Aakash Repeater Courses
Take Aakash iACST and get instant scholarship on coaching programs.
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.
Question1: Differentiate w.r.t. x the function in Exercises 1 to 11.
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.
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.
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.
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
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
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
$f^{''}(x)\left\{\begin{matrix} -6x &x<0 \\ 6x& x>0 \end{matrix}\right.$
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)
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
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 , show that
Answer:
Given function is
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,
The main topics covered in Chapter 5 of continuity and differentiability, miscellaneous exercises are:
Take Aakash iACST and get instant scholarship on coaching programs.
Also, Read,
Below are some useful links for subject-wise NCERT solutions for class 12.
Here are some links to subject-wise solutions for the NCERT exemplar class 12.
Frequently Asked Questions (FAQs)
Yes, the multiplication of two continuous functions is a continuous function.
Yes, subtraction of two continuous functions is a continuous function.
Chemistry holds 25% marks weighatge in the NEET exam.
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.
Biology holds the 50% weightage in the NEET exam.
The JEE main has an equal weightage of three subjects Physics, Chemistry, and Maths.
The maximum marks for JEE Main 2021 is 300 marks.
Over 90% of questions in the board exams are not asked from the miscellaneous exercise.
On Question asked by student community
Hello
Yes, if you’re not satisfied with your marks even after the improvement exam, many education boards allow you to reappear as a private candidate next year to improve your scores. This means you can register independently, study at your own pace, and take the exams without attending regular classes. It’s a good option to improve your results and open up more opportunities for higher studies or careers. Just make sure to check the specific rules and deadlines of your education board so you don’t miss the registration window. Keep your focus, and you will do better next time.
Hello Aspirant,
Yes, in the case that you appeared for the 2025 improvement exam and your roll number is different from what was on the previous year’s marksheet, the board will usually release a new migration certificate. This is because the migration certificate will reflect the most recent exam details, roll number and passing year. You can apply to get it from your board using the process prescribed by them either online or through your school/college.
Yes, if you miss the 1st CBSE exam due to valid reasons, then you can appear for the 2nd CBSE compartment exam.
From the academic year 2026, the board will conduct the CBSE 10th exam twice a year, while the CBSE 12th exam will be held once, as per usual. For class 10th, the second phase exam will act as the supplementary exam. Check out information on w hen the CBSE first exam 2026 will be conducted and changes in 2026 CBSE Board exam by clicking on the link .
If you want to change your stream to humanities after getting a compartment in one subject in the CBSE 12th Board Exam , you actually have limited options to qualify for your board exams. You can prepare effectively and appear in the compartment examination for mathematics again. If you do not wish to continue with the current stream, you can take readmission in the Humanities stream and start from Class 11th again, and continue studying for two more years to qualify for the 12th examination.
The GUJCET Merit List is prepared based on the Class 12th marks and GUJCET marks received by the students. CBSE students who are not from the Gujarat board can definitely compete with GSEB students, as their eligibility is decided based on the combined marks scored by them in GUJCET and the 12th board. The weightage of the GUJCET score is 40% and the weightage of the class 12 scores is 60%.
Take Aakash iACST and get instant scholarship on coaching programs.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE