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

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

Komal MiglaniUpdated on 24 Apr 2025, 07:19 PM IST

If continuity is like a road without any breaks or holes, then differentiability is like when the road is so smooth that there are no bumps or sharp turns, making it easier to measure how steep it is at any given point. In advanced mathematics, continuity and differentiability play a major role in calculus. The miscellaneous exercise of the Continuity and Differentiability chapter combines all the key concepts covered in the chapter, so that the students can enhance their understanding by a comprehensive review of the entire chapter and get better at problem-solving. This article on the NCERT Solutions for Miscellaneous Exercise Chapter 5 Class 12 Maths - Continuity and Differentiability provides clear and step-by-step solutions for the exercise problems given in the exercise and helps the students clear their doubts, so that they can understand the logic behind these solutions and prepare for various examinations. For syllabus, notes, and PDF, refer to this link: NCERT.

Class 12 Maths Chapter 5 Miscellaneous Exercise Solutions: Download PDF

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Continuity and Differentiability Miscellaneous Exercise:

Question1: 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

$\frac{1}{u} \cdot \frac{du}{dx} = \frac{d(x^2 - 3)}{dx} \cdot \log x + (x^2 - 3) \cdot \frac{d(\log x)}{dx}$

$\frac{1}{u} \cdot \frac{du}{dx} = 2x \cdot \log x + (x^2 - 3) \cdot \frac{1}{x}$

$\frac{1}{u} \cdot \frac{du}{dx} = \frac{2x^2 \log x + x^2 - 3}{x}$

$\frac{du}{dx} = u \cdot \left( \frac{2x^2 \log x + x^2 - 3}{x} \right)$

$\frac{du}{dx} = x^{(x^2 - 3)} \cdot \left( \frac{2x^2 \log x + x^2 - 3}{x} \right)$ -(i)
Similarly,
take
Now, take log on both the sides
$\log v=x^2\log (x-3)$
Now, differentiate w.r.t x

$\frac{1}{v} \cdot \frac{dv}{dx} = \frac{d(x^2)}{dx} \cdot \log(x - 3) + x^2 \cdot \frac{d(\log(x - 3))}{dx}$

$\frac{1}{v} \cdot \frac{dv}{dx} = 2x \cdot \log(x - 3) + x^2 \cdot \frac{1}{x - 3}$

$\frac{1}{v} \cdot \frac{dv}{dx} = 2x \log(x - 3) + \frac{x^2}{x - 3}$

$\frac{dv}{dx} = v \cdot \left( 2x \log(x - 3) + \frac{x^2}{x - 3} \right)$

$\frac{dv}{dx} = (x - 3)^{x^2} \cdot \left( 2x \log(x - 3) + \frac{x^2}{x - 3} \right)$ -(ii)
Now

$f(x) = u + v$

$f'(x) = \frac{du}{dx} + \frac{dv}{dx}$

Put the value from equation (i) and (ii):

$f'(x) = x^{(x^2 - 3)} \cdot \left( \frac{2x^2 \log x + x^2 - 3}{x} \right) + (x - 3)^{x^2} \cdot \left( 2x \log(x - 3) + \frac{x^2}{x - 3} \right)$

Therefore, differentiation w.r.t. $x$ is:

$x^{(x^2 - 3)} \cdot \left( \frac{2x^2 \log x + x^2 - 3}{x} \right) + (x - 3)^{x^2} \cdot \left( 2x \log(x - 3) + \frac{x^2}{x - 3} \right)$

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

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{12\sin t}{10(1 - \cos t)} = \frac{6}{5} \cdot \frac{2\sin \frac{t}{2} \cos \frac{t}{2}}{2\sin^2 \frac{t}{2}} = \frac{6}{5} \cdot \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}}$

$(\because \sin 2x = 2\sin x\cos x \ and \ 1-\cos 2x = 2\sin^2x)$
$\frac{dy}{dx}=\frac{6}{5}.\cot \frac{t}{2}$
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
$\frac{d((x - a)^2)}{dx} + \frac{d((y - b)^2)}{dx} = \frac{d(c^2)}{dx} \\ \\
2(x - a) + 2(y - b) \cdot \frac{dy}{dx} = 0 \\ \\
\frac{dy}{dx} = \frac{a - x}{y - b}$ -(ii)
Now, the second derivative

$\frac{d^2y}{dx^2} = \frac{\frac{d(a - x)}{dx} \cdot (y - b) - (a - x) \cdot \frac{d(y - b)}{dx}}{(y - b)^2} \\ \\

\frac{d^2y}{dx^2} = \frac{(-1)(y - b) - (a - x) \cdot \frac{dy}{dx}}{(y - b)^2}$
Now, put values from equation (i) and (ii)
$\frac{d^2y}{dx^2} = \frac{-(y - b) - (a - x) \cdot \frac{a - x}{y - b}}{(y - b)^2} \\ \\
\frac{d^2y}{dx^2} = \frac{-((y - b)^2 + (a - x)^2)}{(y - b)^{\frac{3}{2}}} = \frac{-c^2}{(y - b)^{\frac{3}{2}}}$ $(\because (x - a)^2 + (y - b)^2 = c^2)$
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

$\frac{d(\cos y)}{dx} = \frac{dx}{dx} \cdot \cos(a + y) + x \cdot \frac{d(\cos(a + y))}{dx}$
$-\sin y \cdot \frac{dy}{dx} = \cos(a + y) + x \cdot (-\sin(a + y)) \cdot \frac{dy}{dx}$
$\frac{dy}{dx} \cdot (x \sin(a + y) - \sin y) = \cos(a + y)$
$\frac{dy}{dx} \cdot \left(\frac{\cos y}{\cos(a + b)} \cdot \sin(a + y) - \sin y\right) = \cos(a + b) \quad (\because x = \frac{\cos y}{\cos(a + b)})$
$\frac{dy}{dx} \cdot (\cos y \sin(a + y) - \sin y \cos(a + y)) = \cos^2(a + b)$
$\frac{dy}{dx} \cdot \sin((a + y) - y) = \cos^2(a + b) \quad (\because \cos A \sin B - \sin A \cos B = \sin(A - B))$
$\frac{dy}{dx} = \frac{\cos^2(a + b)}{\sin a}$

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 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 20: 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 21: 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 22: 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
$\frac{dy}{dx} = \frac{d(e^{a\cos^{-1}x})}{dx} \cdot \frac{d(a\cos^{-1}x)}{dx} = e^{a\cos^{-1}x} \cdot \frac{-a}{\sqrt{1 - x^2}} \ \ \ \ \text{-(i)}$
Now, again differentiate w.r.t x
$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{ -a e^{a\cos^{-1}x} \cdot \frac{-a}{\sqrt{1 - x^2}} \cdot \sqrt{1 - x^2} + a e^{a\cos^{-1}x} \cdot \frac{1 \cdot (-2x)}{2\sqrt{1 - x^2}} }{(\sqrt{1 - x^2})^2}$
$= \frac{a^2 e^{a\cos^{-1}x} - \frac{a x e^{a\cos^{-1}x}}{\sqrt{1 - x^2}}}{1 - x^2}$ -(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


Also, Read,

Topics covered in Chapter 5, Continuity and Differentiability: Miscellaneous Exercise

The main topics covered in Chapter 5 of continuity and differentiability, miscellaneous exercises are:

  • Continuity: Mathematically, we can say that a function $f(x)$ is said to be continuous at $x=a$,
    if left-hand limit = right hand limit = function value at $x=a$
    i.e. $\lim\limits_{x→a-}f(x)=\lim\limits_{x→a+}f(x)=f(a)$
  • Differentiability: Differentiability is the property of a function that denotes that the function has a derivative at the given point or interval. Mathematically, we can say that,
    If $\lim\limits_{h \rightarrow 0^{-}} \frac{f(c+h)-f(c)}{h}=\lim\limits _{h \rightarrow 0^{+}} \frac{f(c+h)-f(c)}{h}$ then $f(x)$ is said to be differentiable at $x=c$.
  • Differentiation techniques: There are multiple types of techniques to do differentiation, some of them are derivatives of implicit functions, derivatives of inverse trigonometric functions, logarithmic differentiation, second-order derivatives, etc. Mastering these techniques will make the students more versatile in solving complex problems involving differentiation.

Also, Read,

JEE Main Highest Scoring Chapters & Topics
Just Study 40% Syllabus and Score upto 100%
Download EBook

NCERT Exemplar Solutions Subject Wise

Here are some links to subject-wise solutions for the NCERT exemplar class 12.

Frequently Asked Questions (FAQs)

Q: Does multiplication of two continuous functions is a continuous function ?
A:

Yes, the multiplication of two continuous functions is a continuous function.

Q: Does subtraction of two continuous functions is a continuous function ?
A:

Yes, subtraction of two continuous functions is a continuous function.

Q: What is the weightage of chemistry in NEET ?
A:

Chemistry holds 25% marks weighatge in the NEET exam.

Q: What is the weightage of Continuity and Differentiability in CBSE Class 12 Maths board exams ?
A:

CBSE doesn't provide chapter-wise marks distribution for CBSE Class 12 Maths. A total of 35 marks of questions are asked from the calculus in the CBSE final board exam.

Q: What is the weightage of biology in NEET ?
A:

Biology holds the 50% weightage in the NEET exam.

Q: What is the weightage of maths in JEE Main?
A:

The JEE main has an equal weightage of three subjects Physics, Chemistry, and Maths.

Q: What is the maximum total marks JEE Main?
A:

The maximum marks for JEE Main 2021 is 300 marks.

Q: Does questions form miscellaneous exercise are asked in the board exams?
A:

Over 90% of questions in the board exams are not asked from the miscellaneous exercise.

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Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.



Hello

For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.

Hello,

If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

For the 2025-2026 academic session, the CBSE plans to conduct board exams from 17 February 2026 to 20 May 2026.

You can download it in pdf form from below link

CBSE DATE SHEET 2026

all the best for your exam!!

Hii neeraj!

You can check CBSE class 12th registration number in:

  • Your class 12th board exam admit card. Please do check admit card for registration number, it must be there.
  • You can also check the registration number in your class 12th marksheet in case you have got it.
  • Alternatively you can also visit your school and ask for the same in the administration office they may tell you the registration number.

Hope it helps!