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

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.
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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 ?
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

You can use them people also used problem

Read Complete Answer
Divyan 06 August,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Read Complete Answer
Yuvan 01 September,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer
Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer
kumardurgesh1802 10 July,2024
I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Read Complete Answer
kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

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
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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

 
Read More

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
Read More

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
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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 .
Read More

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
Read More

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.
Read More

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.
Read More

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
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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
Read More

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