NCERT Solutions for Miscellaneous Exercise Chapter 5 Class 12 - Continuity and Differentiability

In the Class 12 Maths chapter 5 miscellaneous exercise solutions, you will get a mixture of questions from all the previous exercises of this Class 12 Maths NCERT textbook chapter. You will get questions related to first-order derivatives of different types of functions, second-order derivatives, mean-value theorem, Rolle's theorem in the miscellaneous exercise chapter 5 Class 12.

This Class 12 NCERT syllabus exercise is a bit difficult as compared to previous exercises, so you may not be able to solve NCERT problems from this exercise at first. You can take help from NCERT solutions for Class 12 Maths chapter 5 miscellaneous exercise to get clarity. There are not many questions asked in the board exams from this exercise, but Class 12 Maths chapter 5 miscellaneous solutions are important for the students who are preparing for competitive exams like JEE main, SRMJEE, VITEEE, MET, etc. You can click on the given link if you are looking for NCERT Solutions for Class 6 to Class 12 at one place.

Also, see

Continuity and Differentiability Exercise 5.1
Continuity and Differentiability Exercise 5.2
Continuity and Differentiability Exercise 5.3
Continuity and Differentiability Exercise 5.4
Continuity and Differentiability Exercise 5.5
Continuity and Differentiability Exercise 5.6
Continuity and Differentiability Exercise 5.7
Continuity and Differentiability Exercise 5.8

Continuity and Differentiability Miscellaneous Exercise:

Question:1 Differentiate w.r.t. x the function in Exercises 1 to 11.

$( 3x^2 - 9x + 5 )^9$

Answer:

Given function is
$f(x)=( 3x^2 - 9x + 5 )^9$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d((3x^2-9x+5)^9)}{dx}= 9(3x^2-9x+5)^8.(6x-9)$
$= 27(2x-3)(3x^2-9x+5)^8$
Therefore, differentiation w.r.t. x is $27(3x^2-9x+5)^8(2x-3)$

Question:2 Differentiate w.r.t. x the function in Exercises 1 to 11.

$\sin ^3 x + \cos ^6 x$

Answer:

Given function is
$f(x)= \sin ^3 x + \cos ^6 x$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^3x +\cos^6x)}{dx}=3\sin^2x.\frac{d(\sin x)}{dx}+6\cos^5x.\frac{d(\cos x)}{dx}$
$=3\sin^2x.\cos x+6\cos^5x.(-\sin x)$
$=3\sin^2x\cos x- 6\cos^5x\sin x = 3\sin x\cos x(\sin x- 2\cos ^4x)$

Therefore, differentiation w.r.t. x is $3\sin x\cos x(\sin x- 2\cos ^4x)$

Question:3 Differentiate w.r.t. x the function in Exercises 1 to 11.

$( 5 x) ^{ 3 \cos 2x }$

Answer:

Given function is
$y=( 5 x) ^{ 3 \cos 2x }$
Take, log on both the sides
$\log y = 3\cos 2x\log 5x$
Now, differentiation w.r.t. x is
By using product rule
$\frac{1}{y}.\frac{dy}{dx} = 3.(-2\sin 2x)\log 5x + 3\cos 2x.\frac{1}{5x}.5= -6\sin2x\log 5x +\frac{3\cos 2x}{x}\\ \frac{dy}{dx} = y.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )\\ \frac{dy}{dx} = (5x)^{3\cos 2x}.\left ( -6\sin2x\log 5x +\frac{3\cos 2x}{x} \right )$

Therefore, differentiation w.r.t. x is $(5x)^{3\cos 2x}.\left ( \frac{3\cos 2x}{x}-6\sin2x\log 5x \right )$

Question:4 Differentiate w.r.t. x the function in Exercises 1 to 11.

$\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1$

Answer:

Given function is
$f(x)=\sin ^ {-1} (x \sqrt x ) , 0 \leq x\leq 1$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\sin^{-1}x\sqrt x)}{dx}=\frac{1}{\sqrt{1-(x\sqrt x)^2}}.\frac{d(x\sqrt x)}{dx}$
$=\frac{1}{\sqrt{1-x^3}}.\left ( 1.\sqrt x+x\frac{1}{2\sqrt x} \right )$
$=\frac{1}{\sqrt{1-x^3}}.\left ( \frac{3\sqrt x}{2} \right )$
$=\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}$

Therefore, differentiation w.r.t. x is $\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}$

Question:5 Differentiate w.r.t. x the function in Exercises 1 to 11.

$\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2$

Answer:

Given function is
$f(x)=\frac{\cos ^{-1}x/2}{\sqrt {2x+7}} , -2 < x < 2$
Now, differentiation w.r.t. x is
By using the Quotient rule
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{\cos^{-1}\frac{x}{2}}{\sqrt{2x+7}})}{dx}=\frac{\frac{d(\cos^{-1}\frac{x}{2})}{dx}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{d(\sqrt{2x+7})}{dx}}{(\sqrt{2x+7})^2}\\ f^{'}(x) = \frac{\frac{-1}{\sqrt{1-(\frac{x}{2})^2}}.\frac{1}{2}.\sqrt{2x+7}-\cos^{-1}\frac{x}{2}.\frac{1}{2.\sqrt{2x+7}}.2}{2x+7}\\ f^{'}(x)= -\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]$

Therefore, differentiation w.r.t. x is $-\left [\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{\cos^{-1}\frac{x}{2}}{(2x+7)^\frac{3}{2}} \right ]$

Question:6 Differentiate w.r.t. x the function in Exercises 1 to 11.

$\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2$

Answer:

Given function is
$f(x)=\cot ^{-1} \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ] , 0 < x < \pi /2$
Now, rationalize the [] part
$\left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} \right ]= \left [ \frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt {1+ \sin x }- \sqrt {1- \sin x }} .\frac{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}{\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x }}\right ]$

$=\frac{(\sqrt { 1+ \sin x }+ \sqrt { 1- \sin x })^2}{(\sqrt{1+\sin x})^2-(\sqrt{1-\sin x})^2} \ \ \ \ \ \ (Using \ (a-b)(a+b)=a^2-b^2)$

$=\frac{((\sqrt { 1+ \sin x })^2+ (\sqrt { 1- \sin x })^2+2(\sqrt { 1+ \sin x })(\sqrt { 1- \sin x }))}{1+\sin x-1+\sin x}$
$(Using \ (a+b)^2=a^2+b^2+2ab)$
$=\frac{1+\sin x+1-\sin x+2\sqrt{1-\sin^2x} }{2\sin x}$

$=\frac{2(1+\cos x)}{2\sin x} = \frac{1+\cos x}{\sin x}$

$=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \ \ \ \ \ (\because 2\cos^2= 1+\cos2x \ and \ \sin2x = 2\sin x\cos x)$

$=\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}} = \cot \frac{x}{2}$
Given function reduces to
$f(x) = \cot^{-1}(\cot \frac{x}{2})\\ f(x) = \frac{x}{2}$
Now, differentiation w.r.t. x is
$f^{'}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{x}{2})}{dx} = \frac{1}{2}$
Therefore, differentiation w.r.t. x is $\frac{1}{2}$

Question:7 Differentiate w.r.t. x the function in Exercises 1 to 11. $( \log x )^{ \log x } , x > 1$

Answer:

Given function is
$y=( \log x )^{ \log x } , x > 1$
Take log on both sides
$\log y=\log x\log( \log x )$
Now, differentiate w.r.t.
$\frac{1}{y}.\frac{dy}{dx}= \frac{1}{x}.\log (\log x)+\log x.\frac{1}{\log x}.\frac{1}{x} = \frac{\log x+1}{x}$
$\frac{dy}{dx} = y.\left ( \frac{\log x+1}{x} \right )\\$
$\frac{dy}{dx} = (\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\$
Therefore, differentiation w.r.t x is $(\log x)^{\log x}.\left ( \frac{\log x+1}{x} \right )\\$

Question:8 $\cos ( a \cos x + b \sin x )$ , for some constant a and b.

Answer:

Given function is
$f(x)=\cos ( a \cos x + b \sin x )$
Now, differentiation w.r.t x
$f^{'}(x)= \frac{d(f(x))}{dx}= \frac{d(\cos(a\cos x+ b \sin x))}{dx}$
$= -\sin(a\cos x+b\sin x).\frac{d(a\cos x+b\sin x)}{dx}$
$= -\sin(a\cos x+b\sin x).(-a\sin x+b\cos x)$
$= (a\sin x-b\cos x)\sin(a\cos x+b\sin x).$
Therefore, differentiation w.r.t x $(a\sin x-b\cos x)\sin(a\cos x+b\sin x)$

Question:9 $(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}$

Answer:

Given function is
$y=(\sin x - \cos x)^{ (\sin x - \cos x), } , \frac{\pi }{4} <x<\frac{3 \pi }{4}$
Take log on both the sides
$\log y=(\sin x - \cos x)\log (\sin x - \cos x)$
Now, differentiate w.r.t. x
$\frac{1}{y}.\frac{dy}{dx} = \frac{d(\sin x-\cos x)}{dx}.\log(\sin x- \cos x)+(\sin x- \cos x).\frac{d(\log(\sin x- \cos x))}{dx}$
$\frac{1}{y}.\frac{dy}{dx} =(\cos x -(-\sin x)).\log(\sin x-\cos x)+(\sin x- \cos x).\frac{(\cos x -(-\sin x))}{(\sin x- \cos x)}$
$\frac{dy}{dx} =y.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )$
$\frac{dy}{dx} =(\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right )$
Therefore, differentiation w.r.t x is $(\sin x-\cos x)^{(\sin x-\cos x)}.(\cos x +\sin x)\left ( \log(\sin x-\cos x)+1 \right ), sinx>cosx$

Question:10 $x ^x + x ^a + a ^x + a ^a$ , for some fixed a > 0 and x > 0

Answer:

Given function is
$f(x)=x ^x + x ^a + a ^x + a ^a$
Lets take
$u = x^x$
Now, take log on both sides
$\log u = x \log x$
Now, differentiate w.r.t x
$\frac{1}{u}.\frac{du}{dx}= \frac{dx}{dx}.\log x+x.\frac{d(\log x)}{dx}\\ \\ \frac{1}{u}.\frac{du}{dx}= 1.\log x+x.\frac{1}{x}\\ \\ \frac{du}{dx}= y.(\log x+1)\\ \\ \frac{du}{dx}= x^x.(\log x+1)$ -(i)
Similarly, take $v = x^a$
take log on both the sides
$\log v = a\log x$
Now, differentiate w.r.t x
$\frac{1}{v}.\frac{dv}{dx}= a.\frac{d(\log x)}{dx}=a.\frac{1}{x}= \frac{a}{x}\\ \\ \frac{dv}{dx}= v.\frac{a}{x}\\ \\ \frac{dv}{dx}= x^a.\frac{a}{x}$ -(ii)

Similarly, take $z = a^x$
take log on both the sides
$\log z = x\log a$
Now, differentiate w.r.t x
$\frac{1}{z}.\frac{dz}{dx}=\log a.\frac{d(x)}{dx}=\log a.1= \log a\\ \\ \frac{dz}{dx}= z.\log a\\ \\ \frac{dz}{dx}= a^x.\log a$ -(iii)

Similarly, take $w = a^a$
take log on both the sides
$\log w = a\log a= \ constant$
Now, differentiate w.r.t x
$\frac{1}{w}.\frac{dw}{dx}= a.\frac{d(a\log a)}{dx}= 0\\ \\ \frac{dw}{dx} = 0$ -(iv)
Now,
$f(x)=u+v+z+w$
$f^{'}(x) = \frac{du}{dx}+\frac{dv}{dx}+\frac{dz}{dx}+\frac{dw}{dx}$
Put values from equation (i) , (ii) ,(iii) and (iv)
$f^{'}(x)= x^x(\log x+1)+ax^{a-1}+a^x\log a$
Therefore, differentiation w.r.t. x is $x^x(\log x+1)+ax^{a-1}+a^x\log a$

Question:11 $x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3$

Answer:

Given function is
$f(x)=x ^{x^2 -3} + ( x-3 ) ^{x^2} , for\: \: x > 3$
take $u=x ^{x^2 -3}$
Now, take log on both the sides
$\log u=(x^2-3)\log x$
Now, differentiate w.r.t x
$\frac{1}{u}.\frac{du}{dx}= \frac{d(x^2-3)}{dx}.\log x+(x^2-3).\frac{d(\log x)}{dx}\\ \\ \frac{1}{u}.\frac{du}{dx} = 2x.\log x+(x^2-3).\frac{1}{x}\\ \\ \frac{1}{u}.\frac{du}{dx} = \frac{2x^2\log x+x^2-3}{x}\\ \\ \frac{du}{dx}= u.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\ \\ \frac{du}{dx}= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )\\ \\$ -(i)
Similarly,
take $v=(x-3)^x^2\\$
Now, take log on both the sides
$\log v=x^2\log (x-3)$
Now, differentiate w.r.t x
$\frac{1}{v}.\frac{dv}{dx}= \frac{d(x^2)}{dx}.\log (x-3)+x^2.\frac{d(\log (x-3))}{dx}\\ \\ \frac{1}{v}.\frac{dv}{dx} = 2x.\log (x-3)+x^2.\frac{1}{(x-3)}\\ \\ \frac{1}{v}.\frac{dv}{dx} = 2x\log(x-3)+\frac{x^2}{x-3}\\ \\ \frac{dv}{dx}= v.\left ( 2x\log(x-3)+\frac{x^2}{x-3} \right )\\ \\ \frac{dv}{dx}= (x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )\\ \\$ -(ii)
Now
$f(x)= u + v$
$f^{'}(x)= \frac{du}{dx}+\frac{dv}{dx}$
Put the value from equation (i) and (ii)
$f^{'}(x)= x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )$
Therefore, differentiation w.r.t x is $x^{(x^2-3)}.\left ( \frac{2x^2\log x+x^2-3}{x} \right )+(x-3)^{x^2}.\left ( 2x\log(x-3)+\frac{x^2}{x-3}\right )$

Question:12 Find dy/dx if $y = 12 (1 - \cos t), x = 10 (t - \sin t),$ $-\frac{\pi }{2} <t< \frac{\pi }{2}$

Answer:

Given equations are
$y = 12 (1 - \cos t), x = 10 (t - \sin t),$
Now, differentiate both y and x w.r.t t independently
$\frac{dy}{dt}=\frac{d(12(1-\cos t))}{dt}= -12(-\sin t)=12\sin t$
And
$\frac{dx}{dt}=\frac{d(10(t-\sin t))}{dt}= 10-10\cos t$
Now
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{12\sin t}{10(1-\cos t)} = \frac{6}{5}.\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{2\sin^2\frac{t}{2}} = \frac{6}{5}.\frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}\\ \\$
$(\because \sin 2x = 2\sin x\cos x \ and \ 1-\cos 2x = 2\sin^2x)$
$\frac{dy}{dx}=\frac{6}{5}.\cot \frac{t}{2}$
Therefore, differentiation w.r.t x is $\frac{6}{5}.\cot \frac{t}{2}$

Question:13 Find dy/dx if $y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1$

Answer:

Given function is
$y = sin ^{-1} x + sin^{-1} \sqrt{1- x^2} , 0 <x< 1$
Now, differentiatiate w.r.t. x
$\frac{dy}{dx}= \frac{d(sin ^{-1} x + sin^{-1} \sqrt{1- x^2})}{dx} = \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-(\sqrt{1-x^2})^2}}.\frac{d(\sqrt{1-x^2})}{dx}\\ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-1+x^2}}.\frac{1}{2\sqrt{1-x^2}}.(-2x)\\ \\ \frac{dy}{dx}= \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\\ \frac{dy}{dx}= 0$
Therefore, differentiatiate w.r.t. x is 0

Question:14 If $x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0 \: \: for \: \: , -1 < x < 1 \: \:prove \: \: that \: \frac{dy}{dx} = -\frac{1}{(1+x)^2}$

Answer:

Given function is
$x \sqrt {1+ y }+ y \sqrt { 1+ x } = 0$
$x\sqrt{1+y} = - y\sqrt{1+x}$
Now, squaring both sides
$(x\sqrt{1+y})^2 = (- y\sqrt{1+x})^2\\ x^2(1+y)=y^2(1+x)\\ x^2+x^2y=y^2x+y^2\\ x^2-y^2=y^2x-x^2y\\ (x-y)(x+y) = -xy(x-y) \\ x+y =-xy\\ y = \frac{-x}{1+x}$
Now, differentiate w.r.t. x is
$\frac{dy}{dx} = \frac{d(\frac{-x}{1+x})}{dx}= \frac{-1.(1+x)-(-x).(1)}{(1+x)^2}= \frac{-1}{(1+x)^2}$
Hence proved

Question:15 If $(x - a)^2 + (y - b)^2 = c^2$ , for some c > 0, prove that $\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}}\:$ is a constant independent of a and b.

Answer:

Given function is
$(x - a)^2 + (y - b)^2 = c^2$
$(y - b)^2 = c^2-(x - a)^2$ - (i)
Now, differentiate w.r.t. x
$\frac{d((x-a)^2)}{dx}+\frac{((y-b)^2)}{dx}=\frac{d(c^2)}{dx}\\ \\ 2(x-a)+2(y-b).\frac{dy}{dx}=0\\ \\ \frac{dy}{dx} = \frac{a-x}{y-b}$ -(ii)
Now, the second derivative
$\frac{d^2y}{dx^2} = \frac{\frac{d(a-x)}{dx}.(y-b) -(a-x).\frac{d(y-b)}{dx}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} =\frac{ (-1).(y-b)-(a-x).\frac{dy}{dx}}{(y-b)^2}\\ \\$
Now, put values from equation (i) and (ii)
$\frac{d^2y}{dx^2} =\frac{-(y-b)-(a-x).\frac{a-x}{y-b}}{(y-b)^2}\\ \\ \frac{d^2y}{dx^2} = \frac{-((y-b)^2+(a-x)^2)}{(y-b)^\frac{3}{2}} = \frac{-c^2}{(y-b)^\frac{3}{2}}$ $(\because (x - a)^2 + (y - b)^2 = c^2)$
Now,
$\frac{\left [ 1+(\frac{dy}{dx} )^2 \right ]^{3/2}}{\frac{d^2 y }{dx^2}} = \frac{\left ( 1+\left ( \frac{x-a}{y-b} \right )^2 \right )^\frac{3}{2}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{\frac{\left ( (y-b)^2 +(x-a)^2\right )^\frac{3}{2}}{(y-b)^\frac{3}{2}}}{\frac{-c^2}{(y-b)^\frac{3}{2}}} = \frac{(c^2)^\frac{3}{2}}{-c^2}= \frac{c^3}{-c^2}= c$ $(\because (x - a)^2 + (y - b)^2 = c^2)$
Which is independent of a and b
Hence proved

Question:16 If $\cos y = x \cos (a + y)$ , with $\cos a \neq \pm 1$ , prove that $\frac{dy}{dx} = \frac{\cos ^2 (a+y )}{\sin a }$

Answer:

Given function is
$\cos y = x \cos (a + y)$
Now, Differentiate w.r.t x
$\frac{d(\cos y)}{dx} = \frac{dx}{dx}.\cos(a+y)+x.\frac{d(\cos (a+y))}{dx}\\ \\ -\sin y \frac{dy}{dx} = 1.\cos (a+y)+x.(-\sin(a+y)).\frac{dy}{dx}\\ \\ \frac{dy}{dx}.(x\sin(a+y)-\sin y)= \cos(a+y)\\ \\ \frac{dy}{dx}.(\frac{\cos y}{\cos (a+b)}.\sin(a+y)-\sin y)= \cos(a+b) \ \ \ \ \ (\because x = \frac{\cos y}{\cos (a+b)})\\ \\ \frac{dy}{dx}.(\cos y\sin(a+y)-\sin y\cos(a+y))=\cos^2(a+b)\\ \\ \frac{dy}{dx}.(\sin(a+y-y))=\cos^2(a+b) \ \ \ \ \ \ \ (\because \cos A\sin B-\sin A\cos B = \sin(A-B))\\ \\ \frac{dy}{dx}= \frac{\cos^2(a+y)}{\sin a}$
Hence proved

Question:17 If $x = a (\cos t + t \sin t)$ and $y = a (\sin t - t \cos t),$ find $\frac{d^2 y }{dx^2 }$

Answer:

Given functions are
$x = a (\cos t + t \sin t)$ and $y = a (\sin t - t \cos t)$
Now, differentiate both the functions w.r.t. t independently
We get
$\frac{dx}{dt} = \frac{d(a(\cos t +t\sin t))}{dt}= a(-\sin t)+a(\sin t+t\cos t)$
$=-a\sin t+a\sin t+at\cos t = at\cos t$
Similarly,
$\frac{dy}{dt} = \frac{d(a(\sin t - t\cos t))}{dt}= a\cos t -a(\cos t+t(-\sin t))$
$= a\cos t -a\cos t+at\sin t =at\sin t$
Now,
$\frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{at\sin t}{at \cos t} = \tan t$
Now, the second derivative
$\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}= \sec^2 t.\frac{dt}{dx}=\frac{\sec^2t.\sec t }{at}=\frac{\sec^3t}{at}$
$(\because \frac{dx}{dt} = at\cos t \Rightarrow \frac{dt}{dx}= \frac{1}{at\cos t}=\frac{\sec t}{at})$
Therefore, $\frac{d^2y}{dx^2}=\frac{\sec^3t}{at}$

Question:18 If $f (x) = |x|^3$ , show that f ''(x) exists for all real x and find it.

Answer:

Given function is
$f (x) = |x|^3$
$f(x)\left\{\begin{matrix} -x^3 & x<0\\ x^3 & x>0 \end{matrix}\right.$
Now, differentiate in both the cases
$f(x)= x^3\\ f^{'}(x)=3x^2\\ f^{''}(x)= 6x$
And
$f(x)= -x^3\\ f^{'}(x)=-3x^2\\ f^{''}(x)= -6x$
In both, the cases f ''(x) exist
Hence, we can say that f ''(x) exists for all real x
and values are
$f^{''}(x)\left\{\begin{matrix} -6x &x<0 \\ 6x& x>0 \end{matrix}\right.$

Question:19 Using mathematical induction prove that $\frac{d}{dx} (x^n) = nx ^{n-1}$ for all positive integers n.

Answer:

Given equation is
$\frac{d}{dx} (x^n) = nx ^{n-1}$
We need to show that $\frac{d}{dx} (x^n) = nx ^{n-1}$ for all positive integers n
Now,
For ( n = 1) $\Rightarrow \frac{d(x^{1})}{dx}= 1.x^{1-1}= 1.x^0=1$
Hence, true for n = 1
For (n = k) $\Rightarrow \frac{d(x^{k})}{dx}= k.x^{k-1}$
Hence, true for n = k
For ( n = k+1) $\Rightarrow \frac{d(x^{k+1})}{dx}= \frac{d(x.x^k)}{dx}$
$= \frac{d(x)}{dx}.x^k+x.\frac{d(x^k)}{dx}$
$= 1.x^k+x.(k.x^{k-1}) = x^k+k.x^k= (k+1)x^k$
Hence, (n = k+1) is true whenever (n = k) is true
Therefore, by the principle of mathematical induction we can say that $\frac{d}{dx} (x^n) = nx ^{n-1}$ is true for all positive integers n

Question:20 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:

Given function is
$\sin (A + B) = \sin A \cos B + \cos A \sin B$
Now, differentiate w.r.t. x
$\frac{d(\sin(A+B))}{dx} = \frac{d\sin A}{dx}.\cos B+\sin A.\frac{d\cos B}{dx}+\frac{d\cos A}{dx}.\sin B+\cos A.\frac{d\sin B}{dx}$
$\cos (A+b)\frac{d(A+B)}{dx}$ $=\frac{dA}{dx}(\cos A\cos B-\sin A\cos B)+\frac{dB}{dx}(\cos A \sin B-\sin A\sin B)$
$=(\cos A \sin B-\sin A\sin B).\frac{d(A+B)}{dx}$
$\cos(A+B)= \cos A\sin B-\sin A\cos B$
Hence, we get the formula by differentiation of sin(A + B)

Question:21 Does there exist a function which is continuous everywhere but not differentiable
at exactly two points? Justify your answer.

Answer:

Consider f(x) = |x| + |x +1|
We know that modulus functions are continuous everywhere and sum of two continuous function is also a continuous function
Therefore, our function f(x) is continuous
Now,
If Lets differentiability of our function at x = 0 and x= -1
L.H.D. at x = 0
$\lim_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^-}\frac{f(h)-f(0)}{h}= \lim_{h\rightarrow 0^-}\frac{|h|+|h+1|-|1|}{h}$
$=\lim_{h\rightarrow 0^-}\frac{-h-(h+1)-1}{h}= 0$ $(|h| = - h \ because\ h < 0)$
R.H.L. at x = 0
$\lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}= \lim_{h\rightarrow 0^+}\frac{|h|+|h+1|-|1|}{h}$
$=\lim_{h\rightarrow 0^+}\frac{h+h+1-1}{h}= \lim_{h\rightarrow 0^+}\frac{2h}{h}= 2$ $(|h| = h \ because \ h > 0)$
R.H.L. is not equal to L.H.L.
Hence.at x = 0 is the function is not differentiable
Now, Similarly
R.H.L. at x = -1
$\lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^+}\frac{f(-1+h)-f(-1)}{h}= \lim_{h\rightarrow 0^+}\frac{|-1+h|+|h|-|-1|}{h}$
$=\lim_{h\rightarrow 0^+}\frac{1-h+h-1}{h}= \lim_{h\rightarrow 0^+}\frac{0}{h}= 0$ $(|h| = h \ because \ h > 0)$
L.H.L. at x = -1
$\lim_{h\rightarrow 0^-}\frac{f(x+h)-f(x)}{h}= \lim_{h\rightarrow 0^-}\frac{f(1+h)-f(1)}{h}= \lim_{h\rightarrow 0^-}\frac{|-1+h|+|h|-|1|}{h}$
$=\lim_{h\rightarrow 1^+}\frac{1-h-h-1}{h}= \lim_{h\rightarrow 0^+}\frac{-2h}{h}= -2$ $(|h| = - h \ because\ h < 0)$
L.H.L. is not equal to R.H.L, so not differentiable at x=-1

Hence, exactly two points where it is not differentiable

Question:22 If $y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}$ , prove that $dy/dx = \begin{vmatrix} f '(x) & g'(x) & h' (x) \\ l& m &n \\ a& b &c \end{vmatrix}$

Answer:

Given that
$y = \begin{vmatrix} f (x) & g(x) & h (x) \\ l& m &n \\ a& b &c \end{vmatrix}$
We can rewrite it as
$y = f(x)(mc-bn)-g(x)(lc-an)+h(x)(lb-am)$
Now, differentiate w.r.t x
we will get
$\frac{dy}{dx} = f^{'}(x)(mc-bn)-g^{'}(x)(lc-an)+h^{'}(x)(lb-am) \Rightarrow \begin{bmatrix} f^{'}(x) &g^{'}(x) &h^{'}(x) \\ l&m &n \\ a& b &c \end{bmatrix}$
Hence proved

Question:23 If $y = e ^{a \cos ^{-1}x} , -1 \leq x \leq 1$ , show that

Answer:

Given function is

$y = e ^{a \cos ^{-1}x} , -1 \leq x \leq 1$

Now, differentiate w.r.t x
we will get
$\frac{dy}{dx}= \frac{d(e^{a\cos^{-1}x})}{dx}.\frac{d(a\cos^{-1}x)}{dx} = e^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}}$ -(i)
Now, again differentiate w.r.t x
$\frac{d^2y}{dx^2}= \frac{d}{dx}\frac{dy}{dx}= \frac{-ae^{a\cos^{-1}x}.\frac{-a}{\sqrt{1-x^2}}.\sqrt{1-x^2}+ae^{a\cos^{-1}x}.\frac{1.(-2x)}{2\sqrt{1-x^2}}}{(\sqrt{1-x^2})^2}$
$=\ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2}$ -(ii)
Now, we need to show that
$( 1- x^2 ) \frac{d^2 y }{dx ^2} - x \frac{dy}{dx} - a ^2 y = 0$
Put the values from equation (i) and (ii)
$(1-x^2).\left ( \ \frac{a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}}{1-x^2} \right )-x.\left ( \frac{-ae^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x}$
$a^2e^{a\cos^{-1}x}-\frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}}+\left ( \frac{axe^{a\cos^{-1}x}}{\sqrt{1-x^2}} \right )-a^2e^{a\cos^{-1}x} = 0$
Hence proved

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Miscellaneous Exercise

Sometimes questions from miscellaneous exercises are asked in the competitive exam so, the Class 12 Maths chapter 5 miscellaneous solutions becomes important.
First, try to solve NCERT problems by yourself, it will check your understanding of the concept
NCERT solutions for Class 12 Maths chapter 5 miscellaneous exercise can be used for reference.

Also see-

NCERT Solutions for Class 12 Maths Chapter 5
NCERT Exemplar Solutions Class 12 Maths Chapter 5

NCERT Solutions of Class 12 Subject Wise

NCERT Solutions for Class 12 Maths
NCERT Solutions for Class 12 Physics
NCERT Solutions for Class 12 Chemistry
NCERT Solutions for Class 12 Biology

Subject Wise NCERT Exampler Solutions

NCERT Exemplar Solutions for Class 12 Maths
NCERT Exemplar Solutions for Class 12 Physics
NCERT Exemplar Solutions for Class 12 Chemistry
NCERT Exemplar Solutions for Class 12 Biology

Happy learning!!!

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NCERT Solutions for Miscellaneous Exercise Chapter 5 Class 12 - Continuity and Differentiability

Continuity and Differentiability Miscellaneous Exercise:

More About NCERT Solutions for Class 12 Maths Chapter 5 Miscellaneous Exercise

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Miscellaneous Exercise

NCERT Solutions of Class 12 Subject Wise

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