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

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
$f^{{}^{\prime }}(x)=\frac{d(f(x))}{dx}=\frac{d({\sin }^{3}x+{\cos }^{6}x)}{dx}=3{\sin }^{2}x.\frac{d(\sin x)}{dx}+6{\cos }^{5}x.\frac{d(\cos x)}{dx}$
$=3{\sin }^{2}x.\cos x+6{\cos }^{5}x.(-\sin x)$
$=3{\sin }^{2}x\cos x-6{\cos }^{5}x\sin x=3\sin x\cos x(\sin x-2{\cos }^{4}x)$

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

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

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

Answer:

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

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

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

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

Answer:

Given function is
$f(x)={\sin }^{-1}(x\sqrt{x}),0\le x\le 1$
Now, differentiation w.r.t. x is
$f^{{}^{\prime }}(x)=\frac{d(f(x))}{dx}=\frac{d({\sin }^{-1}x\sqrt{x})}{dx}=\frac{1}{\sqrt{1-(x\sqrt{x}{)}^2}}.\frac{d(x\sqrt{x})}{dx}$
$=\frac{1}{\sqrt{1-x^3}}.(1.\sqrt{x}+x\frac{1}{2\sqrt{x}})$
$=\frac{1}{\sqrt{1-x^3}}.\left(\frac{3\sqrt{x}}{2}\right)$
$=\frac{3}{2}.\sqrt{\frac{x}{1-x^3}}$

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
$$\begin{gathered} f^{{}^{\prime }}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{{\cos }^{-1}\frac{x}{2}}{\sqrt{2x+7}})}{dx}=\frac{\frac{d({\cos }^{-1}\frac{x}{2})}{dx}.\sqrt{2x+7}-{\cos }^{-1}\frac{x}{2}.\frac{d(\sqrt{2x+7})}{dx}}{(\sqrt{2x+7}{)}^2} \\ {f}^{{}^{\prime }}(x)=\frac{\frac{-1}{\sqrt{1-(\frac{x}{2}{)}^2}}.\frac{1}{2}.\sqrt{2x+7}-{\cos }^{-1}\frac{x}{2}.\frac{1}{2.\sqrt{2x+7}}.2}{2x+7} \\ {f}^{{}^{\prime }}(x)=-[\frac{1}{(\sqrt{4-x^2})(\sqrt{2x+7})}+\frac{{\cos }^{-1}\frac{x}{2}}{(2x+7{)}^{\frac{3}{2}}}] \end{gathered}$$

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

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
$\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]=[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}.\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}]$

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

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

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

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

$=\frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}}=\cot \frac{x}{2}$
Given function reduces to
$$\begin{gathered} f(x)={\cot }^{-1}(\cot \frac{x}{2}) \\ f(x)=\frac{x}{2} \end{gathered}$$
Now, differentiation w.r.t. x is
$f^{{}^{\prime }}(x)=\frac{d(f(x))}{dx}=\frac{d(\frac{x}{2})}{dx}=\frac{1}{2}$
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
$f^{{}^{\prime }}(x)=\frac{d(f(x))}{dx}=\frac{d(\cos (a\cos x+b\sin x))}{dx}$
$=-\sin (a\cos x+b\sin x).\frac{d(a\cos x+b\sin x)}{dx}$
$=-\sin (a\cos x+b\sin x).(-a\sin x+b\cos x)$
$=(a\sin x-b\cos x)\sin (a\cos x+b\sin x).$
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
$\frac{1}{y}.\frac{dy}{dx}=\frac{d(\sin x-\cos x)}{dx}.\log (\sin x-\cos x)+(\sin x-\cos x).\frac{d(\log (\sin x-\cos x))}{dx}$
$\frac{1}{y}.\frac{dy}{dx}=(\cos x-(-\sin x)).\log (\sin x-\cos x)+(\sin x-\cos x).\frac{(\cos x-(-\sin x))}{(\sin x-\cos x)}$
$\frac{dy}{dx}=y.(\cos x+\sin x)(\log (\sin x-\cos x)+1)$
$\frac{dy}{dx}=(\sin x-\cos x{)}^{(\sin x-\cos x)}.(\cos x+\sin x)(\log (\sin x-\cos x)+1)$
Therefore, differentiation w.r.t x is $(\sin x-\cos x{)}^{(\sin x-\cos x)}.(\cos x+\sin x)(\log (\sin x-\cos x)+1),sinx>cosx$

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

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

Similarly, take $w=a^a$
take log on both the sides
$\log w=a\log a=constant$
Now, differentiate w.r.t x
$$\begin{gathered} \frac{1}{w}.\frac{dw}{dx}=a.\frac{d(a\log a)}{dx}=0 \\ \frac{dw}{dx}=0 \end{gathered}$$ -(iv)
Now,
$f(x)=u+v+z+w$
$f^{{}^{\prime }}(x)=\frac{du}{dx}+\frac{dv}{dx}+\frac{dz}{dx}+\frac{dw}{dx}$
Put values from equation (i) , (ii) ,(iii) and (iv)
$f^{{}^{\prime }}(x)=x^x(\log x+1)+a{x}^{a-1}+a^x\log a$
Therefore, differentiation w.r.t. x is $x^x(\log x+1)+a{x}^{a-1}+a^x\log a$

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

Answer:

Given function is
$f(x)=x^{x^2-3}+(x-3{)}^{x^2},forx>3$
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

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

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

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

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

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

$f(x)=u+v$

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

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

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

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

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

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

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

$(\because \sin 2x=2\sin x\cos xand1-\cos 2x=2{\sin }^{2}x)$
$\frac{dy}{dx}=\frac{6}{5}.\cot \frac{t}{2}$
Therefore, differentiation w.r.t x is $\frac{6}{5}.\cot \frac{t}{2}$

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

Answer:

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

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

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
$$\begin{gathered} (x\sqrt{1+y}{)}^2=(-y\sqrt{1+x}{)}^2 \\ {x}^2(1+y)=y^2(1+x) \\ {x}^2+x^{2}y=y^{2}x+y^2 \\ {x}^2-y^2=y^{2}x-x^{2}y \\ (x-y)(x+y)=-xy(x-y) \\ x+y=-xy \\ y=\frac{-x}{1+x} \end{gathered}$$
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$\frac{{[1+(\frac{dy}{dx}{)}^2]}^{3/2}}{\frac{d^{2}y}{d{x}^2}}$ 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
$$\begin{gathered} \frac{d((x-a{)}^2)}{dx}+\frac{d((y-b{)}^2)}{dx}=\frac{d(c^2)}{dx} \\ 2(x-a)+2(y-b)\cdot \frac{dy}{dx}=0 \\ \frac{dy}{dx}=\frac{a-x}{y-b} \end{gathered}$$ -(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}$Now,putvaluesfromequation(i)and(ii)$\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 $\cos a\ne \pm 1$ , 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

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

Hence proved

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

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
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\frac{dy}{dx}={\sec }^{2}t.\frac{dt}{dx}=\frac{{\sec }^{2}t.\sec t}{at}=\frac{{\sec }^{3}t}{at}$
$(\because \frac{dx}{dt}=at\cos t\Rightarrow \frac{dt}{dx}=\frac{1}{at\cos t}=\frac{\sec t}{at})$
Therefore, $\frac{d^{2}y}{d{x}^2}=\frac{{\sec }^{3}t}{at}$

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

Answer:

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

Question 19: Using the fact that $\sin (A+B)=\sin A\cos B+\cos A\sin B$ and the differentiation,
obtain the sum formula for cosines.

Answer:

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

Question 20: Does there exist a function which is continuous everywhere but not differentiable
at exactly two points? Justify your answer.

Answer:

Consider f(x) = |x| + |x +1|
We know that modulus functions are continuous everywhere and sum of two continuous function is also a continuous function
Therefore, our function f(x) is continuous
Now,
If Lets differentiability of our function at x = 0 and x= -1
L.H.D. at x = 0
$\underset{h\to 0^{-}}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0^{-}}{lim}\frac{f(h)-f(0)}{h}=\underset{h\to 0^{-}}{lim}\frac{|h|+|h+1|-|1|}{h}$
$=\underset{h\to 0^{-}}{lim}\frac{-h-(h+1)-1}{h}=0$ $(|h|=-hbecauseh<0)$
R.H.L. at x = 0
$\underset{h\to 0^{+}}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0^{+}}{lim}\frac{f(h)-f(0)}{h}=\underset{h\to 0^{+}}{lim}\frac{|h|+|h+1|-|1|}{h}$
$=\underset{h\to 0^{+}}{lim}\frac{h+h+1-1}{h}=\underset{h\to 0^{+}}{lim}\frac{2h}{h}=2$ $(|h|=hbecauseh>0)$
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
$\underset{h\to 0^{+}}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0^{+}}{lim}\frac{f(-1+h)-f(-1)}{h}=\underset{h\to 0^{+}}{lim}\frac{|-1+h|+|h|-|-1|}{h}$
$=\underset{h\to 0^{+}}{lim}\frac{1-h+h-1}{h}=\underset{h\to 0^{+}}{lim}\frac{0}{h}=0$ $(|h|=hbecauseh>0)$
L.H.L. at x = -1
$\underset{h\to 0^{-}}{lim}\frac{f(x+h)-f(x)}{h}=\underset{h\to 0^{-}}{lim}\frac{f(1+h)-f(1)}{h}=\underset{h\to 0^{-}}{lim}\frac{|-1+h|+|h|-|1|}{h}$
$=\underset{h\to 1^{+}}{lim}\frac{1-h-h-1}{h}=\underset{h\to 0^{+}}{lim}\frac{-2h}{h}=-2$ $(|h|=-hbecauseh<0)$
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