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

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

Question 1.
Is the function defined by f(x) = x² – sin x + 5 continuous at x = π? (NCERT)
Solution:
Given, f(x) = x² – sin x + 5
\(\underset{x \rightarrow \pi^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) f(π + h)
= \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) [(π + h)² – sin (π + h) + 5]
= \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) [(π + h)² + sin h + 5]
= π² + 5
and \(\underset{h \rightarrow 0^{-}}{\ {Lt}}\) f(x) = \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) f(π – h)
= \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) [(π – h)² – sin (π – h) + 5)]
= \(\underset{h \rightarrow 0^{+}}{\ {Lt}}\) [(π – h)² – sin h + 5]
= π² + 5
= f(π)
∴ \(\mathrm{Lt}_{x \rightarrow {\pi}}\) f(x) = f(π)
Thus f(x) is continuous at x = π.

Question 2.
(i) Find the value of k so that f(x) = \(\begin{cases}\frac{\sin k x}{x}, & x<0 \\ 8-3 x & x \geq 0\end{cases}\) may be continuous at x = 0.
(ii) If the function f defined by f(x) = \(\left\{\begin{array}{cc}
\frac{1-\cos 2 x}{2 x^2}, & x \neq 0 \\
k & , x=0
\end{array}\right.\) is continuous at x = 0, find the value of k.
(iii) If the function f is defined by f(x) = \(\left\{\begin{array}{cc}
\frac{1-\tan x}{4 x-\pi}, & 0<x<\frac{\pi}{2}, x \neq \frac{\pi}{4} \\
k, & x=\frac{\pi}{4}
\end{array}\right.\) is continuous at x = \(\frac{\pi}{4}\), find the value of k.
Solution:
(i) L.H.Limit = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{\sin k x}{x}\)
= k . \(\frac{\sin k x}{k x}\)
= k × 1 = k
[∵ \(\underset{\theta \rightarrow 0}{\ {Lt}} \frac{\sin \theta}{\theta}\) = 1]
R.H.Limit = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x)
= \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) 8 – 3x
= 8 – 3 × 0 = 8
and f(0) = 8 – 3 × 0 = 8
Now f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0)

(ii) \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = \(\ {Lt}_{x \rightarrow 0} \frac{1-\cos 2 x}{2 x^2}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{2 \sin ^2 x}{2 x^2}\)
= \(\ {Lt}_{x \rightarrow 0} \frac{\sin ^2 x}{x^2}\)
= \(\left[\ {Lt}_{x \rightarrow 0} \frac{\sin x}{x}\right]^2\)
= 1² = 1
also f(0) = k
since f(x) is continuous at x = 0
∴ \(\underset{x \rightarrow 0}{\mathrm{Lt}}\) f(x) = f(0)
⇒ 1 = k

(iii) \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) f(x) = \(\ {Lt}_{x \rightarrow \frac{\pi}{4}} \frac{1-\tan x}{4 x-\pi}\)
put x = \(\frac{\pi}{4}\) + h
so that as x → \(\frac{\pi}{4}\)
⇒ h → 0

Also, f(\(\frac{\pi}{4}\)) = k
Since f(x) be continuous at x = \(\frac{\pi}{4}\)
∴ \(\mathrm{Lt}_{x \rightarrow \frac{\pi}{4}}\) f(x) = f(\(\frac{\pi}{4}\))
⇒ – \(\frac{1}{2}\) = k
Hence k = – \(\frac{1}{2}\).

Question 3.
Find the points of discontinuity (if any) of the following functions:
(i) f(x) = \(\begin{cases}\frac{x}{|x|}, & \text { if } \quad x<0 \\ -1, & \text { if } x \geq 0\end{cases}\) (NCERT)
(ii) f(x) = |x| – |x + 1| (NCERT)
Solution:
(i) Let c ∈ R be any arbitrary element
Since D_f = R

Case – I:
When c < 0,
f(c) = \(\frac{c}{|c|}\)
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) \(\frac{x}{|x|}\)
= \(\frac{c}{|c|}\)
= f(c)
∴ f is continuous for all c < 0. Case – II: when c > 0 ,
f(c) = – 1
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) – 1
= – 1
= f(c)
∴ f(c) is continuous for all c > 0.

Case – III:
at c = 0
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) – 1 = – 1
and \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) \(\frac{x}{|x|}\)
= \(\ {Lt}_{x \rightarrow 0^{-}} \frac{x}{-x}\) = – 1
[∵ if x < 0 then |x| = – x]
∴ \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = f(0) = 1
Thus f is continuous at x = 0.
Thus, by combining all three cases, f is continuous at everywhere on R and has no points of discontinuity.

(ii) Given f(x) = |x| – |x – 1|
= \(\left\{\begin{array}{ccc}
-x+x-1 & ; & x<0 \\
x+x-1 & ; & 0 \leq x<1 \\
x-(x-1) & ; & x \geq 1
\end{array}\right.\)
= \(\left\{\begin{array}{ccc}
-1 & ; & x<0 \\
2 x-1 & ; & 0 \leq x<1 \\
1 & ; & x \geq 1
\end{array}\right.\)
Hence D_f = R.
So we examine f for continuity at all x ∈ R.
Let c ∈ R be any arbitrary real number.
Thus threee cases arises.

Case – I:
When c < 0 then f(c) = – 1
∴ \(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) – 1
= – 1 = f(c)
∴ f is continuous at all c < 0.

Case – II:
When 0 < c < 1
then f(c) = 2c – 1
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 2x – 1
= 2c – 1 = f(c)
Thus, f is continuous for all c, where 0 < c < 1. Case – III: When c > 1 then f(c) = 1
\(\underset{x \rightarrow c}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow c}{\mathrm{Lt}}\) 1 = 1 = f(c)
∴ f is continuous for all c > 1.

When x = 0:
\(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) – 1 = – 1
\(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) 2x – 1
= 2 × 0 – 1 = – 1
and f(0) = 0 – 1 = – 1
∴ \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) f(x) = f(0)
Thus f(x) is continuous at x = 0.

When x = 1:
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) 2x – 1
= 2 – 1 = 1
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 1 – 1 ;
f(1) = 1
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = f(1)

Thus,f is continuous at x = 1
Hence on combining all cases, f is continuous everywhere in R and has no point of discontinuity.

Question 3 (old).
Discuss the continuity of the function f defined by
f(x) = \(\left\{\begin{array}{l}
3, \text { if } \quad 0 \leq x \leq 1 \\
4, \text { if } 1<x<3 \\
5, \text { if } 3 \leq x \leq 10
\end{array}\right.\). (NCERT)
Solution:
Since constant function is continuous everywhere
when 0 ≤ x < 1 ;
f (x) = 3, which is a constant function
and hence continuous in 0 ≤ x < 1
When 1 < x < 3 ;
f(x) = 4, which is a constant function
and hence continuous for all x ∈ (1, 3)
When 3 f (x) = 5, which is a constant function
and hence continuous is 3 < x ≤ 10.

at x = 1:
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) 3 = 3
and \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 4 = 4
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) ≠ \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x)
Thus f is discontinuous at x = 1.

at x = 3:
\(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) 4 = 4
and \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) 5 = 5
∴ \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) f(x)
Thus f is discontinuous at x = 3.
Hence f is continuous in [0, 10] except at x = 1 and x = 3.

Question 4.
Examine the function f(x) = |x – 1| + |x – 3| for continuity and differentiability at x I and x = 3.
Solution:
Given f(x) = |x – 1| + |x – 3|
= \(\left\{\begin{array}{ccc}
-(x-1)-(x-3) & ; & x<1 \\
(x-1)-(x-3) & ; & 1 \leq x<3 \\
(x-1)+x-3 & ; & x \geq 3
\end{array}\right.\)
= \(\left\{\begin{array}{ccc}
-2 x+4 & ; & x<1 \\
2 & ; & 1 \leq x<3 \\
2 x-4 & ; & x \geq 3
\end{array}\right.\)

Continuity at x = 1:
\(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) (- 2x + 4)
= – 2 + 4 = 2
\(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) 2 = 2
and f(1) = 2
∴ \(\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}\) f(x) = f(1)
∴ f is continuous at x = 1.

Differentiability at x = 1:

∴ Lf'(1) ≠ Rf'(1)
Thus f(x) is not differentiable at x = 1.

Continuity at x = 3:
\(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) 2 = 2
and \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) 2x – 4
= 6 – 4 = 2
Also f(3) = 2 × 3 – 4
= 6 – 4 = 2
Thus, \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) f(x) = \(\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}\) f(x) = f(3)
∴ f is continuous at x = 3.

Differentiability at x = 3:
Lf'(3) = \(\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}\) \(\frac{f(x)-f(3)}{x-3}\)
= \(\ {Lt}_{x \rightarrow 3^{-}} \frac{2-2}{x-3}\)
= 0
Rf'(3) = \(\ {Lt}_{x \rightarrow 3^{+}} \frac{f(x)-f(3)}{x-3}\)
= \(\ {Lt}_{x \rightarrow 3^{+}} \frac{2 x-4-2}{x-3}\)
= \(\frac{2(x-3)}{x-3}\)
∴ Lf'(3) ≠ Rf'(3)
Thus f(x) is not differentiable at x = 3.

Question 5.
Find the derivative of the following functions w.r.t. x:
(i) \(\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}\)
(ii) \(\frac{1-\tan x}{1+\tan x}\)
(iii) sin (\(\sqrt{\sin \sqrt{x}}\)).
Solution:
(i) Let y = \(\frac{\sqrt{a}+\sqrt{x}}{\sqrt{a}-\sqrt{x}}\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(\sqrt{a}-\sqrt{x})\left(\frac{1}{2 \sqrt{x}}\right)-(\sqrt{a}+\sqrt{x})\left(-\frac{1}{2 \sqrt{x}}\right)}{(\sqrt{a}-\sqrt{x})^2}\)
= \(\frac{\sqrt{a}-\sqrt{x}+\sqrt{a}+\sqrt{x}}{2 \sqrt{x}(\sqrt{a}-\sqrt{x})^2}\)
= \(\frac{\sqrt{a}}{\sqrt{x}(\sqrt{a}-\sqrt{x})^2}\)

(ii) Let y = \(\frac{1-\tan x}{1+\tan x}\)
= tan (\(\frac{\pi}{4}\) – x)
\(\frac{d y}{d x}=\sec ^2\left(\frac{\pi}{4}-x\right) \frac{d}{d x}\left(\frac{\pi}{4}-x\right)\)
= \(-\frac{1}{\cos ^2\left(\frac{\pi}{4}-x\right)}\)
= \(\frac{-2}{1+\cos \left(\frac{\pi}{2}-2 x\right)}\)
= \(\frac{-2}{1+\sin 2 x}\)

(iii) Let y = sin (\(\sqrt{\sin \sqrt{x}}\))
Diff. both sides w.r.t. x, we have

Question 6.
Differentiate the following functions w.r.t. x:
(i) \(\frac{\tan 2 x}{1-\cot 2 x}\)
(ii) sin \(\left(\frac{1+x^2}{1-x^2}\right)\)
(iii) \(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\)
Solution:
(i) Let y = \(\frac{\tan 2 x}{1-\cot 2 x}\) ;
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\frac{(1-\cot 2 x) \sec ^2 2 x \cdot 2-2 \tan 2 x \ {cosec}^2 2 x}{(1-\cot 2 x)^2}\)
= \(\frac{2\left[\sec ^2 2 x-\cot 2 x \sec ^2 2 x-\tan 2 x \ {cosec}^2 2 x\right]}{(1-\cot 2 x)^2}\)
= \(\frac{2\left[\sec ^2 2 x-2 \ {cosec} 4 x-2 \ {cosec} 4 x\right]}{(1-\cot 2 x)^2}\)
= \(\frac{2\left[\sec ^2 2 x-4 \ {cosec} 4 x\right]}{(1-\cot 2 x)^2}\)

(ii) Let y = sin \(\left(\frac{1+x^2}{1-x^2}\right)\)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}=\cos \left(\frac{1+x^2}{1-x^2}\right)\left[\frac{\left(1-x^2\right) 2 x-\left(1+x^2\right)(-2 x)}{\left(1-x^2\right)^2}\right]\)
= \(\cos \left(\frac{1+x^2}{1-x^2}\right)\left[\frac{2 x\left(1-x^2+1+x^2\right)}{\left(1-x^2\right)^2}\right]\)
= \(\frac{4 x}{\left(1-x^2\right)^2} \cos \left(\frac{1+x^2}{1-x^2}\right)\)

(iii) Let y = \(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\)

Question 7.
(i) y = log \(\left(\frac{a+b \tan \frac{x}{2}}{a-b \tan \frac{x}{2}}\right)\), prove that \(\frac{d y}{d x}=\frac{a b}{a^2 \cos ^2 \frac{x}{2}-b^2 \sin ^2 \frac{x}{2}}\).
(ii) If 2y = x \(\sqrt{x^2-a^2}\) – a² log (x + \(\sqrt{x^2-a^2}\)), prove that \(\frac{d y}{d x}=\sqrt{x^2-a^2}\).
(iii) If x = 2a sin^-1 \(\sqrt{\frac{y}{2 a}}-\sqrt{2 a y-y^2}\), prove that \(\frac{d y}{d x}=\sqrt{\frac{2 a-y}{y}}\).
Solution:
(i) Given y = log \(\left(\frac{a+b \tan \frac{x}{2}}{a-b \tan \frac{x}{2}}\right)\)
= log (a + b tan \(\frac{x}{2}\)) – log (a – b tan \(\frac{x}{2}\))
Diff. both sides w.r.t. x ; we have

(ii) Given 2y = x \(\sqrt{x^2-a^2}\) – a² log (x + \(\sqrt{x^2-a^2}\))
Diff. both sides w.r.t. x ; we have

(iii) Given x = 2a sin^-1 \(\sqrt{\frac{y}{2 a}}-\sqrt{2 a y-y^2}\)
Diff. both sides w.r.t. y ; we have

Question 8.
Differentiate the following functions w.r.t. x:
(i) sin^-1 \(\left(\frac{a+b \cos x}{b+a \cos x}\right)\)
(ii) cot^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\)
(iii) tan^-1 \(\sqrt{\frac{a-x}{a+x}}\)
Solution:
(i) Let y = sin^-1 \(\left(\frac{a+b \cos x}{b+a \cos x}\right)\)
Diff. both sides w.r.t. x ; we have

(ii) Let y = cot^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) ……….(1)
put x = tan θ
⇒ θ = tan^-1 x
∴ from (1) ;
y = cot^-1 \(\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)\)
= cot^-1 \(\left(\frac{\sec \theta-1}{\tan \theta}\right)\)
⇒ y = cot^-1 \(\left(\frac{1-\cos \theta}{\sin \theta}\right)\)
= cot^-1 \(\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)\)
⇒ y = cot^-1 \(\left(\tan \frac{\theta}{2}\right)\)
= cot^-1 (cot \(\left(\frac{\pi}{2}-\frac{\theta}{2}\right)\))
⇒ y = \(\frac{\pi}{2}-\frac{\theta}{2}\)
= \(\frac{\pi}{2}-\frac{1}{2} \tan ^{-1} x\)

(iii) Let y = tan^-1 \(\sqrt{\frac{a-x}{a+x}}\) ………..(1)
put x = a cos θ
θ = cos^-1 \(\frac{x}{a}\)
∴ from eqn. (1) ; we have
y = tan^-1 \(\sqrt{\frac{a-a \cos \theta}{a+a \cos \theta}}\)
= tan^-1 \(\sqrt{\frac{2 \sin ^2 \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}}}\)
y = tan^-1 (tan \(\frac{\theta}{2}\))
= \(\frac{\theta}{2}\)
= \(\frac{1}{2}\) cos^-1 \(\frac{x}{a}\)
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}=-\frac{1}{2} \frac{1}{\sqrt{1-\frac{x^2}{a^2}}} \frac{1}{a}\)
= – \(\frac{1}{2 \sqrt{a^2-x^2}}\).

Question 9.
If y = 2 tan^-1 \(\sqrt{\frac{x-a}{b-x}}\), a < x < b, prove that \(\left(\frac{d y}{d x}\right)^2+\frac{1}{(x-a)(x-b)}\) = 0.
Solution:
Given y = 2 tan^-1 \(\sqrt{\frac{x-a}{b-x}}\)
put x = a cos² θ + b sin² θ ………..(1)
∴ x – a = a cos² θ + b sin² θ – a
= – a sin² θ + b sin² θ
⇒ x – a = (b – a) sin² θ ………..(2)
and b – x = b – a cos² θ – b sin² θ
= b cos² θ – a cos² θ
= (b – a) cos² θ ………..(3)
∴ y = 2 tan^-1 \(\sqrt{\frac{(b-a) \sin ^2 \theta}{(b-a) \cos ^2 \theta}}\)
= 2 tan^-1 (tan θ)
= 2θ
⇒ \(\frac{d y}{d x}=2 \frac{d \theta}{d x}\)
= \(\frac{2}{(b-a) \sin 2 \theta}\)
= \(\frac{1}{(b-a) \sin \theta \cos \theta}\) ………..(4)
[from (1) ;
\(\frac{d x}{d \theta}\) = 2a cos θ (- sin θ) + 2b sin θ cos θ = (b – a) sin 2θ]
On squaring both sides of eqn. (4) ; we have
\(\left(\frac{d y}{d x}\right)^2=\frac{1}{(b-a)^2 \sin ^2 \theta \cos ^2 \theta}\)
= \(\frac{1}{(b-a)^2 \frac{(x-a)}{b-a}\left(\frac{b-x}{b-a}\right)}\) [using (2) and (3)]
⇒ \(\left(\frac{d y}{d x}\right)^2=\frac{1}{(x-a)(b-x)}\)
⇒ \(\left(\frac{d y}{d x}\right)^2+\frac{1}{(x-a)(x-b)}\) = 0.

Question 10.
Differentiate the following functions w.r.t. x:
(i) tan^-1 \(\left(\frac{\sqrt{x}(3-x)}{1-3 x}\right)\)
(ii) tan^-1 \(\left(\frac{3 a^2 x-x^3}{a\left(a^2-3 x^2\right)}\right)\)
Solution:
(i) Let y = tan^-1 \(\left(\frac{\sqrt{x}(3-x)}{1-3 x}\right)\)
put √x = tan θ
⇒ θ = tan^-1 √x
∴ y = tan^-1 \(\left(\frac{\tan \theta\left(3-\tan ^2 \theta\right)}{1-3 \tan ^2 \theta}\right)\)
= tan^-1 (tan 3θ)
= 3θ
= 3 tan^-1 √x
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{3}{1+(\sqrt{x})^2} \frac{d}{d x} \sqrt{x}\)
= \(\left(\frac{3}{1+x}\right)\left(\frac{1}{2 \sqrt{x}}\right)\)
= \(\frac{3}{2 \sqrt{x}(1+x)}\)

(ii) Let y = tan^-1 \(\left(\frac{3 a^2 x-x^3}{a\left(a^2-3 x^2\right)}\right)\)
put x = a tan θ
θ = tan^-1 \(\frac{x}{a}\)
∴ y = tan^-1 \(\left(\frac{3 a^3 \tan \theta-a^3 \tan ^3 \theta}{a^3-3 a^3 \tan ^2 \theta}\right)\)
= tan^-1 \(\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^3 \theta}\right)\)
= tan^-1 (tan 3θ)
= 3θ
= 3 tan^-1 \(\frac{x}{a}\)
Diff. both sides w.r.t. x ;
\(\frac{d y}{d x}=\frac{3}{1+\frac{x^2}{a^2}}\left(\frac{1}{a}\right)\)
= \(\frac{3 a}{a^2+x^2}\)

Question 11.
If tan y = \(\frac{2 t}{1-t^2}\) and sin x = \(\frac{2 t}{1+t^2}\), find \(\frac{d y}{d x}\). (NCERT Exampler)
Solution:
Given sin x = \(\frac{2 t}{1+t^2}\)
⇒ x = sin^-1 \(\frac{2 t}{1+t^2}\) ………….(1)
and tan y = \(\frac{2 t}{1-t^2}\)
⇒ y = tan^-1 \(\frac{2 t}{1-t^2}\) ………….(2)
putting t = tan θ
and t = tan Φ in eqn. (1) and (2) ; we get
x = sin^-1 \(\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
and y = tan^-1 \(\left(\frac{2 \tan \phi}{1-\tan ^2 \phi}\right)\)
⇒ x = sin^-1 (sin 2θ)
and y = tan^-1 (tan 2Φ)
⇒ x = 2θ
and y = 2Φ
⇒ x = 2 tan^-1 t
⇒ \(\frac{d x}{d t}\) = \(\frac{2}{1+t^2}\)
and y = 2 tan^-1 t
⇒ \(\frac{d y}{d t}=\frac{2}{1+t^2}\)
Thus, \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}\) = 1.

Question 12.
If y = sin^-1 \(\left(\frac{1}{\sqrt{1+x^2}}\right)\) + tan^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\), find \(\frac{d y}{d x}\).
Solution:
Given y = sin^-1 \(\left(\frac{1}{\sqrt{1+x^2}}\right)\) + tan^-1 \(\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) ………..(1)
put x = tan θ
⇒ θ = tan^-1 x
∴ from (1) ;

Question 12 (old).
If y = a^{t + \(\frac{1}{t}\)} and x = (t + \(\frac{1}{t}\))^a, a > 0, find \(\frac{d y}{d x}\). (NCERT)
Solution:
Given y = a^{t + \(\frac{1}{t}\)} …………(1)
and x = (t + \(\frac{1}{t}\))^a …………(2)
Differentiate eqn. (1) and (2) w.r.t. ‘t’ ; we have
\(\frac{d y}{d t}\) = a^{t + \(\frac{1}{t}\)} log a . [1 – \(\frac{1}{t^2}\)]
and \(\frac{d x}{d t}\) = a \(\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right)\)
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}\)
= \(\frac{a^{\left(t+\frac{1}{t}\right)} \log a\left(1-\frac{1}{t^2}\right)}{a\left(t+\frac{1}{t}\right)^{a-1}\left(1-\frac{1}{t^2}\right)}\)
= \(\frac{a^{\left(t+\frac{1}{t}\right)} \log a}{a\left(t+\frac{1}{t}\right)^{a-1}}\).

Question 13.
If y \(\sqrt{1-x^2}\) + x \(\sqrt{1-y^2}\) = c, prove that \(\frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}}\).
Solution:
Given y \(\sqrt{1-x^2}\) + x \(\sqrt{1-y^2}\) = c ……….(1)
put x = sin θ i.e. θ = sin^-1 x
and y = sin Φ i.e. Φ = sin^-1 y
∴ from (1) ;
we have
sin Φ cos θ + sin θ cos Φ = c
⇒ sin (θ + Φ) = c
⇒ θ + Φ = sin^-1 c
⇒ sin^-1 x + sin^-1 y = sin^-1 c
Diff. both sides w.r.t. x ; we have
\(\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\sqrt{\frac{1-y^2}{1-x^2}}\).

Question 14.
Differentiate the following functions w.r.t. x :
(i) tan (x^x)
(ii) (5x)^{3 cos 2x} (NCERT)
(iii) (sin x)^x + sin^-1 √x
(iv) (x + \(\frac{1}{x}\))^x + x^{(1 + \(\frac{1}{x}\))}.
Solution:
(i) Let y = tan (x^x)
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = sec² (x^x) \(\frac{d}{d x}\) (x^x)
= sec² (x^x) \(\frac{d}{d x}\) e^{x log x}
= sec² (x^x) e^{x log x} [x × \(\frac{1}{x}\) + log x – 1]
= x^x sec² (x^x) (1 + log x)

(ii) Let y = (5x)^{3 cos 2x}
Taking logarithm on both sides ; we have
log y = 3 cos 2x log 5x
Diff. both sides w.r.t. x ; we have
\(\frac{1}{y} \frac{d y}{d x}\) = 3 [cos 2x \(\frac{1}{5 x}\) . 5 + log 5x . (- 2 sin 2x)]
⇒ \(\frac{d y}{d x}\) = (5x)^{3 cos 2x} [\(\frac{3 \cos 2 x}{x}\) – 6 sin 2x log 5x]

(iii) Let y = (x + \(\frac{1}{x}\))^x + x^{(1 + \(\frac{1}{x}\))}
⇒ y = u + v …………(1)
∴ \(\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}\) …………(2)
where u = (sin x)^x
⇒ log u = x log sin x
diff. both sides w.r.t. x ; we have
\(\frac{1}{u} \frac{d u}{d x}\) = log sin x + \(\frac{x}{sin x}\) cos x
\(\frac{d u}{d x}\) = (sin x)^x [log sin x + x cot x] ……….(3)
and v = sin^-1 √x
⇒ \(\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^2}} \times \frac{1}{2 \sqrt{x}}\)
\(\frac{d v}{d x}=\frac{1}{2 \sqrt{x} \sqrt{1-x}}=\frac{1}{2 \sqrt{x-x^2}}\) ……….(4)
putting eqn. (3) and eqn. (4) in eqn. (2) ;
∴ \(\frac{d y}{d x}\) = (sin x)^x [log sin x + x cot x] + \(\frac{1}{2 \sqrt{x-x^2}}\)

(iv) Let y = (x + \(\frac{1}{x}\))^x + x^{(1 + \(\frac{1}{x}\))}
⇒ y = u + v ……….(1)
where u = (x + \(\frac{1}{x}\))^x
and v = x^{(1 + \(\frac{1}{x}\))}
Diff. eqn. (1) both sides w.r.t. x ; we have
\(\frac{d y}{d u}=\frac{d u}{d x}+\frac{d v}{d x}\) …………(2)
Since u = (x + \(\frac{1}{x}\))^x ;
Taking logarithm on both sides ; we have
log u = x log (x + \(\frac{1}{x}\)) ;
diff. both sides w.r.t. x,

Question 14 (old).
(iii) (sin x – cos x)^{sin x – cos x}, \(\frac{\pi}{4}\) < x < \(\frac{3 \pi}{4}\) (NCERT).
Solution:
Let y = (sin x – cos x)^{sin x – cos x}, \(\frac{\pi}{4}\) < x < \(\frac{3 \pi}{4}\)
Taking logarithm on both sides ; we have
log y = (sin x – cos x) log (sin x – cos x)
On differentiating both sides ; we have
\(\frac{1}{y} \frac{d y}{d x}\) = \(\frac{(\sin x-\cos x)}{\sin x-\cos x}\) (cos x + sin x) + (cos x + sin x) log (sin x – cos x)
\(\frac{1}{y} \frac{d y}{d x}\) = (cos x + sin x) [1 + log (sin x – cos x)]
⇒ \(\frac{d y}{d x}\) = (sin x – cos x)^{sin x – cos x} (cos x + sin x) [1 + log (sin x – cos x)]

Question 15.
(i) If x^y – y^x = a^b, find \(\frac{d y}{d x}\).
(ii) If y = \((\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}\), prove that \(\frac{d y}{d x}=-\frac{y^2 \tan x}{1-y \log (\cos x)}\). (NCERT Exampler)
Solution:
(i) Given x^y – y^x = a^b
⇒ u – v = a^b …………(1)
where u = x^y ………..(2)
and v = y^x …………..(3)
Diff. eqn. (1) both sides w.r.t. x ;
\(\frac{d u}{d x}-\frac{d v}{d x}\) = 0 ………..(4)
Taking logarithm on both sides of eqn. (2) ;
log u = log x^y = y log x
Diff. both sides w.r.t. x ;
∴ \(\frac{1}{u} \frac{d u}{d x}=\frac{y}{x}+\log x \cdot \frac{d y}{d x}\)
⇒ \(\frac{d u}{d x}=x^y\left[\frac{y}{x}+\log x \cdot \frac{d y}{d x}\right]\) ………..(5)
Taking logarithm on both sides of eqn. (3) ;
log v = log y^x = x log y
Diff. both sides w.r.t. x ; we have
\(\frac{1}{v} \frac{d v}{d x}=\frac{x}{y} \frac{d y}{d x}\) + log y
⇒ \(\frac{d v}{d x}\) = y^x [\(\frac{x}{y} \frac{d y}{d x}\) + log y] ……….(6)
putting eqn. (5) and eqn. (6) in eqn. (4) ; we have
\(x^y\left[\frac{y}{x}+\log x \frac{d y}{d x}\right]-y^x\left[\frac{x}{y} \frac{d y}{d x}+\log y\right]\) = 0
⇒ (x^y log x – xy^{x – 1}) \(\frac{d y}{d x}\) = y^x log y – yx^{y – 1}
⇒ \(\frac{d y}{d x}=\frac{y^x \log y-y x^{y-1}}{x^y \log x-x y^{x-1}}\).

(ii) Given y = (cos x)^y ;
Taking logarithm on both sides ; we have
log y = y log cos x ;
diff. both sides w.r.t. x, we get

Question 16.
If \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\) = 6, prove that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:
Given \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}\) = 6
⇒ x + y = 6 \(\sqrt{xy}\)
Taking logarithm on both sides ; we have
log (x + y) = log 6 + \(\frac{1}{2}\) (log x + log y)
Diff. both sides w.r.t. x ; we have
\(\frac{1}{x+y}\left[1+\frac{d y}{d x}\right]=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y} \frac{d y}{d x}\right)\)
⇒ \(\frac{d y}{d x}\left[\frac{1}{x+y}-\frac{1}{2 y}\right]=\frac{1}{2 x}-\frac{1}{x+y}\)
⇒ \(\frac{d y}{d x}\left(\frac{y-x}{2 y(x+y)}\right)=\frac{y-x}{2 x(x+y)}\)
⇒ \(\frac{d y}{d x}=\frac{y}{x}\).

Question 17.
Find the second derivative of the following functions:
(i) tan² (3x – 2)
(ii) \(\frac{x^2+2 x-1}{x^2-3 x+2}\)
Solution:
(i) Let y = tan² (3x – 2) ;
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = 2 tan (3x – 2) sec² (3x – 2) . 3
Again diff. both sides w.r.t. x : we have
\(\frac{d^2 y}{d x^2}\) = 6 [3 sec² (3x – 2) sec² (3x – 2) + 2 tan² (3x – 2) sec² (3x – 2) × 3]
= 18 sec⁴ (3x – 2) + 36 tan² (3x – 2) sec² (3x – 2)

(ii) Let y = \(\frac{x^2+2 x-1}{x^2-3 x+2}\)
= \(\frac{x^2+2 x-1}{(x-1)(x-2)}\)
= 1 – \(\frac{2}{x-1}+\frac{7}{x-2}\) ………..(1)
[by using partial fractions]
Let \(\frac{x^2+2 x-1}{x^2-3 x+2}=1+\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}}{x-2}\) ………….(2)
Multiply both sides of eqn. (2) ; we have
x² + 2x – 1 = x² – 3x + 2 + A (3x – 2) + B (x – 1)
putting x = 1 in eqn. (3) ; we have
2 = – A
⇒ A = – 2
putting x = 2 in eqn. (3) ; we have
7 = B
Diff. both sides of eqn. (1) w.r.t. x, we have
\(\frac{d y}{d x}=\frac{2}{(x-1)^2}-\frac{7}{(x-2)^2}\)
Again differentiating w.r.t. x, we have
\(\frac{d^2 y}{d x^2}=\frac{-4}{(x-1)^3}+\frac{14}{(x-2)^3}\).

Question 18.
If y = 3x sin 3x + cos 3x, prove that x \(\frac{d^2 y}{d x^2}\) + 9xy = 2 \(\frac{d y}{d x}\).
Solution:
Given y = 3x sin 3x + cos 3x ………..(1)
Diff. eqn. (1) w.r.t. x both sides ; we have
\(\frac{d y}{d x}\) = 3 [3x cos 3x + sin 3x] – 3 sin 3x
= 9 x cos 3x ……….(2)
again differentiating eqn. (2) w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = 9 [- 3x sin 3x + cos 3x]
⇒ x \(\frac{d^2 y}{d x^2}\) = – 27 x² sin 3x + 9x cos 3x
⇒ x \(\frac{d^2 y}{d x^2}\) = – 9x [3x sin 3x + cos 3x] + 18x cos 3x
⇒ x \(\frac{d^2 y}{d x^2}\) = – 9 xy + 2 \(\frac{d y}{d x}\)
[using eqn. (1) and eqn. (2)]
⇒ x \(\frac{d^2 y}{d x^2}\) + 9xy = 2 \(\frac{d y}{d x}\)

Question 19.
If y = cos (3 cos^-1 x), then prove that \(\frac{d^2 y}{d x^2}\) = 24x.
Solution:
Given y = cos (3 cos^-1 x)
put cos^-1 x = θ
⇒ x = cos θ
⇒ y = cos 3θ
= 4 cos³ θ – 3 cos θ
= 4x³ – 3x
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = 12x² – 3 ;
again differentiating w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = 24x.

Question 20.
If sin (x + y) = ky, prove that y₂ + y (1 + y₁)³ = 0.
Solution:
Given sin (x + y) = ky …………(1)
T.P. y₂ + y (1 + y₁)³ = 0
Diff. eqn. (1) both sides w.r.t. x, we have
cos (x + y) (1 + y₁) = ky₁ …………(2)
Differentiating eqn. (2) w.r.t. x ; we have
cos (x + y) y₂ + (1 + y₁) {- sin (x + y) (1 + y₁)} = ky₂
⇒ \(\) y₂ – (1 + y₁)² ky = ky₂
[using eqn. (1) and (2)]
⇒ y₁y₂ – (1 + y₁)² y = y₂ (1 + y₁)
⇒ y₂ (y₁ – 1 – y₁) – (1 + y₁)³ y = 0
⇒ y₂ + y (1 + y₁)³ = 0.

Question 21.
If xy = sin x, prove that \(\frac{d^2 y}{d x^2}\) + \(\frac{2}{x} \frac{d y}{d x}\) + y = 0.
Solution:
Given xy = sin x ……….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
x \(\frac{d y}{d x}\) + y . 1 = cos x ………(2)
Differentiating eqn. (2) w.r.t. x ; we have
x \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) . 1 + \(\frac{d y}{d x}\) = – sin x
⇒ x \(\frac{d^2 y}{d x^2}\) + 2 \(\frac{d y}{d x}\) = – xy
On dividing both sides throughout by x, we have
\(\frac{d^2 y}{d x^2}\) + \(\frac{2}{x} \frac{d y}{d x}\) + y = 0.

Question 22.
If y = \(\frac{2}{\sqrt{a^2-b^2}}\) tan^-1 \(\left(\sqrt{\frac{a \cdot b}{a+b}} \tan \frac{x}{2}\right)\), prove that \(\frac{d^2 y}{d x^2}=\frac{b \sin x}{(a+b \cos x)^2}\).
Solution:
Given y = \(\frac{2}{\sqrt{a^2-b^2}}\) tan^-1 \(\left(\sqrt{\frac{a \cdot b}{a+b}} \tan \frac{x}{2}\right)\) ;
Diff. both sides w.r.t. x ; we have

Question 23.
If x = (a + bt) e^{– nt}, prove that \(\frac{d^2 x}{d t^2}\) + 2n \(\frac{d x}{d t}\) + n²x = 0.
Solution:
Given x = (a + bt) e^{– nt} ……….(1)
Diff. eqn. (1) both sides w.r.t. t ; we have
\(\frac{d x}{d t}\) = (a + bt) e^{– nt} (- n) + e^{– nt} b
⇒ \(\frac{d x}{d t}\) = – nx + be^{– nt} ………..(2)
again differentiating eqn. (2) w.r.t. t, we have
\(\frac{d^2 x}{d t^2}\) = – n \(\frac{d x}{d t}\) + be^{– nt} (- n)
⇒ \(\frac{d^2 x}{d t^2}\) = – n \(\frac{d x}{d t}\) – n [\(\frac{d x}{d t}\) + nx] [using eqn. (2)]
⇒ \(\frac{d^2 x}{d t^2}\) + 2n \(\frac{d x}{d t}\) + n²x = 0

Question 24.
If y = \(\frac{3 a t}{1+t}\) and x = \(\frac{2 a t^2}{1+t}\), find \(\frac{d^2 y}{d x^2}\).
Solution:
Given y = \(\frac{3 a t}{1+t}\) ………….(1)
and x = \(\frac{2 a t^2}{1+t}\) …………(2)
Differentiating eqn. (1) and eqn. (2) w.r.t. t, we have

Question 25.
If x = sin t and y = sin 2t, prove that
(i) (1 – x²) (\(\frac{d y}{d x}\))² = 4 (1 – y²)
(ii) (1 – x²) \(\frac{d^2 y}{d x^2}\) – x \(\frac{d y}{d x}\) + 4y = 0.
Solution:
Given x = sin t ……….(1)
and y = sin 2t ………..(2)
diff. eqns. (1) and (2) ; w.r.t. t ; we have
\(\frac{d x}{d t}\) = cos t ;
\(\frac{d y}{d t}\) = 2 cos 2t
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{2 \cos 2 t}{\cos t}\) ………..(3)
⇒ \(\left(\frac{d y}{d x}\right)^2=4 \frac{\cos ^2 2 t}{\cos ^2 t}\)
= \(\frac{4\left(1-\sin ^2 2 t\right)}{1-\sin ^2 t}\)
= \(\frac{4\left(1-y^2\right)}{1-x^2}\)
[using (1) and (2)]
⇒ (1 – x²) (\(\frac{d y}{d x}\))² = 4 (1 – y²) ………..(4)

(ii) Diff. eqn. (3) both sides w.r.t. x ;
\(\frac{d^2 y}{d x^2}=\frac{2[\cos t(-2 \sin 2 t)-\cos 2 t(-\sin t)]}{\cos ^2 t}\) × \(\frac{d t}{d x}\)
⇒ (1 – sin² t) \(\frac{d^2 y}{d x^2}\) = [- 4 cos t sin 2t + 2 cos 2t sin t] \(\frac{1}{\cos t}\)
= – 4 sin 2t + 2 \(\frac{\cos 2 t}{\cos t}\) sin t
⇒ (1 – x²) \(\frac{d^2 y}{d x^2}\) = – 4y + x \(\frac{d y}{d x}\) [using (1) and (2)]
⇒ (1 – x²) \(\frac{d^2 y}{d x^2}\) – x \(\frac{d y}{d x}\) + 4y = 0.

Question 26.
Verify Rolle’s theorem for the following functions and find point (or points) in the interval where derivative is zero :
f(x) = 2 cos 2 (x – \(\frac{\pi}{4}\)) in [0, π]
Solution:
Given f(x) = 2 cos 2 (x – \(\frac{\pi}{4}\))
= 2 cos (2x – \(\frac{\pi}{2}\))
= 2 cos {- (\(\frac{\pi}{2}\) – 2x)}
= 2 cos (\(\frac{\pi}{2}\) – 2x)
[∵ cos (- θ) = cos θ]
⇒ f(x) = 2 sin 2x ………..(1)
Since, sin 2x is continuous and differentiable everywhere.
∴ f(x) is continuous in [0, π] and f is derivable in (0, π).
also, f(0) = 2 × 0 = 0;
f(π) = 2 sin 2π
= 2 × 0 = 0
∴ f(0) = f(π)
Thus, all the three conditions of Rolle’s theorem are satisfied for function f in [0, π].
Then ∃ atleast one real number c ∈ (0, π)
s.t. f’ (c) = 0
Diff. eqn. (1) w.r.t. x; we have
f’(x) = 4 cos 2x
Now f’ (c) = 0
⇒ 4 cos 2c = 0
⇒ cos 2c = 0
⇒ 2c = (2n + 1) \(\frac{\pi}{2}\), ∀ n ∈ I
⇒ c = (2n + 1) \(\frac{\pi}{4}\), ∀ n ∈ I
∴ c = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \ldots,-\frac{\pi}{4}, \ldots\)
But c ∈ (0, π)
∴ c = \(\frac{\pi}{4}\), \(\frac{3 \pi}{4}\)
Thus, ∃ teo real numbers \(\frac{\pi}{4}\) and \(\frac{3 \pi}{4}\)
s.t. f'(\(\frac{\pi}{4}\)) = 0 = f'(\(\frac{3 \pi}{4}\)).

Question 27.
Verify Lagrange’s mean value theorem for the following function. Also find c of this theorem : f(x) = log x, 1 ≤ x ≤ 2e.
Solution:
Given f(x) = log x, 1 ≤ x ≤ 2e
Since log x is continuous in its domain.
∴ f(x) is continuous in [1, 2e].
Further f’ (x) = \(\frac{1}{x}\), which is defined ∀ x ∈ (1, 2e)
∴ f is derivable in (1, 2e).
Thus, both conditions of lagrange’s mean value theorem are satisfied for f in [1, 2e].
Then ∃ atleast one real number c ∈ (1, 2e) such that
f'(c) = \(\frac{f(2 e)-f(1)}{2 e-1}\) ………….(1)
Diff. eqn. (1) w.r.t. x ; we have
f'(x) = \(\frac{1}{x}\) ;
f(1) = log 1 = 0 ;
f(e) = log 2e
∴ from (2) ; we have
\(\frac{1}{c}=\frac{\log 2 e-0}{2 e-1}\)
⇒ c = \(\frac{2 e-1}{\log 2 e}\)
⇒ c = \(\frac{2 e-1}{\log 2+\log e}\)
= \(\frac{2 e-1}{1+\log 2}\) ∈ (1, 2e)
[∵ e = 2.7183 and log 2 = 0.3010]
[Since 2e – 1 < 2e
⇒ \(\frac{2 e-1}{1+\log 2}\) < \(\frac{2 e}{1+\log 2}\) < 2e and 2e – 1 > 2
⇒ \(\frac{2 e-1}{1+\log 2}\) > \(\frac{2}{1+\log 2}\) > 1]
Thus, ∃ a real number c = \(\frac{2 e-1}{1+\log 2}\) ∈ (1, 2e)
s.t. f'(c) = \(\frac{f(2 e)-f(1)}{2 e-1}\)
Thus, lagrange’s theorem is verified and c = \(\frac{2 e-1}{1+\log 2}\).