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

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

Edited By Ramraj Saini | Updated on Dec 04, 2023 12:20 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Chapter 5 Miscellaneous Exercise

NCERT Solutions for miscellaneous exercise chapter 5 class 12 Continuity and Differentiability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. 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. Miscellaneous exercise class 12 chapter 5 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.

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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:10x ^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:11x ^{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 Iff (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

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

In Class 12 Maths chapter 5 miscellaneous solutions there are 23 questions related to finding derivatives of different types of functions, second-order derivatives, and mixed concepts questions from all the previous exercises of this chapter. Before solving the exercises questions, you can try to solve solved examples given before this exercise. It will help you to get more clarity of the concept and you will be able to solve miscellaneous questions by yourself.

Also Read| Continuity and Differentiability Class 12 Chapter 5 Notes

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.
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Key Features Of NCERT Solutions For Class 12 Chapter 5 Miscellaneous Exercise

  • Comprehensive Coverage: The solutions encompass all the topics covered in miscellaneous exercise class 12 chapter 5, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 chapter 5 maths miscellaneous solutions, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 maths miscellaneous exercise chapter 5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this class 12 maths ch 5 miscellaneous exercise solutions, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for class 12 chapter 5 miscellaneous exercise cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for miscellaneous exercise class 12 chapter 5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Question (FAQs)

1. Does questions form miscellaneous exercise are asked in the board exams?

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

2. Does multiplication of two continuous functions is a continuous function ?

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

3. Does subtraction of two continuous functions is a continuous function ?

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

4. What is the weightage of chemistry in NEET ?

Chemistry holds 25% marks weighatge in the NEET exam.

5. What is the weightage of Continuity and Differentiability in CBSE Class 12 Maths board exams ?

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.

6. What is the weightage of biology in NEET ?

Biology holds the 50% weightage in the NEET exam.

7. What is the weightage of maths in JEE Main?

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

8. What is the maximum total marks JEE Main?

The maximum marks for JEE Main 2021 is 300 marks.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Content Writer

Content writing is meant to speak directly with a particular audience, such as customers, potential customers, investors, employees, or other stakeholders. The main aim of professional content writers is to speak to their targeted audience and if it is not then it is not doing its job. There are numerous kinds of the content present on the website and each is different based on the service or the product it is used for.

2 Jobs Available
Reporter

Individuals who opt for a career as a reporter may often be at work on national holidays and festivities. He or she pitches various story ideas and covers news stories in risky situations. Students can pursue a BMC (Bachelor of Mass Communication), B.M.M. (Bachelor of Mass Media), or MAJMC (MA in Journalism and Mass Communication) to become a reporter. While we sit at home reporters travel to locations to collect information that carries a news value.  

2 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Production Manager
3 Jobs Available
Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness. 

2 Jobs Available
Process Development Engineer

The Process Development Engineers design, implement, manufacture, mine, and other production systems using technical knowledge and expertise in the industry. They use computer modeling software to test technologies and machinery. An individual who is opting career as Process Development Engineer is responsible for developing cost-effective and efficient processes. They also monitor the production process and ensure it functions smoothly and efficiently.

2 Jobs Available
Merchandiser
2 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
ITSM Manager
3 Jobs Available
IT Consultant

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available
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