ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability MCQs

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability MCQs

Choose the correct answer from the given four options in questions (1 to 55) :

Question 1.
The number of points of discontinuity of the rational function f(x) = \(\frac{x^2-3 x+2}{4 x-x^3}\) is
(a) 1
(b) 2
(c) 3
(d) none
Solution:
(c) 3

Given f(x) = \(\frac{x^2-3 x+2}{4 x-x^3}\)
Clearly f(x) is discontinuous
when 4x – x³ = 0
⇒ x (4 – x²) = 0
⇒ x = 0, ± 2
Thus required no. of points of discontinuity of f(x) be three.

Question 2.
The number of points of discontinuity of the function f(x) = |x – 1| + |x – 2| + sin x, x ∈ [0, 4], is
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(d) 0

Given f(x) = |x – 1| + |x – 2| + sin x
When x < 1
∴ |x – 1| = – (x – 1) ;
When 1 ≤ x < 2
∴ |x – 1| = x – 1 ;
When x ≥ 2
∴ |x – 1| = x – 1 ;
|x – 2| = x – 2
∴ f(x) = \(\left\{\begin{array}{ccc}
-(x-1)-(x-2)+\sin x & ; & x<1 \\
x-1-(x-2)+\sin x & ; & 1 \leq x<2 \\
x-1+x-2+\sin x & ; & x \geq 2
\end{array}\right.\)
= \(\left\{\begin{array}{clc}
-2 x+3+\sin x & ; & x<1 \\
1+\sin x & ; & 1 \leq x<2 \\
2 x-3+\sin x & ; & x \geq 2
\end{array}\right.\)

at x = 1
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) – 2x + 3 + sin x
= 1 + sin 1
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 1 + sin x
= 1 + sin 1
∴ f(x) is continous at x = 1.

at x = 2:
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) 1 + sin x
= 1 + sin 2
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) (2x – 3) + sin x
= 1 + sin 2
∴ f(x) is continous at x = 2.

Since every polynomial function and trigonometric function is continuous in its domain.
Thus given function is everywhere continuous.

Question 3.
The functionf(x) = cot x is discontinuous on the set
(a) {x = nπ, n ∈ N}
(b) {x = 2nπ, n ∈ Z}
(c) {x = \(\frac{n \pi}{2}\), n ∈ Z}
(d) {x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z}
Solution:
(d) {x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z}

Given f(x) = cot x
= \(\frac{\cos x}{\sin x}\)
f(x) is not defined when sin x = 0
⇒ x = nπ ∀ n ∈ Z
Thus, f(x) is discontinuous on set {x = nπ, n ∈ N}.

Question 4.
The domain of continuity of the function f(x) = tan x is
(a) R – {nπ, n ∈ Z}
(b) R – {2nπ, n ∈ Z}
(c) R – {(2n + 1)\(\frac{\pi}{2}\), n ∈ Z}
(d) R – {\(\frac{n \pi}{2}\), n ∈ Z}
Solution:
(c) R – {(2n + 1)\(\frac{\pi}{2}\), n ∈ Z}

Given f(x) = tan x
= \(\frac{\sin x}{\cos x}\)
Now f(x) is not defined when cos x = 0
⇒ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
So f(x) is discontinuous at x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
Hence domain of continuity of function f(x) be R – {(2n + 1) \(\frac{\pi}{2}\), n ∈ Z}.

Question 5.
The functionf(x) = [x], where [x] denotes the greatest integer function, is continuous at
(a) – 2
(b) 1
(c) 4
(d) 1.5
Solution:
We know that,
\(\underset{x \rightarrow a^{-}}{\mathrm{Lt}}\) [x] = a – 1
\(\underset{x \rightarrow a^{+}}{\mathrm{Lt}}\) [x] = a, where a ∈ I
Thus, f(x) is discontinuous for all integral points
and \(\underset{x \rightarrow a}{\mathrm{Lt}}\) = [a]
Thus f(x) is continuous for all non-integral points.
Clearly f(x) is continuous at x = 1.5 = \(\frac{3}{2}\) ∉ I.

Question 6.
The function f(x) = \(\left\{\begin{array}{cc}
x-1, & x<2 \\
2 x-3, & x \geq 2
\end{array}\right.\) is continous function
(a) at x = 2 only
(b) for all real values of x
(c) for all real values of x except 2
(d) for all integral values of X only
Solution:
When x < 2 ; f (x) = x – 1 which is a polynomial function and hence continuous everywhere. When x > 2 ;
f (x) = 2x – 3, which is a polynomial function and hence continuous everywhere.
at x = 2
\(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}\) x – 1
= 2 – 1 = 1
\(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}\) 2x – 3
= 4 – 3 = 1
and f(2) = 1
∴ f(x) is continuous at x = 2
Thusf(x) is continuous for all real values of x.

Question 7.
If f(x) = \(\left\{\begin{array}{cc}
5 x-4, & 0 4 x^2+3 a x, & 1<x<2
\end{array}\right.\) is continuous for all x ∈ (0, 2), then the value of a is
(a) 1
(b) 0
(c) – 1
(d) \(\frac{1}{3}\)
Solution:
(c) – 1

Since function f(x) is continuous for all x ∈ (0, 2)
∴ f(x) is continuous at x = 1
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) 5x – 4
= 5 – 4 = 1
and \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) (4x² + 3ax)
= 4 + 3a
⇒ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)
⇒ 1 = 4 + 3a
⇒ – 3 = 3a
⇒ a = – 1.

Question 8.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\sqrt{4+x}-2}{x}, & x \neq 0 \\
k, & x=0
\end{array}\right.\) is continous at x = 0, then the value of k is
(a) 1
(b) 4
(c) \(\frac{1}{4}\)
(d) 0
Solution:
(c) \(\frac{1}{4}\)

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{\sqrt{4+x}-2}{x} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{4+x-4}{x[\sqrt{4+x}+2]}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{1}{\sqrt{4+x}+2}\)
= \(\frac{1}{2+2}=\frac{1}{4}\)
and f(0) = k
Since f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\)

Question 9.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\sqrt{x^2+5}-3}{x+2}, & x \neq-2 \\
k, & x=-2
\end{array}\right.\) is continuous at x = – 2, then the value of k is
(a) – \(\frac{2}{3}\)
(b) 0
(c) \(\frac{2}{3}\)
(d) none of these
Solution:
(a) – \(\frac{2}{3}\)

Question 10.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\sin \pi x}{5 x}, & x \neq 0 \\
k, & x=0
\end{array}\right.\) is continuous at x = 0, then k is equal to
(a) \(\frac{\pi}{5}\)
(b) \(\frac{5}{\pi}\)
(c) 1
(d) 0
Solution:
(b) \(\frac{5}{\pi}\)

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{\sin \pi x}{5 x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{\sin \pi x}{\pi x} \times \frac{\pi}{5}\)
= \(\frac{\pi}{5} \times 1=\frac{\pi}{5}\)
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\sin \theta}{\theta}\) = 1]
Since f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ \(\frac{\pi}{5}\) = k.

Question 11.
The value of function f at x = 0, so that the function f(x) = \(\frac{2^x-2^{-x}}{x}\), x ≠ 0 is continuous at x = 0, is
(a) 0
(b) log 2
(c) log 4
(d) 2⁴
Solution:
(c) log 4

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{2^x-2^{-x}}{x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2^x-1-\left(2^{-x}-1\right)}{x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2^x-1}{x}+\underset{x \rightarrow 0}{\ {Lt}} \frac{2^{-x}-1}{-x}\)
= log 2 + log 2
= 2 lo0g 2
= log 2²
= log 4
Since f(x) is continuous at x = 0
∴ f(0) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = log 4.

Question 12.
If f(x) = \(\frac{2 x+\sin ^{-1} x}{2 x-\tan ^{-1} x}\) is continuous for all x in (- 1, 1), then the value of f(0) is
(a) 2
(b) 3
(c) – 3
(d) \(\frac{3}{2}\)
Solution:
(b) 3

\(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{2 x+\sin ^{-1} x}{2 x-\tan ^{-1} x}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2+\frac{\sin ^{-1} x}{x}}{2-\frac{\tan ^{-1} x}{x}}\)
= \(\frac{2+1}{2-1}\) = 3
[∵ \(\underset{\theta \rightarrow 0}{\mathrm{Lt}} \frac{\sin ^{-1} \theta}{\theta}=\underset{\theta \rightarrow 0}{\mathrm{Lt}} \frac{\tan ^{-1} \theta}{\theta}\) = 1]
Since f(x) is continuous ∀ x ∈ (- 1, 1)
∴ f(x) is continuous at x = 0 ∈ (- 1, 1).
Thus, f(0) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = 3.

Question 13.
If f(x) = \(\left\{\begin{array}{cc}
\frac{1-\tan x}{4 x-\pi}, & x \neq \frac{\pi}{4} \\
k, & x=\frac{\pi}{4}
\end{array}\right.\) is continuous at x = \(\frac{\pi}{4}\), then the value of k is
(a) 1
(b) – 1
(c) \(\frac{1}{2}\)
(d) – \(\frac{1}{2}\)
Solution:
(d) – \(\frac{1}{2}\)

Given f(x) is continuous at x = \(\frac{\pi}{4}\)
∴ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) f(x) = f(latex]\frac{\pi}{4}[/latex])
⇒ – \(\frac{1}{2}\) = k.

Question 14.
If f(x) = \(\left\{\begin{array}{cc}
\tan \left(\frac{\pi}{4}-x\right) & x \neq \frac{\pi}{4} \\
k, & x=\frac{\pi}{4}
\end{array}\right.\) is continuous at x = \(\frac{\pi}{4}\), then the value of k is
(a) 1
(b) 2
(c) \(\frac{1}{2}\)
(d) none of these
Solution:
(c) \(\frac{1}{2}\)

Given f(x) is continuous at x = \(\frac{\pi}{4}\)
∴ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) f(x) = f(\(\frac{\pi}{4}\))
⇒ \(\frac{1}{2}\) = k.

Question 15.
If f(x) = \(\left\{\begin{array}{cc}
\frac{1-\cos p x}{x \sin x}, & x \neq 0 \\
\frac{1}{2}, & x=0
\end{array}\right.\) is continuous at x = 0, then p is equal to
(a) 2
(b) – 2
(c) 1, – 1
(d) none of these
Solution:
(c) 1, – 1

∴ \(\underset{h \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ \(\frac{p^2}{2}=\frac{1}{2}\)
⇒ p = 1.

Question 16.
If f(x) = \(\left\{\begin{array}{cc}
\frac{\sqrt{1-\cos 2 x}}{\sqrt{2} x}, & x \neq 0 \\
k, & x=0
\end{array}\right.\), then which value of k will make function f continuous at x = 0?
(a) 1
(b) – 1
(c) 0
(d) none of these
Solution:
(d) none of these

When x → 0^–
⇒ sin x < 0 as x lies in IVth quadrant
⇒ |sin x| = – sin x
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = – 1
⇒ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
Thus f(x) is discontinuous function.

Question 17.
If f(x) = \(\left\{\begin{array}{cc}
x^2 \sin \frac{1}{x}, & x \neq 0 \\
k, & x=0
\end{array}\right.\) is continuous at x = 0, then the value of k
(a) 1
(b) – 1
(c) 0
(d) none of these
Solution:
(c) 0

Let g(x) = x²
and h(x) = sin \(\frac{1}{x}\)
Here \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) g(x) = \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x² = 0
and |h(x)| = |sin \(\frac{1}{x}\)| ≤ 1 ∀ x ∈ R – {0}
Thus h(x) is bounded in the ngd of o.
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) g(x) h(x) = 0
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) x² sin \(\frac{1}{x}\) = 0
⇒ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = 0
and f(0) = k
Given f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ 0 = k.

Question 18.
If f(x) = \(\left\{\begin{array}{rr}
m x+1, & x \geq \frac{\pi}{2} \\
\sin x+n, & x<\frac{\pi}{2}
\end{array}\right.\) is continuous at x = \(\frac{\pi}{2}\), then
(a) m = 1, n = 0
(b) m = \(\frac{n \pi}{2}\) + 1
(c) n = \(\frac{m \pi}{2}\)
(d) m = n = \(\frac{\pi}{2}\)
Solution:
\(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{+}}{2}
\end{gathered}\) f(x) = \(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{+}}{2}
\end{gathered}\) mx + 1
= \(\frac{m \pi}{2}\) + 1
\(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{-}}{2}
\end{gathered}\) f(x) = \(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{-}}{2}
\end{gathered}\) sin x + n
= sin \(\frac{\pi}{2}\) + n
= 1 + n
Since f(x) is continuous at x = \(\frac{\pi}{2}\)
∴ \(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{-}}{2}
\end{gathered}\) f(x) = \(\begin{gathered}
\mathrm{Lt} \\
x \rightarrow \frac{\pi^{-}}{2}
\end{gathered}\) f(x)
⇒ \(\frac{m \pi}{2}\) + 1 = 1 + n
⇒ n = \(\frac{m \pi}{2}\).

Question 19.
The functionf(x) = | x | at x = 0 is
(a) continuous but not differentiable
(b) differentiable but not continuous
(c) continuous and differentiable
(d) discontinuous and not differentiable
Solution:
(a) continuous but not differentiable

\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) |x|
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x = 0
and \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) |x|
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – x = 0
and f(0) = |0| = 0
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
∴ f(x) is continuous at x = 0
Lf'(0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{|x|-0}{x-0}\)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\)
= – 1
Rf'(0) = \(\underset{x \rightarrow 0^{+}}{\ {Lt}} \frac{|x|-0}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{x}{x}\)
= 1
∴ Lf'(0) ≠ Rf'(0)
⇒ f(x) is not differentiable at x = 0.

Question 20.
The function f(x) = |x| at x = 0 is
(a) continuous but not differentiable
(b) differentiable but not continuous
(c) continuous and differentiable
(d) neither continuous nor differentiable
Solution:
(c) continuous and differentiable

f(x) = x |x|
= \(\left\{\begin{aligned}
-x^2 & ; \quad x<0 \\
x^2 & ; \quad x \geq 0
\end{aligned}\right.\)

Continuity at x = 0:
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x² = 0
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – x² = 0
and f(0) = 0
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
⇒ f(x) is continuous at x = 0.

Differentiability at x = 0:
Lf'(0) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{-x^2-0}{x}\)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x = 0
and Rf'(0) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) \(\frac{x^2-0}{x}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x = 0
∴ Lf'(0) = Rf'(0)
⇒ f(x) is continuous at x = 0.

Question 21.
The derivative of the fiunction f(x) = x |x| at x = 0 is
(a) 1
(b) 0
(c) 2
(d) – 2
Solution:
(c) 2

f(x) = x |x|
= \(\left\{\begin{aligned}
-x^2 & ; \quad x<0 \\
x^2 & ; \quad x \geq 0
\end{aligned}\right.\)

Continuity at x = 0:
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x² = 0
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – x² = 0
and f(0) = 0
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
⇒ f(x) is continuous at x = 0.

Differentiability at x = 0:
Lf'(0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{-x^2-0}{x}\)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) x = 0
and Rf'(0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{x^2-0}{x}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) x = 0
∴ Lf'(0) = Rf'(0)
⇒ f(x) is continuous at x = 0.

Question 22.
If f(x) = \(\left\{\begin{array}{cc}
x, & x \leq 0 \leq 1 \\
2 x-1, & x>1
\end{array}\right.\), then
(a) f is discontinuous at x = 1
(b) f is differentiable at x = 1
(c) f is continuous but not differentiable at x = 1
(d) none of these
Solution:
(c) f is continuous but not differentiable at x = 1

\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 2x – 1
= 2 × 1 – 1 = 1
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) x = 1
and f(1) = 1
∴ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = f(1)
⇒ f(x) is continuous at x = 1.

Differentiability at x = 1:
Lf'(1) = \(\ {Lt}_{x \rightarrow 1^{-}} \frac{x-1}{x-1}\) = 1
Rf'(1) = \(\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{2 x-1-1}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-1}} \frac{2(x-1)}{x-1}\)
= 2
∴ Lf'(1) ≠ Rf'(1)
⇒ f(x) is not differentiable at x = 1.

Question 23.
If f(x) = \(\left\{\begin{array}{cc}
a x^2+1, & x>1 \\
x+a, & x \leq 1
\end{array}\right.\) is derivable at x = 1, then the value of a is
(a) 0
(b) 1
(c) \(\frac{1}{2}\)
(d) 2
Solution:
(c) \(\frac{1}{2}\)

at x = 1
Lf'(1) = \(\ {Lt}_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{x+a-1-a}{x-1}\) = 1
and Rf'(1) = \(\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{a x^2+1-1-a}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{a\left(x^2-1\right)}{x-1}\)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) a(x + 1) = 2a
Since f(x) is derivable at x = 1
∴ Lf'(1) = Rf'(1)
⇒ 1 = 2a
⇒ a = \(\frac{1}{2}\)

Question 24.
The function f(x) = |x| + |x – 1|
(a) differentiable at x = 0 but not at x = 1
(b) differentiable at x = 1 but not at x = 0
(c) neither differentiable at x = 0 nor at x = 1
(d) differentiable at x = 0 as well as at x = 1
Solution:
(c) neither differentiable at x = 0 nor at x = 1

Given f(x) = |x| + |x – 1|
= \(\left\{\begin{array}{ccc}
-x-(x-1) & ; & x<0 \\
x-(x-1) & ; & 0 \leq x<1 \\
x+x-1 & ; & x \geq 1
\end{array}\right.\)
⇒ f(x) = \(\left\{\begin{array}{ccc}
-2 x+1 & ; & x<0 \\
1 & ; & 0 \leq x<1 \\
2 x-1 & ; & x \geq 1
\end{array}\right.\)

at x = 0:
Lf'(0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{-2 x+1-1}{x}\)
= – 2
Rf'(0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{1-1}{x}\) = 0
∴ Lf'(0) ≠ Rf'(0)
∴ f is not differentiable at x = 0.

at x = 1:
Lf'(1) = \(\ e{Lt}_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{1-1}{x-1}\) = 0
and Rf'(1) = \(\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{2 x-1-1}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{2(x-1)}{x-1}\) = 2
∴ Lf'(1) ≠ Rf'(1)
∴ f is not differentiable at x = 1.

Question 25.
The derivative of f(x) = 3 |2 + x| at x = – 3 is
(a) 3
(b) – 3
(c) 0
(d) d0es not exist
Solution:
(b) – 3

Given f(x) = 3 |2 + x|
∴ f'(x) = \(\frac{3(2+x)}{|(2+x)|}\) × 1
⇒ f'(- 3) = \(\frac{3(2-3)}{|2-3|}\) = – 3.

Question 26.
The derivative of f(x) = |x – 1| + |x – 3| at x = 2 is
(a) 2
(b) – 2
(c) 0
(d) does not exist
Solution:
(c) 0

Given f(x) = |x – 1| + |x – 3|
∴ f'(x) = \(\frac{x-1}{|x-1|}+\frac{x-3}{|x-3|}\)
⇒ f'(2) = \(\frac{2-1}{|2-1|}+\frac{2-3}{|2-3|}\)
= \(\frac{1}{|1|}-\frac{1}{|-1|}\)
= 1 – 1 = 0.

Question 27.
The function f(x) = e^|x| is
(a) continuous everywhere but not differentiable at x = 0
(b) continuous and differentiable everywhere
(e) not continuous at x = 0
(d) none of these
Solution:
(a) continuous everywhere but not differentiable at x = 0

\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) e^x
= e⁰ = 1
and \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) e^{– x}
= e⁰ = 1
and f(0) = e^|0|
= e = 1
Thus, \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)
∴ f(x) is continuous at x = 0.
Since exponential function is continuous for all x < 0 and x > 0
Thus f(x) is continuous everywhere.
Further when x < 0 ; f (x) = e^{– x} is differentiable function.
when x > 0 ; f(x) = e^x is differentiable function.

at x = 0:
Lf'(0) = \(\ {Lt}_{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{e^{-x}-1}{x}\)
= – \(\ {Lt}_{x \rightarrow 0^{-}} \frac{e^{-x}-1}{-x}\)
= – 1
and Rf'(0) = \(\ {Lt}_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}\)
= \(\ {Lt}_{x \rightarrow 0^{+}} \frac{e^x-1}{x}\)
= 1
∴ Lf'(0) ≠ Rf'(0)
Thus f(x) is not differentiable at x = 1.

Question 28.
The function f(x) = |sin x|
(a) f is differentiable everywhere
(b) f is continuous everywhere but not differentiable at x = nπ, n ∈ Z
(c) f is continuous everywhere but not differentiable atx=(2n+1) \(\frac{\pi}{2}\), n ∈ Z
(d) none of these
Solution:
(b) f is continuous everywhere but not differentiable at x = nπ, n ∈ Z

f(x) = |sin x|
= \(\left\{\begin{aligned}
-\sin x ; & x<n \pi \\
\sin x ; & x \geq n \pi
\end{aligned}\right.\) ;
x = even
x = nπ (n even)

∴ L.H.D. ≠ R.H.D.
Thus f(x) is not differentiable at x = nπ (n even)

∴ L.H.D. = R.H.D.
Thus f(x) is not differentiable at x = nπ (n = odd)
Hence f(x) is not differentiable at x = nπ, n ∈ Z.

Question 29.
The set of numbers where the function f given by f (x) = | 2x – 1 | cos x is differentiable is
(a) R
(b) R – {\(\frac{1}{2}\)}
(c) (0, ∞)
(d) none of these
Solution:
Refer to MCQ-28 (RD Sharma 10 + 2) Chapter 9.

Question 30.
If y = log (sec e^x2), then \(\frac{d y}{d x}\) =
(a) x² e^x2 tan e^x2
(b) e^x2 tan e^x2
(c) 2x e^x2 tan e^x2
(d) none of these
Solution:
(c) 2x e^x2 tan e^x2

Given y = log (sec e^x2)
∴ \(\frac{d y}{d x}\) = \(\frac{1}{\sec e^{x^2}} \frac{d}{d x}\) sec e^x2
= \(\frac{1}{\sec e^{x^2}}\) sec e^x2 . tan e^x2 . e^x2 . 2x
= e^x2 . 2x tan e^x2.

Question 31.
If y = log \(\left(\frac{1-x^2}{1+x^2}\right)\), |x| < 1, then \(\frac{d y}{d x}\) =
(a) \(\frac{4 x^3}{1-x^4}\)
(b) \(\frac{-4 x}{1-x^4}\)
(c) \(\frac{1}{4-x^4}\)
(d) \(\frac{-4 x^3}{1-x^4}\)
Solution:
(b) \(\frac{-4 x}{1-x^4}\)

Given y = log \(\left(\frac{1-x^2}{1+x^2}\right)\)
= log (1 – x²) – log (1 + x²)
∴ \(\frac{d y}{d x}\) = \(\frac{-2 x}{1-x^2}-\frac{2 x}{1+x^2}\)
= \(\frac{-2 x\left(1+x^2+1-x^2\right)}{1-x^4}\)
= \(\frac{-4 x}{1-x^4}\).

Question 32.
If f(x) = log x, then the derivative of f (log x) w.r.t. x is
(a) \(\frac{\log x}{x}\)
(b) \(\frac{x}{\log x}\)
(c) x log x
(d) \(\frac{1}{x \log x}\)
Solution:
(d) \(\frac{1}{x \log x}\)

Given f(x) = log x
⇒ f(log x) = log (log x)
⇒ \(\frac{d}{d x}\) f (log x) = \(\frac{1}{\log x}\) \(\frac{d}{d x}\) log x
= \(\frac{1}{x \log x}\).

Question 33.
For the curve √x + √y = 4, \(\frac{d y}{d x}\) at (\(\left(\frac{1}{4}, \frac{1}{4}\right)\)) is
(a) – 1
(b) 1
(c) 2
(d) \(\frac{1}{2}\)
Solution:
(a) – 1

Given √x + √y = 4
Diff. both sides w.r.t. x ; we have
\(\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \frac{d y}{d x}\) = 0
\(\frac{d y}{d x}=-\sqrt{\frac{y}{x}}\)
at (\(\left(\frac{1}{4}, \frac{1}{4}\right)\)) ;
\(\frac{d y}{d x}\) = – \(\sqrt{\frac{\frac{1}{4}}{\frac{1}{4}}}\) = – 1

Question 34.
If y = \(\sqrt{\sin x+y}\), then \(\frac{d y}{d x}\) =
(a) \(\frac{\cos x}{2 y-1}\)
(b) \(\frac{\cos x}{1-2 y}\)
(c) \(\frac{\sin x}{2 y-1}\)
(d) \(\frac{\sin x}{1-2 y}\)
Solution:
(a) \(\frac{\cos x}{2 y-1}\)

Given y = \(\sqrt{\sin x+y}\)
⇒ y² = sin x + y ;
diff. both sides w.r.t. x ;
2y \(\frac{d y}{d x}\) = cos x + \(\frac{d y}{d x}\)
⇒ (2y – 1) \(\frac{d y}{d x}\) = cos x
⇒ \(\frac{d y}{d x}\) = \(\frac{\cos x}{2 y-1}\).

Question 35.
The derivative of sec (tan^-1 x) w.r.t. x is
(a) \(\frac{x}{1+x^2}\)
(b) \([\frac{1}{\sqrt{1+x^2}}/latex]
(c) [latex]\frac{x}{\sqrt{1+x^2}}\)
(d) x \(\sqrt{1+x^2}\)
Solution:
(c) \(\frac{x}{\sqrt{1+x^2}}\)

Let y = sec (tan^-1 x)
put tan^-1 x = θ
⇒ x = tan θ

∴ y = sec θ
= \(\sqrt{x^2+1}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{2}\) (x² + 1)^{\(-\frac{1}{2}\)} \(\frac{d}{d x}\) (x² + 1)
= \(\frac{2 x}{2 \sqrt{x^2+1}}\)
= \(\frac{x}{\sqrt{x^2+1}}\)

Question 36.
If f(x) = x tan^-1 x, then f'(1) is equal to
(a) \(\frac{\pi}{4}+\frac{1}{2}\)
(b) \(\frac{\pi}{4}-\frac{1}{2}\)
(c) \(\frac{1}{2}-\frac{\pi}{4}\)
(d) none of these
Solution:
(a) \(\frac{\pi}{4}+\frac{1}{2}\)

Given f(x) = x tan^-1 x
f'(x) = tan^-1 x . 1 + \(\frac{x}{1+x^2}\)
f'(1) = tan^-1 1 + \(\frac{1}{1+1^2}\)
= \(\frac{\pi}{4}+\frac{1}{2}\).

Question 37.
The derivative of tan^-1 \(\left(\frac{3 x-4 x^3}{1-3 x^2}\right)\) w.r.t. x is
(a) \(\frac{1}{1+x^2}\)
(b) \(\frac{3}{1+9 x^2}\)
(c) \(\frac{3}{1+x^2}\)
(d) 3 sec² 3x
Solution:

Let y = tan^-1 \(\left(\frac{3 x-4 x^3}{1-3 x^2}\right)\)
put x = tan θ
⇒ θ = tan^-1 x
⇒ y = tan^-1 \(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)\)
= tan^-1 (tan 3θ)
Thus y = 3θ
= 3 tan^-1 x
⇒ \(\frac{d y}{d x}\) = \(\frac{3}{1+x^2}\).

Question 38.
The derivative of tan^-1 x w.r.t. cot^-1 x is
(a) \(\frac{\pi}{2}\)
(b) – 1
(c) 1
(d) none of these
Solution:
(b) – 1

Let y = tan^-1 x
and z = cot^-1 x
∴ \(\frac{d y}{d x}=\frac{1}{1+x^2}\)
and \(\frac{d z}{d x}=\frac{-1}{1+x^2}\)
we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(– \frac{\frac{1}{1+x^2}}{\frac{1}{1+x^2}}\) = – 1

Question 39.
The derivative of cos^-1 (2x² – 1) w.r.t. cos^-1 x is
(a) 2
(b) \(\frac{2}{x}\)
(c) 1 – x²
(d) \(\frac{-1}{2 \sqrt{1-x^2}}\)
Solution:
(a) 2

Let y = cos^-1 (2x² – 1) ………..(1)
and z = cos^-1 x ………..(2)
we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
putting x = cos θ
∴ θ = cos^-1 x in eqn. (1) ;
y = cos^-1 (2 cos² θ – 1)
= cos^-1 (cos 2θ)
= 2θ
= 2 cos^-1 x
∴ y = 2z
⇒ \(\frac{d y}{d z}\) = 2.

Question 40.
The derivative of sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\) w.r.t. tan^-1 \(\left(\frac{2 x}{1-x^2}\right)\) is
(a) \(\frac{1}{2}\)
(b) 2
(c) \(\frac{1-x^2}{1+x^2}\)
(d) 1
Solution:
(d) 1

Let y = sin^-1 \(\left(\frac{2 x}{1+x^2}\right)\)
= 2 tan^-1 x
and z = tan^-1 \(\left(\frac{2 x}{1-x^2}\right)\)
= 2 tan^-1 x
∴ y = z
⇒ \(\frac{d y}{d z}\) = 1.

Question 41.
The derivative of tan^-1 \(\left(\frac{x}{\sqrt{1-x^2}}\right)\) w.r.t. x is
(a) \(\frac{1}{\sqrt{1-x^2}}\)
(b) \(\frac{1}{\sqrt{1-x^2}}\)
(c) \(– \frac{1}{\sqrt{1-x^2}}\)
(d) none of these
Solution:
(b) \(\frac{1}{\sqrt{1-x^2}}\)

Let y = tan^-1 \(\left(\frac{x}{\sqrt{1-x^2}}\right)\)
put x = sin θ
∴ θ = sin^-1 x
∴ y = tan^-1 \(\left(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\right)\)
= tan^-1 (tan θ)
= θ
= sin^-1 x
⇒ \(\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}\)

Question 42.
The derivative of sin^-1 \(\left(\frac{x}{\sqrt{1+x^2}}\right)\) w.r.t. x is
(a) \(\frac{1}{\sqrt{1-x^2}}\)
(b) \(\frac{1}{\left(1+x^2\right)^{\frac{3}{2}}}\)
(c) \(\frac{1}{1+x^2}\)
(d) none of these
Solution:
(c) \(\frac{1}{1+x^2}\)

Let y = sin^-1 \(\left(\frac{x}{\sqrt{1+x^2}}\right)\)
put x = tan θ
∴ θ = tan^-1 x
⇒ y = sin^-1 \(\left(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\right)\)
= sin^-1 \(\left(\frac{\tan \theta}{\sqrt{1+\tan ^2 \theta}}\right)\)
= sin^-1 \(\left(\frac{\tan \theta}{\sec \theta}\right)\)
⇒ y = θ = tan^-1 x
Thus, \(\frac{d y}{d x}=\frac{1}{1+x^2}\).

Question 43.
If y = cos^-1 \(\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) + cosec^-1 \(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\) then \(\frac{d y}{d x}\) is equal to
(a) \(\frac{\pi}{2}\)
(b) 0
(c) 1
(d) none of these
Solution:
(b) 0

Given y = cos^-1 \(\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) + cosec^-1 \(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)
⇒ y = cos^-1 \( \left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\) + sin^-1 \( \left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
[∵ cosec^-1 x = sin^-1 \(\frac{1}{x}\)]
⇒ y = \(\frac{\pi}{2}\)
[∵ cos^-1 x + sin^-1 x = \(\frac{\pi}{2}\) ∀ x ∈ [- 1, 1]]
∴ \(\frac{d y}{d x}\) = 0.

Question 44.
If y = a (1 + cos t) and x = a (t – sin t) then \(\frac{d y}{d x}\) is equal to
(a) tan \(\frac{t}{2}\)
(b) – tan \(\frac{t}{2}\)
(c) – cot \(\frac{t}{2}\)
(d) none of these
Solution:
(c) – cot \(\frac{t}{2}\)

Given y = a (1 + cos t)
∴ \(\frac{d y}{d t}\) = – a sin t
and x = a (t – sin t)
∴ \(\frac{d x}{d t}\) = a (1 – cos t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{-a \sin t}{a(1-\cos t)}\)
= \(\frac{-2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \sin ^2 \frac{t}{2}}\)
= – cot \(\frac{t}{2}\).

Question 45.
If x = a cos³ t and y = a sin³ t, then \(\frac{d y}{d x}\) is equal to
(a) cos t
(b) sec t
(c) cosec t
(d) – tan t
Solution:
(d) – tan t

Given x = a cos³ t
⇒ \(\frac{d x}{d t}\) = 3a cos² t (- sin t)
and y = a cos³ t
⇒ \(\frac{d y}{d t}\) = 3a sin² t (cos t)
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{3 a \sin ^2 t \cos t}{-3 a \cos ^2 t \sin t}\)
= – tan t.

Question 46.
If x = t² and y = t³, then \(\frac{d^2 y}{d x^2}\) is equal to
(a) \(\frac{3}{2}\)
(b) \(\frac{3}{2}\) t
(c) \(\frac{3}{2 t}\)
(d) \(\frac{3}{4 t}\)
Solution:
(d) \(\frac{3}{4 t}\)

Given x = t²
⇒ \(\frac{d x}{d t}\) = 2t
and y = t³
⇒ \(\frac{d y}{d t}\) = 3t²
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\)
= \(\frac{3 t^2}{2 t}=\frac{3}{2} t\)
⇒ \(\frac{d^2 y}{d x^2}=\frac{d}{d t}\left(\frac{3}{2} t\right) \frac{d t}{d x}\)
= \(\frac{3}{2} \times \frac{1}{2 t}=\frac{3}{4 t}\).

Question 47.
The derivative of log x with respect to \(\frac{1}{x}\) is
(a) – \(\frac{1}{x^3}\)
(b) – \(\frac{1}{x}\)
(c) – x
(d) \(\frac{1}{x^3}\)
Solution:
(c) – x

Let y = log x
⇒ \(\frac{d y}{d x}\) = \(\frac{1}{x}\)
and z = \(\frac{1}{x}\)
⇒ \(\frac{d z}{d x}\) = \(-\frac{1}{x^2}\)
we want to diff. y w.r.t. z i.e. to find \(\frac{d y}{d z}\)
∴ \(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)
= \(\frac{\frac{1}{x}}{-\frac{1}{x^2}}\)
= – x.

Question 48.
The function f : R → R given by f(x) = – |x – 1| is
(a) continuous as well as differentiable at x = 1
(b) not continuous but differentiable at x = 1
(c) continuous but not differentiable at x = 1
(d) neither continuous nor differentiable atx= 1.
Solution:
(c) continuous but not differentiable at x = 1

Given f(x) = – |x – 1|
= \(\left\{\begin{array}{ccc}
(x-1) & ; & x<1 \\
-(x-1) & ; & x \geq 1
\end{array}\right.\)

Continuity at x = 1:
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) – (x – 1)
= – (1 – 1) = 0
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) (x – 1)
= (1 – 1) = 0
and f(1) = – (1 – 1) = 0
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = f(1)
Thus f(x) is continuous at x = 1.

Differentiability at x = 1:
Lf'(1) = \(\ {Lt}_{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{-}} \frac{x-1-0}{x-1}\)
= 1
Rf'(0) = \(\ {Lt}_{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\)
= \(\ {Lt}_{x \rightarrow 1^{+}} \frac{-(x-1)-0}{x-1}\)
= – 1
∴ Lf'(1) ≠ Rf'(1)
Thus, f(x) is not differentiable at x = 1.

Question 49.
If y = f(x²) and f'(x) = e^√x, then \(\frac{d y}{d x}\) is equal to
(a) 2x e^2√x
(b) 2x e^x
(c) 4x e^√x
(d) 4x e^x
Solution:
(b) 2x e^x

Given y = f(x²)
∴ \(\frac{d y}{d x}\) = f'(x²) . 2x
= \(e^{\sqrt{x^2}}\) . 2x
= 2x e^x
[∵ f'(x) = e^√x
⇒ f'(x²) = \(e^{\sqrt{x^2}}\).

Question 50.
If y = log_e \(\left(\frac{x^2}{e^2}\right)\), then \(\frac{d^2 y}{d x^2}\) is equal to
(a) \(-\frac{1}{x}\)
(b) \(-\frac{1}{x^2}\)
(c) \(\frac{2}{x^2}\)
(d) \(-\frac{2}{x^2}\)
Solution:
(d) \(-\frac{2}{x^2}\)

Given y = log_e \(\left(\frac{x^2}{e^2}\right)\)
= log x² – log e²
= 2 log x – log e²
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{2}{x}\)
∴ \(\frac{d^2 y}{d x^2}=-\frac{2}{x^2}\).

Question 51.
In Rolle’s theorem, the value of c for the function f (x) = x³ – 3x in the interval [0, √3] is
(a) 1
(b) – 1
(c) \(\frac{1}{3}\)
(d) \(\frac{2}{3}\)
Solution:
(a) 1

Given f(x) = x³ – 3x
∴ f’(x) = 3x² – 3
which exists for all x ∈ R
Since f (x) is polynomial in x so it is continuous and differentiable everywhere.
f(x) is continuous in [0, π] and f(x) is diff. in (0, √3)
Also, f(0) = 0 ;
f(√3) = (√3)³ – 3√3
= 3√3 – 3√3 = 0
∴ f(0) = f(5)
Thus, all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real number c ∈ (0, √3) s.t f’ (c) = 0
⇒ 3c² – 3 = 0
⇒ c = ± 1
∴ c = 1 ∈ (0, √3).

Question 52.
The value of c is Rolle’s theorem for the function f(x) = e^x sin x, x ∈[0, π], is
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{2}\)
(d) \(\frac{3 \pi}{4}\)
Solution:
(d) \(\frac{3 \pi}{4}\)

Given f(x) = e^x sin x
Since exponential function is continuous and differentiable everywhere.
f’(x) = e^x cos x + sin x e^x
= e^x (cos x + sin x)
Further the product of continuous and differentiable is continuous and differentiable.
∴ f(x) is continuous in [0, π] and differentiable in (0, π).
Now f(0) = e⁰ sin 0 = 0
f(π) = e^π sin π = 0
∴ f(0) = f(π)
Thus all the three conditions of Rolle’s theorem are satisfied so ∃ atleast one real number c ∈ (0, π).
s.t. f'(c) = 0
⇒ e^c (cos c + sin c)= 0
⇒ cos c + sin c = 0
[∵ e^c > 0]
⇒ tan c = – 1
= – tan \(\frac{\pi}{4}\)
= tan (- \(\frac{\pi}{4}\))
⇒ c = nπ – \(\frac{\pi}{4}\), n ∈ Z
⇒ c = \(\frac{3 \pi}{4}\), – \(\frac{\pi}{4}\), ………….. but c ∈ (0, π)
∴ c = \(\frac{3 \pi}{4}\).

Question 53.
Rolle’s theorem is applicable in the interval [- 1, 1] for the function
(a) f (x) = x
(b) f(x) = x²
(c) f(x) = x³ + 2
(d) f(x) = |x|
Solution:
(b) f(x) = x²

Now f(x) = x², which is polynomial in x and hence continuous and differentiable everywhere
∴ f (x) is continuous in [- 1, 1] and differentiable in (- 1, 1).
Further f(1) = 1² = 1 ;
f(- 1) = (- 1)² = 1
∴ f(1) = f(- 1)
So all three conditions of Rolle’s theorem
so ∃ atleastonerealnumber c ∈ (- 1, 1)
s.t f’(c) = 0
⇒ 2c = 0
⇒ c = 0 ∈ (- 1, 1)
Rolle’s theorem is applicable.

Question 54.
The value of c in Lagrange’s Mean Value= theorem for the function f(x) = x (x – 2) in the interval [1, 2] is
(a) \(\frac{3}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{3}{4}\)
(d) \(\frac{5}{4}\)
Solution:
(a) \(\frac{3}{2}\)

Given f(x) = x (x – 2)
which is a polynomial in x and hence continuous and differentiable everywhere.
Now f’ (x) = 2x – 2 which exists ∀ x ∈ R
Thus f (x) is continuous in [1, 2] and differentiable in (1,2).
So all the conditions of Lagrange’s mean value theorem are satisfied so ∃atleast one real no c ∈ (1, 2)
s.t. \(\frac{f(2)-f(1)}{2-1}\) = f’9c) ………….(1)
⇒ Now f(2) = 0;
f(1) = 1 (1 – 2) = – 1
∴ from (1) ;
\(\frac{0-(-1)}{1}\) = 2x – 2
⇒ 2c – 2 = 1
⇒ 2c = 3
⇒ c ∈ (1, 2).

Question 55.
The value of c in Lagrange’s Mean Value theorem for the function f(x) = x + \(\frac{1}{x}\) in the interval [1, 3] is
(a) 1
(b) 2
(c) √3
(d) – √3
Solution:
(c) √3

Given f(x) = x + \(\frac{1}{x}\)
∴ f'(x) = 1 – \(\frac{1}{x^2}\)
which exists for all x ∈ R – {0}
Thus f(x) is continuous in [1, 3] and differentiable in (1, 3).
So all the conditions of Lagrange’s mean value theorem arte satisfied.
So ∃ atleast one real number c ∈ (1, 3).
s.t. \(\frac{f(3)-f(1)}{3-1}\) = f'(3) …………(1)
Now f(1) = 1 + \(\frac{1}{1}\) = 2 ;
f(3) = 3 + \(\frac{1}{3}\) = \(\frac{10}{3}\)
∴ from (1) ;
⇒ \(\frac{\frac{10}{3}-2}{2}=1-\frac{1}{c^2}\)
⇒ \(\frac{4}{6}=1-\frac{1}{c^2}\)
⇒ \(\frac{1}{c^2}=1-\frac{2}{3}=\frac{1}{3}\)
⇒ c² = 3
⇒ c = ± √3
Here c = √3 ∈ (1, 3).