − 1 is,
lim x → − 1 + f(x) = lim x → − 1 + (2x) = 2 × ( − 1) = − 2
∴ lim x → − 1f(x) = f( − 1)
Therefore, f is continuous at x = − 1
Case III:
If − 1 < c < 1, then f(c) = 2c
Therefore, f is continuous at all points of the interval (−1, 1)
lim x → ∞f(x) = lim x → ∞(2x) = 2c
∴ lim f(x) = f(c)
Case IV
If c = 1, then f(c) = f(1) = 2 × 1 = 2
The left hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1(2x) = 2 × 1 = 2
The right hand limit of f at x = 1 is,
lim x → 1f(x) = lim x → 1 12 = 2
∴ lim x → 1f(x) = f(c)
Therefore, f is continuous at x = 2
Case V :
If c > 1, then f(c) = 2 and lim x → cf(x) = lim x → c(2) = 2
lim x → cf(x) = f(c)
Therefore, f is continuous at all points x, such that x > 1
Thus, from the above observations, it can be concluded that f is continuous at all points
of the real line.
Page : 160 , Block Name : Exercise 5.1
Q17 Find the relationship between a and b so that the function f de ned by
f(x) =
{ ax + 1, if x ≤ 3
bx + 3, if x > 3
is continuous at x = 3
Answer. The given function f is
Page 16
f(x) =
{ ax + 1, if x ≤ 3
bx + 3, if x > 3
If f is continuous at x = 3, then
lim x → 3f(x) = lim x → 3 + f(x) = f(3)
Also,
lim x → 3f(x) = lim x → 3(ax + 1) = 3a + 1
lim x → 3f(x) = lim x → 3 3(bx + 3) = 3b + 3
f(3) = 3a + 1
Therefore, from (1), we obtain
3a + 1 = 3b + 3 = 3a + 1
⇒ 3a + 1 = 3b + 3
⇒ 3a = 3b + 2
2
⇒a=b+ 3
2
Therefore, the required relationship is given by, a = b + 3
Page : 160 , Block Name : Exercise 5.1
Q18 For what value of λis the function de ned by
( )
f(x) =
{ λ x 2 − 2x ,
4x + 1, if x > 0
if x ≤ 0
continuous at x = 0? What about continuity at x = 1 ?
Answer. The given function is
f(x) =
{ (
λ x 2 − 2x ,
4x + 1,
)
if x > 0
if x ≤ 0
If f is continuous at x = 0, then
lim x → ∞f(x) = lim x → 0f(x) = f(0)
( ) (
⇒ lim x → ∞λ x 2 − 2x = lim x → 0 − (4x + 1) = λ 0 2 − 2 × 0 )
( )
⇒ λ 02 − 2 × 0 = 4 × 0 + 1 = 0
⇒ 0 = 1 = 0, which is not possible
Therefore, there is no value of λ for which f is continuous at x = 0
{ ax + 1, if x ≤ 3
bx + 3, if x > 3
If f is continuous at x = 3, then
lim x → 3f(x) = lim x → 3 + f(x) = f(3)
Also,
lim x → 3f(x) = lim x → 3(ax + 1) = 3a + 1
lim x → 3f(x) = lim x → 3 3(bx + 3) = 3b + 3
f(3) = 3a + 1
Therefore, from (1), we obtain
3a + 1 = 3b + 3 = 3a + 1
⇒ 3a + 1 = 3b + 3
⇒ 3a = 3b + 2
2
⇒a=b+ 3
2
Therefore, the required relationship is given by, a = b + 3
Page : 160 , Block Name : Exercise 5.1
Q18 For what value of λis the function de ned by
( )
f(x) =
{ λ x 2 − 2x ,
4x + 1, if x > 0
if x ≤ 0
continuous at x = 0? What about continuity at x = 1 ?
Answer. The given function is
f(x) =
{ (
λ x 2 − 2x ,
4x + 1,
)
if x > 0
if x ≤ 0
If f is continuous at x = 0, then
lim x → ∞f(x) = lim x → 0f(x) = f(0)
( ) (
⇒ lim x → ∞λ x 2 − 2x = lim x → 0 − (4x + 1) = λ 0 2 − 2 × 0 )
( )
⇒ λ 02 − 2 × 0 = 4 × 0 + 1 = 0
⇒ 0 = 1 = 0, which is not possible
Therefore, there is no value of λ for which f is continuous at x = 0
Page 17
At x = 1
f(1) = 4x + 1 = 4 × 1 + 1 = 5
lim x → 1(4x + 1) = 4 × 1 + 1 = 5
∴ lim x → 1f(x) = f(1)
Therefore, for any values of λ , f is continuous at x = 1
Page : 160 , Block Name : Exercise 5.1
Q19
Show that the function defined by g(x) = x − [x] is discontinuous at all integral point.
Here [x] denotes the greatest integer less than or equal to x.
Answer.
The given function is g ( x ) = x − [x]
It is evident that g is defined at all integral points.
Let n be an integer.
Then,
g(n) = n − [n] = n − n = 0
The left hand limit of f at x = n is,
lim x → mg(x) = lim x → m(x − [x]) = lim x → 1(x) − lim x → ∞[x] = n − (n − 1) = 1
It is observed that the left and right hand limits of f at x = n do not coincide. Therefore, f is not
continuous at x = n Hence, g is discontinuous at all integral points
Page : 160 , Block Name : Exercise 5.1
Q20 Is the function de ned by f(x) = x 2 − sinx + 5 continuous at x = p ?
Answer. The given function is f(x) = x 2 − sinx + 5
It is evident that f is defined at x = p
At x = π, f(x) = f(π) = π 2 − sinπ + 5 = π 2 − 0 + 5 = π 2 + 5
(
Consider lim x → xf(x) = lim x → x x 2 − sinx + 5 )
Put x = π + h
If x → π, then it is evident that h → 0
(
Consider lim f(x) = lim x → π x 2 − sinx + 5 )
Put x = π + h
If x → π, then it is evident that h → 0
f(1) = 4x + 1 = 4 × 1 + 1 = 5
lim x → 1(4x + 1) = 4 × 1 + 1 = 5
∴ lim x → 1f(x) = f(1)
Therefore, for any values of λ , f is continuous at x = 1
Page : 160 , Block Name : Exercise 5.1
Q19
Show that the function defined by g(x) = x − [x] is discontinuous at all integral point.
Here [x] denotes the greatest integer less than or equal to x.
Answer.
The given function is g ( x ) = x − [x]
It is evident that g is defined at all integral points.
Let n be an integer.
Then,
g(n) = n − [n] = n − n = 0
The left hand limit of f at x = n is,
lim x → mg(x) = lim x → m(x − [x]) = lim x → 1(x) − lim x → ∞[x] = n − (n − 1) = 1
It is observed that the left and right hand limits of f at x = n do not coincide. Therefore, f is not
continuous at x = n Hence, g is discontinuous at all integral points
Page : 160 , Block Name : Exercise 5.1
Q20 Is the function de ned by f(x) = x 2 − sinx + 5 continuous at x = p ?
Answer. The given function is f(x) = x 2 − sinx + 5
It is evident that f is defined at x = p
At x = π, f(x) = f(π) = π 2 − sinπ + 5 = π 2 − 0 + 5 = π 2 + 5
(
Consider lim x → xf(x) = lim x → x x 2 − sinx + 5 )
Put x = π + h
If x → π, then it is evident that h → 0
(
Consider lim f(x) = lim x → π x 2 − sinx + 5 )
Put x = π + h
If x → π, then it is evident that h → 0
Page 18
(
∴ lim f(x) = lim x 2 − sinx + 5
r→π x→π
)
= lim [(π + h) − sin(π + h) + 5 ]
2
h→0
= lim (π + h) 2 − lim sin(π + h) + lim 5
h→0 h→0 h→0
= (π + 0) 2 − lim [sinπcosh + cosπsinh] + 5
h→0
= π 2 − lim sinπcosh − lim cosπsinh + 5
h→0 h→0
= π 2 − sinπcos0 − cosπsin0 + 5
= π 2 − 0 × 1 − ( − 1) × 0 + 5
= π2 + 5
∴ lim x → ∞f(x) = f(π)
Therefore, the given function f is continuous at x = n
Page : 160 , Block Name : Exercise 5.1
Q21 Discuss the continuity of the following functions.
(a) f (x) = sin x + cos x
(b) f (x) = sin x − cos x
(c) f (x) = sin x × cos x
Answer.
It is known that if g and h are two continuous functions, then
g + h, g − h, and g h are also continuous.
It has to proved first that g(x) = sinx and h(x) = cosx are continuous functions.
Let g(x) = sinx
It is evident that g(x) = sinx is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
g(c) = sinc
lim g(x) = lim sinx
x→ϵ
= lim sin(c + h)
= lim [sinccosh + coscsinh]
k→0
= lim (sinccosh) + lim (coscsinh)
h→0 h→0
= sinccos0 + coscsin0
= sinc + 0
= sinc
∴ lim f(x) = lim x 2 − sinx + 5
r→π x→π
)
= lim [(π + h) − sin(π + h) + 5 ]
2
h→0
= lim (π + h) 2 − lim sin(π + h) + lim 5
h→0 h→0 h→0
= (π + 0) 2 − lim [sinπcosh + cosπsinh] + 5
h→0
= π 2 − lim sinπcosh − lim cosπsinh + 5
h→0 h→0
= π 2 − sinπcos0 − cosπsin0 + 5
= π 2 − 0 × 1 − ( − 1) × 0 + 5
= π2 + 5
∴ lim x → ∞f(x) = f(π)
Therefore, the given function f is continuous at x = n
Page : 160 , Block Name : Exercise 5.1
Q21 Discuss the continuity of the following functions.
(a) f (x) = sin x + cos x
(b) f (x) = sin x − cos x
(c) f (x) = sin x × cos x
Answer.
It is known that if g and h are two continuous functions, then
g + h, g − h, and g h are also continuous.
It has to proved first that g(x) = sinx and h(x) = cosx are continuous functions.
Let g(x) = sinx
It is evident that g(x) = sinx is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
g(c) = sinc
lim g(x) = lim sinx
x→ϵ
= lim sin(c + h)
= lim [sinccosh + coscsinh]
k→0
= lim (sinccosh) + lim (coscsinh)
h→0 h→0
= sinccos0 + coscsin0
= sinc + 0
= sinc
Page 19
Let h(x) = cosx
It is evident that h(x) = cosx is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
h(c) = cosc
Therefore, h is a continuous function.
Therefore, it can be concluded that
(a) f(x) = g(x) + h(x) = sinx + cosx is a continuous function
(b) f(x) = g(x) − h(x) = sinx − cosx is a continuous function
(c) f(x) = g(x) × h(x) = sinx × cosx is a continuous function
Page : 160 , Block Name : Exercise 5.1
Q22 Discuss the continuity of the cosine, cosecant, secant and cotangent functions,
Answer. It is known that if g and h are two continuous functions, then
h(x)
(i) g ( x ) , g(x) ≠ 0 is continuous
1
(ii) g ( x ) , g(x) ≠ 0 is continuous
1
(iii) h ( x ) , h(x) ≠ 0 is continuous
It has to be proved first that g(x) = sinx and h(x) = cosx are continuous functions.
Let g(x) = sinx
It is evident that g(x) = sinx is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
g(c) = sinc
lim g(x) = lim sinx
x→c x→c
= lim sin(c + h)
h→0
= sinccosθ + coscsin0
= sinc + 0
= sinc
Therefore, g is a continuous function.
Let h(x) = cosx
It is evident that h(x) = cosx is defined for every real number.
Let c be a real number. Put x = c + h
If x @ c, then h @ 0
h(c) = cosc
lim h(x) = lim cosx
x→c x→c
= math:
Processing lim cos(c
100% + h)
h→0
It is evident that h(x) = cosx is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
h(c) = cosc
Therefore, h is a continuous function.
Therefore, it can be concluded that
(a) f(x) = g(x) + h(x) = sinx + cosx is a continuous function
(b) f(x) = g(x) − h(x) = sinx − cosx is a continuous function
(c) f(x) = g(x) × h(x) = sinx × cosx is a continuous function
Page : 160 , Block Name : Exercise 5.1
Q22 Discuss the continuity of the cosine, cosecant, secant and cotangent functions,
Answer. It is known that if g and h are two continuous functions, then
h(x)
(i) g ( x ) , g(x) ≠ 0 is continuous
1
(ii) g ( x ) , g(x) ≠ 0 is continuous
1
(iii) h ( x ) , h(x) ≠ 0 is continuous
It has to be proved first that g(x) = sinx and h(x) = cosx are continuous functions.
Let g(x) = sinx
It is evident that g(x) = sinx is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
g(c) = sinc
lim g(x) = lim sinx
x→c x→c
= lim sin(c + h)
h→0
= sinccosθ + coscsin0
= sinc + 0
= sinc
Therefore, g is a continuous function.
Let h(x) = cosx
It is evident that h(x) = cosx is defined for every real number.
Let c be a real number. Put x = c + h
If x @ c, then h @ 0
h(c) = cosc
lim h(x) = lim cosx
x→c x→c
= math:
Processing lim cos(c
100% + h)
h→0
Page 20
= lim h → 0[cosccosh − sincsinh]
= lim h → 0cosccosh − lim h → 0sincsinh
= cosccos0 − sincsin0
= cosc × 1 − sinc × 0
= cosc
∴ lim x → ∞h(x) = h(c)
Therefore, h(x) = cosx is continuous function.
It can be concluded that,
It can be concluded that,
1
cscx = sin x , sinx ≠ 0 is continuous
⇒ cscx, x ≠ nπ(n ∈ Z) is continuous
1
cscx = sin x , sinx ≠ 0 is continuous
⇒ cscx, x ≠ nπ(n ∈ Z) is continuous
Therefore, cosecant is continuous except at x = np rn iz
cos x
cotx = sin x , sinx ≠ 0 is continuous
⇒ cotx, x ≠ nπ(n ∈ Z) is continuous
Therefore, cotangent is continuous except at x = np, nˆ z
Page : 160 , Block Name : Exercise 5.1
Q23 Find the points of discontinuity of f, where
{
sin x
x , if x < 0
f(x) =
x + 1, if x ≥ 0
Answer. The given function f is
{
sin x
x , if x < 0
f(x) =
x + 1, if x ≥ 0
It is evident that f is de ned at all points of the real line. Let c be a real number.
Case I:
If c < 0, then f(c) =
sin c
c
and lim x → cf(x) = lim x → c ( )
sin x
x
=
sin c
x
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < 0
= lim h → 0cosccosh − lim h → 0sincsinh
= cosccos0 − sincsin0
= cosc × 1 − sinc × 0
= cosc
∴ lim x → ∞h(x) = h(c)
Therefore, h(x) = cosx is continuous function.
It can be concluded that,
It can be concluded that,
1
cscx = sin x , sinx ≠ 0 is continuous
⇒ cscx, x ≠ nπ(n ∈ Z) is continuous
1
cscx = sin x , sinx ≠ 0 is continuous
⇒ cscx, x ≠ nπ(n ∈ Z) is continuous
Therefore, cosecant is continuous except at x = np rn iz
cos x
cotx = sin x , sinx ≠ 0 is continuous
⇒ cotx, x ≠ nπ(n ∈ Z) is continuous
Therefore, cotangent is continuous except at x = np, nˆ z
Page : 160 , Block Name : Exercise 5.1
Q23 Find the points of discontinuity of f, where
{
sin x
x , if x < 0
f(x) =
x + 1, if x ≥ 0
Answer. The given function f is
{
sin x
x , if x < 0
f(x) =
x + 1, if x ≥ 0
It is evident that f is de ned at all points of the real line. Let c be a real number.
Case I:
If c < 0, then f(c) =
sin c
c
and lim x → cf(x) = lim x → c ( )
sin x
x
=
sin c
x
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x, such that x < 0
Page 21
Case II:
If c > 0, then f(c) = c + 1 and lim x → ∞f(x) = lim x → ∞(x + 1) = c + 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x , such that x ≥ 0
Case III:
If c = 0, then f(c) = f(0) = 0 + 1 = 1
The left hand limit of f at x = 0 is,
sin x
lim x → 0f(x) = lim x → 0 x = 1
The right hand limit of f at x = 0 is,
lim x → 0 + f(x) = lim x → 0 + (x + 1) = 1
∴ lim x → 0f(x) = lim x → 0 + f(x) = f(0)
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at all points of the real line.
Thus, f has no point of discontinuity.
Page : 160 , Block Name : Exercise 5.1
Q24 Determine if f de ned by
{
1
x 2sin x , if x ≠ 0
f(x) = is a continuous function?
0, if x = 0
Answer. The given function f is
{
1
x 2sin x , if x ≠ 0
f(x) =
0, if x = 0
It is evident that f is de ned at all points of the real line. Let c be a real number.
1
If c ≠ 0, then f(c) = c 2sin c
(
lim x → cf(x) = lim x → ∞ x 2sin x
1
) ( lim
= x → ex
2
) ( lim sin ) = d sin
1
x
2
1
c
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x ≠ 0
Case II:
If c = 0, then f(0) = 0
( )
lim x → 0 − f(x) = lim x → 0 x 2sin x
1
( )
= lim x → 0 x 2sin x
1
Processing math: 100% 1
It is known that, − 1 ≤ sin x ≤ 1, x ≠ 0
If c > 0, then f(c) = c + 1 and lim x → ∞f(x) = lim x → ∞(x + 1) = c + 1
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x , such that x ≥ 0
Case III:
If c = 0, then f(c) = f(0) = 0 + 1 = 1
The left hand limit of f at x = 0 is,
sin x
lim x → 0f(x) = lim x → 0 x = 1
The right hand limit of f at x = 0 is,
lim x → 0 + f(x) = lim x → 0 + (x + 1) = 1
∴ lim x → 0f(x) = lim x → 0 + f(x) = f(0)
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at all points of the real line.
Thus, f has no point of discontinuity.
Page : 160 , Block Name : Exercise 5.1
Q24 Determine if f de ned by
{
1
x 2sin x , if x ≠ 0
f(x) = is a continuous function?
0, if x = 0
Answer. The given function f is
{
1
x 2sin x , if x ≠ 0
f(x) =
0, if x = 0
It is evident that f is de ned at all points of the real line. Let c be a real number.
1
If c ≠ 0, then f(c) = c 2sin c
(
lim x → cf(x) = lim x → ∞ x 2sin x
1
) ( lim
= x → ex
2
) ( lim sin ) = d sin
1
x
2
1
c
∴ lim x → ∞f(x) = f(c)
Therefore, f is continuous at all points x ≠ 0
Case II:
If c = 0, then f(0) = 0
( )
lim x → 0 − f(x) = lim x → 0 x 2sin x
1
( )
= lim x → 0 x 2sin x
1
Processing math: 100% 1
It is known that, − 1 ≤ sin x ≤ 1, x ≠ 0
Page 22
1
⇒ − x 2 ≤ sin x ≤ x 2
( )
⇒ lim x → 0 − x 2 ≤ lim x → 0 x 2sin x
( )1
≤ lim x → 0x 2
( )
⇒ 0 ≤ lim x → 0 x 2sin x
1
≤0
( )
⇒ lim x → 0 x 2sin x
1
=0
∴ lim x → 0 + f(x) = 0
( )
Similarly, lim f(x) = lim x → 0 + x 2sin x
1
( ) 1
= lim x → 0 x 2sin x =0
∴ lim x → 0f(x) = f(0) = lim x → 0f(x)
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at every point of
the real line.
Thus, f is a continuous function.
Page : 160 , Block Name : Exercise 5.1
Q25 Examine the continuity of f, where f is de ned by
f(x) = { sinx − cosx,
−1
if x ≠ 0
if x = 0
Answer. The given function f is
f(x) = { sinx − cosx,
−1
if x ≠ 0
if x = 0
If c ≠ 0, then f(c) = sinc − cosc
lim x → cf(x) = lim x → c(sinx − cosx) = sinc − cosc
∴ lim x → ∞f(x) = f(c)
Case II:
If c = 0, then f(0) = − 1
lim x → 0f(x) = lim x → 0(sinx − cosx) = sin0 − cos0 = 0 − 1 = − 1
∴ lim x → 0f(x) = lim x → 0 − f(x) = f(0)
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at every point of
the real line.
Thus, f is a continuous function.
⇒ − x 2 ≤ sin x ≤ x 2
( )
⇒ lim x → 0 − x 2 ≤ lim x → 0 x 2sin x
( )1
≤ lim x → 0x 2
( )
⇒ 0 ≤ lim x → 0 x 2sin x
1
≤0
( )
⇒ lim x → 0 x 2sin x
1
=0
∴ lim x → 0 + f(x) = 0
( )
Similarly, lim f(x) = lim x → 0 + x 2sin x
1
( ) 1
= lim x → 0 x 2sin x =0
∴ lim x → 0f(x) = f(0) = lim x → 0f(x)
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at every point of
the real line.
Thus, f is a continuous function.
Page : 160 , Block Name : Exercise 5.1
Q25 Examine the continuity of f, where f is de ned by
f(x) = { sinx − cosx,
−1
if x ≠ 0
if x = 0
Answer. The given function f is
f(x) = { sinx − cosx,
−1
if x ≠ 0
if x = 0
If c ≠ 0, then f(c) = sinc − cosc
lim x → cf(x) = lim x → c(sinx − cosx) = sinc − cosc
∴ lim x → ∞f(x) = f(c)
Case II:
If c = 0, then f(0) = − 1
lim x → 0f(x) = lim x → 0(sinx − cosx) = sin0 − cos0 = 0 − 1 = − 1
∴ lim x → 0f(x) = lim x → 0 − f(x) = f(0)
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at every point of
the real line.
Thus, f is a continuous function.
Page 23
Page : 161 , Block Name : Exercise 5.1
Q26 Find the values of k so that the function f is continuous at the indicated point.
{
kcos x π
π − 2x
, if x ≠ 2
π
f(x) = π
at x = 2
3, if x = 2
Answer. The given function
{
kcos x π
π − 2x
, if x ≠ 2 π
f(x) = π
at x = 2
3, if x = 2
π π
The given function f is continuous at x = 2 , if f is defined at x = 2 and if the value of the f
π
at x= 2
f is defined at and f 2
() π
=3
kcos x
lim x → π f(x) = lim x → π π − 2x
2 2
π
Put x = 2 + h
π
Then x → 2 ⇒ h → 0
∴ lim f(x) = lim
kcosx
= lim
( ) π
kcos 2 + h
π π − 2x
( )
π b→0 π
x→ 2 x→ 2
π−2 2 +h
− sinh k sinh k k
= k lim = lim = ⋅1=
h→0 − 2h 2h→0 h 2 2
∴ lim x → π f(x) = f 2
2 () π
k
⇒ 2 =3
k
⇒ 2 =6
⇒k=6
Therefore,
Processing the required
math: 100% value of k is 6
Q26 Find the values of k so that the function f is continuous at the indicated point.
{
kcos x π
π − 2x
, if x ≠ 2
π
f(x) = π
at x = 2
3, if x = 2
Answer. The given function
{
kcos x π
π − 2x
, if x ≠ 2 π
f(x) = π
at x = 2
3, if x = 2
π π
The given function f is continuous at x = 2 , if f is defined at x = 2 and if the value of the f
π
at x= 2
f is defined at and f 2
() π
=3
kcos x
lim x → π f(x) = lim x → π π − 2x
2 2
π
Put x = 2 + h
π
Then x → 2 ⇒ h → 0
∴ lim f(x) = lim
kcosx
= lim
( ) π
kcos 2 + h
π π − 2x
( )
π b→0 π
x→ 2 x→ 2
π−2 2 +h
− sinh k sinh k k
= k lim = lim = ⋅1=
h→0 − 2h 2h→0 h 2 2
∴ lim x → π f(x) = f 2
2 () π
k
⇒ 2 =3
k
⇒ 2 =6
⇒k=6
Therefore,
Processing the required
math: 100% value of k is 6
Page 24
Page : 161 , Block Name : Exercise 5.1
Q28 Find the values of k so that the function f is continuous at the indicated point.
f(x) =
{ kx + 1, if x ≤ π
cosx, if x > π
at x = π
Answer. The given function is
f(x) =
{ kx + 1, if x ≤ π
cosx, if x > π
at x = π
The given function f is continuous at x = p, if f is de ned at x = p and if the value of f at x = p
equals the limit of f at x = p
It is evident that f is defined at x = p and f(π) = kπ + 1
lim x → π + f(x) = lim x → π + f(x) = f(π)
⇒ lim x → π − (kx + 1) = lim x → x ∗ cosx = kπ + 1
⇒ kπ + 1 = cosπ = kπ + 1
⇒ kπ + 1 = − 1 = kπ + 1
2
⇒k= − π
Page : 161 , Block Name : Exercise 5.1
Q29 Find the values of k so that the function f is continuous at the indicated point.
f(x) = { kx + 1, if x ≤ 5
3x − 5, if x > 5
at x = 5
Answer. The given function f is
f(x) = { kx + 1, if x ≤ 5
3x − 5, if x > 5
at x = 5
The given function f is continuous at x = 5, if f is de ned at x = 5 and if the value of f at x = 5 equals
the limit of f at x = 5
It is evident that f is defined at x = 5 and f(5) = kx + 1 = 5k + 1
lim x → 5f(x) = lim x → 5f(x) = f(5)
⇒ lim x → 5(kx + 1) = lim x → 1(3x − 5) = 5k + 1
⇒ 5k + 1 = 15 − 5 = 5k + 1
⇒ 5k + 1 = 10
⇒ 5k = 9
9
⇒k= 5
Page : 161 ,math:
Processing Block Name : Exercise 5.1
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Q28 Find the values of k so that the function f is continuous at the indicated point.
f(x) =
{ kx + 1, if x ≤ π
cosx, if x > π
at x = π
Answer. The given function is
f(x) =
{ kx + 1, if x ≤ π
cosx, if x > π
at x = π
The given function f is continuous at x = p, if f is de ned at x = p and if the value of f at x = p
equals the limit of f at x = p
It is evident that f is defined at x = p and f(π) = kπ + 1
lim x → π + f(x) = lim x → π + f(x) = f(π)
⇒ lim x → π − (kx + 1) = lim x → x ∗ cosx = kπ + 1
⇒ kπ + 1 = cosπ = kπ + 1
⇒ kπ + 1 = − 1 = kπ + 1
2
⇒k= − π
Page : 161 , Block Name : Exercise 5.1
Q29 Find the values of k so that the function f is continuous at the indicated point.
f(x) = { kx + 1, if x ≤ 5
3x − 5, if x > 5
at x = 5
Answer. The given function f is
f(x) = { kx + 1, if x ≤ 5
3x − 5, if x > 5
at x = 5
The given function f is continuous at x = 5, if f is de ned at x = 5 and if the value of f at x = 5 equals
the limit of f at x = 5
It is evident that f is defined at x = 5 and f(5) = kx + 1 = 5k + 1
lim x → 5f(x) = lim x → 5f(x) = f(5)
⇒ lim x → 5(kx + 1) = lim x → 1(3x − 5) = 5k + 1
⇒ 5k + 1 = 15 − 5 = 5k + 1
⇒ 5k + 1 = 10
⇒ 5k = 9
9
⇒k= 5
Page : 161 ,math:
Processing Block Name : Exercise 5.1
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Page 25
Q30 Find the values of a and b such that the function de ned by
{
5, if x ≤ 2
f(x) = ax + b, if 2 < x < 10
21, if x ≥ 10
is a continuous function.
Answer. The given function f is
{
5, if x ≤ 2
f(x) = ax + b, if 2 < x < 10
21, if x ≥ 10
It is evident that the given function f is de ned at all points of the real line. If f is a continuous
function, then f is continuous at all real numbers. In particular, f is continuous at x = 2 and x = 10
Since f is continuous at x = 2, we obtain
lim x → 2 − f(x) = lim x → 2 + f(x) = f(2)
⇒ lim x → 2(5) = lim x → 2 + (ax + b) = 5
⇒ 5 = 2a + b = 5
⇒ 2a + b = 5
lim x → 10 − f(x) = lim x → 10 + f(x) = f(10)
⇒ lim x → 10(ax + b) = lim x → 10 − (21) = 21
⇒ 10a + b = 21 = 21
⇒ 10a + b = 21
On subtracting equation (1) from equation (2), we obtain 8a = 16 ⇒ a = 2 By putting a = 2 in
equation (1), we obtain 2 × 2 + b = 5 ⇒ 4 + b = 5
⇒b=1
Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.
Page : 161 , Block Name : Exercise 5.1
Q31 Show that the function de ned by
( )
f(x) = cos x 2 is a continuous function.
Answer.
The given function is f(x) = cos x 2 ( )
This function f is defined for every real number and f can be written as the composition
of two functions as,
f = g ∘ h, where g(x) = cosx and h(x) = x 2
[ ∵ (goh)(x) = g(h(x)) = g (x ) = cos (x ) = f(x) ]
Processing math: 100% 2 2
{
5, if x ≤ 2
f(x) = ax + b, if 2 < x < 10
21, if x ≥ 10
is a continuous function.
Answer. The given function f is
{
5, if x ≤ 2
f(x) = ax + b, if 2 < x < 10
21, if x ≥ 10
It is evident that the given function f is de ned at all points of the real line. If f is a continuous
function, then f is continuous at all real numbers. In particular, f is continuous at x = 2 and x = 10
Since f is continuous at x = 2, we obtain
lim x → 2 − f(x) = lim x → 2 + f(x) = f(2)
⇒ lim x → 2(5) = lim x → 2 + (ax + b) = 5
⇒ 5 = 2a + b = 5
⇒ 2a + b = 5
lim x → 10 − f(x) = lim x → 10 + f(x) = f(10)
⇒ lim x → 10(ax + b) = lim x → 10 − (21) = 21
⇒ 10a + b = 21 = 21
⇒ 10a + b = 21
On subtracting equation (1) from equation (2), we obtain 8a = 16 ⇒ a = 2 By putting a = 2 in
equation (1), we obtain 2 × 2 + b = 5 ⇒ 4 + b = 5
⇒b=1
Therefore, the values of a and b for which f is a continuous function are 2 and 1 respectively.
Page : 161 , Block Name : Exercise 5.1
Q31 Show that the function de ned by
( )
f(x) = cos x 2 is a continuous function.
Answer.
The given function is f(x) = cos x 2 ( )
This function f is defined for every real number and f can be written as the composition
of two functions as,
f = g ∘ h, where g(x) = cosx and h(x) = x 2
[ ∵ (goh)(x) = g(h(x)) = g (x ) = cos (x ) = f(x) ]
Processing math: 100% 2 2
Page 26
It has to be first proved that g(x) = cosx and h(x) = x 2 are continuous functions.
It is evident that g is defined for every real number.
Let c be a real number.
Then, g(c) = cosc
Put x = c + h
If x → c, then h → 0
lim g(x) = lim cosx
x→ϵ x→c
= lim cos(c + h)
h→0
= lim [cosccosh − sincsinh]
h→0
= lim cosccosh − lim sincsinh
h→0 h→0
= cosccos0 − sincsin0
= cosc × 1 − sinc × 0
= cosc
∴ lim x → ∞g(x) = g(c)
Therefore, g(x) = cosx is continuous function.
h(x) = x 2
Clearly, h is defined for every real number.
Let k be a real number, then h(k) = k 2
lim x → ih(x) = lim x → kx 2 = k 2
∴ lim x → kh(x) = h(k)
Therefore, h is a continuous function. It is known that for real valued functions g and h,such that
(g o h) is de ned at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous
at c.
( )
f(x) = (goh)(x) = cos x 2 is a continuous function.
Page : 161 , Block Name : Exercise 5.1
Q32 Show that the function de ned by f(x) = | cosx | is a continuous function.
Answer. The given function is f(x) = | cosx |
This function f is de ned for every real number and f can be written as the composition of two
functions as,
f = g ∘ h, where g(x) = | x | and h(x) = cosx
[ ∵ (goh)(x) = g(h(x)) − g(cosx) = | cosx | = f(x)]
It has to be rst proved that g(x) = | x | and h(x) = cosx are continuous functions.
g(x) = | x | can be written as
g(x) =
{ − x, if x < 0
x, math:if100%
Processing x≥0
It is evident that g is defined for every real number.
Let c be a real number.
Then, g(c) = cosc
Put x = c + h
If x → c, then h → 0
lim g(x) = lim cosx
x→ϵ x→c
= lim cos(c + h)
h→0
= lim [cosccosh − sincsinh]
h→0
= lim cosccosh − lim sincsinh
h→0 h→0
= cosccos0 − sincsin0
= cosc × 1 − sinc × 0
= cosc
∴ lim x → ∞g(x) = g(c)
Therefore, g(x) = cosx is continuous function.
h(x) = x 2
Clearly, h is defined for every real number.
Let k be a real number, then h(k) = k 2
lim x → ih(x) = lim x → kx 2 = k 2
∴ lim x → kh(x) = h(k)
Therefore, h is a continuous function. It is known that for real valued functions g and h,such that
(g o h) is de ned at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is continuous
at c.
( )
f(x) = (goh)(x) = cos x 2 is a continuous function.
Page : 161 , Block Name : Exercise 5.1
Q32 Show that the function de ned by f(x) = | cosx | is a continuous function.
Answer. The given function is f(x) = | cosx |
This function f is de ned for every real number and f can be written as the composition of two
functions as,
f = g ∘ h, where g(x) = | x | and h(x) = cosx
[ ∵ (goh)(x) = g(h(x)) − g(cosx) = | cosx | = f(x)]
It has to be rst proved that g(x) = | x | and h(x) = cosx are continuous functions.
g(x) = | x | can be written as
g(x) =
{ − x, if x < 0
x, math:if100%
Processing x≥0
Page 27
Clearly, g is de ned for all real numbers. Let c be a real number.
Case I:
If c < 0, then g(c) = − c and lim x → ϵg(x) = lim x → c( − x) = − c
∴ lim x → eg(x) = g(c)
Therefore, g is continuous at all points x, such that x < 0
case II:
If c > 0, then g(c) = c and lim x → cg(x) = lim x → cx = c
∴ lim x → eg(x) = g(c)
Case III:
If c = 0, then g(c) = g(0) = 0
lim x → 0g(x) = lim x → 0( − x) = 0
lim x → 0 + g(x) = lim x → 0 + (x) = 0
∴ lim x → 0g(x) = lim x → 0 0(x) = g(0)
Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is
continuous at all points. h (x) = cos x
It is evident that h(x) = cosx is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
lim h(x) = lim cosx
x→c
= lim cos(c + h)
h→∞
= lim [cosccosh − sincsinh]
h→0
= cosccos0 − sincsin0
= cosc × 1 − sinc × 0
= cosc
∴ lim x → ch(x) = h(c)
Therefore, h (x) = cos x is a continuous function. It is known that for real valued functions g and
h,such that (g o h) is de ned at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is
continuous at c.
f(x) = (goh)(x) = g(h(x)) = g(cosx) = | cosx | is a continuous function.
Page : 161 , Block Name : Exercise 5.1
Q33 Examine that sin | x | is a continuous function. .
Answer. f(x) = sin | x |
This function f is de ned for every real number and f can be written as the composition of two
functions as,
f = g o h, where g(x) = | x | and h(x) = sinx
[ ∵ (goh)(x) = g(h(x)) = g(sinx) = | sinx | = f(x)]
It has to be proved rst that g(x) = | x | and h(x) = sinx are continuous functions.
Case I:
If c < 0, then g(c) = − c and lim x → ϵg(x) = lim x → c( − x) = − c
∴ lim x → eg(x) = g(c)
Therefore, g is continuous at all points x, such that x < 0
case II:
If c > 0, then g(c) = c and lim x → cg(x) = lim x → cx = c
∴ lim x → eg(x) = g(c)
Case III:
If c = 0, then g(c) = g(0) = 0
lim x → 0g(x) = lim x → 0( − x) = 0
lim x → 0 + g(x) = lim x → 0 + (x) = 0
∴ lim x → 0g(x) = lim x → 0 0(x) = g(0)
Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is
continuous at all points. h (x) = cos x
It is evident that h(x) = cosx is defined for every real number.
Let c be a real number. Put x = c + h
If x → c, then h → 0
lim h(x) = lim cosx
x→c
= lim cos(c + h)
h→∞
= lim [cosccosh − sincsinh]
h→0
= cosccos0 − sincsin0
= cosc × 1 − sinc × 0
= cosc
∴ lim x → ch(x) = h(c)
Therefore, h (x) = cos x is a continuous function. It is known that for real valued functions g and
h,such that (g o h) is de ned at c, if g is continuous at c and if f is continuous at g (c), then (f o g) is
continuous at c.
f(x) = (goh)(x) = g(h(x)) = g(cosx) = | cosx | is a continuous function.
Page : 161 , Block Name : Exercise 5.1
Q33 Examine that sin | x | is a continuous function. .
Answer. f(x) = sin | x |
This function f is de ned for every real number and f can be written as the composition of two
functions as,
f = g o h, where g(x) = | x | and h(x) = sinx
[ ∵ (goh)(x) = g(h(x)) = g(sinx) = | sinx | = f(x)]
It has to be proved rst that g(x) = | x | and h(x) = sinx are continuous functions.
Page 28
g(x) = | x | can be written as
g(x) =
{ − x, if x < 0
x, if x ≥ 0
Let c be a real number.
Case I:
If c < 0, then g(c) = 4c and lim t → tg(x) = lim x → c( − x) = − c
∴ lim x → cg(x) = g(c)
Therefore, g is continuous at all points x, such that x < 0
Case II:
If c > 0, then g(c) = c and lim x → rg(x) = lim x → cx = c
∴ lim x → ϵg(x) = g(c)
Therefore, g is continuous at all points x, such that x > 0
Case III:
If c = 0, then g(c) = g(0) = 0
lim x → ∞g(x) = lim x → 0( − x) = 0
lim x → 0 + g(x) = lim x → 0 ′(x) = 0
∴ lim x → 0g(x) = lim x → 0 ( (x) = g(0)
Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is
continuous at all points.
h (x) = sin x
It is evident that h (x) = sin x is de ned for every real number. Let c be a real number. Put x = c + k
If x → c, then k → 0 h (c) = sin c
lim h(x) = lim sinx
x→c
= lim sin(c + k)
x→c
= lim [sinccosk + coscsink]
k→0
= lim (sinccosk) + lim (coscsink)
k→0 h→0
= sinccos0 + coscsin0
= sinc + 0
= sinc
∴ lim x → ch(x) = g(c)
Therefore, h is a continuous function.
It is known that for real valued functions g and h,such that (g o h) is de ned at c, if g is continuous
at c and if f is continuous at g (c), then (f o g) is continuous at c.
f(x) = (goh)(x) = g(h(x)) = g(sinx) = | sinx | is a continuous function.
Page : 161 , Block Name : Exercise 5.1
Q34 Find all
Processing the 100%
math: points of discontinuity of f defined by f(x) = | x | − | x + 1 |
g(x) =
{ − x, if x < 0
x, if x ≥ 0
Let c be a real number.
Case I:
If c < 0, then g(c) = 4c and lim t → tg(x) = lim x → c( − x) = − c
∴ lim x → cg(x) = g(c)
Therefore, g is continuous at all points x, such that x < 0
Case II:
If c > 0, then g(c) = c and lim x → rg(x) = lim x → cx = c
∴ lim x → ϵg(x) = g(c)
Therefore, g is continuous at all points x, such that x > 0
Case III:
If c = 0, then g(c) = g(0) = 0
lim x → ∞g(x) = lim x → 0( − x) = 0
lim x → 0 + g(x) = lim x → 0 ′(x) = 0
∴ lim x → 0g(x) = lim x → 0 ( (x) = g(0)
Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is
continuous at all points.
h (x) = sin x
It is evident that h (x) = sin x is de ned for every real number. Let c be a real number. Put x = c + k
If x → c, then k → 0 h (c) = sin c
lim h(x) = lim sinx
x→c
= lim sin(c + k)
x→c
= lim [sinccosk + coscsink]
k→0
= lim (sinccosk) + lim (coscsink)
k→0 h→0
= sinccos0 + coscsin0
= sinc + 0
= sinc
∴ lim x → ch(x) = g(c)
Therefore, h is a continuous function.
It is known that for real valued functions g and h,such that (g o h) is de ned at c, if g is continuous
at c and if f is continuous at g (c), then (f o g) is continuous at c.
f(x) = (goh)(x) = g(h(x)) = g(sinx) = | sinx | is a continuous function.
Page : 161 , Block Name : Exercise 5.1
Q34 Find all
Processing the 100%
math: points of discontinuity of f defined by f(x) = | x | − | x + 1 |
Page 29
Answer. The given function is f(x) = | x | − | x + 1 |
The two functions, g and h, are defined as
g(x) = | x | and h(x) = | x + 1 |
Then, f = g − h
The continuity of g and h is examined first.
g(x) = | x | Can be written as
g(x) = { − x, if x < 0
x, if x ≥ 0
Clearly, g is de ned for all real numbers. Let c be a real number.
Case I:
If c < 0, then g(c) = − c and lim x → ϵg(x) = lim x → c( − x) = − c
∴ lim x → cg(x) = g(c)
Therefore, g is continuous at all points x, such that x < 0
Case II:
If c > 0, then g(c) = c and lim g(x) = lim x = c
∴ lim x → cg(x) = g(c)
Case III:
If c = 0, then g(c) = g(0) = 0
lim x → 0g(x) = lim x → 0( − x) = 0
lim x → 0 + g(x) = lim x → 0 ′(x) = 0
∴ lim x → 0g(x) = lim x → 0 + (x) = g(0)
Therefore, g is continuous at x = 0
Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is
continuous at all points.
h(x) = | x + 1 | can be written as
Clearly, h is de ned for every real number.
Let c be a real number.
Case I:
If c < − 1, then h(c) = − (c + 1) and lim x → ϵh(x) = lim x → ∞[ − (x + 1)] = − (c + 1)
∴ lim x → ch(x) = h(c)
Therefore, h is continuous at all points x, such that x < − 1
Case II:
If c > − 1, then h(c) = c + 1 and lim x → ch(x) = lim x → c(x + 1) = c + 1
∴ lim x → ∞h(x) = h(c)
Therefore, h is continuous at all points x, such that x > − 1
The two functions, g and h, are defined as
g(x) = | x | and h(x) = | x + 1 |
Then, f = g − h
The continuity of g and h is examined first.
g(x) = | x | Can be written as
g(x) = { − x, if x < 0
x, if x ≥ 0
Clearly, g is de ned for all real numbers. Let c be a real number.
Case I:
If c < 0, then g(c) = − c and lim x → ϵg(x) = lim x → c( − x) = − c
∴ lim x → cg(x) = g(c)
Therefore, g is continuous at all points x, such that x < 0
Case II:
If c > 0, then g(c) = c and lim g(x) = lim x = c
∴ lim x → cg(x) = g(c)
Case III:
If c = 0, then g(c) = g(0) = 0
lim x → 0g(x) = lim x → 0( − x) = 0
lim x → 0 + g(x) = lim x → 0 ′(x) = 0
∴ lim x → 0g(x) = lim x → 0 + (x) = g(0)
Therefore, g is continuous at x = 0
Therefore, g is continuous at x = 0 From the above three observations, it can be concluded that g is
continuous at all points.
h(x) = | x + 1 | can be written as
Clearly, h is de ned for every real number.
Let c be a real number.
Case I:
If c < − 1, then h(c) = − (c + 1) and lim x → ϵh(x) = lim x → ∞[ − (x + 1)] = − (c + 1)
∴ lim x → ch(x) = h(c)
Therefore, h is continuous at all points x, such that x < − 1
Case II:
If c > − 1, then h(c) = c + 1 and lim x → ch(x) = lim x → c(x + 1) = c + 1
∴ lim x → ∞h(x) = h(c)
Therefore, h is continuous at all points x, such that x > − 1
Page 30
Case III:
If c = − 1, then h(c) = h( − 1) = − 1 + 1 = 0
lim x → − 1h(x) = lim x → − 1 + [ − (x + 1)] = − ( − 1 + 1) = 0
lim x → − 1 − h(x) = lim x → − 1 + (x + 1) = ( − 1 + 1) = 0
∴ lim x → − 1 − h(x) = lim x → − 1 − h(x) = h( − 1)
Therefore, h is continuous at x = −1 From the above three observations, it can be concluded that h
is continuous at all points of the real line. g and h are continuous functions. Therefore, f = g − h is
also a continuous function. Therefore, f has no point of discontinuity.
Page : 161 , Block Name : Exercise 5.1
Q1 Differentiate the functions with respect to x.
(
sin x 2 + 5 )
Answer.
( )
Let f(x) = sin x 2 + 5 , u(x) = x 2 + 5, and v(t) = sint
(
Then, ( wou )(x) = v(u(x)) = v x 2 + 5 = tan x 2 + 5 = f(x)) ( )
Thus, f is a composite of two functions.
Put t = u(x) = x 2 + 5
Then, we obtain
t = u(x) = x 2 + 5
Then , we obtain
dv d
dt dt ( )
= (sint) = cost = cos x 2 + 5
dt d d d
dx
=
dx ( ) ( )
x2 + 5 =
dx
x2 +
dx
(5) = 2x + 0 = 2x
= cos (x + 5 ) × 2x = 2xcos (x + 5 )
df dv dt
2 2
dx
= dt ⋅dx
Alternate method
d d
dx [ ( )] ( ) dx (x + 5 )
sin x 2 + 5 = cos x 2 + 5 ⋅ 2
( ) [ dx (x ) + dx (5) ]
d d
= cos x 2 + 5 ⋅ 2
= cos (x + 5 ) ⋅ [2x + 0]
2
= 2xcos (x + 5 )
2
Page : 166 ,math:
Processing Block Name : Exercise 5.2
100%
If c = − 1, then h(c) = h( − 1) = − 1 + 1 = 0
lim x → − 1h(x) = lim x → − 1 + [ − (x + 1)] = − ( − 1 + 1) = 0
lim x → − 1 − h(x) = lim x → − 1 + (x + 1) = ( − 1 + 1) = 0
∴ lim x → − 1 − h(x) = lim x → − 1 − h(x) = h( − 1)
Therefore, h is continuous at x = −1 From the above three observations, it can be concluded that h
is continuous at all points of the real line. g and h are continuous functions. Therefore, f = g − h is
also a continuous function. Therefore, f has no point of discontinuity.
Page : 161 , Block Name : Exercise 5.1
Q1 Differentiate the functions with respect to x.
(
sin x 2 + 5 )
Answer.
( )
Let f(x) = sin x 2 + 5 , u(x) = x 2 + 5, and v(t) = sint
(
Then, ( wou )(x) = v(u(x)) = v x 2 + 5 = tan x 2 + 5 = f(x)) ( )
Thus, f is a composite of two functions.
Put t = u(x) = x 2 + 5
Then, we obtain
t = u(x) = x 2 + 5
Then , we obtain
dv d
dt dt ( )
= (sint) = cost = cos x 2 + 5
dt d d d
dx
=
dx ( ) ( )
x2 + 5 =
dx
x2 +
dx
(5) = 2x + 0 = 2x
= cos (x + 5 ) × 2x = 2xcos (x + 5 )
df dv dt
2 2
dx
= dt ⋅dx
Alternate method
d d
dx [ ( )] ( ) dx (x + 5 )
sin x 2 + 5 = cos x 2 + 5 ⋅ 2
( ) [ dx (x ) + dx (5) ]
d d
= cos x 2 + 5 ⋅ 2
= cos (x + 5 ) ⋅ [2x + 0]
2
= 2xcos (x + 5 )
2
Page : 166 ,math:
Processing Block Name : Exercise 5.2
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Page 31
Q2 Differentiate the functions with respect to x.
Cos (Sin x)
Answer.
Let f(x) = cos(sinx), u(x) = sinx, and v(t) = cost
Then, (vou)(x) = v(u(x)) = v(sinx) = cos(sinx) = f(x)
Thus, f is a composite function of two functions.
Put t = u(x) = sinx
dv d
∴ dt = dt [cost] = − sint = − sin(sinx)
dt d
dx
= dx (sinx) = cosx
df dv dt
By chain rule dx = dt ⋅ dx = − sin(sinx) ⋅ cosx = − cosxsin(sinx)
Alternate Method
d d
dx
[cos(sinx)] = − sin(sinx) ⋅ dx (sinx) = − sin(sinx) ⋅ cosx = − cosxsin(sinx)
Page : 166 , Block Name : Exercise 5.2
Q3 Differentiate the functions with respect to x. sin (ax + b)
Answer.
Let f(x) = sin(ax + b), u(x) = ax + b, and v(t) = sint
Then, ( wou )(x) = v(u(x)) = v(ax + b) = sin(ax + b) = f(x)
Thus, f is a composite function of two functions, u and v.
Put t = u(x) = ax + b
Therefore,
dv d
dt
= dt (sint) = cost = cos(ax + b)
dt d d d
dx
= dx (ax + b) = dx (ax) + dx (b) = a + 0 = a
Hence, by chain rule, we obtain
df dv dt
dx
= dt ⋅ dx = cos(ax + b) ⋅ a = acos(ax + b)
Alternate method
d d
[sin(ax + b)] = cos(ax + b) ⋅ (ax + b)
dx dx
= cos(ax + b) ⋅
[ d
dx
(ax) +
d
dx
(b)
]
= cos(ax + b) ⋅ (a + 0)
= acos(ax + b)
Processing
Page : 166 ,math:
Block100%
Name : Exercise 5.2
Cos (Sin x)
Answer.
Let f(x) = cos(sinx), u(x) = sinx, and v(t) = cost
Then, (vou)(x) = v(u(x)) = v(sinx) = cos(sinx) = f(x)
Thus, f is a composite function of two functions.
Put t = u(x) = sinx
dv d
∴ dt = dt [cost] = − sint = − sin(sinx)
dt d
dx
= dx (sinx) = cosx
df dv dt
By chain rule dx = dt ⋅ dx = − sin(sinx) ⋅ cosx = − cosxsin(sinx)
Alternate Method
d d
dx
[cos(sinx)] = − sin(sinx) ⋅ dx (sinx) = − sin(sinx) ⋅ cosx = − cosxsin(sinx)
Page : 166 , Block Name : Exercise 5.2
Q3 Differentiate the functions with respect to x. sin (ax + b)
Answer.
Let f(x) = sin(ax + b), u(x) = ax + b, and v(t) = sint
Then, ( wou )(x) = v(u(x)) = v(ax + b) = sin(ax + b) = f(x)
Thus, f is a composite function of two functions, u and v.
Put t = u(x) = ax + b
Therefore,
dv d
dt
= dt (sint) = cost = cos(ax + b)
dt d d d
dx
= dx (ax + b) = dx (ax) + dx (b) = a + 0 = a
Hence, by chain rule, we obtain
df dv dt
dx
= dt ⋅ dx = cos(ax + b) ⋅ a = acos(ax + b)
Alternate method
d d
[sin(ax + b)] = cos(ax + b) ⋅ (ax + b)
dx dx
= cos(ax + b) ⋅
[ d
dx
(ax) +
d
dx
(b)
]
= cos(ax + b) ⋅ (a + 0)
= acos(ax + b)
Processing
Page : 166 ,math:
Block100%
Name : Exercise 5.2
Page 32
Q4 Differentiate the functions with respect to x. Sec(tan(√x))
Answer.
Let f(x) = sec(tan√x), u(x) = √x, v(t) = tant, and w(s) = secs
Then, (wovou)(x) = w[v(u(x))] = w[v(√x)] = w(tan√x) = sec(tan√x) = f(x)
Thus, f is a composite function of three functions, u, v, and w .
Put s = v(t) = tant and t = u(x) = √x
dw d
ds
= ds (secs) = secsinx = sec(tant) ⋅ tan(tant) [s = tant]
= sec(tan√x) ⋅ tan(tan√x)
ds d
dt
= dt (tant) = sec 2t = sec 2√x
dt
dx
d
= dx (√x) = dx x 2
d
()1 1 1
= 2 ⋅ x2 =
1
2√ x
Hence, by chain rule, we obtain
dt dw ds dt
= ⋅ ⋅
dx ds dt dx
1
= sec(tan√x) ⋅ tan(tan√x) × sec 2√x ×
2√x
1
= sec 2√xsec(tan√x)tan(tan√x)
2√x
sec 2√xsec(tan√x)tan(tan√x)
=
2√x
Alternate method
d d
[sec(tan√x)] = sec(tan√x) ⋅ tan(tan√x) ⋅ (tan√x)
dx dx
d
= sec(tan√x) ⋅ tan(tan√x) ⋅ sec 2(√x) ⋅
dx √
( x)
1
= sec(tan√x) ⋅ tan(tan√x) ⋅ sec 2(√x) ⋅
2√x
sec(tan√x) ⋅ tan(tan√x)sec 2(√x)
=
2√x
Page : 166 , Block Name : Exercise 5.2
Q5 Differentiate the functions with respect to x
sin ( ax + b )
cos ( cx + d )
Answer.
Let f(x) = sec(tan√x), u(x) = √x, v(t) = tant, and w(s) = secs
Then, (wovou)(x) = w[v(u(x))] = w[v(√x)] = w(tan√x) = sec(tan√x) = f(x)
Thus, f is a composite function of three functions, u, v, and w .
Put s = v(t) = tant and t = u(x) = √x
dw d
ds
= ds (secs) = secsinx = sec(tant) ⋅ tan(tant) [s = tant]
= sec(tan√x) ⋅ tan(tan√x)
ds d
dt
= dt (tant) = sec 2t = sec 2√x
dt
dx
d
= dx (√x) = dx x 2
d
()1 1 1
= 2 ⋅ x2 =
1
2√ x
Hence, by chain rule, we obtain
dt dw ds dt
= ⋅ ⋅
dx ds dt dx
1
= sec(tan√x) ⋅ tan(tan√x) × sec 2√x ×
2√x
1
= sec 2√xsec(tan√x)tan(tan√x)
2√x
sec 2√xsec(tan√x)tan(tan√x)
=
2√x
Alternate method
d d
[sec(tan√x)] = sec(tan√x) ⋅ tan(tan√x) ⋅ (tan√x)
dx dx
d
= sec(tan√x) ⋅ tan(tan√x) ⋅ sec 2(√x) ⋅
dx √
( x)
1
= sec(tan√x) ⋅ tan(tan√x) ⋅ sec 2(√x) ⋅
2√x
sec(tan√x) ⋅ tan(tan√x)sec 2(√x)
=
2√x
Page : 166 , Block Name : Exercise 5.2
Q5 Differentiate the functions with respect to x
sin ( ax + b )
cos ( cx + d )
Page 33
sin ( ax + b ) g(x)
Answer. The given function is f(x) = cos ( cx + d ) = h ( x ) where g (x) = sin (ax + b) and
h(x) = cos(cx + d)
g ′h − gh ′
∴ f′ =
h2
Consider g(x) = sin(ax + b)
Let u(x) = ax + b, v(t) = sint
Then, (wou)(x) = v(u(x)) = v(ax + b) = sin(ax + b) = g(x)
∴ g is a composite function of two functions, u and v
Put t = u(x) = ax + b
dv d
dt
= dt (sint) = cost = cos(ax + b)
dt d d d
dx
= dx (αx + b) = dx (ax) + dx (b) = a + 0 = a
Therefore, by chain rule, we obtain
dg dv dt
g ′ = dx = dt ⋅ dx = cos(ax + b) ⋅ a = acos(ax + b)
Consider h(x) = cos(cx + d)
Let p(x) = cx + d, q(y) = cosy
Then, (qop)(x) = q(p(x)) = q(cx + d) = cos(cx + d) = h(x)
Put y = p(x) = cx + d
dq d
dy
= dy (cosy) = − siny = − sin(cx + d)
dy d d d
dx
= dx (cx + d) = dx (cx) + dx (d) = c
Therefore, by chain rule, we obtain
dh dq dy
h ′ = dx = dy ⋅ dx = − sin(cx + d) × c = − csin(cx + d)
acos(ax + b) ⋅ cos(cx + d) − sin(ax + b){ − csin(cx + d)}
∴ f′ =
[cos(cx + d)] 2
acos(ax + b) sin(cx + d) 1
= + csin(ax + b) ⋅ ×
cos(cx + d) cos(cx + d) cos(cx + d)
= acos(ax + b)sec(cx + d) + csin(ax + b)tan(cx + d)sec(cx + d)
Page : 166 , Block Name : Exercise 5.2
Q6 Differentiate the functions with respect to x. cosx 3 ⋅ sin 2 x 5 ( )
Answer.
The given function is
[cosx ⋅ sin (x )] = sin (x ) × (cosx ) + cosx × [sin (x )]
d d d
3 2 5 2 5 3 3 2 4
dx dx dx
Answer. The given function is f(x) = cos ( cx + d ) = h ( x ) where g (x) = sin (ax + b) and
h(x) = cos(cx + d)
g ′h − gh ′
∴ f′ =
h2
Consider g(x) = sin(ax + b)
Let u(x) = ax + b, v(t) = sint
Then, (wou)(x) = v(u(x)) = v(ax + b) = sin(ax + b) = g(x)
∴ g is a composite function of two functions, u and v
Put t = u(x) = ax + b
dv d
dt
= dt (sint) = cost = cos(ax + b)
dt d d d
dx
= dx (αx + b) = dx (ax) + dx (b) = a + 0 = a
Therefore, by chain rule, we obtain
dg dv dt
g ′ = dx = dt ⋅ dx = cos(ax + b) ⋅ a = acos(ax + b)
Consider h(x) = cos(cx + d)
Let p(x) = cx + d, q(y) = cosy
Then, (qop)(x) = q(p(x)) = q(cx + d) = cos(cx + d) = h(x)
Put y = p(x) = cx + d
dq d
dy
= dy (cosy) = − siny = − sin(cx + d)
dy d d d
dx
= dx (cx + d) = dx (cx) + dx (d) = c
Therefore, by chain rule, we obtain
dh dq dy
h ′ = dx = dy ⋅ dx = − sin(cx + d) × c = − csin(cx + d)
acos(ax + b) ⋅ cos(cx + d) − sin(ax + b){ − csin(cx + d)}
∴ f′ =
[cos(cx + d)] 2
acos(ax + b) sin(cx + d) 1
= + csin(ax + b) ⋅ ×
cos(cx + d) cos(cx + d) cos(cx + d)
= acos(ax + b)sec(cx + d) + csin(ax + b)tan(cx + d)sec(cx + d)
Page : 166 , Block Name : Exercise 5.2
Q6 Differentiate the functions with respect to x. cosx 3 ⋅ sin 2 x 5 ( )
Answer.
The given function is
[cosx ⋅ sin (x )] = sin (x ) × (cosx ) + cosx × [sin (x )]
d d d
3 2 5 2 5 3 3 2 4
dx dx dx
Page 34
( ) ( )
d
= − sinx 3sin 2 x 5 × 3x 2 + 2sinx 5cosx 3 ⋅ cosx 5 × dx x 5
( )
= − 3x 2sinx 3 − sin 2 x 5 + 2sinx 5cosx 3cosx 3 ⋅ x5x 4
= 10x 4sinx 5cosx 5cosx 2 − 3x 2sinx 3sin 2 x 5 ( )
Page : 166 , Block Name : Exercise 5.2
Q7 Differentiate the functions with respect to x. 2 cot x 2 √ ( )
Answer.
d
dx [ √ ( )]
2 cot x 2
1 d
=2⋅ ×
dx [cot (x )]
2
√ ( )
2 cot x 2
√
sin (x ) 2
d
× − csc (x ) ×
dx ( )
= 2 x2 2
cos (x ) 2
√
sin (x ) 2
1
= − × × (2x)
cos (x ) sin (x )
2 2 2
− 2x
=
√cos x2√sin x2sin x2
− 2√2x
=
√2sin x2cos x2sin x2
− 2√2x
=
√
sin x 2 sin 2x 2
Page : 166 , Block Name : Exercise 5.2
Q8 Differentiate the functions with respect to x. cos(√x)
Answer.
d
= − sinx 3sin 2 x 5 × 3x 2 + 2sinx 5cosx 3 ⋅ cosx 5 × dx x 5
( )
= − 3x 2sinx 3 − sin 2 x 5 + 2sinx 5cosx 3cosx 3 ⋅ x5x 4
= 10x 4sinx 5cosx 5cosx 2 − 3x 2sinx 3sin 2 x 5 ( )
Page : 166 , Block Name : Exercise 5.2
Q7 Differentiate the functions with respect to x. 2 cot x 2 √ ( )
Answer.
d
dx [ √ ( )]
2 cot x 2
1 d
=2⋅ ×
dx [cot (x )]
2
√ ( )
2 cot x 2
√
sin (x ) 2
d
× − csc (x ) ×
dx ( )
= 2 x2 2
cos (x ) 2
√
sin (x ) 2
1
= − × × (2x)
cos (x ) sin (x )
2 2 2
− 2x
=
√cos x2√sin x2sin x2
− 2√2x
=
√2sin x2cos x2sin x2
− 2√2x
=
√
sin x 2 sin 2x 2
Page : 166 , Block Name : Exercise 5.2
Q8 Differentiate the functions with respect to x. cos(√x)
Answer.
Page 35
Let f(x) = cos(√x)
Also, let u(x) = √x
And, v(t) = cost
Then, (vou)(x) = v(u(x))
= v(√x)
= cos√x
= f(x)
Clearly, f is a composite function of two functions, u and v, such that
t = u(x) = √x
Then,
dt
dx
=
d
dx
(√x) =
d
dx ()
1
x2 =
1 −1
2
x 2
1
=
2√x
By using chain rule, we obtain
dt dv dt
dx
= dt ⋅ dx
1
= − sin(√x) ⋅
2√ x
1
= − sin(√x)
2√ x
sin ( √x )
= −
2√ x
Alternate method
d d
[cos(√x)] = − sin(√x) ⋅ (√x)
dx dx
d 2
= − sin(√x) ×
dx
x ( )
1 1
= − sin√x × x 2
2
− sin√x
=
2√x
Page : 166 , Block Name : Exercise 5.2
Q9
Prove that the function f given by
f(x) = | x − 1 | , x ∈ R is not differentiable at x = 1
Answer. The given function is f(x) = | x − 1 | , x ∈ R
Also, let u(x) = √x
And, v(t) = cost
Then, (vou)(x) = v(u(x))
= v(√x)
= cos√x
= f(x)
Clearly, f is a composite function of two functions, u and v, such that
t = u(x) = √x
Then,
dt
dx
=
d
dx
(√x) =
d
dx ()
1
x2 =
1 −1
2
x 2
1
=
2√x
By using chain rule, we obtain
dt dv dt
dx
= dt ⋅ dx
1
= − sin(√x) ⋅
2√ x
1
= − sin(√x)
2√ x
sin ( √x )
= −
2√ x
Alternate method
d d
[cos(√x)] = − sin(√x) ⋅ (√x)
dx dx
d 2
= − sin(√x) ×
dx
x ( )
1 1
= − sin√x × x 2
2
− sin√x
=
2√x
Page : 166 , Block Name : Exercise 5.2
Q9
Prove that the function f given by
f(x) = | x − 1 | , x ∈ R is not differentiable at x = 1
Answer. The given function is f(x) = | x − 1 | , x ∈ R
Page 36
It is known that a function f is differentiable at a point x = c in its domain if both
f(c+h) −f(c) f(c+h) −f(c)
lim h → ∞ h
and lim h
are finite and equal.
To check the differentiability of the given function at x = 1 ,
consider the left hand limit of f at x = 1
f(1+h) −f(1) |1+h−1| − |1−1|
lim h → ∞ h
= lim h → ∞ h
|h−0 −h
= lim h → 0 h = lim h → ∞ h (h < 0 ⇒ | h | = − h)
= −1
Consider the right hand limit of f at x = 1
f(1+h) −f(1) |1+h−1| − |1−1|
lim h → ∞ h
= lim h → ∞ h
|h| −0 h
= lim h → ∞ h
= lim h → ∞ h (h > 0 ⇒ | h | = h)
=1
since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x
=1
Page : 166 , Block Name : Exercise 5.2
Q10
Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not
differentiable at x = 1 and x = 2 .
Answer.
The given function f is f(x) = [x], 0 < x < 3
It is known that a function f is differentiable at a point x = c in its domain if both
f(c+h) −f(c) f(c+h) −f(c)
lim h → 0 h
and lim h → 0 h
are finite and equal.
To check the differentiability of the given function at x = 1, consider the left hand limit of
f at x = 1
f(1+h) −f(1) [1+h] − [1]
lim h → ∞ h
= lim h → 0 h
1−1
= lim h → 0 + h = lim h → 0 ′0 = 0
Since the left and right hand limits of f are not equal, f is not differentiable at
x=1
f(2+h) −f(2) [2+h] − [2]
lim h → 0 h
= lim h → ∞ h
1−2 −1
= lim h → 0 h = lim h → ∞ h = ∞
f(c+h) −f(c) f(c+h) −f(c)
lim h → ∞ h
and lim h
are finite and equal.
To check the differentiability of the given function at x = 1 ,
consider the left hand limit of f at x = 1
f(1+h) −f(1) |1+h−1| − |1−1|
lim h → ∞ h
= lim h → ∞ h
|h−0 −h
= lim h → 0 h = lim h → ∞ h (h < 0 ⇒ | h | = − h)
= −1
Consider the right hand limit of f at x = 1
f(1+h) −f(1) |1+h−1| − |1−1|
lim h → ∞ h
= lim h → ∞ h
|h| −0 h
= lim h → ∞ h
= lim h → ∞ h (h > 0 ⇒ | h | = h)
=1
since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x
=1
Page : 166 , Block Name : Exercise 5.2
Q10
Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not
differentiable at x = 1 and x = 2 .
Answer.
The given function f is f(x) = [x], 0 < x < 3
It is known that a function f is differentiable at a point x = c in its domain if both
f(c+h) −f(c) f(c+h) −f(c)
lim h → 0 h
and lim h → 0 h
are finite and equal.
To check the differentiability of the given function at x = 1, consider the left hand limit of
f at x = 1
f(1+h) −f(1) [1+h] − [1]
lim h → ∞ h
= lim h → 0 h
1−1
= lim h → 0 + h = lim h → 0 ′0 = 0
Since the left and right hand limits of f are not equal, f is not differentiable at
x=1
f(2+h) −f(2) [2+h] − [2]
lim h → 0 h
= lim h → ∞ h
1−2 −1
= lim h → 0 h = lim h → ∞ h = ∞
Page 37
Consider the right hand limit of f at x = 1
f(2+h) −f(2) [2+h] − [2]
lim h → ∞ h
= lim t → ∞ h
2−2
= lim h → 0 + h = lim k → 00 = 0
since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x
=2
Page : 166 , Block Name : Exercise 5.2
dy
Q1 Find dx : in 2x + 3y = sinx
Answer.
The given relationship is 2x + 3y = sinx
Differentiating this relationship with respect to x, we obtain
d d
dx
(2x + 3y) = dx (sinx)
d d
⇒ dx (2x) + dx (3y) = cosx
dy
⇒ 2 + 3 dx = cosx
dy
⇒ 3 dx = cosx − 2
dy cos x − 2
∴ dx = 3
Page : 166 , Block Name : Exercise 5.3
dy
Q2 Find dx : in 2x + 3y = siny
Answer.
The given relationship is 2x + 3y = siny
Differentiating this relationship with respect to x, we obtain
d d d
dx
(2x) + dx (3y) = dx (siny)
dy dy
⇒ 2 + 3 dx = cosy dx [ By using chain rule ]
dy
⇒ 2 = (cosy − 3) dx
dy 2
∴ dx = cos y − 3
Page : 166 , Block Name : Exercise 5.3
dy
Q3Processing
Find dx math: 100% 2
: in ax + by = cosy
f(2+h) −f(2) [2+h] − [2]
lim h → ∞ h
= lim t → ∞ h
2−2
= lim h → 0 + h = lim k → 00 = 0
since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x
=2
Page : 166 , Block Name : Exercise 5.2
dy
Q1 Find dx : in 2x + 3y = sinx
Answer.
The given relationship is 2x + 3y = sinx
Differentiating this relationship with respect to x, we obtain
d d
dx
(2x + 3y) = dx (sinx)
d d
⇒ dx (2x) + dx (3y) = cosx
dy
⇒ 2 + 3 dx = cosx
dy
⇒ 3 dx = cosx − 2
dy cos x − 2
∴ dx = 3
Page : 166 , Block Name : Exercise 5.3
dy
Q2 Find dx : in 2x + 3y = siny
Answer.
The given relationship is 2x + 3y = siny
Differentiating this relationship with respect to x, we obtain
d d d
dx
(2x) + dx (3y) = dx (siny)
dy dy
⇒ 2 + 3 dx = cosy dx [ By using chain rule ]
dy
⇒ 2 = (cosy − 3) dx
dy 2
∴ dx = cos y − 3
Page : 166 , Block Name : Exercise 5.3
dy
Q3Processing
Find dx math: 100% 2
: in ax + by = cosy
Page 38
Answer.
The given relationship is ax + by 2 = cosy
Differentiating this relationship with respect to x, we obtain
( )
d d d
dx
(ax) + dx by 2 = dx (cosy)
⇒ a + b (y ) = (cosy)
d d
2
dx dx
( )
d dy d dy
Using chain rule, we obtain dx y 2 = 2y dx and dx (cosy) = − siny dx
From (1) and (2), we obtain
dy dy
a + b × 2y dx = − siny dx
dy
⇒ (2by + siny) dx = − a
dy −a
∴ dx = 2by + sin y
Page : 166 , Block Name : Exercise 5.3
dy
Q4 Find dx : in xy + y 2 = tanx + y
Answer.
The given relationship is xy + y 2 = tanx + y
Differentiating this relationship with respect to x, we obtain
( )
d d
dx
xy + y 2 = dx (tanx + y)
( )
d d d dy
⇒ dx (xy) + dx y 2 = dx (tanx) + dx
[ d
⇒ y ⋅ dx (x) + x ⋅ dx
dy
] dy dy
+ 2y dx = sec 2x + dx
dy dy dy
⇒ y ⋅ 1 + x ⋅ dx + 2y dx = sec 2x + dx
dy
⇒ (x + 2y − 1) dx = sec 2x − y
dy sec 2 x − y
∴ dx = ( x + 2y − 1 )
Page : 166 , Block Name : Exercise 5.3
dy
Q5 Find dx : in x 2 + xy + y 2 = 100
Answer.
The given relationship is ax + by 2 = cosy
Differentiating this relationship with respect to x, we obtain
( )
d d d
dx
(ax) + dx by 2 = dx (cosy)
⇒ a + b (y ) = (cosy)
d d
2
dx dx
( )
d dy d dy
Using chain rule, we obtain dx y 2 = 2y dx and dx (cosy) = − siny dx
From (1) and (2), we obtain
dy dy
a + b × 2y dx = − siny dx
dy
⇒ (2by + siny) dx = − a
dy −a
∴ dx = 2by + sin y
Page : 166 , Block Name : Exercise 5.3
dy
Q4 Find dx : in xy + y 2 = tanx + y
Answer.
The given relationship is xy + y 2 = tanx + y
Differentiating this relationship with respect to x, we obtain
( )
d d
dx
xy + y 2 = dx (tanx + y)
( )
d d d dy
⇒ dx (xy) + dx y 2 = dx (tanx) + dx
[ d
⇒ y ⋅ dx (x) + x ⋅ dx
dy
] dy dy
+ 2y dx = sec 2x + dx
dy dy dy
⇒ y ⋅ 1 + x ⋅ dx + 2y dx = sec 2x + dx
dy
⇒ (x + 2y − 1) dx = sec 2x − y
dy sec 2 x − y
∴ dx = ( x + 2y − 1 )
Page : 166 , Block Name : Exercise 5.3
dy
Q5 Find dx : in x 2 + xy + y 2 = 100
Answer.
Page 39
2 2
The given relationship is x + xy + y = 100
Differentiating this relationship with respect to x, we obtain
( )
d d
dx
x 2 + xy + y 2 = dx (100)
( ) ( )
d d d
⇒ dx x 2 + dx (xy) + dx y 2 = 0
[
⇒ 2x + y ⋅ dx (x) + x ⋅ dx
d dy
] dy
+ 2y dx = 0
dy dy
⇒ 2x + y ⋅ 1 + x ⋅ dx + 2y dx = 0
dy
⇒ 2x + y + (x + 2y) dx = 0
dy 2x + y
∴ dx = − x + 2y
Page : 166 , Block Name : Exercise 5.3
dy
Q6 Find dx : in x 3 + x 2y + xy 2 + y 3 = 81
Answer.
The given relationship is x 3 + x 2y + xy 2 + y 3 = 81
Differentiating this relationship with respect to x , we obtain
( )
d d
dx
x 3 + x 2y + xy 2 + y 3 = dx (81)
( )
( ) ( ) ( )
d d d d
⇒ dx x 3 + dx x 2y + dx xy 2 + dx y 3 = 0
⇒ 3x + [y (x ) + x ] + [y (x) + x (y ) ] + 3y
d dy d d dy
2 2 2 2 2 2
dx dx dx dx dx = 0
[
⇒ 3x 2 + y ⋅ 2x + x 2 dx
dy
] [
+ y 2 ⋅ 1 + x ⋅ 2y ⋅ dx
dy
] dy
+ 3y 2 dx = 0
( ) ( )
dy
⇒ x 2 + 2xy + 3y 2 dx + 3x 2 + 2xy + y 2 = 0
dy (
− 3x 2 + 2xy + y 2 )
∴ dx =
( x + 2xy + 3y )
2 2
Page : 166 , Block Name : Exercise 5.3
dy
Q7 Find dx : in sin 2y + cosxy = π
Answer.
The given relationship is x + xy + y = 100
Differentiating this relationship with respect to x, we obtain
( )
d d
dx
x 2 + xy + y 2 = dx (100)
( ) ( )
d d d
⇒ dx x 2 + dx (xy) + dx y 2 = 0
[
⇒ 2x + y ⋅ dx (x) + x ⋅ dx
d dy
] dy
+ 2y dx = 0
dy dy
⇒ 2x + y ⋅ 1 + x ⋅ dx + 2y dx = 0
dy
⇒ 2x + y + (x + 2y) dx = 0
dy 2x + y
∴ dx = − x + 2y
Page : 166 , Block Name : Exercise 5.3
dy
Q6 Find dx : in x 3 + x 2y + xy 2 + y 3 = 81
Answer.
The given relationship is x 3 + x 2y + xy 2 + y 3 = 81
Differentiating this relationship with respect to x , we obtain
( )
d d
dx
x 3 + x 2y + xy 2 + y 3 = dx (81)
( )
( ) ( ) ( )
d d d d
⇒ dx x 3 + dx x 2y + dx xy 2 + dx y 3 = 0
⇒ 3x + [y (x ) + x ] + [y (x) + x (y ) ] + 3y
d dy d d dy
2 2 2 2 2 2
dx dx dx dx dx = 0
[
⇒ 3x 2 + y ⋅ 2x + x 2 dx
dy
] [
+ y 2 ⋅ 1 + x ⋅ 2y ⋅ dx
dy
] dy
+ 3y 2 dx = 0
( ) ( )
dy
⇒ x 2 + 2xy + 3y 2 dx + 3x 2 + 2xy + y 2 = 0
dy (
− 3x 2 + 2xy + y 2 )
∴ dx =
( x + 2xy + 3y )
2 2
Page : 166 , Block Name : Exercise 5.3
dy
Q7 Find dx : in sin 2y + cosxy = π
Answer.
Page 40
The given relationship is sin 2y + cosxy = π
Differentiating this relationship with respect to x, we obtain
(sin y + cosxy ) = (π)
d d
2
dx dx
⇒ (sin y ) + (cosxy) = 0
d d
2
dx dx
Using chain rule, we obtain
(sin y ) = 2siny (siny) = 2sinycosy
d d dy
2
dx dx dx
d
dx
d
(cosxy) = − sinxy dx (xy) = − sinxy y dx (x) + x dx [ d dy
]
= − sinxy y.1 + x dx
[ dy
] = − ysinxy − xsinxy dx
dy
From above results we get
dy dy
2sinycosy dx − ysinxy − xsinxy dx = 0
dy
⇒ (2sinycosy − xsinxy) dx = ysinxy
dy
⇒ (sin2y − xsinxy) dx = ysinxy
dy ysin xy
∴ dx = sin 2y − xsin xy
Page : 166 , Block Name : Exercise 5.3
dy
Q8 Find dx : in sin 2x + cos 2y = 1
Answer.
The given relationship is sin 2x + cos 2y = 1
Differentiating this relationship with respect to x, we obtain
( )
d d
dx
sin 2x + cos 2y = dx (1)
( ) ( )
d d
⇒ dx sin 2x + dx cos 2y = 0
d d
⇒ 2sinx ⋅ dx (sinx) + 2cosy ⋅ dx (cosy) = 0
dy
⇒ 2sinxcosx + 2cosy( − siny) ⋅ dx = 0
dy
⇒ sin2xsin2y dx = 0
dy sin 2x
∴ dx = sin 2y
Differentiating this relationship with respect to x, we obtain
(sin y + cosxy ) = (π)
d d
2
dx dx
⇒ (sin y ) + (cosxy) = 0
d d
2
dx dx
Using chain rule, we obtain
(sin y ) = 2siny (siny) = 2sinycosy
d d dy
2
dx dx dx
d
dx
d
(cosxy) = − sinxy dx (xy) = − sinxy y dx (x) + x dx [ d dy
]
= − sinxy y.1 + x dx
[ dy
] = − ysinxy − xsinxy dx
dy
From above results we get
dy dy
2sinycosy dx − ysinxy − xsinxy dx = 0
dy
⇒ (2sinycosy − xsinxy) dx = ysinxy
dy
⇒ (sin2y − xsinxy) dx = ysinxy
dy ysin xy
∴ dx = sin 2y − xsin xy
Page : 166 , Block Name : Exercise 5.3
dy
Q8 Find dx : in sin 2x + cos 2y = 1
Answer.
The given relationship is sin 2x + cos 2y = 1
Differentiating this relationship with respect to x, we obtain
( )
d d
dx
sin 2x + cos 2y = dx (1)
( ) ( )
d d
⇒ dx sin 2x + dx cos 2y = 0
d d
⇒ 2sinx ⋅ dx (sinx) + 2cosy ⋅ dx (cosy) = 0
dy
⇒ 2sinxcosx + 2cosy( − siny) ⋅ dx = 0
dy
⇒ sin2xsin2y dx = 0
dy sin 2x
∴ dx = sin 2y
Page 41
Page : 166 , Block Name : Exercise 5.3
dy
Q9 Find dx : in y = sin − 1
( )
2x
1 + x2
Answer. The given relationship is
y = sin − 1
( ) 2x
1 + x2
y = sin − 1
( ) 2x
1 + x2
2x
⇒ siny =
1 + x2
Differentiating this relationship with respect to x, we obtain
d
dx
(siny) = dx
d
( ) 2x
1 + x2
dy
⇒ cosy dx = dx
2x
( )
d 2x
1 + x2
, is of the form of v
1 + x2
Therefore, by quotient rule, we obtain
( ) ( 1 + x ) ⋅ ( 2x ) − 2x ⋅ ( 1 + x )
d d
2 2
d 2x dx dx
dx
=
1 + x2
(1 + x ) 2 2
( 1 + x ) ⋅ 2 − 2x ⋅ [ 0 + 2x ] 2 + 2x − 4x 2 ( 1 − x )
2
2 2
2
= = =
(1+x ) 2 2
(1+x ) (1+x ) 2 2 2 2
√ ( ) √ (1 + x ) − 4x
2 2 2
2x 2
⇒ cosy = √1 − sin 2y = 1−
1 + x2
=
(1 + x )
2 2
√
(1 − x ) 1 − x 2 2
2
= = 2
(1 + x ) 1 + x 2 2
dy
Q9 Find dx : in y = sin − 1
( )
2x
1 + x2
Answer. The given relationship is
y = sin − 1
( ) 2x
1 + x2
y = sin − 1
( ) 2x
1 + x2
2x
⇒ siny =
1 + x2
Differentiating this relationship with respect to x, we obtain
d
dx
(siny) = dx
d
( ) 2x
1 + x2
dy
⇒ cosy dx = dx
2x
( )
d 2x
1 + x2
, is of the form of v
1 + x2
Therefore, by quotient rule, we obtain
( ) ( 1 + x ) ⋅ ( 2x ) − 2x ⋅ ( 1 + x )
d d
2 2
d 2x dx dx
dx
=
1 + x2
(1 + x ) 2 2
( 1 + x ) ⋅ 2 − 2x ⋅ [ 0 + 2x ] 2 + 2x − 4x 2 ( 1 − x )
2
2 2
2
= = =
(1+x ) 2 2
(1+x ) (1+x ) 2 2 2 2
√ ( ) √ (1 + x ) − 4x
2 2 2
2x 2
⇒ cosy = √1 − sin 2y = 1−
1 + x2
=
(1 + x )
2 2
√
(1 − x ) 1 − x 2 2
2
= = 2
(1 + x ) 1 + x 2 2
Page 42
1 − x2 dy (
2 1 − x2 )
2 × dx =
1+x
(1+x ) 2 2
dy 2
⇒ dx =
1 + x2
Page : 166 , Block Name : Exercise 5.3
Q10 Find dx
dy
: in y = tan − 1
( )
3x − x 3
1 − 3x 2
, −
1
√3
<x<
1
√3
Answer. The given relationship is
y = tan − 1
( )3x − x 3
1 − 3x 2
y = tan − 1
( ) 3x − x 3
1 − 3x 2
3x − x 3
⇒ tany =
1 − 3x 2
y y
3tan 3 − tan 3 3
It is know that ntany = y
1 − 3tan 2 3
on comparing above two results
d
dx
d
(x) = dx tan 3( ) y
⇒ 1 = sec 2 3 ⋅ dx 3
y d
() y
y 1 dy
⇒ 1 = sec 2 3 ⋅ 3 ⋅ dx
dy 3 3
⇒ dx = y = y
sec 2 3 1 + tan 2 3
dy 3
∴ dx =
1 + x2
Page : 166 , Block Name : Exercise 5.3
dy
Q11 Find dx : in y = cos − 1
( )
1 − x2
1 + x2
,0 < x < 1
2 1 − x2 )
2 × dx =
1+x
(1+x ) 2 2
dy 2
⇒ dx =
1 + x2
Page : 166 , Block Name : Exercise 5.3
Q10 Find dx
dy
: in y = tan − 1
( )
3x − x 3
1 − 3x 2
, −
1
√3
<x<
1
√3
Answer. The given relationship is
y = tan − 1
( )3x − x 3
1 − 3x 2
y = tan − 1
( ) 3x − x 3
1 − 3x 2
3x − x 3
⇒ tany =
1 − 3x 2
y y
3tan 3 − tan 3 3
It is know that ntany = y
1 − 3tan 2 3
on comparing above two results
d
dx
d
(x) = dx tan 3( ) y
⇒ 1 = sec 2 3 ⋅ dx 3
y d
() y
y 1 dy
⇒ 1 = sec 2 3 ⋅ 3 ⋅ dx
dy 3 3
⇒ dx = y = y
sec 2 3 1 + tan 2 3
dy 3
∴ dx =
1 + x2
Page : 166 , Block Name : Exercise 5.3
dy
Q11 Find dx : in y = cos − 1
( )
1 − x2
1 + x2
,0 < x < 1
Page 43
Answer. The given relationship is,
y = cos − 1
( ) 1 − x2
1 + x2
1 − x2
⇒ cosy =
1 + x2
y
1 − tan 2 2 1 − x2
⇒ y =
1 + tan 2 2 1 + x2
On comparing L.H.S. and R.H.S. of the above relationship, we obtain
y
tan 2 = x
Differentiating this relationship with respect to x, we obtain
y
sec 2 2 ⋅ dx 2
d
() y d
= dx (x)
y 1 dy
⇒ sec 2 2 × 2 dx = 1
dy 2
⇒ dx = y
sec 2 2
dy 2
⇒ dx = y
1 + tan 2 2
dy 1
⇒ dx =
1 + x2
Page : 166 , Block Name : Exercise 5.3
dy
Q12 Find dx : in y = sin −1
( )
1 − x2
1 + x2
,0 < x < 1
Answer. The given relationship is
y = sin − 1
( )
1 − x2
1 + x2
y = sin −1
( )
1 − x2
1 + x2
1 − x2
⇒ siny =
1 + x2
Differentiating this relationship with respect to x, we obtain
y = cos − 1
( ) 1 − x2
1 + x2
1 − x2
⇒ cosy =
1 + x2
y
1 − tan 2 2 1 − x2
⇒ y =
1 + tan 2 2 1 + x2
On comparing L.H.S. and R.H.S. of the above relationship, we obtain
y
tan 2 = x
Differentiating this relationship with respect to x, we obtain
y
sec 2 2 ⋅ dx 2
d
() y d
= dx (x)
y 1 dy
⇒ sec 2 2 × 2 dx = 1
dy 2
⇒ dx = y
sec 2 2
dy 2
⇒ dx = y
1 + tan 2 2
dy 1
⇒ dx =
1 + x2
Page : 166 , Block Name : Exercise 5.3
dy
Q12 Find dx : in y = sin −1
( )
1 − x2
1 + x2
,0 < x < 1
Answer. The given relationship is
y = sin − 1
( )
1 − x2
1 + x2
y = sin −1
( )
1 − x2
1 + x2
1 − x2
⇒ siny =
1 + x2
Differentiating this relationship with respect to x, we obtain
Page 44
d
dx
(siny) = dx
d
( ) 1 − x2
1 + x2
Using chain rule, we obtain
d dy
dx
(siny) = cosy ⋅ dx
cosy = √1 − sin y = 2
√ ( )
1−
1 − x2 2
1 + x2
√
(1 + x ) − (1 − x ) 2 2 2 2
2
√
64x 2x
= = = 2
(1 + x ) (1 + x ) 1 + x 2 2 2 2
d 2x dy
∴ dx (siny) =
1 + x 2 dx
( ) (1 + x ) ⋅ (1 − x ) − ( 1 − x ) ⋅ ( 1 + x )
2 2 ′ 2 2 ′
d 1 − x2
dx 1 + x
= 2
(1+x ) 2 2
( 1 + x ) ( − 2x ) − ( 1 − x ) ⋅ ( 2x )
2 2
=
(1+x ) 2 2
( 1 + x ) ( − 2x ) − ( 1 − x ) ⋅ ( 2x )
2 2
=
(1+x ) 2 2
− 2x − 2x 3 − 2x + 2x 3
= From above results
( 1 + x2 ) 2
− 4x
=
(1+x ) 2 2
2x dy − 4x
2 dx =
1+x
(1+x ) 2 2
dy −2
⇒ dx =
1 + x2
Alternate method
dx
(siny) = dx
d
( ) 1 − x2
1 + x2
Using chain rule, we obtain
d dy
dx
(siny) = cosy ⋅ dx
cosy = √1 − sin y = 2
√ ( )
1−
1 − x2 2
1 + x2
√
(1 + x ) − (1 − x ) 2 2 2 2
2
√
64x 2x
= = = 2
(1 + x ) (1 + x ) 1 + x 2 2 2 2
d 2x dy
∴ dx (siny) =
1 + x 2 dx
( ) (1 + x ) ⋅ (1 − x ) − ( 1 − x ) ⋅ ( 1 + x )
2 2 ′ 2 2 ′
d 1 − x2
dx 1 + x
= 2
(1+x ) 2 2
( 1 + x ) ( − 2x ) − ( 1 − x ) ⋅ ( 2x )
2 2
=
(1+x ) 2 2
( 1 + x ) ( − 2x ) − ( 1 − x ) ⋅ ( 2x )
2 2
=
(1+x ) 2 2
− 2x − 2x 3 − 2x + 2x 3
= From above results
( 1 + x2 ) 2
− 4x
=
(1+x ) 2 2
2x dy − 4x
2 dx =
1+x
(1+x ) 2 2
dy −2
⇒ dx =
1 + x2
Alternate method
Page 45
y = sin − 1
( ) 1 − x2
1 + x2
1 − x2
siny =
1 + x2
( )
⇒ 1 + x 2 siny = 1 − x 2
⇒ (1 + x )siny = 1 − x
2 2
⇒ (1 + siny)x 2 = 1 − x 2
1 − sin y
⇒ x 2 = 1 + sin y
( 2
y
cos 2 − sin 2
y
) 2
⇒x =
( y
cos 2 + sin 2
y
) 2
y y
cos 2 − sin 2
⇒x= y y
cos 2 + sin 2
y
1 − tan 2
⇒x= y
1 + tan 2
y
1 − tan 2
⇒x= y
1 + tan 2
⇒ x = tan 4 − 2 ( ) π y
Differentiating this relationship with respect to x, we obtain
d
dx
d
[ ( )]
(x) = dx ⋅ tan 4 − 2
π y
( ) ( )
⇒ 1 = sec 2 4 − 2
π y d
⋅ dx 4 − 2
π y
[ ( )] ( )
⇒ 1 = 1 = 1 + tan 2 4 − 2
π y
⋅
1 dy
− 2 dx
( )( )
1 dy
⇒ 1 = 1 + x2 − 2 dx
dy −2
⇒ dx =
1 + x2
( ) 1 − x2
1 + x2
1 − x2
siny =
1 + x2
( )
⇒ 1 + x 2 siny = 1 − x 2
⇒ (1 + x )siny = 1 − x
2 2
⇒ (1 + siny)x 2 = 1 − x 2
1 − sin y
⇒ x 2 = 1 + sin y
( 2
y
cos 2 − sin 2
y
) 2
⇒x =
( y
cos 2 + sin 2
y
) 2
y y
cos 2 − sin 2
⇒x= y y
cos 2 + sin 2
y
1 − tan 2
⇒x= y
1 + tan 2
y
1 − tan 2
⇒x= y
1 + tan 2
⇒ x = tan 4 − 2 ( ) π y
Differentiating this relationship with respect to x, we obtain
d
dx
d
[ ( )]
(x) = dx ⋅ tan 4 − 2
π y
( ) ( )
⇒ 1 = sec 2 4 − 2
π y d
⋅ dx 4 − 2
π y
[ ( )] ( )
⇒ 1 = 1 = 1 + tan 2 4 − 2
π y
⋅
1 dy
− 2 dx
( )( )
1 dy
⇒ 1 = 1 + x2 − 2 dx
dy −2
⇒ dx =
1 + x2
Page 46
Page : 166 , Block Name : Exercise 5.3
dy
Q13 Find dx : in y = cos − 1
( ) 2x
1 + x2
, −1<x<1
Answer. The given relationship is
y = cos − 1
( ) 2x
1 + x2
y = cos − 1
( ) 2x
1 + x2
2x
⇒ cosy =
1 + x2
Differentiating this relationship with respect to x, we obtain
d
dx
(cosy) = dx ⋅
d
( ) 2x
1 + x2
( 1 + x ) ⋅ ( 2x ) − 2x ⋅ ( 1 + x )
d d
2 2
dy dx dx
⇒ − siny ⋅ dx =
(1+x ) 2 2
dy ( 1 + x ) × 2 − 2x ⋅ 2x 2
⇒ − √1 − cos y dx =
2
(1+x ) 2 2
[√ ( ) ] [ ( ) ] ( )
2 2 1 − x2
2x dy
⇒ 1− dx
= −
1 + x2 2
1 + x2
( ) ( )
√( ) ( )
2
1 − x2 dy − 2 1 − x2
⇒ =
2 dx 2
1 + x2 1 + x2
1 − x2 dy (
− 2 1 − x2 )
⇒ 2 ⋅ dx =
1+x
(1+x ) 2 2
dy −2
⇒ dx =
1 + x2
Page : 166 , Block Name : Exercise 5.3
Processing
Q14
dy
Find dxmath: y = sin − 1 2x 1 − x 2 , −
: in 100% (√ ) √ 1
2
<x<
1
√2
dy
Q13 Find dx : in y = cos − 1
( ) 2x
1 + x2
, −1<x<1
Answer. The given relationship is
y = cos − 1
( ) 2x
1 + x2
y = cos − 1
( ) 2x
1 + x2
2x
⇒ cosy =
1 + x2
Differentiating this relationship with respect to x, we obtain
d
dx
(cosy) = dx ⋅
d
( ) 2x
1 + x2
( 1 + x ) ⋅ ( 2x ) − 2x ⋅ ( 1 + x )
d d
2 2
dy dx dx
⇒ − siny ⋅ dx =
(1+x ) 2 2
dy ( 1 + x ) × 2 − 2x ⋅ 2x 2
⇒ − √1 − cos y dx =
2
(1+x ) 2 2
[√ ( ) ] [ ( ) ] ( )
2 2 1 − x2
2x dy
⇒ 1− dx
= −
1 + x2 2
1 + x2
( ) ( )
√( ) ( )
2
1 − x2 dy − 2 1 − x2
⇒ =
2 dx 2
1 + x2 1 + x2
1 − x2 dy (
− 2 1 − x2 )
⇒ 2 ⋅ dx =
1+x
(1+x ) 2 2
dy −2
⇒ dx =
1 + x2
Page : 166 , Block Name : Exercise 5.3
Processing
Q14
dy
Find dxmath: y = sin − 1 2x 1 − x 2 , −
: in 100% (√ ) √ 1
2
<x<
1
√2
Page 47
Answer.
(√
y = sin − 1 2x 1 − x 2 )
(√
y = sin − 1 2x 1 − x 2 )
√
⇒ siny = 2x 1 − x 2
Differentiating this relationship with respect to x, we obtain
dy
[ (√ ) √
cosy dx = 2 x dx
d
1 − x2 + 1 − x 2 dx
dx
]
⇒ √
dy
[
1 − sin 2y dx = 2 2 ⋅
x − 2x
√ 1 − x2
+ √1 − x 2 ]
⇒ √1 − (2x√1 − x ) 2 2
dy
dx = 2 [ − x2 + 1 − x2
√1 − x 2 ]
⇒ √1 − 4x 2 (1 − x 2 ) dx = 2
dy 1 − 2x
[ ] √1 − x 2
2
dy 2
⇒ dx =
√1 − x 2
Page : 166 , Block Name : Exercise 5.3
dy
Q15 Find dx : in y = sec − 1
( ) 1
2x 2 − 1
,0 < x <
√2
1
Answer. The given relationship is
y = sec − 1
( ) 1
2x 2 − 1
y = sec − 1
( ) 1
2x 2 − 1
(√
y = sin − 1 2x 1 − x 2 )
(√
y = sin − 1 2x 1 − x 2 )
√
⇒ siny = 2x 1 − x 2
Differentiating this relationship with respect to x, we obtain
dy
[ (√ ) √
cosy dx = 2 x dx
d
1 − x2 + 1 − x 2 dx
dx
]
⇒ √
dy
[
1 − sin 2y dx = 2 2 ⋅
x − 2x
√ 1 − x2
+ √1 − x 2 ]
⇒ √1 − (2x√1 − x ) 2 2
dy
dx = 2 [ − x2 + 1 − x2
√1 − x 2 ]
⇒ √1 − 4x 2 (1 − x 2 ) dx = 2
dy 1 − 2x
[ ] √1 − x 2
2
dy 2
⇒ dx =
√1 − x 2
Page : 166 , Block Name : Exercise 5.3
dy
Q15 Find dx : in y = sec − 1
( ) 1
2x 2 − 1
,0 < x <
√2
1
Answer. The given relationship is
y = sec − 1
( ) 1
2x 2 − 1
y = sec − 1
( ) 1
2x 2 − 1
Page 48
1
⇒ secy =
2x 2 − 1
⇒ cosy = 2x 2 − 1
⇒ 2x 2 = 1 + cosy
y
⇒ 2x 2 = 2cos 2 2
y
⇒ x = cos 2
Differentiating this relationship with respect to x, we obtain
d
dx
d
(x) = dx cos 2 ( ) y
⇒ 1 = − sin 2 ⋅ dx 2
y d
() y
−1 1 dy
⇒ y = 2 dx
sin 2
dy −2 −2
⇒ dx = y = y
sin 2
√ 1 − cos 2 2
dy −2
⇒ dx =
√1 − x 2
Page : 166 , Block Name : Exercise 5.3
ex
Q1 Differentiate the following w.r.t. x: sin x
ex
Answer. Lety = sin x
By using the quotient rule, we obtain
( )
d d
dy sin x dx e x − e x dx ( sin x )
dx
=
sin 2 x
( )
sinx ⋅ e x − e x ⋅ (cosx)
=
sin 2x
e x(sinx − cosx)
= , x ≠ nπ, n ∈ Z
sin 2x
Page : 174 , Block Name : Exercise 5.4
Q2 Differentiate the following w.r.t. x:
−1
e sin x
⇒ secy =
2x 2 − 1
⇒ cosy = 2x 2 − 1
⇒ 2x 2 = 1 + cosy
y
⇒ 2x 2 = 2cos 2 2
y
⇒ x = cos 2
Differentiating this relationship with respect to x, we obtain
d
dx
d
(x) = dx cos 2 ( ) y
⇒ 1 = − sin 2 ⋅ dx 2
y d
() y
−1 1 dy
⇒ y = 2 dx
sin 2
dy −2 −2
⇒ dx = y = y
sin 2
√ 1 − cos 2 2
dy −2
⇒ dx =
√1 − x 2
Page : 166 , Block Name : Exercise 5.3
ex
Q1 Differentiate the following w.r.t. x: sin x
ex
Answer. Lety = sin x
By using the quotient rule, we obtain
( )
d d
dy sin x dx e x − e x dx ( sin x )
dx
=
sin 2 x
( )
sinx ⋅ e x − e x ⋅ (cosx)
=
sin 2x
e x(sinx − cosx)
= , x ≠ nπ, n ∈ Z
sin 2x
Page : 174 , Block Name : Exercise 5.4
Q2 Differentiate the following w.r.t. x:
−1
e sin x
Page 49
−1
Answer. Let y = e sin x
By using the chain rule, we obtain
dy
dx = dx e
d
sin
( −1 x
)
(sin x )
dy −1 d
⇒ dx = e sin x⋅
dx
−1
−1 x 1
= e sin ⋅
√1 − x 2
−1 x
e sin
=
√1 − x 2
−1 x
dy e sin
∴ = , x ∈ ( − 1, 1)
dx 1 − x2
√
Page : 174 , Block Name : Exercise 5.4
3
Q3 Differentiate the following w.r.t. x: e x
Answer.
3
Let y = e x
By using the chain rule, we obtain
dy
dx
d
= dx e x ( ) 3 3 d
( ) 3
= e x ⋅ dx x 3 = e x ⋅ 3x 2 = 3x 2e x
3
Page : 174 , Block Name : Exercise 5.4
Q4 Differentiate the following w.r.t. x: sin tan − 1e − x ( )
Answer.
(
Let y = sin tan − 1e − x )
By using the chain rule, we obtain
[ ( )]
dy d
dx
= dx sin tan − 1e − x
Answer. Let y = e sin x
By using the chain rule, we obtain
dy
dx = dx e
d
sin
( −1 x
)
(sin x )
dy −1 d
⇒ dx = e sin x⋅
dx
−1
−1 x 1
= e sin ⋅
√1 − x 2
−1 x
e sin
=
√1 − x 2
−1 x
dy e sin
∴ = , x ∈ ( − 1, 1)
dx 1 − x2
√
Page : 174 , Block Name : Exercise 5.4
3
Q3 Differentiate the following w.r.t. x: e x
Answer.
3
Let y = e x
By using the chain rule, we obtain
dy
dx
d
= dx e x ( ) 3 3 d
( ) 3
= e x ⋅ dx x 3 = e x ⋅ 3x 2 = 3x 2e x
3
Page : 174 , Block Name : Exercise 5.4
Q4 Differentiate the following w.r.t. x: sin tan − 1e − x ( )
Answer.
(
Let y = sin tan − 1e − x )
By using the chain rule, we obtain
[ ( )]
dy d
dx
= dx sin tan − 1e − x
Page 50
1 d
(
= cos tan − 1e − x ⋅ ) ⋅
dx (
e )
−x
1 + e −x ( ) 2
(
cos tan − 1e − x ) d
= ⋅ e −x ⋅ ( − x)
1 + e − 2x dx
(
e − xcos tan − 1e − x )
= × ( − 1)
1 + e − 2x
(
− e − xcos tan − 1e − x )
=
1 + e − 2x
Page : 174 , Block Name : Exercise 5.4
Q5 Differentiate the following w.r.t. x:log cose x ( )
Answer. y = log cose x ( )
By using the chain rule, we obtain
dy d
dx
=
dx [ (
log cose x )]
1 d
= ⋅
cose x dx
cose x ) (
1 d
⋅ ( − sine ) ⋅
dx ( )
x x
= x
e
cose
− sine x
= ⋅ ex
cose x
π
= − e xtane x, e x ≠ (2n + 1) 2 , n ∈ N
Page : 174 , Block Name : Exercise 5.4
2 5
Q6 Differentiate the following w.r.t. X: e x + e x + … + e x
Answer.
d
dx ( 2
ex + ex + … + ex
3
)
d
( )
= dx e x + dx e x
d
( ) ( ) ( ) ( )
2
+ dx e x
d 3 d
+ dx e x
4 d
+ dx e x
3
(
= cos tan − 1e − x ⋅ ) ⋅
dx (
e )
−x
1 + e −x ( ) 2
(
cos tan − 1e − x ) d
= ⋅ e −x ⋅ ( − x)
1 + e − 2x dx
(
e − xcos tan − 1e − x )
= × ( − 1)
1 + e − 2x
(
− e − xcos tan − 1e − x )
=
1 + e − 2x
Page : 174 , Block Name : Exercise 5.4
Q5 Differentiate the following w.r.t. x:log cose x ( )
Answer. y = log cose x ( )
By using the chain rule, we obtain
dy d
dx
=
dx [ (
log cose x )]
1 d
= ⋅
cose x dx
cose x ) (
1 d
⋅ ( − sine ) ⋅
dx ( )
x x
= x
e
cose
− sine x
= ⋅ ex
cose x
π
= − e xtane x, e x ≠ (2n + 1) 2 , n ∈ N
Page : 174 , Block Name : Exercise 5.4
2 5
Q6 Differentiate the following w.r.t. X: e x + e x + … + e x
Answer.
d
dx ( 2
ex + ex + … + ex
3
)
d
( )
= dx e x + dx e x
d
( ) ( ) ( ) ( )
2
+ dx e x
d 3 d
+ dx e x
4 d
+ dx e x
3
Page 51
[ 2
= e x + e x × dx x 2
d
( ) ] + [e ⋅ (x ) ] + [e ⋅ (x ) ] + [e ⋅ (x ) ]
x3
d
dx
3 xx
d
dx
4 x3
d
dx
5
( 2
) ( 3
) (
= e x + e x × 2x + e x × 3x 2 + e x × 4x 3 + e x × 5x 4
4
) ( 3
)
2 3 4 3
= e x + 2xe x + 3x 2e x + 4x 3e x + 5x 4e x
Page : 174 , Block Name : Exercise 5.4
Q7 Differentiate the following w.r.t. X: √ e √x , x > 0
Answer. y = √e v
y 2 = e √x
By differentiating this relationship with respect to x, we obtain
y 2 = e √x
dy d
⇒ 2y dx = e √x dx (√x)
dy 1 1
⇒ 2y dx = e √x 2 ⋅
√x
dy e √x
⇒ dx =
4y√x
dy e √x
⇒ dx =
√
4 e √x √ x
dy e √x
⇒ dx = ,x > 0
√
4 xe √x
Page : 174 , Block Name : Exercise 5.4
Q8 Differentiate the following w.r.t. X: log(logx), x > 1
Answer.
Let y = log(logx)
By using the chaln rule, we obtain
dy d
= [log(logx)]
dx dx
1 d
= ⋅ (logx)
logx dx
1 1
= ⋅
logx x
1
=Processing
, x math:
> 1 100%
xlogx
= e x + e x × dx x 2
d
( ) ] + [e ⋅ (x ) ] + [e ⋅ (x ) ] + [e ⋅ (x ) ]
x3
d
dx
3 xx
d
dx
4 x3
d
dx
5
( 2
) ( 3
) (
= e x + e x × 2x + e x × 3x 2 + e x × 4x 3 + e x × 5x 4
4
) ( 3
)
2 3 4 3
= e x + 2xe x + 3x 2e x + 4x 3e x + 5x 4e x
Page : 174 , Block Name : Exercise 5.4
Q7 Differentiate the following w.r.t. X: √ e √x , x > 0
Answer. y = √e v
y 2 = e √x
By differentiating this relationship with respect to x, we obtain
y 2 = e √x
dy d
⇒ 2y dx = e √x dx (√x)
dy 1 1
⇒ 2y dx = e √x 2 ⋅
√x
dy e √x
⇒ dx =
4y√x
dy e √x
⇒ dx =
√
4 e √x √ x
dy e √x
⇒ dx = ,x > 0
√
4 xe √x
Page : 174 , Block Name : Exercise 5.4
Q8 Differentiate the following w.r.t. X: log(logx), x > 1
Answer.
Let y = log(logx)
By using the chaln rule, we obtain
dy d
= [log(logx)]
dx dx
1 d
= ⋅ (logx)
logx dx
1 1
= ⋅
logx x
1
=Processing
, x math:
> 1 100%
xlogx
Page 52
Page : 174 , Block Name : Exercise 5.4
cos x
Q9 Differentiate the following w.r.t. X: log x , x > 0
Answer.
cos x
y = log x
By using the quotient rule, we obtain
d d
dy dx ( cos x ) × log x − cos x × dx ( log x )
dx
=
( log x ) 2
1
− sin xlog x − cos x × x
=
( log x ) 2
=
−
[ xlog x − cos x × x
1
( log x ) 2
Page : 174 , Block Name : Exercise 5.4
Q10 Differentiate the following w.r.t. X: cos logx + e x , x > 0 ( )
Answer. y = cos log x + e ( x
)
dy d
dx (
= − sin logx + e x ⋅
dx
logx + e x ) ( )
) [ dx (logx) + dx (e ) ]
d d
(
= − sin logx + e x ⋅ x
) (x + e )
1
(
= − sin logx + e x ⋅ x
= −
( ) (
1
x
+ e x sin logx + e x , x > 0 )
Page : 174 , Block Name : Exercise 5.4
Q1 Differentiate the function with respect to x.
cosx ⋅ cos2x ⋅ cos3x
Answer. y = cosx ⋅ cos2x ⋅ cos3x
Taking logarithm on both the sides, we obtain
cos x
Q9 Differentiate the following w.r.t. X: log x , x > 0
Answer.
cos x
y = log x
By using the quotient rule, we obtain
d d
dy dx ( cos x ) × log x − cos x × dx ( log x )
dx
=
( log x ) 2
1
− sin xlog x − cos x × x
=
( log x ) 2
=
−
[ xlog x − cos x × x
1
( log x ) 2
Page : 174 , Block Name : Exercise 5.4
Q10 Differentiate the following w.r.t. X: cos logx + e x , x > 0 ( )
Answer. y = cos log x + e ( x
)
dy d
dx (
= − sin logx + e x ⋅
dx
logx + e x ) ( )
) [ dx (logx) + dx (e ) ]
d d
(
= − sin logx + e x ⋅ x
) (x + e )
1
(
= − sin logx + e x ⋅ x
= −
( ) (
1
x
+ e x sin logx + e x , x > 0 )
Page : 174 , Block Name : Exercise 5.4
Q1 Differentiate the function with respect to x.
cosx ⋅ cos2x ⋅ cos3x
Answer. y = cosx ⋅ cos2x ⋅ cos3x
Taking logarithm on both the sides, we obtain
Page 53
logy = log(cosx ⋅ cos2x ⋅ cos3x)
⇒ logy = log(cosx) + log(cos2x) + log(cos3x)
Differentiating both sides with respect to x, we obtain
1 dy 1 d 1 d 1 d
y dx
= cos x ⋅ dx (cosx) + cos 2x ⋅ dx (cos2x) + cos 3x ⋅ dx (cos3x)
dy
[ sin x sin 2x
⇒ dx = y − cos x − cos 2x ⋅ dx (2x) − cos 3x ⋅ dx (3x)
dy
d sin 3x d
]
∴ dx = − cosx ⋅ cos2x ⋅ cos3x[tanx + 2tan2x + 3tan3x]
Page : 178 , Block Name : Exercise 5.5
Q2 Differentiate the function with respect to x.
(x−1) (x−2)
√ (x−3) (x−4) (x−5)
Answer. Let
(x−1) (x−2)
y=
√ (x−3) (x−4) (x−5)
Taking logarithm on both the sides, we obtain
(x−1) (x−2)
logy = log
√ (x−3) (x−4) (x−5)
1
[ (x−1) (x−2)
⇒ logy = 2 log ( x − 3 ) ( x − 4 ) ( x − 5 )
1
]
⇒ logy = 2 [log{(x − 1)(x − 2)} − log{(x − 3)(x − 4)(x − 5)}]
1
⇒ logy = 2 [log(x − 1) + log(x − 2) − log(x − 3) − log(x − 4) − log(x − 5)]
Differentiating both sides with respect to x, we obtain
[
1 d 1 d 1 d
1 dy
⋅ (x − 1) + x − 2 ⋅ dx (x − 2) − x − 3 ⋅ dx (x − 3)
1 x − 1 dx
y dx
= 2 1 d 1 d
− x − 4 ⋅ dx (x − 4) − x − 5 , dx (x − 5)
dy y
( 1 1
⇒ dx = 2 x − 1 + x − 2 − x − 3 − x − 4 − x − 5
1 1 1
)
dy
∴ dx = 2
1
√
(x−1) (x−2)
(x−3) (x−2) [ 1 1
x−1 + x−2 − x−3 − x−4 − x−5
1 1 1
]
Processing
Page : 178 ,math:
Block100%
Name : Exercise 5.5
⇒ logy = log(cosx) + log(cos2x) + log(cos3x)
Differentiating both sides with respect to x, we obtain
1 dy 1 d 1 d 1 d
y dx
= cos x ⋅ dx (cosx) + cos 2x ⋅ dx (cos2x) + cos 3x ⋅ dx (cos3x)
dy
[ sin x sin 2x
⇒ dx = y − cos x − cos 2x ⋅ dx (2x) − cos 3x ⋅ dx (3x)
dy
d sin 3x d
]
∴ dx = − cosx ⋅ cos2x ⋅ cos3x[tanx + 2tan2x + 3tan3x]
Page : 178 , Block Name : Exercise 5.5
Q2 Differentiate the function with respect to x.
(x−1) (x−2)
√ (x−3) (x−4) (x−5)
Answer. Let
(x−1) (x−2)
y=
√ (x−3) (x−4) (x−5)
Taking logarithm on both the sides, we obtain
(x−1) (x−2)
logy = log
√ (x−3) (x−4) (x−5)
1
[ (x−1) (x−2)
⇒ logy = 2 log ( x − 3 ) ( x − 4 ) ( x − 5 )
1
]
⇒ logy = 2 [log{(x − 1)(x − 2)} − log{(x − 3)(x − 4)(x − 5)}]
1
⇒ logy = 2 [log(x − 1) + log(x − 2) − log(x − 3) − log(x − 4) − log(x − 5)]
Differentiating both sides with respect to x, we obtain
[
1 d 1 d 1 d
1 dy
⋅ (x − 1) + x − 2 ⋅ dx (x − 2) − x − 3 ⋅ dx (x − 3)
1 x − 1 dx
y dx
= 2 1 d 1 d
− x − 4 ⋅ dx (x − 4) − x − 5 , dx (x − 5)
dy y
( 1 1
⇒ dx = 2 x − 1 + x − 2 − x − 3 − x − 4 − x − 5
1 1 1
)
dy
∴ dx = 2
1
√
(x−1) (x−2)
(x−3) (x−2) [ 1 1
x−1 + x−2 − x−3 − x−4 − x−5
1 1 1
]
Processing
Page : 178 ,math:
Block100%
Name : Exercise 5.5
Page 54
Q3 Differentiate the function with respect to x.
(logx) cos x
Answer.
Let y = (logx) cos x
Taking logarithm on both the sides, we obtain
logy = cosx ⋅ log(logx)
Differentiating both sides with respect to x, we obtain
1 dy d d
⋅
y dx
= dx (cosx) × log(logx) + cosx × dx [log(logx)]
1 dy 1 d
⇒ y ⋅ dx = − sinxlog(logx) + cosx × log x ⋅ dx (logx)
dy
[
⇒ dx = y − sinxlog(logx) + log x × x
cos x 1
]
dy
[ cos x
∴ dx = (logx) cos x xlog x − sinxlog(logx) ]
Page : 178 , Block Name : Exercise 5.5
Q4 Differentiate the function with respect to x.
x x − 2 ln x
Answer.
Let y = x x − 2 in x
Also, let x x = u and 2 din x = v
∴y=u−v
dy du dv
⇒ dx = dx − dx
u = xx
Taking logarithm on both the sides, we obtain
logu = xlogx
1 du
u dx
= [ d
dx
(x) × logx + x × dx (logx)
d
]
du
[
⇒ dx = u 1 × logx + x × x
1
]
du
⇒ dx = x x(logx + 1)
du
⇒ dx = x x(1 + logx)
v = 2 sin x
(logx) cos x
Answer.
Let y = (logx) cos x
Taking logarithm on both the sides, we obtain
logy = cosx ⋅ log(logx)
Differentiating both sides with respect to x, we obtain
1 dy d d
⋅
y dx
= dx (cosx) × log(logx) + cosx × dx [log(logx)]
1 dy 1 d
⇒ y ⋅ dx = − sinxlog(logx) + cosx × log x ⋅ dx (logx)
dy
[
⇒ dx = y − sinxlog(logx) + log x × x
cos x 1
]
dy
[ cos x
∴ dx = (logx) cos x xlog x − sinxlog(logx) ]
Page : 178 , Block Name : Exercise 5.5
Q4 Differentiate the function with respect to x.
x x − 2 ln x
Answer.
Let y = x x − 2 in x
Also, let x x = u and 2 din x = v
∴y=u−v
dy du dv
⇒ dx = dx − dx
u = xx
Taking logarithm on both the sides, we obtain
logu = xlogx
1 du
u dx
= [ d
dx
(x) × logx + x × dx (logx)
d
]
du
[
⇒ dx = u 1 × logx + x × x
1
]
du
⇒ dx = x x(logx + 1)
du
⇒ dx = x x(1 + logx)
v = 2 sin x
Page 55
Taking logarithm on both the sides with respect to x, we obtain
logv = sinx ⋅ log2
Differentiating both sides with respect to x, we obtain
1 dv d
⋅
v dx
= log2 ⋅ dx (sinx)
dv
⇒ dx = vlog2cosx
dv
⇒ dx = 2 sin xcosxlog2
dy
∴ dx = x x(1 + logx) − 2 din xcosxlog2
Page : 178 , Block Name : Exercise 5.5
Q5 Differentiate the function with respect to x.
(x + 3) 2 ⋅ (x + 4) 3 ⋅ (x + 5) 4
Answer.
Let y = (x + 3) 2 ⋅ (x + 4) 3 ⋅ (x + 5) 4
Taking logarithm on both the sides, we obtain
logy = log(x + 3) 2 + log(x + 4) 3 + log(x + 5) 4
⇒ logy = 2log(x + 3) + 3log(x + 4) + 4log(x + 5)
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
1 dy 1 d 1 d 1 d
⋅
y dx
= 2 ⋅ x + 3 ⋅ dx (x + 3) + 3 ⋅ x + 4 ⋅ dx (x + 4) + 4 ⋅ x + 5 ⋅ dx (x + 5)
dy
[ 2 3
⇒ dx = y x + 3 + x + 4 + x + 5
4
]
dy
⇒ dx = (x + 3) 2(x + 4) 3(x + 5) 4 ⋅ [ 2(x+4) (x+5) +3(x+3) (x+5) +4(x+3) (x+4)
(x+3) (x+5) ]
[( ) ( ) ( )]
dy
⇒ dx = (x + 3)(x + 4) 2(x + 5) 3 ⋅ 2 x 2 + 9x + 20 + 3 x 2 + 8x + 15 + 4 x 2 + 7x + 12
( )
dy
∴ dx = (x + 3)(x + 4) 2(x + 5) 3 9x 2 + 70x + 133
Page : 178 , Block Name : Exercise 5.5
Q6 Differentiate the function with respect to x.
( ) 1 x
( ) 1
x+ x + x 1+ x
Answer.
logv = sinx ⋅ log2
Differentiating both sides with respect to x, we obtain
1 dv d
⋅
v dx
= log2 ⋅ dx (sinx)
dv
⇒ dx = vlog2cosx
dv
⇒ dx = 2 sin xcosxlog2
dy
∴ dx = x x(1 + logx) − 2 din xcosxlog2
Page : 178 , Block Name : Exercise 5.5
Q5 Differentiate the function with respect to x.
(x + 3) 2 ⋅ (x + 4) 3 ⋅ (x + 5) 4
Answer.
Let y = (x + 3) 2 ⋅ (x + 4) 3 ⋅ (x + 5) 4
Taking logarithm on both the sides, we obtain
logy = log(x + 3) 2 + log(x + 4) 3 + log(x + 5) 4
⇒ logy = 2log(x + 3) + 3log(x + 4) + 4log(x + 5)
Differentiating both sides with respect to x, we obtain
Differentiating both sides with respect to x, we obtain
1 dy 1 d 1 d 1 d
⋅
y dx
= 2 ⋅ x + 3 ⋅ dx (x + 3) + 3 ⋅ x + 4 ⋅ dx (x + 4) + 4 ⋅ x + 5 ⋅ dx (x + 5)
dy
[ 2 3
⇒ dx = y x + 3 + x + 4 + x + 5
4
]
dy
⇒ dx = (x + 3) 2(x + 4) 3(x + 5) 4 ⋅ [ 2(x+4) (x+5) +3(x+3) (x+5) +4(x+3) (x+4)
(x+3) (x+5) ]
[( ) ( ) ( )]
dy
⇒ dx = (x + 3)(x + 4) 2(x + 5) 3 ⋅ 2 x 2 + 9x + 20 + 3 x 2 + 8x + 15 + 4 x 2 + 7x + 12
( )
dy
∴ dx = (x + 3)(x + 4) 2(x + 5) 3 9x 2 + 70x + 133
Page : 178 , Block Name : Exercise 5.5
Q6 Differentiate the function with respect to x.
( ) 1 x
( ) 1
x+ x + x 1+ x
Answer.
Page 56
( ) 1 x
( ) 1
Let y = x + x + x 1+ x
( ) 1 x
( ) 1
Also, let u = x + x and v = x 1 + x
∴y=u+v
dy du dv
⇒ dx = dx + dx
Then u = x + x
( ) 1 x
( )
⇒ logu = log x + x
1 x
( )
⇒ logu = xlog x + x
1
Differentiating both sides with respect to x, we obtain
1
⋅
u dx
du d
= dx (x) × log x + x( ) [ ( )] 1 d
+ x × dx log x + x
1
1 du
⇒ u dx = 1 × log x + x
( ) ( ) ( )
1
+x×
1
x+ x
1
d
⋅ dx x + x
1
[ ( )
]
1
x− x
= (x + ) log (x + ) +
du 1 x 1
⇒ dx x x
( ) x+ x
1
du
⇒ dx = x + x
( )[ ( )
1 x
log x + x
1
+
]x2 − 1
x2 + 1
du
⇒ dx = x + x
( )[ 1 x x2 − 1
x2 + 1 ( )]
+ log x + x
1
( )
v = x 1+ x
1
( ) 1
Let y = x + x + x 1+ x
( ) 1 x
( ) 1
Also, let u = x + x and v = x 1 + x
∴y=u+v
dy du dv
⇒ dx = dx + dx
Then u = x + x
( ) 1 x
( )
⇒ logu = log x + x
1 x
( )
⇒ logu = xlog x + x
1
Differentiating both sides with respect to x, we obtain
1
⋅
u dx
du d
= dx (x) × log x + x( ) [ ( )] 1 d
+ x × dx log x + x
1
1 du
⇒ u dx = 1 × log x + x
( ) ( ) ( )
1
+x×
1
x+ x
1
d
⋅ dx x + x
1
[ ( )
]
1
x− x
= (x + ) log (x + ) +
du 1 x 1
⇒ dx x x
( ) x+ x
1
du
⇒ dx = x + x
( )[ ( )
1 x
log x + x
1
+
]x2 − 1
x2 + 1
du
⇒ dx = x + x
( )[ 1 x x2 − 1
x2 + 1 ( )]
+ log x + x
1
( )
v = x 1+ x
1
Page 57
( )
y = x 1+ x
1
⇒ logv = log x
[ ( )] 1+ x
1
( )
⇒ logv = 1 + x logx
1
Differentiating both sides with respect to x, we obtain
1
⋅
v dx
=
dv
[ ( )]
d
dx
1+ x
1
( )
× logx + 1 + x
1 d
⋅ dx logx
1 dv
⇒ v dx =
( ) −
1
x2
logx + 1 + x
( ) 1 1
⋅ x
1 dv log x 1 1
⇒ v dx = − + x +
x2 x2
dv
⇒ dx = v
[ − log x + x + 1
x2 ]
dv
⇒ dx = x 1 + x ( )( 1 x + 1 − log x
x2 )
Therefore, from (1), (2), and (3), we obtain
( )[ 1 x x2 − 1
( )] ( )( )
dy 1 1 x + 1 − log x
dx
= x+ x + log x + x + x 1+ x
x2 + 1 x2
Page : 178 , Block Name : Exercise 5.5
Q7 Differentiate the function with respect to x.
(logx) x + x log x
Answer.
Let y = (logx) x + x ln x
Also, let u = (logx) x and v = x log x
∴y=u+v
y = x 1+ x
1
⇒ logv = log x
[ ( )] 1+ x
1
( )
⇒ logv = 1 + x logx
1
Differentiating both sides with respect to x, we obtain
1
⋅
v dx
=
dv
[ ( )]
d
dx
1+ x
1
( )
× logx + 1 + x
1 d
⋅ dx logx
1 dv
⇒ v dx =
( ) −
1
x2
logx + 1 + x
( ) 1 1
⋅ x
1 dv log x 1 1
⇒ v dx = − + x +
x2 x2
dv
⇒ dx = v
[ − log x + x + 1
x2 ]
dv
⇒ dx = x 1 + x ( )( 1 x + 1 − log x
x2 )
Therefore, from (1), (2), and (3), we obtain
( )[ 1 x x2 − 1
( )] ( )( )
dy 1 1 x + 1 − log x
dx
= x+ x + log x + x + x 1+ x
x2 + 1 x2
Page : 178 , Block Name : Exercise 5.5
Q7 Differentiate the function with respect to x.
(logx) x + x log x
Answer.
Let y = (logx) x + x ln x
Also, let u = (logx) x and v = x log x
∴y=u+v
Page 58
dy du dv
⇒ dx = dx + dx
u = (logx) x
Differentiating both sides with respect to x, we obtain
⇒ logu = log (logx) x [ ]
⇒ logu = xlog(logx)
1 du d d
u dx
= dx (x) × log(logx) + x ⋅ dx [log(logx)]
du
[
⇒ dx = u 1 × log(logx) + x ⋅ log x ⋅ dx (logx)
1 d
]
du
[
⇒ dx = (logx) x log(logx) + log x ⋅ x
x 1
]
du
[
⇒ dx = (logx) x log(logx) + log x
1
]
du
⇒ dx = (logx) x [ log ( log x ) ⋅ log x + 1
log x ]
du
⇒ dx = (logx) x − 1[1 + logx ⋅ log(logx)]
y = x log x
( )
⇒ logv = log x log x
⇒ logv = log (x ) = (logx) log x 2
Differentiating both sides with respect to x, we obtain
[ ]
1 dv d
⋅
v dx
= dx (logx) 2
1 dv d
⇒ v ⋅ dx = 2(logx) ⋅ dx (logx)
dv 1
⇒ dx = 2v(logx) ⋅ x
dv
⇒ dx = 2x log x − 1 ⋅ logx
Therefore, from (1), (2), and (3), we obtain
dy
dx
= (logx) n − 1[1 + logx ⋅ log(logx)] + 2x log x − 1 ⋅ logx
Page : 178 , Block Name : Exercise 5.5
Q8 Differentiate the function with respect to x.
(sinx) x + sin − 1√x
Answer.
⇒ dx = dx + dx
u = (logx) x
Differentiating both sides with respect to x, we obtain
⇒ logu = log (logx) x [ ]
⇒ logu = xlog(logx)
1 du d d
u dx
= dx (x) × log(logx) + x ⋅ dx [log(logx)]
du
[
⇒ dx = u 1 × log(logx) + x ⋅ log x ⋅ dx (logx)
1 d
]
du
[
⇒ dx = (logx) x log(logx) + log x ⋅ x
x 1
]
du
[
⇒ dx = (logx) x log(logx) + log x
1
]
du
⇒ dx = (logx) x [ log ( log x ) ⋅ log x + 1
log x ]
du
⇒ dx = (logx) x − 1[1 + logx ⋅ log(logx)]
y = x log x
( )
⇒ logv = log x log x
⇒ logv = log (x ) = (logx) log x 2
Differentiating both sides with respect to x, we obtain
[ ]
1 dv d
⋅
v dx
= dx (logx) 2
1 dv d
⇒ v ⋅ dx = 2(logx) ⋅ dx (logx)
dv 1
⇒ dx = 2v(logx) ⋅ x
dv
⇒ dx = 2x log x − 1 ⋅ logx
Therefore, from (1), (2), and (3), we obtain
dy
dx
= (logx) n − 1[1 + logx ⋅ log(logx)] + 2x log x − 1 ⋅ logx
Page : 178 , Block Name : Exercise 5.5
Q8 Differentiate the function with respect to x.
(sinx) x + sin − 1√x
Answer.
Page 59
Let y = (sinx) x + sin − 1√x
Also, let u = (sinx) x and v = sin − 1√x
∴y=u+v
dy du dv
⇒ dx = dx + dx
u = (sinx) x
⇒ logu = log(sinx) x
⇒ logu = xlog(sinx)
Differentiating both sides with respect to x, we obtain
du d d
⇒ dx = dx (x) × log(sinx) + x × dx [log(sinx)]
du
[ 1
⇒ dx = u 1 ⋅ log(sinx) + x ⋅ sin x ⋅ dx (sinx)
d
]
du
[
⇒ dx = (sinx) x log(sinx) + sin x ⋅ cosx
x
]
du
⇒ dx = (sinx) ⊤(xcotx + logsinx)
v = sin − 1√x
Differentiating both sides with respect to x, we obtain
du
[
⇒ dx = (sinx) x log(sinx) + sin x ⋅ cosx
x
]
du
⇒ dx = (sinx) T(xcotx + logsinx)
v = sin − 1√x
Differentiating both sides with respect to x, we obtain
dv 1 d
dx
= ⋅ dx (√x)
√ 1 − ( √x ) 2
du
⇒ dx = (sinx) ′(xcotx + logsinx)
v = sin − 1√x
Differentiating both sides with respect to x, we obtain
dv 1 d
dx
= ⋅ dx (√x)
√ 1 − ( √x ) 2
dv 1 1
⇒ dx = ⋅
√1 − x 2√ x
dv 1
⇒ dx =
√
2 x − x2
Also, let u = (sinx) x and v = sin − 1√x
∴y=u+v
dy du dv
⇒ dx = dx + dx
u = (sinx) x
⇒ logu = log(sinx) x
⇒ logu = xlog(sinx)
Differentiating both sides with respect to x, we obtain
du d d
⇒ dx = dx (x) × log(sinx) + x × dx [log(sinx)]
du
[ 1
⇒ dx = u 1 ⋅ log(sinx) + x ⋅ sin x ⋅ dx (sinx)
d
]
du
[
⇒ dx = (sinx) x log(sinx) + sin x ⋅ cosx
x
]
du
⇒ dx = (sinx) ⊤(xcotx + logsinx)
v = sin − 1√x
Differentiating both sides with respect to x, we obtain
du
[
⇒ dx = (sinx) x log(sinx) + sin x ⋅ cosx
x
]
du
⇒ dx = (sinx) T(xcotx + logsinx)
v = sin − 1√x
Differentiating both sides with respect to x, we obtain
dv 1 d
dx
= ⋅ dx (√x)
√ 1 − ( √x ) 2
du
⇒ dx = (sinx) ′(xcotx + logsinx)
v = sin − 1√x
Differentiating both sides with respect to x, we obtain
dv 1 d
dx
= ⋅ dx (√x)
√ 1 − ( √x ) 2
dv 1 1
⇒ dx = ⋅
√1 − x 2√ x
dv 1
⇒ dx =
√
2 x − x2
Page 60
Therefore, from (1), (2), and (3), we obtain
dy 1
dx
= (sinx) x(xcotx + logsinx) +
√
2 x − x2
Page : 178 , Block Name : Exercise 5.5
Q9 Differentiate the function with respect to x.
x sin x + (sinx) cos x
Answer.
Let y = x sin x + (sinx) cos x
Also, let u = x sin x and v = (sinx) cos x
∴y=u+v
u = x sin x
( )
⇒ logu = log x sin x
⇒ logu = sinxlogx
Differentiating both sides with respect to x, we obtain
1 du d d
u dx
= dx (sinx) ⋅ logx + sinx ⋅ dx (logx)
du
[
⇒ dx = u cosxlogx + sinx ⋅ x
1
]
du
[
⇒ dx = x sin x cosxlogx + x
sin x
]
du
[
⇒ dx = x sin x cosxlogx + x
sin x
]
v = (sinx) cos x
⇒ logv = log(sinx) cos x
⇒ logv = cosxlog(sinx)
Differentiating both sides with respect to x, we obtain
1 dv d d
v dx
= dx (cosx) × log(sinx) + cosx × dx [log(sinx)]
dv
[
⇒ dx = v − sinx ⋅ log(sinx) + cosx ⋅ sin x ⋅ dx (sinx)
1 d
]
dv
[
⇒ dx = (sinx) cos x − sinxlogsinx + sin x cosx
cos x
]
dy 1
dx
= (sinx) x(xcotx + logsinx) +
√
2 x − x2
Page : 178 , Block Name : Exercise 5.5
Q9 Differentiate the function with respect to x.
x sin x + (sinx) cos x
Answer.
Let y = x sin x + (sinx) cos x
Also, let u = x sin x and v = (sinx) cos x
∴y=u+v
u = x sin x
( )
⇒ logu = log x sin x
⇒ logu = sinxlogx
Differentiating both sides with respect to x, we obtain
1 du d d
u dx
= dx (sinx) ⋅ logx + sinx ⋅ dx (logx)
du
[
⇒ dx = u cosxlogx + sinx ⋅ x
1
]
du
[
⇒ dx = x sin x cosxlogx + x
sin x
]
du
[
⇒ dx = x sin x cosxlogx + x
sin x
]
v = (sinx) cos x
⇒ logv = log(sinx) cos x
⇒ logv = cosxlog(sinx)
Differentiating both sides with respect to x, we obtain
1 dv d d
v dx
= dx (cosx) × log(sinx) + cosx × dx [log(sinx)]
dv
[
⇒ dx = v − sinx ⋅ log(sinx) + cosx ⋅ sin x ⋅ dx (sinx)
1 d
]
dv
[
⇒ dx = (sinx) cos x − sinxlogsinx + sin x cosx
cos x
]
Page 61
dv
⇒ dx = (sinx) cos x[ − sinxlogsinx + cotxcosx]
dv
⇒ dx = (sinx) cos x[cotxcosx − sinxlogsinx]
From (1), (2), and (3), we obtain
dy
dx (
= x sin x cosxlogx + x
sin x
) + (sinx) cos x[cosxcotx − sinxlogsinx]
Page : 178 , Block Name : Exercise 5.5
Q10 Differentiate the function with respect to x.
x2 + 1
x xson x +
x2 − 1
Answer.
x2 + 1
Let y = x xcos x +
x2 − 1
x2 + 1
Also, let u = x sen x and v =
x2 − 1
∴y=u+v
dy du dv
⇒ dx = dx + dx
u = x rog x
⇒ logu = log x xan x ( )
⇒ logu = xcosxlogx
Differentiating both sides with respect to x, we obtain
1 du d d d
u dx = dx (x) ⋅ cosx ⋅ logx + x ⋅ dx (cosx) ⋅ logx + xcosx ⋅ dx (logx)
du
[
⇒ dx = u 1 ⋅ cosx ⋅ logx + x ⋅ ( − sinx)logx + xcosx ⋅ x
1
]
du
⇒ dx = x ros x(cosxlogx − xsinxlogx + cosx)
x2 + 1
v=
x2 − 1
(
⇒ logv = log x 2 + 1 − log x 2 − 1) ( )
Differentiating both sides with respect to x, we obtain
⇒ dx = (sinx) cos x[ − sinxlogsinx + cotxcosx]
dv
⇒ dx = (sinx) cos x[cotxcosx − sinxlogsinx]
From (1), (2), and (3), we obtain
dy
dx (
= x sin x cosxlogx + x
sin x
) + (sinx) cos x[cosxcotx − sinxlogsinx]
Page : 178 , Block Name : Exercise 5.5
Q10 Differentiate the function with respect to x.
x2 + 1
x xson x +
x2 − 1
Answer.
x2 + 1
Let y = x xcos x +
x2 − 1
x2 + 1
Also, let u = x sen x and v =
x2 − 1
∴y=u+v
dy du dv
⇒ dx = dx + dx
u = x rog x
⇒ logu = log x xan x ( )
⇒ logu = xcosxlogx
Differentiating both sides with respect to x, we obtain
1 du d d d
u dx = dx (x) ⋅ cosx ⋅ logx + x ⋅ dx (cosx) ⋅ logx + xcosx ⋅ dx (logx)
du
[
⇒ dx = u 1 ⋅ cosx ⋅ logx + x ⋅ ( − sinx)logx + xcosx ⋅ x
1
]
du
⇒ dx = x ros x(cosxlogx − xsinxlogx + cosx)
x2 + 1
v=
x2 − 1
(
⇒ logv = log x 2 + 1 − log x 2 − 1) ( )
Differentiating both sides with respect to x, we obtain
Page 62
1 dv 2x 2x
v dx
= −
x2 + 1 x2 − 1
[ ( ) ( )
]
dv 2x x 2 − 1 − 2x x 2 + 1
⇒ dx = v
(x + 1 ) ( (x − 1 )
2 2
⇒ dx =
dv x2 + 1
x2 − 1
×
[( )( )]
x2 + 1
− 4x
x2 − 1
dv − 4x
⇒ dx =
(x −1 )
2 2
From (1), (2), and (3), we obtain
dy 4x
dx
= x sen x[cosx(1 + logx) − xsinxlogx] −
(x −1 )
2 2
Page : 178 , Block Name : Exercise 5.5
Q11 Differentiate the function with respect to x.
1
(xcosx) x + (xsinx) x
Answer.
1
Let y = (xcosx) x + (xsinx) x
1
x
Also, let u = (xcosx) and v = (xsinx) x
⇒y=u+v
dy du dv
⇒ dx = dx + dx
u = (xcosx) x
⇒ logu = log(xcosx) x
⇒ logu = xlog(xcosx)
⇒ logu = x[logx + logcosx]
⇒ logu = xlogx + xlogcosx
Differentiating both sides with respect to x, we obtain
1 du d d
u dx
= dx (xlogx) + dx (xlogcosx)
du
⇒ dx = u [{ d
logx ⋅ dx (x) + x ⋅ dx (logx)
d
}{ +
d d
logcosx ⋅ dx (x) + x ⋅ dx (logcosx) }]
du
⇒ dx = (xcosx) x [( logx ⋅ 1 + x ⋅ x
1
){
+
1 d
logcosx ⋅ 1 + x ⋅ cos x ⋅ dx (cosx) }]
v dx
= −
x2 + 1 x2 − 1
[ ( ) ( )
]
dv 2x x 2 − 1 − 2x x 2 + 1
⇒ dx = v
(x + 1 ) ( (x − 1 )
2 2
⇒ dx =
dv x2 + 1
x2 − 1
×
[( )( )]
x2 + 1
− 4x
x2 − 1
dv − 4x
⇒ dx =
(x −1 )
2 2
From (1), (2), and (3), we obtain
dy 4x
dx
= x sen x[cosx(1 + logx) − xsinxlogx] −
(x −1 )
2 2
Page : 178 , Block Name : Exercise 5.5
Q11 Differentiate the function with respect to x.
1
(xcosx) x + (xsinx) x
Answer.
1
Let y = (xcosx) x + (xsinx) x
1
x
Also, let u = (xcosx) and v = (xsinx) x
⇒y=u+v
dy du dv
⇒ dx = dx + dx
u = (xcosx) x
⇒ logu = log(xcosx) x
⇒ logu = xlog(xcosx)
⇒ logu = x[logx + logcosx]
⇒ logu = xlogx + xlogcosx
Differentiating both sides with respect to x, we obtain
1 du d d
u dx
= dx (xlogx) + dx (xlogcosx)
du
⇒ dx = u [{ d
logx ⋅ dx (x) + x ⋅ dx (logx)
d
}{ +
d d
logcosx ⋅ dx (x) + x ⋅ dx (logcosx) }]
du
⇒ dx = (xcosx) x [( logx ⋅ 1 + x ⋅ x
1
){
+
1 d
logcosx ⋅ 1 + x ⋅ cos x ⋅ dx (cosx) }]
Page 63
du
⇒ dx = (xcosx) x (logx + 1) +
[ { x
logcosx + cos x ⋅ ( − sinx)
}]
du
⇒ dx = (xcosx) x[(1 + logx) + (logcosx − xtanx)]
du
⇒ dx = (xcosx) x[1 − xtanx + (logx + logcosx)]
du
⇒ dx = (xcosx) x[1 − xtanx + log(xcosx)]
1
v = (xsinx) x
1
⇒ logv = log(xsinx) x
1
⇒ logv = x log(xsinx)
1
⇒ logv = x (logx + logsinx)
1 1
⇒ logv = x logx + x logsinx
1 dv
v dx
d
( ) [
1
= dx x logx + dx x log(sinx)
d 1
]
1 dv
[ ()
⇒ v dx = logx ⋅ dx x
d 1 1 d
][
+ x ⋅ dx (logx) + log(sinx) ⋅ dx x
d
() 1 1 d
+ x ⋅ dx {log(sinx)} ]
1 dv
⇒ v dx = logx ⋅
[ ( ) ][ () −
1
x2
1
+ x ⋅ x
1
+ log(sinx) ⋅
1
x2
+
1
x2
1
+ xsin x ⋅ dx (sinx)
d
]
1 dv
⇒ v dx =
1
x [ ]
2 (1 − logx) +
log ( sin x )
x2
1
+ xsin x + xsin x
1
dv
⇒ dx = (xsinx) x
[ 1
] 1 − log x − log ( sin x ) + xcot x
x2
dv
⇒ dx = (xsinx) x
[ ]1 1 − log ( xsin x ) + xcot x
x2
From (1), (2), and (3), we obtain
From (1), (2), and (3), we obtain
dy
dx
= (xcosx) x[1 − xtanx + log(xcosx)] + (xsinx) x
1
[ xcot x + 1 − log ( xsin x )
x2 ]
Page : 178 , Block Name : Exercise 5.5
⇒ dx = (xcosx) x (logx + 1) +
[ { x
logcosx + cos x ⋅ ( − sinx)
}]
du
⇒ dx = (xcosx) x[(1 + logx) + (logcosx − xtanx)]
du
⇒ dx = (xcosx) x[1 − xtanx + (logx + logcosx)]
du
⇒ dx = (xcosx) x[1 − xtanx + log(xcosx)]
1
v = (xsinx) x
1
⇒ logv = log(xsinx) x
1
⇒ logv = x log(xsinx)
1
⇒ logv = x (logx + logsinx)
1 1
⇒ logv = x logx + x logsinx
1 dv
v dx
d
( ) [
1
= dx x logx + dx x log(sinx)
d 1
]
1 dv
[ ()
⇒ v dx = logx ⋅ dx x
d 1 1 d
][
+ x ⋅ dx (logx) + log(sinx) ⋅ dx x
d
() 1 1 d
+ x ⋅ dx {log(sinx)} ]
1 dv
⇒ v dx = logx ⋅
[ ( ) ][ () −
1
x2
1
+ x ⋅ x
1
+ log(sinx) ⋅
1
x2
+
1
x2
1
+ xsin x ⋅ dx (sinx)
d
]
1 dv
⇒ v dx =
1
x [ ]
2 (1 − logx) +
log ( sin x )
x2
1
+ xsin x + xsin x
1
dv
⇒ dx = (xsinx) x
[ 1
] 1 − log x − log ( sin x ) + xcot x
x2
dv
⇒ dx = (xsinx) x
[ ]1 1 − log ( xsin x ) + xcot x
x2
From (1), (2), and (3), we obtain
From (1), (2), and (3), we obtain
dy
dx
= (xcosx) x[1 − xtanx + log(xcosx)] + (xsinx) x
1
[ xcot x + 1 − log ( xsin x )
x2 ]
Page : 178 , Block Name : Exercise 5.5
Page 64
dy
Q12 Find dx of function.
xy + yx = 1
Answer.
The given function is x y + y x = 1
Let x y = u and y x = v
Then, the function becomes u + v = 1
du dv
∴ dx + dx = 0
u = x′
⇒ logu = log x r ( )
⇒ logu = ylogx
1 du dy d
u dx
= logx dx + y ⋅ dx (logx)
du
[
⇒ dx = u logx dx + y ⋅ x
dy 1
]
du
(
⇒ dx = x y logx dx + x
dy y
)
y = yx
⇒ logv = log y x ( )
⇒ logv = xlogy
Differentiating both sides with respect to x, we obtain
1 dv d d
⋅
v dx
= logy ⋅ dx (x) + x ⋅ dx (logy)
dv
(
⇒ dx = v logy ⋅ 1 + x ⋅ y ⋅ dx
1 dy
)
dv
(
⇒ dx = y x logy + y dx
x dy
)
From (1), (2), and (3), we obtain
( dy
x y logx dx + x
y
) (
+ y x logy + y dx
x dy
) =0
( ) ( )
dy
⇒ x ylogx + xy x − 1 dx = − yx y − 1 + y xlogy
dy yx y − 1 + y xlog y
∴ dx = −
x ylog x + xy x − 1
Processing
Page : 178 ,math:
Block100%
Name : Exercise 5.5
Q12 Find dx of function.
xy + yx = 1
Answer.
The given function is x y + y x = 1
Let x y = u and y x = v
Then, the function becomes u + v = 1
du dv
∴ dx + dx = 0
u = x′
⇒ logu = log x r ( )
⇒ logu = ylogx
1 du dy d
u dx
= logx dx + y ⋅ dx (logx)
du
[
⇒ dx = u logx dx + y ⋅ x
dy 1
]
du
(
⇒ dx = x y logx dx + x
dy y
)
y = yx
⇒ logv = log y x ( )
⇒ logv = xlogy
Differentiating both sides with respect to x, we obtain
1 dv d d
⋅
v dx
= logy ⋅ dx (x) + x ⋅ dx (logy)
dv
(
⇒ dx = v logy ⋅ 1 + x ⋅ y ⋅ dx
1 dy
)
dv
(
⇒ dx = y x logy + y dx
x dy
)
From (1), (2), and (3), we obtain
( dy
x y logx dx + x
y
) (
+ y x logy + y dx
x dy
) =0
( ) ( )
dy
⇒ x ylogx + xy x − 1 dx = − yx y − 1 + y xlogy
dy yx y − 1 + y xlog y
∴ dx = −
x ylog x + xy x − 1
Processing
Page : 178 ,math:
Block100%
Name : Exercise 5.5
Page 65
dy
Q13 Find dx of function.
yx = xy
Answer.
The given function is x y + y x = 1
Let x ′ = u and y x = v
Then, the function becomes u + v = 1
du dv
∴ dx + dx = 0
u = xr
⇒ logu = log x x ( )
⇒ logu = ylogx
1 du dy d
u dx
= logx dx + y ⋅ dx (logx)
du
[
⇒ dx = u logx dx + y ⋅ x
dy 1
]
du
(
⇒ dx = x y logx dx + x
dy y
)
v = yk
1 dv d d
⋅
v dx
= logy ⋅ dx (x) + x ⋅ dx (logy)
dv
(
⇒ dx = y x logy + y dx
x dy
)
From (1), (2), and (3), we obtain
( dy
x y logx dx + x
y
) (
+ y x logy + y dx
x dy
) =0
( ) ( )
dy
⇒ x ylogx + xy x − 1 dx = − yx y − 1 + y xlogy
dy yx y − 1 + y ′log y
∴ dx = −
x ylog x + xy x − 1
Page : 178 , Block Name : Exercise 5.5
dy
Q14 Find dx of function.
(cosx) x = (cosy) x
Answer.
Q13 Find dx of function.
yx = xy
Answer.
The given function is x y + y x = 1
Let x ′ = u and y x = v
Then, the function becomes u + v = 1
du dv
∴ dx + dx = 0
u = xr
⇒ logu = log x x ( )
⇒ logu = ylogx
1 du dy d
u dx
= logx dx + y ⋅ dx (logx)
du
[
⇒ dx = u logx dx + y ⋅ x
dy 1
]
du
(
⇒ dx = x y logx dx + x
dy y
)
v = yk
1 dv d d
⋅
v dx
= logy ⋅ dx (x) + x ⋅ dx (logy)
dv
(
⇒ dx = y x logy + y dx
x dy
)
From (1), (2), and (3), we obtain
( dy
x y logx dx + x
y
) (
+ y x logy + y dx
x dy
) =0
( ) ( )
dy
⇒ x ylogx + xy x − 1 dx = − yx y − 1 + y xlogy
dy yx y − 1 + y ′log y
∴ dx = −
x ylog x + xy x − 1
Page : 178 , Block Name : Exercise 5.5
dy
Q14 Find dx of function.
(cosx) x = (cosy) x
Answer.
Page 66
The given function is (cosx) y = (cosy) x
Taking logarithm on both the sides, we obtain
ylogcosx = xlogcosy
Differentiating both sides, we obtain
dy d d d
logcosx ⋅ dx + y ⋅ dx (logcosx) = logcosy ⋅ dx (x) + x ⋅ dx (logcosy)
dy 1 d 1 d
⇒ logcosx dx + y ⋅ cos x ⋅ dx (cosx) = logcosy ⋅ 1 + x ⋅ cos y ⋅ dx (cosy)
dy y x dy
⇒ logcosx dx + cos x ⋅ ( − sinx) = logcosy + cos y ( − siny) ⋅ dx
dy dy
⇒ logcosx dx − ytanx = logcosy − xtany dx
dy
⇒ (logcosx + xtany) dx = ytanx + logcosy
dy ytan x + log cos y
∴ dx = xtan y + log cos x
dy ytan x + log cos y
∴ dx = xtan y + log cos x
Page : 178 , Block Name : Exercise 5.5
dy
Q15 Find dx of function.
xy = e ( x − y )
Answer.
The given function is xy = e ( x − y )
Taking logarithm on both the sides, we obtain
log(xy) = log e x − y ( )
⇒ logx + logy = (x − y)loge
⇒ logx + logy = (x − y) × 1
⇒ logx + logy = (x − y) × 1
⇒ logx + logy = x − y
d d d dy
dx
(logx) + dx (logy) = dx (x) − dx
1 1 dy dy
⇒ x + y dx = 1 − dx
( ) 1
⇒ 1 + y dx = 1 − x
dy 1
⇒ ( ) y+1
y
dy
dx =
x−1
x
dy y(x−1)
∴Processing
dx
= x ( ymath:
+ 1 ) 100%
Taking logarithm on both the sides, we obtain
ylogcosx = xlogcosy
Differentiating both sides, we obtain
dy d d d
logcosx ⋅ dx + y ⋅ dx (logcosx) = logcosy ⋅ dx (x) + x ⋅ dx (logcosy)
dy 1 d 1 d
⇒ logcosx dx + y ⋅ cos x ⋅ dx (cosx) = logcosy ⋅ 1 + x ⋅ cos y ⋅ dx (cosy)
dy y x dy
⇒ logcosx dx + cos x ⋅ ( − sinx) = logcosy + cos y ( − siny) ⋅ dx
dy dy
⇒ logcosx dx − ytanx = logcosy − xtany dx
dy
⇒ (logcosx + xtany) dx = ytanx + logcosy
dy ytan x + log cos y
∴ dx = xtan y + log cos x
dy ytan x + log cos y
∴ dx = xtan y + log cos x
Page : 178 , Block Name : Exercise 5.5
dy
Q15 Find dx of function.
xy = e ( x − y )
Answer.
The given function is xy = e ( x − y )
Taking logarithm on both the sides, we obtain
log(xy) = log e x − y ( )
⇒ logx + logy = (x − y)loge
⇒ logx + logy = (x − y) × 1
⇒ logx + logy = (x − y) × 1
⇒ logx + logy = x − y
d d d dy
dx
(logx) + dx (logy) = dx (x) − dx
1 1 dy dy
⇒ x + y dx = 1 − dx
( ) 1
⇒ 1 + y dx = 1 − x
dy 1
⇒ ( ) y+1
y
dy
dx =
x−1
x
dy y(x−1)
∴Processing
dx
= x ( ymath:
+ 1 ) 100%
Page 67
Page : 178 , Block Name : Exercise 5.5
Q16 Find the derivative of the function given by f(x) = (1 + x) 1 + x 2 ( )(1 + x )(1 + x ) and hence
4 3
nd f ′(1)
Answer. The given relationship is f(x) = (1 + x) 1 + x 2 ( )(1 + x )(1 + x )Taking logarithm on both
4 3
the sides, we obtain
( ) ( )
logf(x) = log(1 + x) + log 1 + x 2 + log 1 + x 4 + log 1 + x 8 Differentiating both sides with respect ( )
to x, we obtain
( ) ( ) ( )
1 d d d d d
f(x)
⋅ dx [f(x)] = dx log(1 + x) + dx log 1 + x 2 + dx log 1 + x 4 + dx log 1 + x 8
+ (1 + x ) + (1 + x ) + (1 + x )
1 1 d 1 d 1 d 1 d
⇒ f ( x ) ⋅ f ′(x) = 1 + x ⋅ dx (1 + x) + dx
2
⋅ 4
⋅ 8
1 + x2 1 + x 4 dx 1 + x 8 dx
[
⇒ f ′(x) = f(x) 1 + x +
1
1 + x2
⋅ 2x +
1
1 + x4
⋅ 4x 3 +
1 + x3
⋅ 8x 7
1 1
]
′
∴ f (x) = (1 + x) 1 + x ( 2
)(1 + x )(1 + x )
4 8
[ 1 2x 4x 3
1 + x + 1 + x2 + 1 + x4 + 1 + x8
8x 7
]
′
Hence, f (1) = (1 + 1) 1 + 1 ( 2 4 ′
[
)(1 + 1 )(1 + 1 ) 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
1 2×1
2
4 × 13
4
8 × 17
2 ]
=2×2×2×2
[ 1
2
+
2
2
+
4
2
+
8
2 ]
15
= 16 × ( 1+2+4+8
2 )
= 16 × 2 = 120
Page : 178 , Block Name : Exercise 5.5
Q17 Differentiate x 4 − 5x + 8 ( )(x + 7x + 9 ) in three ways mentioned below
3
(i) By using product rule.
(ii) By expanding the product to obtain a single polynomial.
(iii) By logarithmic differentiation. Do they all give the same answer?
Answer. Letmath:
Processing y = 100% (
x 5 − 5x + 8 )(x + 7x + 9 )
3
Q16 Find the derivative of the function given by f(x) = (1 + x) 1 + x 2 ( )(1 + x )(1 + x ) and hence
4 3
nd f ′(1)
Answer. The given relationship is f(x) = (1 + x) 1 + x 2 ( )(1 + x )(1 + x )Taking logarithm on both
4 3
the sides, we obtain
( ) ( )
logf(x) = log(1 + x) + log 1 + x 2 + log 1 + x 4 + log 1 + x 8 Differentiating both sides with respect ( )
to x, we obtain
( ) ( ) ( )
1 d d d d d
f(x)
⋅ dx [f(x)] = dx log(1 + x) + dx log 1 + x 2 + dx log 1 + x 4 + dx log 1 + x 8
+ (1 + x ) + (1 + x ) + (1 + x )
1 1 d 1 d 1 d 1 d
⇒ f ( x ) ⋅ f ′(x) = 1 + x ⋅ dx (1 + x) + dx
2
⋅ 4
⋅ 8
1 + x2 1 + x 4 dx 1 + x 8 dx
[
⇒ f ′(x) = f(x) 1 + x +
1
1 + x2
⋅ 2x +
1
1 + x4
⋅ 4x 3 +
1 + x3
⋅ 8x 7
1 1
]
′
∴ f (x) = (1 + x) 1 + x ( 2
)(1 + x )(1 + x )
4 8
[ 1 2x 4x 3
1 + x + 1 + x2 + 1 + x4 + 1 + x8
8x 7
]
′
Hence, f (1) = (1 + 1) 1 + 1 ( 2 4 ′
[
)(1 + 1 )(1 + 1 ) 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
1 2×1
2
4 × 13
4
8 × 17
2 ]
=2×2×2×2
[ 1
2
+
2
2
+
4
2
+
8
2 ]
15
= 16 × ( 1+2+4+8
2 )
= 16 × 2 = 120
Page : 178 , Block Name : Exercise 5.5
Q17 Differentiate x 4 − 5x + 8 ( )(x + 7x + 9 ) in three ways mentioned below
3
(i) By using product rule.
(ii) By expanding the product to obtain a single polynomial.
(iii) By logarithmic differentiation. Do they all give the same answer?
Answer. Letmath:
Processing y = 100% (
x 5 − 5x + 8 )(x + 7x + 9 )
3
Page 68
(i)
Let x 2 − 5x + 8 = u and x 3 + 7x + 9 = v
∴ y = mv
dy du dv
⇒ dx = dx ⋅ v + u ⋅ dx
( ) ( ) ( ) ( )
dy d d
⇒ dx = dx x 2 − 5x + 8 ⋅ x 3 + 7x + 9 + x 2 − 5x + 8 ⋅ dx x 3 + 7x + 9
= (2x − 5) (x + 7x + 9 ) + (x − 5x + 8 )(3x + 7 )
dy
⇒ dx 3 2 2
= 2x (x + 7x + 9 ) − 5 (x + 7x + 9 ) + x (3x + 7 ) − 5x (3x + 7 ) + 8 (3x + 7 )
dy
⇒ dx 3 3 2 2 2 2
= (2x + 14x + 18x ) − 5x − 35x − 45 + (3x + 7x ) − 15x − 35x + 24x + 56
dy
⇒ dx 4 2 3 4 2 3 2
dy
∴ dx = 5x 4 − 20x 5 + 45x 2 − 52x + 11
(ii)
(
y = x 2 − 5x + 8)(x + 7x + 9 ) 3
= x (x + 7x + 9 ) − 5x (x + 7x + 9 ) + 8 (x + 7x + 9 )
2 3 3 3
= x 5 + 7x 3 + 9x 2 − 5x 4 − 35x 2 − 45x + 8x 3 + 56x + 72
= x 5 − 5x 4 + 15x 3 − 26x 2 + 11x + 72
dy d 2
∴
dx
=
dx (
x − 5x 4 + 15x 2 − 26x 2 + 11x + 72 )
d d d d d d
dx ( ) dx ( ) dx ( ) dx ( )
= x − 5 2 x + 15 x4 − 26 x + 11 5
(x) + (72) 2
dx dx
= 5x 4 − 5 × 4x 3 + 15 × 3x 2 − 26 × 2x + 11 × 1 + 0
= 5x 4 − 20x 3 + 45x 2 − 52x + 11
(iii)
(
y = x 2 − 5x + 8 )(x + 7x + 9 )
3
Taking logarithm on both the sides, we obtain
( )
logy = log x 2 − 5x + 8 + log x 3 + 7x + 9 ( )
Differentiating both sides with respect to x, we obtain
Let x 2 − 5x + 8 = u and x 3 + 7x + 9 = v
∴ y = mv
dy du dv
⇒ dx = dx ⋅ v + u ⋅ dx
( ) ( ) ( ) ( )
dy d d
⇒ dx = dx x 2 − 5x + 8 ⋅ x 3 + 7x + 9 + x 2 − 5x + 8 ⋅ dx x 3 + 7x + 9
= (2x − 5) (x + 7x + 9 ) + (x − 5x + 8 )(3x + 7 )
dy
⇒ dx 3 2 2
= 2x (x + 7x + 9 ) − 5 (x + 7x + 9 ) + x (3x + 7 ) − 5x (3x + 7 ) + 8 (3x + 7 )
dy
⇒ dx 3 3 2 2 2 2
= (2x + 14x + 18x ) − 5x − 35x − 45 + (3x + 7x ) − 15x − 35x + 24x + 56
dy
⇒ dx 4 2 3 4 2 3 2
dy
∴ dx = 5x 4 − 20x 5 + 45x 2 − 52x + 11
(ii)
(
y = x 2 − 5x + 8)(x + 7x + 9 ) 3
= x (x + 7x + 9 ) − 5x (x + 7x + 9 ) + 8 (x + 7x + 9 )
2 3 3 3
= x 5 + 7x 3 + 9x 2 − 5x 4 − 35x 2 − 45x + 8x 3 + 56x + 72
= x 5 − 5x 4 + 15x 3 − 26x 2 + 11x + 72
dy d 2
∴
dx
=
dx (
x − 5x 4 + 15x 2 − 26x 2 + 11x + 72 )
d d d d d d
dx ( ) dx ( ) dx ( ) dx ( )
= x − 5 2 x + 15 x4 − 26 x + 11 5
(x) + (72) 2
dx dx
= 5x 4 − 5 × 4x 3 + 15 × 3x 2 − 26 × 2x + 11 × 1 + 0
= 5x 4 − 20x 3 + 45x 2 − 52x + 11
(iii)
(
y = x 2 − 5x + 8 )(x + 7x + 9 )
3
Taking logarithm on both the sides, we obtain
( )
logy = log x 2 − 5x + 8 + log x 3 + 7x + 9 ( )
Differentiating both sides with respect to x, we obtain
Page 69
( ) ( )
1 dy d d
y dx
= dx log x 2 − 5x + 8 + dx log x 3 + 7x + 9
⋅ (x − 5x + 8 ) + (x + 7x + 9 )
1 dy 1 d 1 d
⇒ y dx = 2 ⋅ 3
x 2 − 5x + 8 dx x 3 + 7x + 9 dx
dy
⇒ dx = y
[ 1
x 2 − 5x + 8
× (2x − 5) + 3
x + 7x + 9
×
1
(3x + 7 )
2
]
⇒ dx =
dy
(x 2 − 5x + 8 )( x 3 + 7x + 9 ) [ 2x − 5
x 2 − 5x + 8
+
3x 2 + 7
x 3 + 7x + 9 ]
() ( ) ( ) ( )
dy
⇒ dx = 2x x 3 + 7x + 9 − 5 x 3 + 7x + 9 + 3x 2 x 2 − 5x + 8 + 7 x 2 − 5x + 8
= (2x + 14x + 18x ) − 5x − 35x − 45 + (3x − 15x + 24x ) + (7x − 35x + 56 )
dy
⇒ dx 4 2 3 4 4 2 2
dy
⇒ dx = 5x 4 − 20x 3 + 45x 2 − 52x + 4
dy
From the above three observations, it can be concluded that all the results of dx are same .
Page : 178 , Block Name : Exercise 5.5
Q18 If u, v and w are functions of x, then show that
d du dv dv
dx
(u, v, w) = dx vw + u. dx , w + u, v, dx
in two ways- rst by repeated application of product rule, second by logarithmic differentiation.
Answer.
Let y = u, v, w = u. (v, w)
By applying product rule, we obtain
dy du d
dx
= dx , (v, w) + u, dx (v − w)
dy du dv dw
⇒ dx = dx ⋅ v ⋅ w + u ⋅ dx ⋅ w + u ⋅ v ⋅ dx
By taking logarithm on both sides of the equation y = u. v, w, we obtain
logy = logu + logv + logw
Differentiating both sides with respect to x, we obtain
1 dy d d d
⋅
y dx
= dx (logu) + dx (logv) + dx (logw)
1 dy 1 du 1 dv 1 dw
⇒ y ⋅ dx = u dx + v dx + w dx
dy
(
⇒ dx = y u dx + v dx + w dx
1 du 1 dv 1 dw
)
1 dy d d
y dx
= dx log x 2 − 5x + 8 + dx log x 3 + 7x + 9
⋅ (x − 5x + 8 ) + (x + 7x + 9 )
1 dy 1 d 1 d
⇒ y dx = 2 ⋅ 3
x 2 − 5x + 8 dx x 3 + 7x + 9 dx
dy
⇒ dx = y
[ 1
x 2 − 5x + 8
× (2x − 5) + 3
x + 7x + 9
×
1
(3x + 7 )
2
]
⇒ dx =
dy
(x 2 − 5x + 8 )( x 3 + 7x + 9 ) [ 2x − 5
x 2 − 5x + 8
+
3x 2 + 7
x 3 + 7x + 9 ]
() ( ) ( ) ( )
dy
⇒ dx = 2x x 3 + 7x + 9 − 5 x 3 + 7x + 9 + 3x 2 x 2 − 5x + 8 + 7 x 2 − 5x + 8
= (2x + 14x + 18x ) − 5x − 35x − 45 + (3x − 15x + 24x ) + (7x − 35x + 56 )
dy
⇒ dx 4 2 3 4 4 2 2
dy
⇒ dx = 5x 4 − 20x 3 + 45x 2 − 52x + 4
dy
From the above three observations, it can be concluded that all the results of dx are same .
Page : 178 , Block Name : Exercise 5.5
Q18 If u, v and w are functions of x, then show that
d du dv dv
dx
(u, v, w) = dx vw + u. dx , w + u, v, dx
in two ways- rst by repeated application of product rule, second by logarithmic differentiation.
Answer.
Let y = u, v, w = u. (v, w)
By applying product rule, we obtain
dy du d
dx
= dx , (v, w) + u, dx (v − w)
dy du dv dw
⇒ dx = dx ⋅ v ⋅ w + u ⋅ dx ⋅ w + u ⋅ v ⋅ dx
By taking logarithm on both sides of the equation y = u. v, w, we obtain
logy = logu + logv + logw
Differentiating both sides with respect to x, we obtain
1 dy d d d
⋅
y dx
= dx (logu) + dx (logv) + dx (logw)
1 dy 1 du 1 dv 1 dw
⇒ y ⋅ dx = u dx + v dx + w dx
dy
(
⇒ dx = y u dx + v dx + w dx
1 du 1 dv 1 dw
)
Page 70
dy
⇒ dx = u, v, w ⋅ ( 1 du
u dx
1 dv
+ v dx + w dx
1 dw
)
dy du dv dv
∴ dx = dx ⋅ v ⋅ w + u ⋅ dx ⋅ w + u ⋅ v ⋅ dx
Page : 179 , Block Name : Exercise 5.5
Q1 If x and y are connected parametrically by the equation, without eliminating the
dy
parameter, find dx
x = 2at 2, y = at 4
Answer.
The given equations are x = 2at 2 and y = at 4
( ) ()
dx d d
Then, dt = dt 2at 2 = 2a ⋅ dt t 2 = 2a ⋅ 2t = 4at
( ) ()
dy d d
dt
= dt at 4 = a ⋅ dt t 4 = a ⋅ 4 ⋅ t 3 = 4at 3
dy ( ) dy
dt
4at 3
∴ dx = = 4at
= t2
( ) dx
dt
Page : 181 , Block Name : Exercise 5.6
Q2 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
x = acosθ, y = bcosθ
Answer.
The given equations are x = acosθ and y = bcosθ
dx d
Then, dθ = dθ (acosθ) = a( − sinθ) = − asinθ
dy d
dθ
= dθ (bcosθ) = b( − sinθ) = − bsinθ
∴ dx
dy
=
( ) =
dy
dθ
− bsin θ
= a
b
− asin θ
( ) dy
dθ
Page : 181 , Block Name : Exercise 5.6
⇒ dx = u, v, w ⋅ ( 1 du
u dx
1 dv
+ v dx + w dx
1 dw
)
dy du dv dv
∴ dx = dx ⋅ v ⋅ w + u ⋅ dx ⋅ w + u ⋅ v ⋅ dx
Page : 179 , Block Name : Exercise 5.5
Q1 If x and y are connected parametrically by the equation, without eliminating the
dy
parameter, find dx
x = 2at 2, y = at 4
Answer.
The given equations are x = 2at 2 and y = at 4
( ) ()
dx d d
Then, dt = dt 2at 2 = 2a ⋅ dt t 2 = 2a ⋅ 2t = 4at
( ) ()
dy d d
dt
= dt at 4 = a ⋅ dt t 4 = a ⋅ 4 ⋅ t 3 = 4at 3
dy ( ) dy
dt
4at 3
∴ dx = = 4at
= t2
( ) dx
dt
Page : 181 , Block Name : Exercise 5.6
Q2 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
x = acosθ, y = bcosθ
Answer.
The given equations are x = acosθ and y = bcosθ
dx d
Then, dθ = dθ (acosθ) = a( − sinθ) = − asinθ
dy d
dθ
= dθ (bcosθ) = b( − sinθ) = − bsinθ
∴ dx
dy
=
( ) =
dy
dθ
− bsin θ
= a
b
− asin θ
( ) dy
dθ
Page : 181 , Block Name : Exercise 5.6
Page 71
Q3 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
x = sint, y = cos2t
Answer.
The given equations are x = sint and y = cos2t
dx d
Then, dt = dt (sint) = cost
dy d d
dt
= dt (cos2t) = − sin2t ⋅ dt (2t) = − 2sin2t
∴ dx
dy
=
( )= dy
dt
− 2sin 2t
=
− 2 ⋅ 2sin tcos t
= − 4sint
cos t cos t
( ) dx
dt
Page : 181 , Block Name : Exercise 5.6
Q4 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
4
x = 4t, y = y
4
Answer. The given equations are x = 4t and y = t
dx d
dt
= dt (4t) = 4
dy
dt
= dt t
d
() 4
() d
= 4 ⋅ dt t
1
=4⋅
()
−1
t2
=
−4
t2
( ) ( )
−4
dy
dt t2
dy −1
∴ dx = dx
= 4
=
t2
Page : 181 , Block Name : Exercise 5.6
Q5 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
x = cosθ − cos2θ, y = sinθ − sin2l
Answer.
dy
nd dx
x = sint, y = cos2t
Answer.
The given equations are x = sint and y = cos2t
dx d
Then, dt = dt (sint) = cost
dy d d
dt
= dt (cos2t) = − sin2t ⋅ dt (2t) = − 2sin2t
∴ dx
dy
=
( )= dy
dt
− 2sin 2t
=
− 2 ⋅ 2sin tcos t
= − 4sint
cos t cos t
( ) dx
dt
Page : 181 , Block Name : Exercise 5.6
Q4 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
4
x = 4t, y = y
4
Answer. The given equations are x = 4t and y = t
dx d
dt
= dt (4t) = 4
dy
dt
= dt t
d
() 4
() d
= 4 ⋅ dt t
1
=4⋅
()
−1
t2
=
−4
t2
( ) ( )
−4
dy
dt t2
dy −1
∴ dx = dx
= 4
=
t2
Page : 181 , Block Name : Exercise 5.6
Q5 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
x = cosθ − cos2θ, y = sinθ − sin2l
Answer.
Page 72
The given equations are x = cosθ − cos2θ and y = sinθ − sin2θ
dx d d d
Then, = (cosθ − cos2θ) = (cosθ) = (cos2θ)
dθ dθ dθ dθ
= − sinθ − ( − 2sin2θ) = 2sin2θ − sinθ
dy d d d
dθ
= dθ (sinθ − sin2θ) = dθ (sinθ) − dθ (sin2θ)
= cosθ − 2cos2θ
dy ( ) dy
dθ
cos θ − 2cos 2θ
∴ dx = = 2sin 2θ − sin θ
( ) dx
dθ
Page : 181 , Block Name : Exercise 5.6
Q6 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
x = a(θ − sinθ), y = a(1 + cosθ)
Answer.
The given equations are x = a ( θ − sin θ ) and y = a(1 + cosθ)
dx
[ d d
Then, dθ = a dθ (θ) − dθ (sinθ) = a(1 − cosθ) ]
dy
[ d d
]
dθ = a dθ (1) + dθ (cosθ) = a[0 + ( − sinθ)] = − asinθ
∴ dx
dy
=
( ) =
dy
dθ
− asin θ
=
θ
− 2sin 2 cos 2
θ
=
− cos 2
θ
= − cot 2
θ
dx a ( 1 − cos θ ) θ θ
2sin 2 2 sin 2
Page : 181 , Block Name : Exercise 5.6
Q7 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
sin 3 t cos 3 t
x= ,y =
√cos 2t √cos 2t
sin 3 t cos 3 t
Answer. The given equation are and y =
√cos 2t √cos 2t
dx d d d
Then, = (cosθ − cos2θ) = (cosθ) = (cos2θ)
dθ dθ dθ dθ
= − sinθ − ( − 2sin2θ) = 2sin2θ − sinθ
dy d d d
dθ
= dθ (sinθ − sin2θ) = dθ (sinθ) − dθ (sin2θ)
= cosθ − 2cos2θ
dy ( ) dy
dθ
cos θ − 2cos 2θ
∴ dx = = 2sin 2θ − sin θ
( ) dx
dθ
Page : 181 , Block Name : Exercise 5.6
Q6 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
x = a(θ − sinθ), y = a(1 + cosθ)
Answer.
The given equations are x = a ( θ − sin θ ) and y = a(1 + cosθ)
dx
[ d d
Then, dθ = a dθ (θ) − dθ (sinθ) = a(1 − cosθ) ]
dy
[ d d
]
dθ = a dθ (1) + dθ (cosθ) = a[0 + ( − sinθ)] = − asinθ
∴ dx
dy
=
( ) =
dy
dθ
− asin θ
=
θ
− 2sin 2 cos 2
θ
=
− cos 2
θ
= − cot 2
θ
dx a ( 1 − cos θ ) θ θ
2sin 2 2 sin 2
Page : 181 , Block Name : Exercise 5.6
Q7 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
sin 3 t cos 3 t
x= ,y =
√cos 2t √cos 2t
sin 3 t cos 3 t
Answer. The given equation are and y =
√cos 2t √cos 2t
Page 73
Then,
dx
dt
=
dt
d
[ ]
sin 3t
√cos2t
( )
d d
√cos2t ⋅ dt sin 3t − sin 3t ⋅ dt √cos2t
=
cos2t
d 1
√cos2t ⋅ 3sin 2t ⋅ dt (sint) − sin 3t × d
2√cos2t ⋅ dt (cos2t)
sin 3 t
3√cos2t ⋅ sin 2tcost − ⋅ ( − 2sin2t)
2√cos 2t
cos2t
3cos 2tsin 2 tcos t + sin 3 tsin 2t
=
cos 2t√cos 2t
d d d
√cos 2t ⋅ 3cos 2 t ⋅ dr ( cos t ) − cos 3 t ⋅ 2√cos 2t ⋅ dt ( cos 2t )
= cos 2t
1
3√cos 2tcos 2 t ( − sin t ) − cos 3 t ⋅
2√cos 2t ⋅ ( − 2sin 2t )
= cos 2t
− 3cos 2t ⋅ cos 2 t ⋅ sin t + cos 3 tsin 2t
=
cos 2t ⋅ √cos 2t
∴ dx
dy
=
( ) =
dy
dt
− 3cos 2t ⋅ cos 2 tsin t + cos 3 tsin 2t
3cos 2tsin 2 tcos t + sin 3 tsin 2t
( ) dx
dt
− 3cos2t ⋅ cos 2t ⋅ sint + cos 3t(2sintcost)
=
3cos2tsin 2tcost + sin 3t(2sintcost)
[
sintcost − 3cos2t ⋅ cost + 2cos 3t ]
= ]
[
sintcost 3cos2tsint + 2sin 3t
[ − 3 ( 2cos t − 1 ) cos t + 2cos t ]
2 3
=
[ 3 ( 1 − 2sin t ) sin t + 2sin t ]
2 3
− 4cos 2t + 3cost
=
3sint − 4sin 3t
− cos3t
=
sin3
= − cot3t
Page : 181 , Block Name : Exercise 5.6
Q8 If x and y are connected parametrically by the equation, without eliminating the parameter,
dx
dt
=
dt
d
[ ]
sin 3t
√cos2t
( )
d d
√cos2t ⋅ dt sin 3t − sin 3t ⋅ dt √cos2t
=
cos2t
d 1
√cos2t ⋅ 3sin 2t ⋅ dt (sint) − sin 3t × d
2√cos2t ⋅ dt (cos2t)
sin 3 t
3√cos2t ⋅ sin 2tcost − ⋅ ( − 2sin2t)
2√cos 2t
cos2t
3cos 2tsin 2 tcos t + sin 3 tsin 2t
=
cos 2t√cos 2t
d d d
√cos 2t ⋅ 3cos 2 t ⋅ dr ( cos t ) − cos 3 t ⋅ 2√cos 2t ⋅ dt ( cos 2t )
= cos 2t
1
3√cos 2tcos 2 t ( − sin t ) − cos 3 t ⋅
2√cos 2t ⋅ ( − 2sin 2t )
= cos 2t
− 3cos 2t ⋅ cos 2 t ⋅ sin t + cos 3 tsin 2t
=
cos 2t ⋅ √cos 2t
∴ dx
dy
=
( ) =
dy
dt
− 3cos 2t ⋅ cos 2 tsin t + cos 3 tsin 2t
3cos 2tsin 2 tcos t + sin 3 tsin 2t
( ) dx
dt
− 3cos2t ⋅ cos 2t ⋅ sint + cos 3t(2sintcost)
=
3cos2tsin 2tcost + sin 3t(2sintcost)
[
sintcost − 3cos2t ⋅ cost + 2cos 3t ]
= ]
[
sintcost 3cos2tsint + 2sin 3t
[ − 3 ( 2cos t − 1 ) cos t + 2cos t ]
2 3
=
[ 3 ( 1 − 2sin t ) sin t + 2sin t ]
2 3
− 4cos 2t + 3cost
=
3sint − 4sin 3t
− cos3t
=
sin3
= − cot3t
Page : 181 , Block Name : Exercise 5.6
Q8 If x and y are connected parametrically by the equation, without eliminating the parameter,
Page 74
dy
(
nd dx x = a cost + logtan 2 , y = asint
t
)
Answer. The given equations are x = a cost + logtan 2
( t
) and y = asint
Then,
dx
dt
=a⋅ [ d
dt
(cost) +
d
dr ( logtan
t
2 )]
[
= a − sint +
tan 2
1
l
,
d
dt ( )tan
t
2
]
[
= a − sint + cot
t
2
⋅ sec 2
t
2
⋅
d
dt 2( )] t
[
= a − sint +
1
t
2sin 2 cos 2
t
]
(
= a − sint +
1
sint )
=a
( − sin 2t + 1
sint )
cos 2t
=a
sint
dy d
dt
= a dt (sint) = acost
dy ( ) dy
dt
acos t sin t
∴ dx = = = cos t = tant
( ) ( dx
dt
cos 2 t
a sin t
)
Page : 181 , Block Name : Exercise 5.6
Q9 If x and y are connected parametrically by the equation, without eliminating the parameter,
nd
dy
Processing
x = asecθ,math:
y = 100%
btanθ
dx
(
nd dx x = a cost + logtan 2 , y = asint
t
)
Answer. The given equations are x = a cost + logtan 2
( t
) and y = asint
Then,
dx
dt
=a⋅ [ d
dt
(cost) +
d
dr ( logtan
t
2 )]
[
= a − sint +
tan 2
1
l
,
d
dt ( )tan
t
2
]
[
= a − sint + cot
t
2
⋅ sec 2
t
2
⋅
d
dt 2( )] t
[
= a − sint +
1
t
2sin 2 cos 2
t
]
(
= a − sint +
1
sint )
=a
( − sin 2t + 1
sint )
cos 2t
=a
sint
dy d
dt
= a dt (sint) = acost
dy ( ) dy
dt
acos t sin t
∴ dx = = = cos t = tant
( ) ( dx
dt
cos 2 t
a sin t
)
Page : 181 , Block Name : Exercise 5.6
Q9 If x and y are connected parametrically by the equation, without eliminating the parameter,
nd
dy
Processing
x = asecθ,math:
y = 100%
btanθ
dx
Page 75
Answer.
The given equations are x = asecθ and y = btanθ
dx d
Then, dθ = a ⋅ dθ (secθ) = asecθtanθ
dy d
dθ
= b ⋅ dθ (tanθ) = bsec 2θ
dy ( ) dy
dθ
bsec 2 θ b bcos θ b 1 b
∴ dx = = asec θtan θ
= a secθcotθ = acos θsin θ = a × sin θ = a cscθ
( ) dx
dθ
Page : 181 , Block Name : Exercise 5.6
Q10 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
x = a(cosθ + θsinθ), y = a(sinθ − θcosθ)
Answer.
The given equations are (x = a(cosθ + θsinθ) and y = a(sinθ − θcosθ)
dx
[ d d
] [
Then, dθ = a dθ cosθ + dθ (θsinθ) = a − sinθ + θ dθ (sinθ) + sinθ dθ (θ)
d d
]
= a[ − sinθ + θcosθ + sinθ] = aθcosθ
dy
dθ
=a
[ d
dθ
(sinθ) −
d
dθ ] [
(θcosθ) = a cosθ −
{ θ
d
dθ
(cosθ) + cosθ ⋅
d
dθ
(θ)
}]
= a[cosθ + θsinθ − cosθ]
= aθsinθ
dy ( ) dy
dθ
aθsin θ
∴ dx = = aθcos θ
= tanθ
( ) dx
dθ
Page : 181 , Block Name : Exercise 5.6
dy y
√ √
−1 t −1 y
Q11 x = a sin ,y = a cos , show that dx = − x
√asin and y = √acos t
−1 −1
Answer. The given equations are x =
The given equations are x = asecθ and y = btanθ
dx d
Then, dθ = a ⋅ dθ (secθ) = asecθtanθ
dy d
dθ
= b ⋅ dθ (tanθ) = bsec 2θ
dy ( ) dy
dθ
bsec 2 θ b bcos θ b 1 b
∴ dx = = asec θtan θ
= a secθcotθ = acos θsin θ = a × sin θ = a cscθ
( ) dx
dθ
Page : 181 , Block Name : Exercise 5.6
Q10 If x and y are connected parametrically by the equation, without eliminating the parameter,
dy
nd dx
x = a(cosθ + θsinθ), y = a(sinθ − θcosθ)
Answer.
The given equations are (x = a(cosθ + θsinθ) and y = a(sinθ − θcosθ)
dx
[ d d
] [
Then, dθ = a dθ cosθ + dθ (θsinθ) = a − sinθ + θ dθ (sinθ) + sinθ dθ (θ)
d d
]
= a[ − sinθ + θcosθ + sinθ] = aθcosθ
dy
dθ
=a
[ d
dθ
(sinθ) −
d
dθ ] [
(θcosθ) = a cosθ −
{ θ
d
dθ
(cosθ) + cosθ ⋅
d
dθ
(θ)
}]
= a[cosθ + θsinθ − cosθ]
= aθsinθ
dy ( ) dy
dθ
aθsin θ
∴ dx = = aθcos θ
= tanθ
( ) dx
dθ
Page : 181 , Block Name : Exercise 5.6
dy y
√ √
−1 t −1 y
Q11 x = a sin ,y = a cos , show that dx = − x
√asin and y = √acos t
−1 −1
Answer. The given equations are x =
Page 76
√asin t and y = √acos t
−1 −1
x=
1 1
⇒x= a ( sin − 1 t
) 2
and y = a ( cos − 1 t
) 2
1 1
−1 t −1 t
⇒ x = a 2 sin and y = a 2 cos
1
−1 t
Consider x = a 2 sin
Taking logarithm on both the sides, we obtain
1
logx = 2 sin − 1tloga
( )
1 dx 1 d
∴ x ⋅ dt = 2 loga ⋅ dt sin − 1t
dx x 1
⇒ dt = 2 loga ⋅
√1 − t 2
dx xlog a
⇒ dt =
√
2 1 − t2
1
logx = 2 sin − 1tloga
( )
1 dx 1 d
∴ x ⋅ dt = 2 loga ⋅ dt sin − 1t
dx x 1
⇒ dt = 2 loga ⋅
√1 − t 2
dx xlog a
⇒ dt =
√
2 1 − t2
1
logy = 2 cos − 1tloga
( )
1 dy 1 d
∴ y ⋅ dx = 2 loga ⋅ dt cos − 1t
dy
⇒ dt =
ylog a
2 ⋅
( ) −1
√1 − t 2
dy − ylog a
⇒ dt =
√
2 1 − t2
( ) (√ )
− ylog a
dy
dt 2 1 − t2
dy y
∴ dx = = = − x
( ) (√ )
dx
dt
xlog a
2 1 − t2
Hence proved
Page : 181 , Block Name : Exercise 5.6
−1 −1
x=
1 1
⇒x= a ( sin − 1 t
) 2
and y = a ( cos − 1 t
) 2
1 1
−1 t −1 t
⇒ x = a 2 sin and y = a 2 cos
1
−1 t
Consider x = a 2 sin
Taking logarithm on both the sides, we obtain
1
logx = 2 sin − 1tloga
( )
1 dx 1 d
∴ x ⋅ dt = 2 loga ⋅ dt sin − 1t
dx x 1
⇒ dt = 2 loga ⋅
√1 − t 2
dx xlog a
⇒ dt =
√
2 1 − t2
1
logx = 2 sin − 1tloga
( )
1 dx 1 d
∴ x ⋅ dt = 2 loga ⋅ dt sin − 1t
dx x 1
⇒ dt = 2 loga ⋅
√1 − t 2
dx xlog a
⇒ dt =
√
2 1 − t2
1
logy = 2 cos − 1tloga
( )
1 dy 1 d
∴ y ⋅ dx = 2 loga ⋅ dt cos − 1t
dy
⇒ dt =
ylog a
2 ⋅
( ) −1
√1 − t 2
dy − ylog a
⇒ dt =
√
2 1 − t2
( ) (√ )
− ylog a
dy
dt 2 1 − t2
dy y
∴ dx = = = − x
( ) (√ )
dx
dt
xlog a
2 1 − t2
Hence proved
Page : 181 , Block Name : Exercise 5.6
Page 77
Q1 Find the second order derivatives of the function
x 2 + 3x + 2
Answer.
Let y = x 2 + 3x + 2
Then,
( )
dy d d d
dx
= dx x 2 + dx (3x) + dx (2) = 2x + 3 + 0 = 2x + 3
d 2y d d d
∴ 2 = dx (2x + 3) = dx (2x) + dx (3) = 2 + 0 = 2
dx
Page : 183 , Block Name : Exercise 5.7
Q2 Find the second order derivatives of the function
x 20
Answer.
Let y = x 20
Then,
( )
dy d
dx
= dx x 20 = 20x 19
d 2y
= (20x ) = 20 (x ) = 20 ⋅ 19 ⋅ x = 380x
d d
19 19 18 18
∴ dx dx
dx 2
Page : 183 , Block Name : Exercise 5.7
Q3 Find the second order derivatives of the function.
x ⋅ cosx
Answer.
Let y = x ⋅ cosx
Then
dy d d d
dx = dx (x ⋅ cosx) = cosx ⋅ dx (x) + x dx (cosx) = cosx ⋅ 1 + x( − sinx) = cosx − xsinx
d 2y d d d
∴ = dx [cosx − xsinx] = dx (cosx) − dx (xsinx)
dx 2
d 2y d d d
∴ 2
= [cosx − xsinx] = (cosx) − (xsinx)
dx dx dx dx
[
= − sinx − sinx ⋅
d
dx
(x) + x ⋅
d
dx
(sinx)
]
= − sinx − (sinx + xcosx)
= − (xcosx + 2sinx)
x 2 + 3x + 2
Answer.
Let y = x 2 + 3x + 2
Then,
( )
dy d d d
dx
= dx x 2 + dx (3x) + dx (2) = 2x + 3 + 0 = 2x + 3
d 2y d d d
∴ 2 = dx (2x + 3) = dx (2x) + dx (3) = 2 + 0 = 2
dx
Page : 183 , Block Name : Exercise 5.7
Q2 Find the second order derivatives of the function
x 20
Answer.
Let y = x 20
Then,
( )
dy d
dx
= dx x 20 = 20x 19
d 2y
= (20x ) = 20 (x ) = 20 ⋅ 19 ⋅ x = 380x
d d
19 19 18 18
∴ dx dx
dx 2
Page : 183 , Block Name : Exercise 5.7
Q3 Find the second order derivatives of the function.
x ⋅ cosx
Answer.
Let y = x ⋅ cosx
Then
dy d d d
dx = dx (x ⋅ cosx) = cosx ⋅ dx (x) + x dx (cosx) = cosx ⋅ 1 + x( − sinx) = cosx − xsinx
d 2y d d d
∴ = dx [cosx − xsinx] = dx (cosx) − dx (xsinx)
dx 2
d 2y d d d
∴ 2
= [cosx − xsinx] = (cosx) − (xsinx)
dx dx dx dx
[
= − sinx − sinx ⋅
d
dx
(x) + x ⋅
d
dx
(sinx)
]
= − sinx − (sinx + xcosx)
= − (xcosx + 2sinx)
Page 78
Page : 183 , Block Name : Exercise 5.7
Q4 Find the second order derivatives of the function.
logx
Answer.
Let y = logx
Then,
dy d 1
dx
= dx (logx) = x
∴
d 2y
dx 2
d
= dx x ()
1
=
−1
x2
Page : 183 , Block Name : Exercise 5.7
Q5 Find the second order derivatives of the function.
x 3logx
Answer.
Let y = x 3logx
Then
dy d 3 d 3 d
dx
=
dx [
x logx = logx ⋅ ] dx
x + x3 ⋅ ( )
dx
(logx)
1
= logx ⋅ 3x 2 + x 3 ⋅ = logx ⋅ 3x 2 + x 2
x
= x 2(1 + 3logx)
( )
d d
= (1 + 3logx) ⋅ dx x 2 + x 2 dx (1 + 3logx)
3
= (1 + 3logx) ⋅ 2x + x 2 ⋅ x
= 2x + 6xlogx + 3x
= 5x + 6xlogx
= x(5 + 6logx)
Page : 183 , Block Name : Exercise 5.7
Q6 Find the second order derivatives of the function
y = e xsin5x
Answer.
Q4 Find the second order derivatives of the function.
logx
Answer.
Let y = logx
Then,
dy d 1
dx
= dx (logx) = x
∴
d 2y
dx 2
d
= dx x ()
1
=
−1
x2
Page : 183 , Block Name : Exercise 5.7
Q5 Find the second order derivatives of the function.
x 3logx
Answer.
Let y = x 3logx
Then
dy d 3 d 3 d
dx
=
dx [
x logx = logx ⋅ ] dx
x + x3 ⋅ ( )
dx
(logx)
1
= logx ⋅ 3x 2 + x 3 ⋅ = logx ⋅ 3x 2 + x 2
x
= x 2(1 + 3logx)
( )
d d
= (1 + 3logx) ⋅ dx x 2 + x 2 dx (1 + 3logx)
3
= (1 + 3logx) ⋅ 2x + x 2 ⋅ x
= 2x + 6xlogx + 3x
= 5x + 6xlogx
= x(5 + 6logx)
Page : 183 , Block Name : Exercise 5.7
Q6 Find the second order derivatives of the function
y = e xsin5x
Answer.
Page 79
dy d d d
dx (
= e sin5x ) = sin5x (e ) + e
x
(sin5x)x x
dx dx dx
d
= sin5x ⋅ e x + e x ⋅ cos5x ⋅ (5x) = e xsin5x + e xcos5x ⋅ 5
dx
= e x(sin5x + 5cos5x)
d 2y d
∴
dx 2
= [e (sin5x + 5cos5x) ]
dx
x
d d
= (sin5x + 5cos5x) ⋅
dx (e )+e ⋅
dx
x
(sin5x + 5cos5x) x
= (sin5x + 5cos5x) ⋅ (e ) + e ⋅ (sin5x + 5cos5x)
d d
x x
dx dx
[ d
= (sin5x + 5cos5x)e x + e x cos5x ⋅ dx (5x) + 5( − sin5x) ⋅ dx (5x)
d
]
= e x(sin5x + 5cos5x) + e x(5cos5x − 25sin5x)
= e x(10cos5x − 24sin5x) = 2e x(5cos5x − 12sin5x)
Page : 183 , Block Name : Exercise 5.7
Q7 Find the second order derivatives of the function.
e 6xcos3x
Answer.
Let y = e 6xcos3x
Then
dy d 6x d 6x d
dx
=
dx (
e ⋅ cos3x = cos3x ⋅ ) dx
e + e 6x ⋅
dx ( )
(cos3x)
d d
= cos3x ⋅ e 6x ⋅ (6x) + e 6x ⋅ ( − sin3x) ⋅ (3x)
dx dx
= 6e 6xcos3x − 3e 6xsin3x
d 2y d d d
∴ 2 =
dx dx (
6e 6xcos3x − 3e 6xsin3x = 6 ⋅
dx )
e 6xcos3x − 3 ⋅
dx
e 6xsin3x ( ) ( )
[
= 6 ⋅ 6e 6xcos3x − 3e 6xsin3x − 3 ⋅ sin3x ⋅ ] [ d
dx ( )
e 6x
+e ⋅
dx
(sin3x) ]
4x
d
= 36e 6xcos3x − 18e 4xsin3x − 18e 6πsin3x − 9e 6xcos3x
= 27e 6xcos3x − 36e 6xsin3x
= 9e 6x(3cos3x − 4sin3x)
Page : 183 , Block Name : Exercise 5.7
Q8 Find the second order derivatives of the function.
−1
tanProcessing
x math: 100%
dx (
= e sin5x ) = sin5x (e ) + e
x
(sin5x)x x
dx dx dx
d
= sin5x ⋅ e x + e x ⋅ cos5x ⋅ (5x) = e xsin5x + e xcos5x ⋅ 5
dx
= e x(sin5x + 5cos5x)
d 2y d
∴
dx 2
= [e (sin5x + 5cos5x) ]
dx
x
d d
= (sin5x + 5cos5x) ⋅
dx (e )+e ⋅
dx
x
(sin5x + 5cos5x) x
= (sin5x + 5cos5x) ⋅ (e ) + e ⋅ (sin5x + 5cos5x)
d d
x x
dx dx
[ d
= (sin5x + 5cos5x)e x + e x cos5x ⋅ dx (5x) + 5( − sin5x) ⋅ dx (5x)
d
]
= e x(sin5x + 5cos5x) + e x(5cos5x − 25sin5x)
= e x(10cos5x − 24sin5x) = 2e x(5cos5x − 12sin5x)
Page : 183 , Block Name : Exercise 5.7
Q7 Find the second order derivatives of the function.
e 6xcos3x
Answer.
Let y = e 6xcos3x
Then
dy d 6x d 6x d
dx
=
dx (
e ⋅ cos3x = cos3x ⋅ ) dx
e + e 6x ⋅
dx ( )
(cos3x)
d d
= cos3x ⋅ e 6x ⋅ (6x) + e 6x ⋅ ( − sin3x) ⋅ (3x)
dx dx
= 6e 6xcos3x − 3e 6xsin3x
d 2y d d d
∴ 2 =
dx dx (
6e 6xcos3x − 3e 6xsin3x = 6 ⋅
dx )
e 6xcos3x − 3 ⋅
dx
e 6xsin3x ( ) ( )
[
= 6 ⋅ 6e 6xcos3x − 3e 6xsin3x − 3 ⋅ sin3x ⋅ ] [ d
dx ( )
e 6x
+e ⋅
dx
(sin3x) ]
4x
d
= 36e 6xcos3x − 18e 4xsin3x − 18e 6πsin3x − 9e 6xcos3x
= 27e 6xcos3x − 36e 6xsin3x
= 9e 6x(3cos3x − 4sin3x)
Page : 183 , Block Name : Exercise 5.7
Q8 Find the second order derivatives of the function.
−1
tanProcessing
x math: 100%
Page 80
Answer.
Let y = tan − 1x
Then,
( )
dy d 1
dx
= dx tan − 1x =
1 + x2
∴
d 2y
dx 2
= dx
d
( ) 1
1 + x2
d
= dx 1 + x 2( ) = ( − 1) ⋅ (1 + x ) ⋅ (1 + x )
−1 2 −2
d
dx
2
−1 − 2x
= × 2x =
(1+x ) 2 2
(1 +x ) 2 2
Page : 183 , Block Name : Exercise 5.7
Q9 Find the second order derivatives of the function.
log(logx)
Answer.
Let y = log(logx)
Then
dy d 1 d 1
dx
= dx [log(logx)] = log x ⋅ dx (logx) = xlog x = (xlogx) − 1
=
−1
( xlog x ) 2 [ d
logx ⋅ dx (x) + x ⋅ dx (logx)
d
]
=
−1
( xlog x ) 2 ⋅ [ logx ⋅ 1 + x ⋅ x
1
] =
− ( 1 + log x )
( xlog x ) 2
Page : 183 , Block Name : Exercise 5.7
Q10
Find the second order derivatives of the function.
sin(logx)
dy d d cos ( log x )
Answer. dx = dx [sin(logx)] = cos(logx) ⋅ dx (logx) = x
∴
d 2y
dx 2 = dx
d
[ cos ( log x )
x ]
Let y = tan − 1x
Then,
( )
dy d 1
dx
= dx tan − 1x =
1 + x2
∴
d 2y
dx 2
= dx
d
( ) 1
1 + x2
d
= dx 1 + x 2( ) = ( − 1) ⋅ (1 + x ) ⋅ (1 + x )
−1 2 −2
d
dx
2
−1 − 2x
= × 2x =
(1+x ) 2 2
(1 +x ) 2 2
Page : 183 , Block Name : Exercise 5.7
Q9 Find the second order derivatives of the function.
log(logx)
Answer.
Let y = log(logx)
Then
dy d 1 d 1
dx
= dx [log(logx)] = log x ⋅ dx (logx) = xlog x = (xlogx) − 1
=
−1
( xlog x ) 2 [ d
logx ⋅ dx (x) + x ⋅ dx (logx)
d
]
=
−1
( xlog x ) 2 ⋅ [ logx ⋅ 1 + x ⋅ x
1
] =
− ( 1 + log x )
( xlog x ) 2
Page : 183 , Block Name : Exercise 5.7
Q10
Find the second order derivatives of the function.
sin(logx)
dy d d cos ( log x )
Answer. dx = dx [sin(logx)] = cos(logx) ⋅ dx (logx) = x
∴
d 2y
dx 2 = dx
d
[ cos ( log x )
x ]
Page 81
d d
x ⋅ dx [cos(logx)] − cos(logx) ⋅ dx (x)
=
x2
=
x⋅
[ d
]
− sin(logx) ⋅ dx (logx) − cos(logx) ⋅ 1
x2
1
− xsin(logx) ⋅ x − cos(logx)
=
x2
− [sin(logx) + cos(logx)]
=
x2
Page : 183 , Block Name : Exercise 5.7
d 2y
Q11 If y = 5cosx − 3sinx, prove that +y=0
dx 2
Answer.
It is given that, y = 5cosx − 3sinx
Then,
dy d d d d
= (5cosx) − (3sinx) = 5 (cosx) − 3 (sinx)
dx dx dx dx dx
= 5( − sinx) − 3cosx = − (5sinx + 3cosx)
d 2y d
∴ 2 = [ − (5sinx + 3cosx)]
dx dx
= − 5⋅ [ dx
d
(sinx) + 3 ⋅
d
dx
(cosx) ]
= − [5cosx + 3( − sinx)]
= − [5cosx − 3sinx]
d 2y
∴ +y=0
dx 2
Hence proved
Page : 183 , Block Name : Exercise 5.7
d 2y
Q12 y = cos − 1x, find in terms of y
dx 2
Answer.
It is given that, y = cos − 1x
Then,
x ⋅ dx [cos(logx)] − cos(logx) ⋅ dx (x)
=
x2
=
x⋅
[ d
]
− sin(logx) ⋅ dx (logx) − cos(logx) ⋅ 1
x2
1
− xsin(logx) ⋅ x − cos(logx)
=
x2
− [sin(logx) + cos(logx)]
=
x2
Page : 183 , Block Name : Exercise 5.7
d 2y
Q11 If y = 5cosx − 3sinx, prove that +y=0
dx 2
Answer.
It is given that, y = 5cosx − 3sinx
Then,
dy d d d d
= (5cosx) − (3sinx) = 5 (cosx) − 3 (sinx)
dx dx dx dx dx
= 5( − sinx) − 3cosx = − (5sinx + 3cosx)
d 2y d
∴ 2 = [ − (5sinx + 3cosx)]
dx dx
= − 5⋅ [ dx
d
(sinx) + 3 ⋅
d
dx
(cosx) ]
= − [5cosx + 3( − sinx)]
= − [5cosx − 3sinx]
d 2y
∴ +y=0
dx 2
Hence proved
Page : 183 , Block Name : Exercise 5.7
d 2y
Q12 y = cos − 1x, find in terms of y
dx 2
Answer.
It is given that, y = cos − 1x
Then,
Page 82
−1
( )= √ ( )
dy d −1
dx
= dx cos − 1x = − 1 − x2 2
1 − x2
[( )]
−1
d 2y d
2
= dx − 1−x 2
dx 2
( )(
−3
) ( )
1 d
= − − 2 ⋅ 1 − x2 2 ⋅ dx 1 − x 2
1
= × ( − 2x)
2 √( 1 − x2 ) 3
d 2y −x
2 =
dx
√ ( 1 − x2 ) 3
d 2y −x
⇒ 2 =
dx
√ ( 1 − x2 ) 3
y = cos − 1x ⇒ x = cosy
Putting x = cosy in equation (i), we obtain
d 2y − cosy
=
dx 2
1 − cos 2y √(
3
)
d 2y − cosy
⇒ =
dx 2
√ (sin2y )3
− cosy
=
sin 3y
− cosy 1
= ×
siny sin 2y
d 2y
⇒ = − coty ⋅ coscc 2y
dx 2
Page : 184 , Block Name : Exercise 5.7
Q13 y = 3cos(logx) + 4sin(logx), show that x 2y 2 + xy 1 + y = 0
Answer.
It is given that, y = 3cos(logx) + 4sin(logx)
Then,
( )= √ ( )
dy d −1
dx
= dx cos − 1x = − 1 − x2 2
1 − x2
[( )]
−1
d 2y d
2
= dx − 1−x 2
dx 2
( )(
−3
) ( )
1 d
= − − 2 ⋅ 1 − x2 2 ⋅ dx 1 − x 2
1
= × ( − 2x)
2 √( 1 − x2 ) 3
d 2y −x
2 =
dx
√ ( 1 − x2 ) 3
d 2y −x
⇒ 2 =
dx
√ ( 1 − x2 ) 3
y = cos − 1x ⇒ x = cosy
Putting x = cosy in equation (i), we obtain
d 2y − cosy
=
dx 2
1 − cos 2y √(
3
)
d 2y − cosy
⇒ =
dx 2
√ (sin2y )3
− cosy
=
sin 3y
− cosy 1
= ×
siny sin 2y
d 2y
⇒ = − coty ⋅ coscc 2y
dx 2
Page : 184 , Block Name : Exercise 5.7
Q13 y = 3cos(logx) + 4sin(logx), show that x 2y 2 + xy 1 + y = 0
Answer.
It is given that, y = 3cos(logx) + 4sin(logx)
Then,
Page 83
d d
y1 = 3 ⋅ [cos(logx)] + 4 ⋅ [sin(logx)]
dx dx
=3⋅ [ − sin(logx) ⋅
− 3sin ( log x )
d
dx ] [
(logx) + 4 ⋅ cos(logx) ⋅
4cos ( log x )
d
dx
(logx)
4cos ( log x ) − 3sin ( log x )
]
∴ y1 = x
+ x
= x
∴ y 2 = dx
d
( 4cos ( log x ) − 3sin ( log x )
x )
x{4cos(logx) − 3sin(logx)} ′ − {4cos(logx) − 3sin(logx)}(x) ′
=
x2
[ } ′
x 4{cos(logx)} ′ − 3sin(logx) ] − {4cos(logx) − 3sin(logx)} !
=
x2
[
x − 4sin(logx) ⋅ (logx) ′ − 3cos(logx)(logx) ′ − 4cos(logx) + 3sin(logx)]
=
x2
[
x − 4sin(logx) ⋅ (logx) ′ − 3cos(logx)(logx) ′ − 4cos(logx) + 3sin(logx)]
=
x2
=
[ 1 1
x − 4sin(logx) ⋅ x − 3cos(logx) 2 − 4cos(logx) + 3sin(logx)
x
x2
− 4sin(logx) − 3cos(logx) − 4cos(logx) + 3sin(logx)
=
x2
− sin ( log x ) − 7cos ( log x )
=
x2
x 2y 2 + xy 1 + y
= x2
( − sin ( log x ) − 7cos ( log x )
x2 ) (
+x
4cos ( log x ) − 3sin ( log x )
x ) + 3cos(logx) + 4sin(logx)
= − sin(logx) − 7cos(logx) + 4cos(logx) − 3sin(logx) + 3cos(logx) + 4sin(logx)
Hence, proved.
Page : 184 , Block Name : Exercise 5.7
d 2y dy
Q14 If y = Ae mx + Be m Show that 2 − (m + n) dx + mny = 0
dx
Answer. It is given that y = Ae mx + Be α
y1 = 3 ⋅ [cos(logx)] + 4 ⋅ [sin(logx)]
dx dx
=3⋅ [ − sin(logx) ⋅
− 3sin ( log x )
d
dx ] [
(logx) + 4 ⋅ cos(logx) ⋅
4cos ( log x )
d
dx
(logx)
4cos ( log x ) − 3sin ( log x )
]
∴ y1 = x
+ x
= x
∴ y 2 = dx
d
( 4cos ( log x ) − 3sin ( log x )
x )
x{4cos(logx) − 3sin(logx)} ′ − {4cos(logx) − 3sin(logx)}(x) ′
=
x2
[ } ′
x 4{cos(logx)} ′ − 3sin(logx) ] − {4cos(logx) − 3sin(logx)} !
=
x2
[
x − 4sin(logx) ⋅ (logx) ′ − 3cos(logx)(logx) ′ − 4cos(logx) + 3sin(logx)]
=
x2
[
x − 4sin(logx) ⋅ (logx) ′ − 3cos(logx)(logx) ′ − 4cos(logx) + 3sin(logx)]
=
x2
=
[ 1 1
x − 4sin(logx) ⋅ x − 3cos(logx) 2 − 4cos(logx) + 3sin(logx)
x
x2
− 4sin(logx) − 3cos(logx) − 4cos(logx) + 3sin(logx)
=
x2
− sin ( log x ) − 7cos ( log x )
=
x2
x 2y 2 + xy 1 + y
= x2
( − sin ( log x ) − 7cos ( log x )
x2 ) (
+x
4cos ( log x ) − 3sin ( log x )
x ) + 3cos(logx) + 4sin(logx)
= − sin(logx) − 7cos(logx) + 4cos(logx) − 3sin(logx) + 3cos(logx) + 4sin(logx)
Hence, proved.
Page : 184 , Block Name : Exercise 5.7
d 2y dy
Q14 If y = Ae mx + Be m Show that 2 − (m + n) dx + mny = 0
dx
Answer. It is given that y = Ae mx + Be α
Page 84
( ) ( )
dy d d d d
dx
= A ⋅ dx e mx + B ⋅ dx e nx = A ⋅ e mx ⋅ dx (mx) + B ⋅ e nx ⋅ dx (nx) = Ame mx + Bne m
d 2y
( ) ( ) ( )
d d d
= dx Ame mx + Bne nx = Am ⋅ dx e mx + Bn ⋅ dx e nx
dx 2
d d
= Am ⋅ e mx ⋅ dx (mx) + Bn ⋅ e mx ⋅ dx (nx) = Am 2e nx + Bn 2e nx
(
= Am 2e mx + Bn 2e m − (m + n) ⋅ Ame mx + Bne m + mn Ae mx + Be m ) ( )
= 0 Hence proved
Page : 184 , Block Name : Exercise 5.7
d 2y
Q15 If y = 500e 7x + 600e − 7x Show that = 49y
dx 2
Answer.
It is given that, y = 500e 7x + 600e − 7x
Then,
dy d 7x d
dx
= 500 ⋅
dx
e ( )
+ 600 ⋅
dx
e − 7x ( )
d d
= 500 ⋅ e 7x ⋅ (7x) + 600 ⋅ e − 2x ⋅ ( − 7x)
dx dx
= 3500e 7x − 4200e − 7x
d d
= 3500 ⋅ e 7x ⋅ (7x) − 420 ⋅ e − 7x ⋅ ( − 7x)
dx dx
= 7 × 3500 ⋅ e 7x + 7 × 4200 ⋅ e − 7x
= 49 × 500e 7x + 49 × 600 ⋅ e − 7x
(
= 49 500e 7x + 400e − 7x )
= 49y
Hence proved.
Page : 184 , Block Name : Exercise 5.7
Q16 e y(x + 1) = 1 Show that
d 2y
dx 2 = () dy
dx
2
Answer.
2
The given relationship is e ( x + 1 ) = 1
e y(x + 1) = 1
1
⇒ ey = x + 1
dy d d d d
dx
= A ⋅ dx e mx + B ⋅ dx e nx = A ⋅ e mx ⋅ dx (mx) + B ⋅ e nx ⋅ dx (nx) = Ame mx + Bne m
d 2y
( ) ( ) ( )
d d d
= dx Ame mx + Bne nx = Am ⋅ dx e mx + Bn ⋅ dx e nx
dx 2
d d
= Am ⋅ e mx ⋅ dx (mx) + Bn ⋅ e mx ⋅ dx (nx) = Am 2e nx + Bn 2e nx
(
= Am 2e mx + Bn 2e m − (m + n) ⋅ Ame mx + Bne m + mn Ae mx + Be m ) ( )
= 0 Hence proved
Page : 184 , Block Name : Exercise 5.7
d 2y
Q15 If y = 500e 7x + 600e − 7x Show that = 49y
dx 2
Answer.
It is given that, y = 500e 7x + 600e − 7x
Then,
dy d 7x d
dx
= 500 ⋅
dx
e ( )
+ 600 ⋅
dx
e − 7x ( )
d d
= 500 ⋅ e 7x ⋅ (7x) + 600 ⋅ e − 2x ⋅ ( − 7x)
dx dx
= 3500e 7x − 4200e − 7x
d d
= 3500 ⋅ e 7x ⋅ (7x) − 420 ⋅ e − 7x ⋅ ( − 7x)
dx dx
= 7 × 3500 ⋅ e 7x + 7 × 4200 ⋅ e − 7x
= 49 × 500e 7x + 49 × 600 ⋅ e − 7x
(
= 49 500e 7x + 400e − 7x )
= 49y
Hence proved.
Page : 184 , Block Name : Exercise 5.7
Q16 e y(x + 1) = 1 Show that
d 2y
dx 2 = () dy
dx
2
Answer.
2
The given relationship is e ( x + 1 ) = 1
e y(x + 1) = 1
1
⇒ ey = x + 1
Page 85
e y(x + 1) = 1
1
⇒ ey = x + 1
Taking logarithm on both the sides, we obtain
1
y = log ( x + 1 )
Differentiating this relationship with respect to x, we obtain
dy
dx ( )
= (x + 1) dx x + 1
d 1
= (x + 1) ⋅
−1
(x+1) 2 = x+1
−1
∴
d 2y
dx 2 ( )
= − dx x + 1
d 1
= −
( ) −1
( x + 1 )2
=
1
( x + 1 )2
⇒
d 2y
dx 2 = ( ) x+1
−1 2
⇒
d 2y
dx 2 = () dy
dx
2
Hence proved
Page : 184 , Block Name : Exercise 5.7
−1
Q17 if y = tan x ( ) , show that (x2 + 1 )2y2 + 2x (x2 + 1 )y1 = 2
2
Answer.
The given relationship is y = tan − 1x ( ) 2
Then,
( )
d
y 1 = 2tan − 1x dx tan − 1x
1
⇒ y 1 = 2tan − 1x ⋅
1 + x2
( )
⇒ 1 + x 2 y 1 = 2tan − 1x
Again differentiating with respect to x on both the sides, we obtain
( )
1 + x 2 y 2 + 2xy 1 = 2
( ) 1
1 + x2
( ) 2
⇒ 1 + x 2 y 2 + 2x 1 + x 2 y 1 = 2 ( )
Hence, proved.
1
⇒ ey = x + 1
Taking logarithm on both the sides, we obtain
1
y = log ( x + 1 )
Differentiating this relationship with respect to x, we obtain
dy
dx ( )
= (x + 1) dx x + 1
d 1
= (x + 1) ⋅
−1
(x+1) 2 = x+1
−1
∴
d 2y
dx 2 ( )
= − dx x + 1
d 1
= −
( ) −1
( x + 1 )2
=
1
( x + 1 )2
⇒
d 2y
dx 2 = ( ) x+1
−1 2
⇒
d 2y
dx 2 = () dy
dx
2
Hence proved
Page : 184 , Block Name : Exercise 5.7
−1
Q17 if y = tan x ( ) , show that (x2 + 1 )2y2 + 2x (x2 + 1 )y1 = 2
2
Answer.
The given relationship is y = tan − 1x ( ) 2
Then,
( )
d
y 1 = 2tan − 1x dx tan − 1x
1
⇒ y 1 = 2tan − 1x ⋅
1 + x2
( )
⇒ 1 + x 2 y 1 = 2tan − 1x
Again differentiating with respect to x on both the sides, we obtain
( )
1 + x 2 y 2 + 2xy 1 = 2
( ) 1
1 + x2
( ) 2
⇒ 1 + x 2 y 2 + 2x 1 + x 2 y 1 = 2 ( )
Hence, proved.
Page 86
Page : 184 , Block Name : Exercise 5.7
Q1 Verify Rolle’s Theorem for the function f(x) = x 2 + 2x − 8, x ∈ [ − 4, 2]
Answer. The given function f(x) = x 2 + 2x − 8, being a polynomial function, is continuous in [−4, 2]
and is differentiable in (−4, 2).
f( − 4) = ( − 4) 2 + 2 × ( − 4) − 8 = 16 − 8 − 8 = 0
f(2) = (2) 2 + 2 × 2 − 8 = 4 + 4 − 8 = 0
∴ f( − 4) = f(2) = 0
⇒ Thevalueoff(x)at − 4and2coincides.
Rolle’s Theorem states that there is a point c ∈ ( − 4, 2) such that f ′(c) = 0
f(x) = x 2 + 2x − 8
⇒ f ′(x) = 2x + 2
∴ f ′(c) = 0
⇒ 2c + 2 = 0
⇒ c = − 1, where c = − 1 ∈ ( − 4, 2)
Hence, Rolle's Theorem is verified for the given function.
Page : 186 , Block Name : Exercise 5.8
Q2 Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say
something about the converse of Rolle’s Theorem from these examples?
f(x) = [x] for x ∈ [5, 9]
f(x) = [x] for x ∈ [ − 2, 2]
(iii) f(x) = x 2 − 1 for x ∈ [1, 2]
Answer.
By Rolle's Theorem, for a function f : [a, b] → R , if
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
(c) f(a) = f(b)
then, there exists some c ∈ (a, b) such that f ′(c) = 0
Therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of
the three conditions of the hypothesis.
(i) f(x) = [x] for x ∈ [5, 9]
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9
⇒ f(x) is not continuous in [5, 9]
Also, f(5) = [5] = 5 and f(9) = [9] = 9
∴ f(5) ≠ f(9)
The differentiability of f in (5, 9) is checked as follows.
Q1 Verify Rolle’s Theorem for the function f(x) = x 2 + 2x − 8, x ∈ [ − 4, 2]
Answer. The given function f(x) = x 2 + 2x − 8, being a polynomial function, is continuous in [−4, 2]
and is differentiable in (−4, 2).
f( − 4) = ( − 4) 2 + 2 × ( − 4) − 8 = 16 − 8 − 8 = 0
f(2) = (2) 2 + 2 × 2 − 8 = 4 + 4 − 8 = 0
∴ f( − 4) = f(2) = 0
⇒ Thevalueoff(x)at − 4and2coincides.
Rolle’s Theorem states that there is a point c ∈ ( − 4, 2) such that f ′(c) = 0
f(x) = x 2 + 2x − 8
⇒ f ′(x) = 2x + 2
∴ f ′(c) = 0
⇒ 2c + 2 = 0
⇒ c = − 1, where c = − 1 ∈ ( − 4, 2)
Hence, Rolle's Theorem is verified for the given function.
Page : 186 , Block Name : Exercise 5.8
Q2 Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say
something about the converse of Rolle’s Theorem from these examples?
f(x) = [x] for x ∈ [5, 9]
f(x) = [x] for x ∈ [ − 2, 2]
(iii) f(x) = x 2 − 1 for x ∈ [1, 2]
Answer.
By Rolle's Theorem, for a function f : [a, b] → R , if
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
(c) f(a) = f(b)
then, there exists some c ∈ (a, b) such that f ′(c) = 0
Therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of
the three conditions of the hypothesis.
(i) f(x) = [x] for x ∈ [5, 9]
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9
⇒ f(x) is not continuous in [5, 9]
Also, f(5) = [5] = 5 and f(9) = [9] = 9
∴ f(5) ≠ f(9)
The differentiability of f in (5, 9) is checked as follows.
Page 87
The left hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−1−n −1
lim h → 0 h
= lim h → ∞ h
= lim h → 0 h
= lim h → ∞ h = ∞
The right hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−n
lim h → ∞ h
= lim h → 0 + h
= lim h → 0 h = lim h → ∞0 = 0
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle's
Theorem.
Hence, Rolle's Theorem is not applicable for f(x) = [x] for x ∈ [5, 9]
(ii) f(x) = [x] for x ∈ [ − 2, 2]
It is evident that the given function f(x) is not continuous at every integral point
In particular, f(x) is not continuous at x = − 2 and x = 2
⇒ f(x) is not continuous in [ − 2, 2] .
Also, f( − 2) = [ − 2] = − 2 and f(2) = [2] = 2
∴ f( − 2) ≠ f(2)
∴ f( − 2) ≠ f(2)
The differentiability of f in ( − 2, 2) is checked as follows.
The left hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−1−n −1
lim h → 0 h
= lim h → 0 h
= lim h → 0 h
= lim h → ∞ h = ∞
The right hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−n
lim h → 0 h
= lim h → 0 − h
= lim h → 0 h = lim h → ∞0 = 0
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x
=n
∴ f is not differentiable in ( − 2, 2)
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle's
Theorem.
Hence, Rolles Theorem is not applicable for f(x) = [x] for x ∈ [ − 2, 2]
(iii) f(x) = x 2 − 1 for x ∈ [1, 2]
It is evident that f, being a polynomial function, is continuous in [1, 2] and is
differentiable in (1, 2) .
f(1) = (1) 2 − 1 = 0
f(2) = (2) 2 − 1 = 3
af (1) ≠ f(2)
It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem.
f(x) = x 2 − 1 for x ∈ [1, 2]
Page : 186 , Block Name : Exercise 5.8
Q3Processing math: 100%
f(n+h) −f(n) [n+h] − [n] n−1−n −1
lim h → 0 h
= lim h → ∞ h
= lim h → 0 h
= lim h → ∞ h = ∞
The right hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−n
lim h → ∞ h
= lim h → 0 + h
= lim h → 0 h = lim h → ∞0 = 0
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle's
Theorem.
Hence, Rolle's Theorem is not applicable for f(x) = [x] for x ∈ [5, 9]
(ii) f(x) = [x] for x ∈ [ − 2, 2]
It is evident that the given function f(x) is not continuous at every integral point
In particular, f(x) is not continuous at x = − 2 and x = 2
⇒ f(x) is not continuous in [ − 2, 2] .
Also, f( − 2) = [ − 2] = − 2 and f(2) = [2] = 2
∴ f( − 2) ≠ f(2)
∴ f( − 2) ≠ f(2)
The differentiability of f in ( − 2, 2) is checked as follows.
The left hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−1−n −1
lim h → 0 h
= lim h → 0 h
= lim h → 0 h
= lim h → ∞ h = ∞
The right hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−n
lim h → 0 h
= lim h → 0 − h
= lim h → 0 h = lim h → ∞0 = 0
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x
=n
∴ f is not differentiable in ( − 2, 2)
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle's
Theorem.
Hence, Rolles Theorem is not applicable for f(x) = [x] for x ∈ [ − 2, 2]
(iii) f(x) = x 2 − 1 for x ∈ [1, 2]
It is evident that f, being a polynomial function, is continuous in [1, 2] and is
differentiable in (1, 2) .
f(1) = (1) 2 − 1 = 0
f(2) = (2) 2 − 1 = 3
af (1) ≠ f(2)
It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem.
f(x) = x 2 − 1 for x ∈ [1, 2]
Page : 186 , Block Name : Exercise 5.8
Q3Processing math: 100%
Page 88
If f : [ − 5, 5] → R is a differentiable function and if f ′(x) does not vanish anywhere, then
prove that f( − 5) ≠ f(5)
Answer.
It is given that f : [ − 5, 5] → R is a differentiable function.
since every differentiable function is a continuous function, we obtain
(a) f is continuous on [ − 5, 5].
(b) f is differentiable on ( − 5, 5)
f(5) −f( −5)
f ′(c) = 5− ( −5)
⇒ 10f ′(c) = f(5) − f( − 5)
It is also given that f ′(x) does not vanish anywhere.
⇒ f ′(c) ≠ 0
⇒ 10f ′(c) ≠ 0
⇒ f(5) − f( − 5) ≠ 0
⇒ f(5) ≠ f( − 5)
Hence Proved .
Page : 186 , Block Name : Exercise 5.8
Q4
Verify Mean value Theorem, if f(x) = x 2 − 4x − 3 in the interval [a, b], where
a = 1 and b = 4 .
Answer.
The given function is f(x) = x 2 − 4x − 3
f being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4)
whose derivative is 2x − 4.
f(1) = 1 2 − 4 × 1 − 3 = − 6, f(4) = 4 2 − 4 × 4 − 3 = − 3
f(b) −f(a) f(4) −f(1) −3− ( −6) 3
∴ b−a
= 4−1
= 3
= 3 =1
Mean Value Theorem states that there is a point c ∈ (1, 4) such that f ′(c) = 1
f ′(c) = 1
⇒ 2c − 4 = 1
5 5
⇒ c = 2 , where c = 2 ϵ(1, 4)
Hence, Mean Value Theorem is verified for the given function.
Page : 186 , Block Name : Exercise 5.8
Q5
prove that f( − 5) ≠ f(5)
Answer.
It is given that f : [ − 5, 5] → R is a differentiable function.
since every differentiable function is a continuous function, we obtain
(a) f is continuous on [ − 5, 5].
(b) f is differentiable on ( − 5, 5)
f(5) −f( −5)
f ′(c) = 5− ( −5)
⇒ 10f ′(c) = f(5) − f( − 5)
It is also given that f ′(x) does not vanish anywhere.
⇒ f ′(c) ≠ 0
⇒ 10f ′(c) ≠ 0
⇒ f(5) − f( − 5) ≠ 0
⇒ f(5) ≠ f( − 5)
Hence Proved .
Page : 186 , Block Name : Exercise 5.8
Q4
Verify Mean value Theorem, if f(x) = x 2 − 4x − 3 in the interval [a, b], where
a = 1 and b = 4 .
Answer.
The given function is f(x) = x 2 − 4x − 3
f being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4)
whose derivative is 2x − 4.
f(1) = 1 2 − 4 × 1 − 3 = − 6, f(4) = 4 2 − 4 × 4 − 3 = − 3
f(b) −f(a) f(4) −f(1) −3− ( −6) 3
∴ b−a
= 4−1
= 3
= 3 =1
Mean Value Theorem states that there is a point c ∈ (1, 4) such that f ′(c) = 1
f ′(c) = 1
⇒ 2c − 4 = 1
5 5
⇒ c = 2 , where c = 2 ϵ(1, 4)
Hence, Mean Value Theorem is verified for the given function.
Page : 186 , Block Name : Exercise 5.8
Q5
Page 89
Verify Mean Value Theorem, if f(x) = x 3 − 5x 2 − 3x in the interval [a, b], where a = 1 and
b = 3. Find all c ∈ ( 1 , 3 ) for which f ′(c) = 0
Answer.
The given function f is f(x) = x 3 − 5x 2 − 3x
f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3)
whose derivative is 3x 2 − 10x − 3 .
f(1) = 1 3 − 5 × 1 2 − 3 × 1 = − 7, f(3) = 3 3 − 5 × 3 2 − 3 × 3 = − 27
f(b) −f(a) f(3) −f(1) − 27 − ( − 7 )
∴ b−a
= 3−1
= 3−1
= − 10
f ′(c) = − 10
⇒ 3c 2 − 10c − 3 = 10
⇒ 3c 2 − 10c + 7 = 0
⇒ 3c 2 − 3c − 7c + 7 = 0
⇒ 3c(c − 1) − 7(c − 1) = 0
⇒ (c − 1)(3c − 7) = 0
7 7
⇒ c = 1, 3 , where c = 3 ϵ(1, 3)
7
Hence, Mean Value Theorem is verified for the given function and c = 3 ∈ (1, 3) is the
only point for which f ′(c) = 0
Page : 186 , Block Name : Exercise 5.8
Q6 Examine the applicability of Mean Value Theorem for all three functions given in the above Q2.
Answer.
Mean Value Theorem states that for a function f : [a, b] → R , if
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b)
f(b) −f(a)
then, there exists some c ∈ (a, b) such that f ′(c) = b−a
Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the
two conditions of the hypothesis.
(i) f(x) = [x] for x ∈ [5, 9]
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9
⇒ f(x) is not continuous in [5, 9].
The left hand limit of f at x = n is
f(n+h) −f(n) [n+h] − [n] n−1−n −1
lim h → 0 h
= lim h → 0 h
= lim k → 0 h
= lim h → ∞ h =
The right hand limit of f at x = n is,
b = 3. Find all c ∈ ( 1 , 3 ) for which f ′(c) = 0
Answer.
The given function f is f(x) = x 3 − 5x 2 − 3x
f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3)
whose derivative is 3x 2 − 10x − 3 .
f(1) = 1 3 − 5 × 1 2 − 3 × 1 = − 7, f(3) = 3 3 − 5 × 3 2 − 3 × 3 = − 27
f(b) −f(a) f(3) −f(1) − 27 − ( − 7 )
∴ b−a
= 3−1
= 3−1
= − 10
f ′(c) = − 10
⇒ 3c 2 − 10c − 3 = 10
⇒ 3c 2 − 10c + 7 = 0
⇒ 3c 2 − 3c − 7c + 7 = 0
⇒ 3c(c − 1) − 7(c − 1) = 0
⇒ (c − 1)(3c − 7) = 0
7 7
⇒ c = 1, 3 , where c = 3 ϵ(1, 3)
7
Hence, Mean Value Theorem is verified for the given function and c = 3 ∈ (1, 3) is the
only point for which f ′(c) = 0
Page : 186 , Block Name : Exercise 5.8
Q6 Examine the applicability of Mean Value Theorem for all three functions given in the above Q2.
Answer.
Mean Value Theorem states that for a function f : [a, b] → R , if
(i) f is continuous on [a, b]
(ii) f is differentiable on (a, b)
f(b) −f(a)
then, there exists some c ∈ (a, b) such that f ′(c) = b−a
Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the
two conditions of the hypothesis.
(i) f(x) = [x] for x ∈ [5, 9]
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9
⇒ f(x) is not continuous in [5, 9].
The left hand limit of f at x = n is
f(n+h) −f(n) [n+h] − [n] n−1−n −1
lim h → 0 h
= lim h → 0 h
= lim k → 0 h
= lim h → ∞ h =
The right hand limit of f at x = n is,
Page 90
since the left and right hand limits of f at x = n are not equal, f is not differentiable at x
=n
if is not differentiable in (5, 9) .
It is observed that f does not satisfy all the conditions of the hypothesis of Mean value
Theorem.
Hence, Mean Value Theorem is not applicable for f(x) = [x] for x ∈ [5, 9]
(ii) f(x) = [x] for x ∈ [ − 2, 2]
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = − 2 and x = 2
⇒ f(x) is not continuous in [ − 2, 2] .
The differentiability of f in ( − 2, 2) is checked as follows.
Let n be an integer such that n ∈ ( − 2, 2) .
The left hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−1−n −1
lim h → 0 h
= lim h → 0 h
= lim h → 0 h
= lim h → 0 h = ∞
The right hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−n
lim h → 0 h
= lim h → 0 h
= lim b → ∞ h = lim h → 00 = 0
It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem
f(x) = x 2 − 1 for x ∈ [1, 2]
It is evident that f , being a polynomial function, is continuous in [1, 2] and is
differentiable in (1, 2) .
It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean value Theorem is applicable for f(x) = x 2 − 1 for x ∈ [1, 2]
It can be proved as follows.
f(1) = 1 2 − 1 = 0, f(2) = 2 2 − 1 = 3
f(b) −f(a) f(2) −f(1) 3−0
∴ b−a
= 2−1
= 1
=3
∴ f ′(x) = 2x
∴ f ′(c) = 3
3
⇒ c = 2 = 1.5, where 1.5 ∈ [1, 2]
Page : 186 , Block Name : Exercise 5.8
Q1 Differentiate w.r.t. x the function in 3x 2 − 9x + 5( ) 9
Answer.
(
Let y = 3x 2 − 9x + 5 ) 9
Using chain rule, we obtain
( )
dyProcessing
d 9
dx
= dx
3xmath:
2 − 9x100%
+5
=n
if is not differentiable in (5, 9) .
It is observed that f does not satisfy all the conditions of the hypothesis of Mean value
Theorem.
Hence, Mean Value Theorem is not applicable for f(x) = [x] for x ∈ [5, 9]
(ii) f(x) = [x] for x ∈ [ − 2, 2]
It is evident that the given function f(x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = − 2 and x = 2
⇒ f(x) is not continuous in [ − 2, 2] .
The differentiability of f in ( − 2, 2) is checked as follows.
Let n be an integer such that n ∈ ( − 2, 2) .
The left hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−1−n −1
lim h → 0 h
= lim h → 0 h
= lim h → 0 h
= lim h → 0 h = ∞
The right hand limit of f at x = n is,
f(n+h) −f(n) [n+h] − [n] n−n
lim h → 0 h
= lim h → 0 h
= lim b → ∞ h = lim h → 00 = 0
It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem
f(x) = x 2 − 1 for x ∈ [1, 2]
It is evident that f , being a polynomial function, is continuous in [1, 2] and is
differentiable in (1, 2) .
It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean value Theorem is applicable for f(x) = x 2 − 1 for x ∈ [1, 2]
It can be proved as follows.
f(1) = 1 2 − 1 = 0, f(2) = 2 2 − 1 = 3
f(b) −f(a) f(2) −f(1) 3−0
∴ b−a
= 2−1
= 1
=3
∴ f ′(x) = 2x
∴ f ′(c) = 3
3
⇒ c = 2 = 1.5, where 1.5 ∈ [1, 2]
Page : 186 , Block Name : Exercise 5.8
Q1 Differentiate w.r.t. x the function in 3x 2 − 9x + 5( ) 9
Answer.
(
Let y = 3x 2 − 9x + 5 ) 9
Using chain rule, we obtain
( )
dyProcessing
d 9
dx
= dx
3xmath:
2 − 9x100%
+5
Page 91
( ) ⋅ (3x − 9x + 5 )
8 d
= 9 3x 2 − 9x + 5 dx
2
= 9 (3x − 9x + 5 ) ⋅ (6x − 9)
2 8
= 9 (3x − 9x + 5 ) ⋅ 3(2x − 3)
2 8
= 27 (3x − 9x + 5 ) (2x − 3)
2 8
Page : 191 , Block Name : Miscellaneous Exercise
Q2 Differentiate w.r.t. x the function in sin 3x + cos 6x
Answer.
Let y = sin 3x + cos 6x
dy d d
∴
dx
=
dx ( )
sin 3x +
dx
cos 6x ( )
d d
= 3sin 2x ⋅ (sinx) + 6cos 5x ⋅ (cosx)
dx dx
= 3sin 2x ⋅ cosx + 6cos 5x ⋅ ( − sinx)
(
= 3sinxcosx sinx − 2cos 4x )
Page : 191 , Block Name : Miscellaneous Exercise
Q3 Differentiate w.r.t. x the function in (5x) 3cos 2x
Answer.
Let y = (5x) sen 2x
Taking logarithm on both the sides, we obtain
logy = 3cos2xlog5x
Differentiating both sides with respect to x, we obtain
8 d
= 9 3x 2 − 9x + 5 dx
2
= 9 (3x − 9x + 5 ) ⋅ (6x − 9)
2 8
= 9 (3x − 9x + 5 ) ⋅ 3(2x − 3)
2 8
= 27 (3x − 9x + 5 ) (2x − 3)
2 8
Page : 191 , Block Name : Miscellaneous Exercise
Q2 Differentiate w.r.t. x the function in sin 3x + cos 6x
Answer.
Let y = sin 3x + cos 6x
dy d d
∴
dx
=
dx ( )
sin 3x +
dx
cos 6x ( )
d d
= 3sin 2x ⋅ (sinx) + 6cos 5x ⋅ (cosx)
dx dx
= 3sin 2x ⋅ cosx + 6cos 5x ⋅ ( − sinx)
(
= 3sinxcosx sinx − 2cos 4x )
Page : 191 , Block Name : Miscellaneous Exercise
Q3 Differentiate w.r.t. x the function in (5x) 3cos 2x
Answer.
Let y = (5x) sen 2x
Taking logarithm on both the sides, we obtain
logy = 3cos2xlog5x
Differentiating both sides with respect to x, we obtain
Page 92
1 dy
y dx [ d
= 3 log5x ⋅ dx (cos2x) + cos2x ⋅ dx (log5x)
d
]
dy
[ d
⇒ dx = 3y log5x( − sin2x) ⋅ dx (2x) + cos2x ⋅ 5x ⋅ dx (5x)
1 d
]
dy
[
⇒ dx = 3y − 2sin2xlog5x +
cos 2x
x ]
dy
⇒ dx = 3y
[ 3cos 2x
x
− 6sin2xlog5x
]
dy
∴ dx = (5x) 3cos 2x
[ 3cos 2x
x
− 6sin2xlog5x
]
Page : 191 , Block Name : Miscellaneous Exercise
Q4 Differentiate w.r.t. x the function in sin − 1(x√x), 0 ≤ x ≤ 1
Answer.
Let y = sin − 1(x√x)
Using chain rule, we obtain
dy d
= sin − 1(x√x)
dx dx
1 d
= × (x√x)
1 − (x√x) 2 dx
√
=
1
√1 − x 3 dx
⋅
d
()
x2
3
1 3 1
= × ⋅ x2
2
√1 − x 3
3√x
=
√1 − x 3
3 x
=
2 √ 1 − x3
Page : 191 , Block Name : Miscellaneous Exercise
y dx [ d
= 3 log5x ⋅ dx (cos2x) + cos2x ⋅ dx (log5x)
d
]
dy
[ d
⇒ dx = 3y log5x( − sin2x) ⋅ dx (2x) + cos2x ⋅ 5x ⋅ dx (5x)
1 d
]
dy
[
⇒ dx = 3y − 2sin2xlog5x +
cos 2x
x ]
dy
⇒ dx = 3y
[ 3cos 2x
x
− 6sin2xlog5x
]
dy
∴ dx = (5x) 3cos 2x
[ 3cos 2x
x
− 6sin2xlog5x
]
Page : 191 , Block Name : Miscellaneous Exercise
Q4 Differentiate w.r.t. x the function in sin − 1(x√x), 0 ≤ x ≤ 1
Answer.
Let y = sin − 1(x√x)
Using chain rule, we obtain
dy d
= sin − 1(x√x)
dx dx
1 d
= × (x√x)
1 − (x√x) 2 dx
√
=
1
√1 − x 3 dx
⋅
d
()
x2
3
1 3 1
= × ⋅ x2
2
√1 − x 3
3√x
=
√1 − x 3
3 x
=
2 √ 1 − x3
Page : 191 , Block Name : Miscellaneous Exercise
Page 93
x
cos − 1 2
Q5 Differentiate w.r.t. x the function in , −2<x<2
√2x + 7
Answer.
x
cos − 1 2
By quotient rule, we obtain
dy
√2x + 7 dx
d
( cos − 1 2
x
) (
− cos − 1 2
x
) √
d
dx ( 2x + 7 )
dx
=
( √2x + 7 ) 2
=
√2x + 7
[ √ ()
1−
−1
x
2
2
d
⋅ dx
() (x
2
] −
dx
cos − 1 2
x
) 1
2√2x + 7
d
⋅ dx ( 2x + 7 )
(2x + 7)
x
− √2x + 7 cos − 1 2
= −
√ 4 − x 2 × (2x + 7) (√2x + 7)(2x + 7)
= −
[√ 1
4 − x 2√2x + 7
+
2
(2x + 7) 2 ]
Page : 191 , Block Name : Miscellaneous Exercise
Q6 Differentiate w.r.t. x the function in cot − 1
[ √1 + sin x + √1 − sin x
√1 + sin x − √1 − sin x ] 1
,0 < x < 2
Answer.
Let y = cot − 1
[ √1 + sin x + √1 − sin x
√1 + sin x − √1 − sin x ]
√1 + sin x + √1 − sin x
Then,
√1 + sin x + √1 − sin x
( √1 + sin x + √1 − sin x ) 2
=
( √1 + sin x − √1 − sin x + √1 − sin x ) 2
( 1 + sin x ) + ( 1 − sin x ) + 2√ ( 1 − sin x ) ( 1 − sin x )
= ( 1 + sin x ) − ( 1 − sin x )
cos − 1 2
Q5 Differentiate w.r.t. x the function in , −2<x<2
√2x + 7
Answer.
x
cos − 1 2
By quotient rule, we obtain
dy
√2x + 7 dx
d
( cos − 1 2
x
) (
− cos − 1 2
x
) √
d
dx ( 2x + 7 )
dx
=
( √2x + 7 ) 2
=
√2x + 7
[ √ ()
1−
−1
x
2
2
d
⋅ dx
() (x
2
] −
dx
cos − 1 2
x
) 1
2√2x + 7
d
⋅ dx ( 2x + 7 )
(2x + 7)
x
− √2x + 7 cos − 1 2
= −
√ 4 − x 2 × (2x + 7) (√2x + 7)(2x + 7)
= −
[√ 1
4 − x 2√2x + 7
+
2
(2x + 7) 2 ]
Page : 191 , Block Name : Miscellaneous Exercise
Q6 Differentiate w.r.t. x the function in cot − 1
[ √1 + sin x + √1 − sin x
√1 + sin x − √1 − sin x ] 1
,0 < x < 2
Answer.
Let y = cot − 1
[ √1 + sin x + √1 − sin x
√1 + sin x − √1 − sin x ]
√1 + sin x + √1 − sin x
Then,
√1 + sin x + √1 − sin x
( √1 + sin x + √1 − sin x ) 2
=
( √1 + sin x − √1 − sin x + √1 − sin x ) 2
( 1 + sin x ) + ( 1 − sin x ) + 2√ ( 1 − sin x ) ( 1 − sin x )
= ( 1 + sin x ) − ( 1 − sin x )
Page 94
√
2 + 2 1 − sin 2x
=
2sinx
1 + cosx
=
sinx
x
2cos 2 2
= x x
2sin 2 cos 2
x x
= cot cos
2 2
x
= cot
2
Therefore, equation (1) becomes
y = cot − 1 cot 2
( ) x
x
⇒y= 2
dy 1 d
∴ dx = 2 dx (x)
dy 1
⇒ dx = 2
Page : 191 , Block Name : Miscellaneous Exercise
Q7 Differentiate w.r.t. x the function in (logx) tax x, x > 1
Answer. y = (logx) ln x
Taking logarithm on both the sides, we obtain
logy = logx ⋅ log(logx)
Differentiating both sides with respect to x, we obtain
1 dy d
y dx = dx [logx ⋅ log(logx)]
1 dy d d
⇒ y dx = log(logx) ⋅ dx (logx) + logx ⋅ dx [log(logx)]
dy
[ 1
⇒ dx = y log(logx) ⋅ x + logx ⋅ log x ⋅ dx (logx)
1 d
]
dy
[
⇒ dx = y x log(logx) + x
1 1
]
dy
∴ dx = (logx) log x x + [ 1 log ( log x )
x ]
2 + 2 1 − sin 2x
=
2sinx
1 + cosx
=
sinx
x
2cos 2 2
= x x
2sin 2 cos 2
x x
= cot cos
2 2
x
= cot
2
Therefore, equation (1) becomes
y = cot − 1 cot 2
( ) x
x
⇒y= 2
dy 1 d
∴ dx = 2 dx (x)
dy 1
⇒ dx = 2
Page : 191 , Block Name : Miscellaneous Exercise
Q7 Differentiate w.r.t. x the function in (logx) tax x, x > 1
Answer. y = (logx) ln x
Taking logarithm on both the sides, we obtain
logy = logx ⋅ log(logx)
Differentiating both sides with respect to x, we obtain
1 dy d
y dx = dx [logx ⋅ log(logx)]
1 dy d d
⇒ y dx = log(logx) ⋅ dx (logx) + logx ⋅ dx [log(logx)]
dy
[ 1
⇒ dx = y log(logx) ⋅ x + logx ⋅ log x ⋅ dx (logx)
1 d
]
dy
[
⇒ dx = y x log(logx) + x
1 1
]
dy
∴ dx = (logx) log x x + [ 1 log ( log x )
x ]
Page 95
Page : 191 , Block Name : Miscellaneous Exercise
Q8 Differentiate w.r.t. x the function in cos(acosx + bsinx), for some constant a and b
Answer.
Let y = cos(acosx + bsinx)
By using chain rule, we obtaln
dy d
dx
= dx cos(acosx + bsinx)
dy d
⇒ = − sin(acosx + bsinx) ⋅ (acosx + bsinx)
dx dx
= − sin(acosx + bsinx) ⋅ [a( − sinx) + bcosx]
= (asinx − bcosx) ⋅ sin(acosx + bsinx)
Page : 191 , Block Name : Miscellaneous Exercise
π 3π
Q9 Differentiate w.r.t. x the function in (sinx − cosx) ( sin x − cos x ) , 4 < x < 4
Answer.
Let y = (sinx − cosx) ( sin x − cos x )
Taking logarithm on both the sides, we obtain
[
logy = log (sinx − cosx) ( sin x − cos x ) ]
⇒ logy = (sinx − cosx) ⋅ log(sinx − cosx)
Differentiating both sides with respect to x, we obtain
1 dy d
y dx
= dx [(sinx − cosx)log(sinx − cosx)]
1 dy d d
⇒ y dx = log(sinx − cosx) ⋅ dx (sinx − cosx) + (sinx − cosx) ⋅ dx log(sinx − cosx)
1 dy 1 d
⇒ y dx = log(sinx − cosx) ⋅ (cosx + sinx) + (sinx − cosx) ⋅ ( sin x − cos x ) ⋅ dx (sinx − cosx)
dy
⇒ dx = (sinx − cosx) ( sin x − cos x ) [(cosx + sinx) ⋅ log(sinx − cosx) + (cosx + sinx)]
dy
∴ dx = (sinx − cosx) ( ln x − cos x ) (cosx + sinx)[1 + log(sinx − cosx)]
Page : 191 , Block Name : Miscellaneous Exercise
Q10 x x + x a + a x + a a, for some fixed a > 0 and x > 0
Answer. y = x x + x ′′ + a x + a a
Q8 Differentiate w.r.t. x the function in cos(acosx + bsinx), for some constant a and b
Answer.
Let y = cos(acosx + bsinx)
By using chain rule, we obtaln
dy d
dx
= dx cos(acosx + bsinx)
dy d
⇒ = − sin(acosx + bsinx) ⋅ (acosx + bsinx)
dx dx
= − sin(acosx + bsinx) ⋅ [a( − sinx) + bcosx]
= (asinx − bcosx) ⋅ sin(acosx + bsinx)
Page : 191 , Block Name : Miscellaneous Exercise
π 3π
Q9 Differentiate w.r.t. x the function in (sinx − cosx) ( sin x − cos x ) , 4 < x < 4
Answer.
Let y = (sinx − cosx) ( sin x − cos x )
Taking logarithm on both the sides, we obtain
[
logy = log (sinx − cosx) ( sin x − cos x ) ]
⇒ logy = (sinx − cosx) ⋅ log(sinx − cosx)
Differentiating both sides with respect to x, we obtain
1 dy d
y dx
= dx [(sinx − cosx)log(sinx − cosx)]
1 dy d d
⇒ y dx = log(sinx − cosx) ⋅ dx (sinx − cosx) + (sinx − cosx) ⋅ dx log(sinx − cosx)
1 dy 1 d
⇒ y dx = log(sinx − cosx) ⋅ (cosx + sinx) + (sinx − cosx) ⋅ ( sin x − cos x ) ⋅ dx (sinx − cosx)
dy
⇒ dx = (sinx − cosx) ( sin x − cos x ) [(cosx + sinx) ⋅ log(sinx − cosx) + (cosx + sinx)]
dy
∴ dx = (sinx − cosx) ( ln x − cos x ) (cosx + sinx)[1 + log(sinx − cosx)]
Page : 191 , Block Name : Miscellaneous Exercise
Q10 x x + x a + a x + a a, for some fixed a > 0 and x > 0
Answer. y = x x + x ′′ + a x + a a
Page 96
Also, let x x = u, x a = v, a x = w, and a a = s
∴y=u+v+w+s
dy du dv dw ds
⇒ dx = dx + dx + dx + dx
u = xx
⇒ logu = logx x
⇒ logu = xlogx
1 du d d
u dx
= logx ⋅ dx (x) + x ⋅ dx (logx)
du
[
⇒ dx = u logx ⋅ 1 + x ⋅ x
1
]
du
⇒ dx = x x[logx + 1] = x x(1 + logx)
v = xa
( )
dv d
∴ dx = dx x a
dv
⇒ dx = ax a − 1
w = ax
⇒ logw = loga x
⇒ logw = xloga
Differentiating both sides with respect to x, we obtain
1 dw d
w
⋅ dx = loga ⋅ dx (x)
dw
⇒ dx = wloga
dw
⇒ dx = a xloga
s = aa
Since a is constant, a a is also a constant.
ds
dx
=0
From (1), (2), (3), (4), and (5), we obtain
dy
dx
= x x(1 + logx) + ax a − 1 + a xloga + 0
= x x(1 + logx) + ax n − 1 + a xloga
Page : 191 , Block Name : Miscellaneous Exercise
Q11 Differentiate w.r.t. x the function in
2 2
x x − 3 + (x − 3) x , for x > 3
Answer.
∴y=u+v+w+s
dy du dv dw ds
⇒ dx = dx + dx + dx + dx
u = xx
⇒ logu = logx x
⇒ logu = xlogx
1 du d d
u dx
= logx ⋅ dx (x) + x ⋅ dx (logx)
du
[
⇒ dx = u logx ⋅ 1 + x ⋅ x
1
]
du
⇒ dx = x x[logx + 1] = x x(1 + logx)
v = xa
( )
dv d
∴ dx = dx x a
dv
⇒ dx = ax a − 1
w = ax
⇒ logw = loga x
⇒ logw = xloga
Differentiating both sides with respect to x, we obtain
1 dw d
w
⋅ dx = loga ⋅ dx (x)
dw
⇒ dx = wloga
dw
⇒ dx = a xloga
s = aa
Since a is constant, a a is also a constant.
ds
dx
=0
From (1), (2), (3), (4), and (5), we obtain
dy
dx
= x x(1 + logx) + ax a − 1 + a xloga + 0
= x x(1 + logx) + ax n − 1 + a xloga
Page : 191 , Block Name : Miscellaneous Exercise
Q11 Differentiate w.r.t. x the function in
2 2
x x − 3 + (x − 3) x , for x > 3
Answer.
Page 97
2 2
Let y = x x − 3 + (x − 3) x
2 2
Also, let u = x x − 3 and v = (x − 3) x
∴y=u+v
Differentiating both sides with respect to x, we obtain
dy du dv
dx
= dx + dx
2
u = xx − 3
∴ logu = log x x − 3 ( ) 2
(
logu = x 2 − 3 logx )
2
u = xx − 3
∴ logu = log x x − 3 ( ) 2
(
logu = x 2 − 3 logx )
Differentiating with respect to x, we obtain
(
) ( )
1 du d d
u
⋅ dx = logx ⋅ dx x 2 − 3 + x 2 − 3 ⋅ dx (logx)
= logx ⋅ 2x + (x − 3 ) ⋅
1 du 1
⇒ u dx 2
x
⇒ dx
du
= xx
2
( x2 − 3
x
+ 2xlogx
]
Also,
2
v = (x − 3) x
2
∴ logv = log(x − 3) x
⇒ logv = x 2log(x − 3)
Differentiating both sides with respect to x, we obtain
( )
1 dv d d
⋅
v dx
= log(x − 3) ⋅ dx x 2 + x 2 ⋅ dx [log(x − 3)]
1 dv 1 d
⇒ v dx = log(x − 3) ⋅ 2x + x 2 ⋅ x − 3 ⋅ dx (x − 3)
dv
[
⇒ dx = v 2xlog(x − 3) + x − 3 ⋅ 1
x2
]
dv
⇒ dx = (x − 3) x
2
[ x2
x−3
+ 2xlog(x − 3)
]
Let y = x x − 3 + (x − 3) x
2 2
Also, let u = x x − 3 and v = (x − 3) x
∴y=u+v
Differentiating both sides with respect to x, we obtain
dy du dv
dx
= dx + dx
2
u = xx − 3
∴ logu = log x x − 3 ( ) 2
(
logu = x 2 − 3 logx )
2
u = xx − 3
∴ logu = log x x − 3 ( ) 2
(
logu = x 2 − 3 logx )
Differentiating with respect to x, we obtain
(
) ( )
1 du d d
u
⋅ dx = logx ⋅ dx x 2 − 3 + x 2 − 3 ⋅ dx (logx)
= logx ⋅ 2x + (x − 3 ) ⋅
1 du 1
⇒ u dx 2
x
⇒ dx
du
= xx
2
( x2 − 3
x
+ 2xlogx
]
Also,
2
v = (x − 3) x
2
∴ logv = log(x − 3) x
⇒ logv = x 2log(x − 3)
Differentiating both sides with respect to x, we obtain
( )
1 dv d d
⋅
v dx
= log(x − 3) ⋅ dx x 2 + x 2 ⋅ dx [log(x − 3)]
1 dv 1 d
⇒ v dx = log(x − 3) ⋅ 2x + x 2 ⋅ x − 3 ⋅ dx (x − 3)
dv
[
⇒ dx = v 2xlog(x − 3) + x − 3 ⋅ 1
x2
]
dv
⇒ dx = (x − 3) x
2
[ x2
x−3
+ 2xlog(x − 3)
]
Page 98
dv
Substituting the expressions of dx and dx equation (1), we obtain
dy
dx
= xx − 3
2
[ x2 − 3
x ]
+ 2xlogx + (x − 3) x x − 3 + 2xlog(x − 3)
[ x2
]
Page : 191 , Block Name : Miscellaneous Exercise
dy π π
Q12 dx , if y = 12(1 − cost), x = 10(t − sint), − 2 < t < 2
Answer.
It is given that, y = 12(1 − cost), x = 10(t − sint)
dx d d
∴ dt = dt [10(t − sint)] = 10 ⋅ dt (t − sint) = 10(1 − cost)
dy d d
dt
= dt [12(1 − cost)] = 12 ⋅ dt (1 − cost) = 12 ⋅ [0 − ( − sint)] = 12sint
∴ dx
dy
=
( )= dy
dt
t
12 ⋅ 2sin 2 ⋅ cos 2
t
6
= 5 cot 2
t
t
( ) dx
dt
10 ⋅ 2sin 2 2
Page : 191 , Block Name : Miscellaneous Exercise
dy
Q13 Find dx , if y = sin − 1x + sin − 1 1 − x 2, − 1 ≤ x ≤ 1 √
Answer.
It is given that, y = sin − 1x + sin − 1 1 − x 2 √
dy d
[
∴ dx = dx sin − 1x + sin − 1 1 − x 2 √ ]
dy d
(
⇒ dx = dx sin − 1x + dx sin − 1 1 − x 2)
d
( √ )
( )
dy 1 1 1 d
⇒ dx = 2
+ x ⋅ 2
⋅ dx 1 − x 2
√1 − x √
2 1−x
dy 1 1
⇒ dx = + ( − 2x)
√ 1 − x2 √
2x 1 − x 2
dy 1 1
⇒ dx = −
√1 − x 2
√1 − x 2
dy
∴ dx = 0
Page : 191 , Block Name : Miscellaneous Exercise
Substituting the expressions of dx and dx equation (1), we obtain
dy
dx
= xx − 3
2
[ x2 − 3
x ]
+ 2xlogx + (x − 3) x x − 3 + 2xlog(x − 3)
[ x2
]
Page : 191 , Block Name : Miscellaneous Exercise
dy π π
Q12 dx , if y = 12(1 − cost), x = 10(t − sint), − 2 < t < 2
Answer.
It is given that, y = 12(1 − cost), x = 10(t − sint)
dx d d
∴ dt = dt [10(t − sint)] = 10 ⋅ dt (t − sint) = 10(1 − cost)
dy d d
dt
= dt [12(1 − cost)] = 12 ⋅ dt (1 − cost) = 12 ⋅ [0 − ( − sint)] = 12sint
∴ dx
dy
=
( )= dy
dt
t
12 ⋅ 2sin 2 ⋅ cos 2
t
6
= 5 cot 2
t
t
( ) dx
dt
10 ⋅ 2sin 2 2
Page : 191 , Block Name : Miscellaneous Exercise
dy
Q13 Find dx , if y = sin − 1x + sin − 1 1 − x 2, − 1 ≤ x ≤ 1 √
Answer.
It is given that, y = sin − 1x + sin − 1 1 − x 2 √
dy d
[
∴ dx = dx sin − 1x + sin − 1 1 − x 2 √ ]
dy d
(
⇒ dx = dx sin − 1x + dx sin − 1 1 − x 2)
d
( √ )
( )
dy 1 1 1 d
⇒ dx = 2
+ x ⋅ 2
⋅ dx 1 − x 2
√1 − x √
2 1−x
dy 1 1
⇒ dx = + ( − 2x)
√ 1 − x2 √
2x 1 − x 2
dy 1 1
⇒ dx = −
√1 − x 2
√1 − x 2
dy
∴ dx = 0
Page : 191 , Block Name : Miscellaneous Exercise
Page 99
Q14
If x√1 + y + y√1 + x = 0 , for, − 1 < x < 1 , prove that
dy 1
dx
= −
( 1 + x )2
Answer.
It is given that,
x√ 1 + y + y√ 1 + x = 0
Squaring both sides, we obtain
x 2(1 + y) = y 2(1 + x)
⇒ x 2 + x 2y = y 2 + xy 2
⇒ x 2 − y 2 = xy 2 − x 2y
⇒ x 2 − y 2 = xy(y − x)
⇒ (x + y)(x − y) = xy(y − x)
∴ x + y = − xy
⇒ (1 + x)y = − x
−x
⇒ y = (1+x)
Differentiating both sides with respect to x, we obtain
−x
y = (1+x)
d d
dy ( 1 + x ) dx ( x ) − x dx ( 1 + x ) (1+x) −x 1
dx
= − = − = −
( 1 + x )2 ( 1 + x )2 ( 1 + x )2
Hence, rroved.
Page : 191 , Block Name : Miscellaneous Exercise
2 2 2
Q15 if ( x − a ) + ( y − b ) = c , for some c > 0 prove that
[ ( )]
3
dy 2 2
1+ dx
d 2y
is a constant independent of a and b.
dx 2
Answer.
It is given that, (x − a) 2 + (y − b) 2 = c 2
Differentiating both sides with respect to x, we obtain
If x√1 + y + y√1 + x = 0 , for, − 1 < x < 1 , prove that
dy 1
dx
= −
( 1 + x )2
Answer.
It is given that,
x√ 1 + y + y√ 1 + x = 0
Squaring both sides, we obtain
x 2(1 + y) = y 2(1 + x)
⇒ x 2 + x 2y = y 2 + xy 2
⇒ x 2 − y 2 = xy 2 − x 2y
⇒ x 2 − y 2 = xy(y − x)
⇒ (x + y)(x − y) = xy(y − x)
∴ x + y = − xy
⇒ (1 + x)y = − x
−x
⇒ y = (1+x)
Differentiating both sides with respect to x, we obtain
−x
y = (1+x)
d d
dy ( 1 + x ) dx ( x ) − x dx ( 1 + x ) (1+x) −x 1
dx
= − = − = −
( 1 + x )2 ( 1 + x )2 ( 1 + x )2
Hence, rroved.
Page : 191 , Block Name : Miscellaneous Exercise
2 2 2
Q15 if ( x − a ) + ( y − b ) = c , for some c > 0 prove that
[ ( )]
3
dy 2 2
1+ dx
d 2y
is a constant independent of a and b.
dx 2
Answer.
It is given that, (x − a) 2 + (y − b) 2 = c 2
Differentiating both sides with respect to x, we obtain
Page 100
[(x − a) ] + [(y − b) ] = (c )
d d d
2 2 2
dx dx dx
d d
⇒ 2(x − a) ⋅ dx (x − a) + 2(y − b) ⋅ dx (y − b) = 0
dy
⇒ 2(x − a) ⋅ 1 + 2(y − b) ⋅ dx = 0
dy − (x−a)
⇒ dx = y−b
∴
d 2y
dx 2 = dx
d
[ − (x−a)
y−b ]
[ ]
d d
(y − b) ⋅ dx (x − a) − (x − a) ⋅ dx (y − b)
= −
(y − b) 2
[ ]
dy
(y − b) − (x − a) ⋅ dx ]
= −
(y − b) 2
[ { }
]
− (x−a)
(y − b) − (x − a) ⋅ y−b
]
= −
(y − b) 2
= −
[ (y − b) 2 + (x − b) 2
(y − b) 3 ]
[ ]
3
c2 2
c3
( y − b )2
( y − b )3
= c2
= c2
− −
( y − b )3 ( y − b )3
= − c, which is constant and is independent of a and b
Hence, proved.
Page : 191 , Block Name : Miscellaneous Exercise
dy cos 2 ( a + y )
Q16 cosy = xcos(a + y), with cosa ≠ ± 1, prove that dx = sin a
Answer.
d d d
2 2 2
dx dx dx
d d
⇒ 2(x − a) ⋅ dx (x − a) + 2(y − b) ⋅ dx (y − b) = 0
dy
⇒ 2(x − a) ⋅ 1 + 2(y − b) ⋅ dx = 0
dy − (x−a)
⇒ dx = y−b
∴
d 2y
dx 2 = dx
d
[ − (x−a)
y−b ]
[ ]
d d
(y − b) ⋅ dx (x − a) − (x − a) ⋅ dx (y − b)
= −
(y − b) 2
[ ]
dy
(y − b) − (x − a) ⋅ dx ]
= −
(y − b) 2
[ { }
]
− (x−a)
(y − b) − (x − a) ⋅ y−b
]
= −
(y − b) 2
= −
[ (y − b) 2 + (x − b) 2
(y − b) 3 ]
[ ]
3
c2 2
c3
( y − b )2
( y − b )3
= c2
= c2
− −
( y − b )3 ( y − b )3
= − c, which is constant and is independent of a and b
Hence, proved.
Page : 191 , Block Name : Miscellaneous Exercise
dy cos 2 ( a + y )
Q16 cosy = xcos(a + y), with cosa ≠ ± 1, prove that dx = sin a
Answer.
Page 101
It is given that, cosy = xcos(a + y)
d d
∴ dx [cosy] = dx [xcos(a + y)]
dy d d
⇒ − siny dx = cos(a + y) ⋅ dx (x) + x ⋅ dx [cos(a + y)]
dy dy
⇒ − siny dx = cos(a + y) + x ⋅ [ − sin(a + y)] dx
dy
⇒ [xsin(a + y) − siny] dx = cos(a + y)
cos y
since cosy = xcos(a + y), x = cos ( a + y )
Then, equation (1) reduces to
[ cos y
cos ( a + y )
⋅ sin(a + y) − siny dx = cos(a + y)
dy
] dy
⇒ sin(a + y − y) dx = cos 2(a + b)
dy cos 2 ( a + b )
⇒ dx = sin a
Hence, proved.
Page : 192 , Block Name : Miscellaneous Exercise
d 2y
Q17 If x = a(cost + tsint) and y = a(sint − tcost), find
dx 2
Answer.
It is given that, x = a(cost + tsint) and y = a(sint − tcost)
dx d
∴ = a ⋅ (cost + tsint)
dt dt
[
= a − sint + sint ⋅
d
dx
(t) + t ⋅
d
dt
(sint) ]
= a[ − sint + sint + tcost] = atcost
[ {
= a cost −
d
cost ⋅ dt (t) + t ⋅ dt (cost)
d
}]
= a[cost − {cost − tsint}] = atsint
dy ( ) dy
dt
atsin t
∴ dx = = atcos t = tant
( ) dy
dt
Then,
d 2y
dx 2
d
= dx dx
() dy d
= dx (tant) = sec 2t ⋅ dx
dt
d d
∴ dx [cosy] = dx [xcos(a + y)]
dy d d
⇒ − siny dx = cos(a + y) ⋅ dx (x) + x ⋅ dx [cos(a + y)]
dy dy
⇒ − siny dx = cos(a + y) + x ⋅ [ − sin(a + y)] dx
dy
⇒ [xsin(a + y) − siny] dx = cos(a + y)
cos y
since cosy = xcos(a + y), x = cos ( a + y )
Then, equation (1) reduces to
[ cos y
cos ( a + y )
⋅ sin(a + y) − siny dx = cos(a + y)
dy
] dy
⇒ sin(a + y − y) dx = cos 2(a + b)
dy cos 2 ( a + b )
⇒ dx = sin a
Hence, proved.
Page : 192 , Block Name : Miscellaneous Exercise
d 2y
Q17 If x = a(cost + tsint) and y = a(sint − tcost), find
dx 2
Answer.
It is given that, x = a(cost + tsint) and y = a(sint − tcost)
dx d
∴ = a ⋅ (cost + tsint)
dt dt
[
= a − sint + sint ⋅
d
dx
(t) + t ⋅
d
dt
(sint) ]
= a[ − sint + sint + tcost] = atcost
[ {
= a cost −
d
cost ⋅ dt (t) + t ⋅ dt (cost)
d
}]
= a[cost − {cost − tsint}] = atsint
dy ( ) dy
dt
atsin t
∴ dx = = atcos t = tant
( ) dy
dt
Then,
d 2y
dx 2
d
= dx dx
() dy d
= dx (tant) = sec 2t ⋅ dx
dt
Page 102
= sec 2t ⋅ atcos t
1
[ dx
dt
dt 1
= atcost ⇒ dx = atcos t ]
sec 3 t π
= at
,0 < t < 2
Page : 192 , Block Name : Miscellaneous Exercise
Q18 f(x) = | x | 3, show that f ′′(x) exists for all real x , and find it.
Answer. It is known that
|x| =
{ x,
− x,
if x ≥ 0
if x < 0
Therefore, when x ≥ 0, f(x) = | x | 3 = x 3
In this case, f ′(x) = 3x 2 and hence, f ′(x) = 6x
When x < 0, f(x) = | x | 3 = ( − x) 3 = − x 3
In this case, f ′(x) = − 3x 2 and hence, f ′(x) = − 6x
Thus, for f(x) = | x | 3, f ′′(x)
f ′′(x) =
{ 6x,
− 6x,
if x ≥ 0
if x < 0
Page : 192 , Block Name : Miscellaneous Exercise
Q19 Using mathematical induction prove that
(x ) = nx
d
n n − 1for all positive integer n.
dx
Answer.
( )
d
P(n) : dx x n = nx n − 1 for all positive integers n
d
P(1) : dx (x) = 1 = 1 ⋅ x 1 − 1
∴ P(n) is true for n = 1
Let P(k) is true for some positive integer k .
Let P(k) is true for some positive integer k
( )
d
That is, P(k) : dx x T = kx k − 1
It has to be proved that P(k + 1) is also true.
( ) ( )
d d
Consider dx x k + 1 = dx x ⋅ x k
1
[ dx
dt
dt 1
= atcost ⇒ dx = atcos t ]
sec 3 t π
= at
,0 < t < 2
Page : 192 , Block Name : Miscellaneous Exercise
Q18 f(x) = | x | 3, show that f ′′(x) exists for all real x , and find it.
Answer. It is known that
|x| =
{ x,
− x,
if x ≥ 0
if x < 0
Therefore, when x ≥ 0, f(x) = | x | 3 = x 3
In this case, f ′(x) = 3x 2 and hence, f ′(x) = 6x
When x < 0, f(x) = | x | 3 = ( − x) 3 = − x 3
In this case, f ′(x) = − 3x 2 and hence, f ′(x) = − 6x
Thus, for f(x) = | x | 3, f ′′(x)
f ′′(x) =
{ 6x,
− 6x,
if x ≥ 0
if x < 0
Page : 192 , Block Name : Miscellaneous Exercise
Q19 Using mathematical induction prove that
(x ) = nx
d
n n − 1for all positive integer n.
dx
Answer.
( )
d
P(n) : dx x n = nx n − 1 for all positive integers n
d
P(1) : dx (x) = 1 = 1 ⋅ x 1 − 1
∴ P(n) is true for n = 1
Let P(k) is true for some positive integer k .
Let P(k) is true for some positive integer k
( )
d
That is, P(k) : dx x T = kx k − 1
It has to be proved that P(k + 1) is also true.
( ) ( )
d d
Consider dx x k + 1 = dx x ⋅ x k
Page 103
d d
dx (
x )=
dx (
Consider i+1
x⋅x ) k
d d
dx ( )
= xk ⋅ (x) + x ⋅ x k
dx
= xk ⋅ 1 + x ⋅ k ⋅ xk − 1
= x k + kx k
= (k + 1) ⋅ x k
= (k + 1) ⋅ x ( k + 1 ) − 1
Thus, P(k + 1) is true whenever P (k) is true. Therefore, by the principle of mathematical induction,
the statement P(n) is true for every positive integer n . Hence proved
Page : 192 , Block Name : Miscellaneous Exercise
Q20 Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the
sum formula for cosines.
Answer.
sin(A + B) = sinAcosB + cosAsinB
Differentiating both sides with respect to x, we obtain
d d d
dx
[sin(A + B)] = dx (sinAcosB) + dx (cosAsinB)
d d d
⇒ cos(A + B) ⋅ (A + B) = cosB = cosB (sinA) + sinα ⋅
(cosB)
dx dx dx
d d
+ sinB ⋅ (cosA) + cosA ⋅ (sinB)
dx dx
d dλ dB
⇒ cos(A + B) ⋅ (A + B) = cosB ⋅ cosA + sinA( − sinB)
dx dx dx
dA dB
+ sinB( − sinA) ⋅ + cosAcosB
dx dx
⇒ cos(A + B) ⋅ [ dA
dx
+ dx
dB
] = (cosAcosB − sinAsinB) ⋅ [ dA
dx
+ dx
dB
]
∴ cos(A + B) = cosAcosB − sinAsinB
Page : 192 , Block Name : Miscellaneous Exercise
Q21 Does there exist a function which is continuous everywhere but not differentiable at exactly
two points? Justify your answer.
Answer.
dx (
x )=
dx (
Consider i+1
x⋅x ) k
d d
dx ( )
= xk ⋅ (x) + x ⋅ x k
dx
= xk ⋅ 1 + x ⋅ k ⋅ xk − 1
= x k + kx k
= (k + 1) ⋅ x k
= (k + 1) ⋅ x ( k + 1 ) − 1
Thus, P(k + 1) is true whenever P (k) is true. Therefore, by the principle of mathematical induction,
the statement P(n) is true for every positive integer n . Hence proved
Page : 192 , Block Name : Miscellaneous Exercise
Q20 Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the
sum formula for cosines.
Answer.
sin(A + B) = sinAcosB + cosAsinB
Differentiating both sides with respect to x, we obtain
d d d
dx
[sin(A + B)] = dx (sinAcosB) + dx (cosAsinB)
d d d
⇒ cos(A + B) ⋅ (A + B) = cosB = cosB (sinA) + sinα ⋅
(cosB)
dx dx dx
d d
+ sinB ⋅ (cosA) + cosA ⋅ (sinB)
dx dx
d dλ dB
⇒ cos(A + B) ⋅ (A + B) = cosB ⋅ cosA + sinA( − sinB)
dx dx dx
dA dB
+ sinB( − sinA) ⋅ + cosAcosB
dx dx
⇒ cos(A + B) ⋅ [ dA
dx
+ dx
dB
] = (cosAcosB − sinAsinB) ⋅ [ dA
dx
+ dx
dB
]
∴ cos(A + B) = cosAcosB − sinAsinB
Page : 192 , Block Name : Miscellaneous Exercise
Q21 Does there exist a function which is continuous everywhere but not differentiable at exactly
two points? Justify your answer.
Answer.
Page 104
y=x
-∞ < x ≤ 1
2-x
1≤x≤∞
It can be seen from the above graph that, the given function is continuos everywhere but not
differentiable at exactly two points which are 0 and 1.
Page : 192 , Block Name : Miscellaneous Exercise
| | | |
f(x) g(x) h(x) f ′(x) g ′(x) h ′(x)
dy
Q22 if y = l m n prove that dx = l m n
a b c a b c
Answer.
| |
f(x) g(x) h(x)
y= l m n
a b c
⇒ y = (mc − nb)f(x) − (lc − na)g(x) + (lb − ma)h(x)
dy d d d
Then, dx = dx [(mc − nb)f(x)] − dx [(lc − na)g(x)] + dx [(lb − ma)h(x)]
= (mc − nb)f ′(x) − (lc − na)g ′(x) + (lb − ma)h ′(x)
| |
f ′(x) g ′(x) h ′(x)
= l m n
a b c
| |
f ′(x) g ′(x) h ′(x)
dy
dx
= l m n
Processing
a math:
b 100%c
-∞ < x ≤ 1
2-x
1≤x≤∞
It can be seen from the above graph that, the given function is continuos everywhere but not
differentiable at exactly two points which are 0 and 1.
Page : 192 , Block Name : Miscellaneous Exercise
| | | |
f(x) g(x) h(x) f ′(x) g ′(x) h ′(x)
dy
Q22 if y = l m n prove that dx = l m n
a b c a b c
Answer.
| |
f(x) g(x) h(x)
y= l m n
a b c
⇒ y = (mc − nb)f(x) − (lc − na)g(x) + (lb − ma)h(x)
dy d d d
Then, dx = dx [(mc − nb)f(x)] − dx [(lc − na)g(x)] + dx [(lb − ma)h(x)]
= (mc − nb)f ′(x) − (lc − na)g ′(x) + (lb − ma)h ′(x)
| |
f ′(x) g ′(x) h ′(x)
= l m n
a b c
| |
f ′(x) g ′(x) h ′(x)
dy
dx
= l m n
Processing
a math:
b 100%c
Page 105
Page : 192 , Block Name : Miscellaneous Exercise
−1 x
Q23 if y = e acos , − 1 ≤ x ≤ 1 show that
d 2y
(1 − x )
dy
2 − x dx − a 2y = 0
dx 2
−1
Answer. It is given that,y = e acos x
Taking logarithm on both the sides, we obtain
logy = acos − 1xloge
logy = acos − 1x
Differentiating both sides with respect to x, we obtain
1 dy −1
y dx = a ×
√1 − x 2
dy − ay
⇒ dx =
√1 − x 2
By squaring both the sides, we obtain
() a 2y 2
dy 2
dx
=
1 − x2
)( ) = a y
2
(
dy
⇒ 1 − x2 dx
2 2
)( ) = a y
2
(
dy
1 − x2 dx
2 2
Again differentiating both sides with respect to x, we obtain
() dy
dx
2 d
dx (1 − x ) + (1 − x ) ×
2 2
d
dx [( ) ]dy
dx
2 d
( )
= a 2 dx y 2
() dy d 2y
2
( )×2 ,
dy dy
2 2
⇒ dx
( − 2x) + 1 − x dx dx 2 = a ⋅ 2y ⋅ dx
() d 2y
2
( )×2 ⋅
dy dy dy
2
⇒ dx
( − 2x) + 1 − x dx
= a 2 ⋅ 2y ⋅ dx
dx 2
d 2y
⇒ − x + (1 − x )
dy
2 2
=a ⋅y
dx dx 2
d 2y
⇒ (1 − x )
dy
2 2
−x −a y=0 2 dx
dx
Hence, proved.
Page : 192 , Block Name : Miscellaneous Exercise
−1 x
Q23 if y = e acos , − 1 ≤ x ≤ 1 show that
d 2y
(1 − x )
dy
2 − x dx − a 2y = 0
dx 2
−1
Answer. It is given that,y = e acos x
Taking logarithm on both the sides, we obtain
logy = acos − 1xloge
logy = acos − 1x
Differentiating both sides with respect to x, we obtain
1 dy −1
y dx = a ×
√1 − x 2
dy − ay
⇒ dx =
√1 − x 2
By squaring both the sides, we obtain
() a 2y 2
dy 2
dx
=
1 − x2
)( ) = a y
2
(
dy
⇒ 1 − x2 dx
2 2
)( ) = a y
2
(
dy
1 − x2 dx
2 2
Again differentiating both sides with respect to x, we obtain
() dy
dx
2 d
dx (1 − x ) + (1 − x ) ×
2 2
d
dx [( ) ]dy
dx
2 d
( )
= a 2 dx y 2
() dy d 2y
2
( )×2 ,
dy dy
2 2
⇒ dx
( − 2x) + 1 − x dx dx 2 = a ⋅ 2y ⋅ dx
() d 2y
2
( )×2 ⋅
dy dy dy
2
⇒ dx
( − 2x) + 1 − x dx
= a 2 ⋅ 2y ⋅ dx
dx 2
d 2y
⇒ − x + (1 − x )
dy
2 2
=a ⋅y
dx dx 2
d 2y
⇒ (1 − x )
dy
2 2
−x −a y=0 2 dx
dx
Hence, proved.
Page : 192 , Block Name : Miscellaneous Exercise
Document Details
| Board / Org | NCERT |
| Exam | Class 12 |
| Type | Solution |
| Pages | 92 |
| Updated | 30 Apr 2026 |