Miscellaneous Exercise (Revised) - Chapter 5 - Continuity & Differentiability - Ncert Solutions class 12 - Maths

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Chapter 5: Continuity & Differentiability NCERT Solutions Class 12 Maths

Miscellaneous Exercise Question 1:

$\left(3 x^2-9 x+5\right)^9$

Answer:
Let $y=\left(3 x^2-9 x+5\right)^9$
Using chain rule, we obtain
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left(3 x^2-9 x+5\right)^9 \\
& =9\left(3 x^2-9 x+5\right)^8 \cdot \frac{d}{d x}\left(3 x^2-9 x+5\right) \\
& =9\left(3 x^2-9 x+5\right)^8 \cdot(6 x-9) \\
& =9\left(3 x^2-9 x+5\right)^8 \cdot 3(2 x-3) \\
& =27\left(3 x^2-9 x+5\right)^8(2 x-3)
\end{aligned}
$

Miscellaneous Exercise Question 2

$\sin ^3 x+\cos ^6 x$

Answer:
Let $y=\sin ^3 x+\cos ^6 x$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d}{d x}\left(\sin ^3 x\right)+\frac{d}{d x}\left(\cos ^6 x\right) \\
& =3 \sin ^2 x \cdot \frac{d}{d x}(\sin x)+6 \cos ^3 x \cdot \frac{d}{d x}(\cos x) \\
& =3 \sin ^2 x \cdot \cos x+6 \cos ^5 x \cdot(-\sin x) \\
& =3 \sin x \cos x\left(\sin x-2 \cos ^4 x\right)
\end{aligned}
$

Miscellaneous Exercise Question 3 :

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

Answer :
Let $y=(5 x)^{3 \cos 2 x}$
Taking logarithm on both the sides, we obtain
$
\log y=3 \cos 2 x \log 5 x
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=3\left[\log 5 x \cdot \frac{d}{d x}(\cos 2 x)+\cos 2 x \cdot \frac{d}{d x}(\log 5 x)\right] \\
& \Rightarrow \frac{d y}{d x}=3 y\left[\log 5 x(-\sin 2 x) \cdot \frac{d}{d x}(2 x)+\cos 2 x \cdot \frac{1}{5 x} \cdot \frac{d}{d x}(5 x)\right] \\
& \Rightarrow \frac{d y}{d x}=3 y\left[-2 \sin 2 x \log 5 x+\frac{\cos 2 x}{x}\right] \\
& \Rightarrow \frac{d y}{d x}=3 y\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right] \\
& \therefore \frac{d y}{d x}=(5 x)^{3 \cos 2 x}\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]
\end{aligned}
$

Miscellaneous Exercise Question 4:

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

Answer:
Let $y=\sin ^{-1}(x \sqrt{x})$
Using chain rule, we obtain
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x} \sin ^{-1}(x \sqrt{x}) \\
& =\frac{1}{\sqrt{1-(x \sqrt{x})^2}} \times \frac{d}{d x}(x \sqrt{x}) \\
& =\frac{1}{\sqrt{1-x^3}} \cdot \frac{d}{d x}\left(x^{\frac{3}{2}}\right) \\
& =\frac{1}{\sqrt{1-x^3}} \times \frac{3}{2} \cdot x^{\frac{1}{2}} \\
& =\frac{3 \sqrt{x}}{2 \sqrt{1-x^3}} \\
& =\frac{3}{2} \sqrt{\frac{x}{1-x^3}}
\end{aligned}
$

Miscellaneous Exercise Question 5:

$\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}},-2$

Answer :
Let $y=\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}$
By quotient rule, we obtain
$
\begin{aligned}
& \frac{d y}{d x}=\frac{\sqrt{2 x+7} \frac{d}{d x}\left(\cos ^{-1} \frac{x}{2}\right)-\left(\cos ^{-1} \frac{x}{2}\right) \frac{d}{d x}(\sqrt{2 x+7})}{(\sqrt{2 x+7})^2} \\
& =\frac{\sqrt{2 x+7}\left[\frac{-1}{\sqrt{1-\left(\frac{x}{2}\right)^2}} \cdot \frac{d}{d x}\left(\frac{x}{2}\right)\right]-\left(\cos ^{-1} \frac{x}{2}\right) \frac{1}{2 \sqrt{2 x+7}} \cdot \frac{d}{d x}(2 x+7)}{2 x+7} \\
& =\frac{\sqrt{2 x+7} \frac{-1}{\sqrt{4-x^2}}-\left(\cos ^{-1} \frac{x}{2}\right) \frac{2}{2 \sqrt{2 x+7}}}{2 x+7} \\
& =\frac{-\sqrt{2 x+7}}{\sqrt{4-x^2} \times(2 x+7)}-\frac{\cos ^{-1} \frac{x}{2}}{(\sqrt{2 x+7})(2 x+7)} \\
& =-\left[\frac{1}{\sqrt{4-x^2} \sqrt{2 x+7}}+\frac{\cos ^{-1} \frac{x}{2}}{(2 x+7)^{\frac{3}{2}}}\right] \\
&
\end{aligned}
$

Miscellaneous Exercise Question 6 :
$
\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0 $

Answer:
Let $y=\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$
$
\begin{aligned}
& \text { Then, } \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}-\sqrt{1-\sin x})(\sqrt{1+\sin x}+\sqrt{1-\sin x})} \\
& =\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1-\sin x)(1+\sin x)}}{(1+\sin x)-(1-\sin x)} \\
& =\frac{2+2 \sqrt{1-\sin ^2 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}} \\
& =\cot \frac{x}{2}
\end{aligned}
$

Therefore, equation (1) becomes

$\begin{aligned}
& y=\cot ^{-1}\left(\cot \frac{x}{2}\right) \\
& \Rightarrow y=\frac{x}{2} \\
& \therefore \frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x) \\
& \Rightarrow \frac{d y}{d x}=\frac{1}{2}
\end{aligned}$

Miscellaneous Exercise Question 7:

$(\log x)^{\log x}, x>1$

Answer :
Let $y=(\log x)^{\log x}$
Taking logarithm on both the sides, we obtain
$
\log y=\log x \cdot \log (\log x)
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[\log x \cdot \log (\log x)] \\
& \Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\log x) \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}[\log (\log x)] \\
& \Rightarrow \frac{d y}{d x}=y\left[\log (\log x) \cdot \frac{1}{x}+\log x \cdot \frac{1}{\log x} \cdot \frac{d}{d x}(\log x)\right] \\
& \Rightarrow \frac{d y}{d x}=y\left[\frac{1}{x} \log (\log x)+\frac{1}{x}\right] \\
& \therefore \frac{d y}{d x}=(\log x)^{\log x}\left[\frac{1}{x}+\frac{\log (\log x)}{x}\right]
\end{aligned}
$

Miscellaneous Exercise Question 8:

$\cos (a \cos x+b \sin x)$, for some constant $a$ and $b$.

Answer:
Let $y=\cos (a \cos x+b \sin x)$
By using chain rule, we obtain
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x} \cos (a \cos x+b \sin x) \\
& \Rightarrow \frac{d y}{d x}=-\sin (a \cos x+b \sin x) \cdot \frac{d}{d x}(a \cos x+b \sin x) \\
& =-\sin (a \cos x+b \sin x) \cdot[a(-\sin x)+b \cos x] \\
& =(a \sin x-b \cos x) \cdot \sin (a \cos x+b \sin x) \\
&
\end{aligned}
$

Miscellaneous Exercise Question 9:
$
(\sin x-\cos x)^{(\sin x-\cos x)}, \frac{\pi}{4} $

Answer:
Let $y=(\sin x-\cos x)^{(\sin x-\cos x)}$
Taking logarithm on both the sides, we obtain
$
\begin{aligned}
& \log y=\log \left[(\sin x-\cos x)^{(\sin x-\cos x)}\right] \\
& \Rightarrow \log y=(\sin x-\cos x) \cdot \log (\sin x-\cos x)
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=\frac{d}{d x}[(\sin x-\cos x) \log (\sin x-\cos x)] \\
& \Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \cdot \frac{d}{d x}(\sin x-\cos x)+(\sin x-\cos x) \cdot \frac{d}{d x} \log (\sin x-\cos x) \\
& \Rightarrow \frac{1}{y} \frac{d y}{d x}=\log (\sin x-\cos x) \cdot(\cos x+\sin x)+(\sin x-\cos x) \cdot \frac{1}{(\sin x-\cos x)} \cdot \frac{d}{d x}(\sin x-\cos x) \\
& \Rightarrow \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}[(\cos x+\sin x) \cdot \log (\sin x-\cos x)+(\cos x+\sin x)] \\
& \therefore \frac{d y}{d x}=(\sin x-\cos x)^{(\sin x-\cos x)}(\cos x+\sin x)[1+\log (\sin x-\cos x)]
\end{aligned}
$

Miscellaneous Exercise Question 10:

$x^x+x^a+a^x+a^a$, for some fixed $a>0$ and $x>0$

Answer :
Let $y=x^x+x^a+a^x+a^a$
Also, let $x^x=u, x^a=v, a^x=w$, and $a^a=s$
$
\begin{aligned}
& \therefore y=u+v+w+s \\
& \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}+\frac{d w}{d x}+\frac{d s}{d x} \\
& u=x^x \\
& \Rightarrow \log u=\log x^x \\
& \Rightarrow \log u=x \log x
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{u} \frac{d u}{d x}=\log x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\log x) \\
& \Rightarrow \frac{d u}{d x}=u\left[\log x \cdot 1+x \cdot \frac{1}{x}\right] \\
& \Rightarrow \frac{d u}{d x}=x^x[\log x+1]=x^x(1+\log x) \\
& v=x^a
\end{aligned}
$

$
\begin{aligned}
& \therefore \frac{d v}{d x}=\frac{d}{d x}\left(x^a\right) \\
& \Rightarrow \frac{d v}{d x}=a x^{a-1} \\
& w=a^x \\
& \Rightarrow \log w=\log a^x \\
& \Rightarrow \log w=x \log a
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{w} \cdot \frac{d w}{d x}=\log a \cdot \frac{d}{d x}(x) \\
& \Rightarrow \frac{d w}{d x}=w \log a \\
& \Rightarrow \frac{d w}{d x}=a^x \log a \\
& S=a^a
\end{aligned}
$

Since $a$ is constant, $a^a$ is also a constant.
$
\therefore \frac{d s}{d x}=0
$

From (1), (2), (3), (4), and (5), we obtain
$
\begin{aligned}
\frac{d y}{d x} & =x^x(1+\log x)+a x^{a-1}+a^x \log a+0 \\
& =x^x(1+\log x)+a x^{a-1}+a^x \log a
\end{aligned}
$

Miscellaneous Exercise Question 11:

$x^{x^2-3}+(x-3)^{x^2}$, for $x>3$

Answer :
Let $y=x^{x^2-3}+(x-3)^{x^2}$
Also, let $u=x^{x^2-3}$ and $v=(x-3)^{x^2}$
$
\therefore y=u+v
$

Differentiating both sides with respect to $x$,we obtain
$
\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}
$
$
\begin{aligned}
& u=x^{x^2-3} \\
& \therefore \log u=\log \left(x^{x^2-3}\right) \\
& \log u=\left(x^2-3\right) \log x
\end{aligned}
$

Differentiating with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{u} \cdot \frac{d u}{d x}=\log x \cdot \frac{d}{d x}\left(x^2-3\right)+\left(x^2-3\right) \cdot \frac{d}{d x}(\log x) \\
& \Rightarrow \frac{1}{u} \frac{d u}{d x}=\log x \cdot 2 x+\left(x^2-3\right) \cdot \frac{1}{x} \\
& \Rightarrow \frac{d u}{d x}=x^{x^2-3} \cdot\left[\frac{x^2-3}{x}+2 x \log x\right]
\end{aligned}
$

Also,

$
\begin{aligned}
& v=(x-3)^{x^2} \\
& \therefore \log v=\log (x-3)^{x^2} \\
& \Rightarrow \log v=x^2 \log (x-3)
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{v} \cdot \frac{d v}{d x}=\log (x-3) \cdot \frac{d}{d x}\left(x^2\right)+x^2 \cdot \frac{d}{d x}[\log (x-3)] \\
& \Rightarrow \frac{1}{v} \frac{d v}{d x}=\log (x-3) \cdot 2 x+x^2 \cdot \frac{1}{x-3} \cdot \frac{d}{d x}(x-3) \\
& \Rightarrow \frac{d v}{d x}=v\left[2 x \log (x-3)+\frac{x^2}{x-3} \cdot 1\right] \\
& \Rightarrow \frac{d v}{d x}=(x-3)^{x^2}\left[\frac{x^2}{x-3}+2 x \log (x-3)\right]
\end{aligned}
$

Substituting the expressions of $\frac{d u}{d x}$ and $\frac{d v}{d x}$ in equation (1), we obtain
$
\frac{d y}{d x}=x^{x^2-3}\left[\frac{x^2-3}{x}+2 x \log x\right]+(x-3)^{x^2}\left[\frac{x^2}{x-3}+2 x \log (x-3)\right]
$

Miscellaneous Exercise Question 12:

Find $\frac{d y}{d x}$, if $y=12(1-\cos t), x=10(t-\sin t),-\frac{\pi}{2}$

Answer :
It is given that, $y=12(1-\cos t), x=10(t-\sin t)$
$
\begin{aligned}
& \therefore \frac{d x}{d t}=\frac{d}{d t}[10(t-\sin t)]=10 \cdot \frac{d}{d t}(t-\sin t)=10(1-\cos t) \\
& \frac{d y}{d t}=\frac{d}{d t}[12(1-\cos t)]=12 \cdot \frac{d}{d t}(1-\cos t)=12 \cdot[0-(-\sin t)]=12 \sin t \\
& \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{12 \sin t}{10(1-\cos t)}=\frac{12 \cdot 2 \sin \frac{t}{2} \cdot \cos \frac{t}{2}}{10 \cdot 2 \sin ^2 \frac{t}{2}}=\frac{6}{5} \cot \frac{t}{2}
\end{aligned}
$

Miscellaneous Exercise Question 13:

Find $\frac{d y}{d x}$, if $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^2},-1 \leq x \leq 1$

Answer :
It is given that, $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^2}$
$
\begin{aligned}
& \therefore \frac{d y}{d x}=\frac{d}{d x}\left[\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^2}\right] \\
& \Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{-1} x\right)+\frac{d}{d x}\left(\sin ^{-1} \sqrt{1-x^2}\right) \\
& \Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-\left(\sqrt{1-x^2}\right)^2}} \cdot \frac{d}{d x}\left(\sqrt{1-x^2}\right) \\
& \Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{x} \cdot \frac{1}{2 \sqrt{1-x^2}} \cdot \frac{d}{d x}\left(1-x^2\right) \\
& \Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{2 x \sqrt{1-x^2}}(-2 x) \\
& \Rightarrow \frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \\
& \therefore \frac{d y}{d x}=0 \\
&
\end{aligned}
$

Miscellaneous Exercise Question 14:

If $x \sqrt{1+y}+y \sqrt{1+x}=0$, for, -1

$
\frac{d y}{d x}=-\frac{1}{(1+x)^2}
$

Answer :
It is given that,
$
\begin{aligned}
& x \sqrt{1+y}+y \sqrt{1+x}=0 \\
& \Rightarrow x \sqrt{1+y}=-y \sqrt{1+x}
\end{aligned}
$

Squaring both sides, we obtain
$
\begin{aligned}
& x^2(1+y)=y^2(1+x) \\
& \Rightarrow x^2+x^2 y=y^2+x y^2 \\
& \Rightarrow x^2-y^2=x y^2-x^2 y \\
& \Rightarrow x^2-y^2=x y(y-x) \\
& \Rightarrow(x+y)(x-y)=x y(y-x) \\
& \therefore x+y=-x y \\
& \Rightarrow(1+x) y=-x \\
& \Rightarrow y=\frac{-x}{(1+x)}
\end{aligned}
$

Differentiating both sides with respect to $x$, we obtain

$
\begin{aligned}
& y=\frac{-x}{(1+x)} \\
& \frac{d y}{d x}=-\frac{(1+x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+x)}{(1+x)^2}=-\frac{(1+x)-x}{(1+x)^2}=-\frac{1}{(1+x)^2}
\end{aligned}
$

Hence, proved.

Miscellaneous Exercise Question 15:

If $(x-a)^2+(y-b)^2=c^2$, for some $c>0$, prove that
$
\frac{\left[1+\left(\frac{d y}{d x}\right)^2\right]^{\frac{3}{2}}}{\frac{d^2 y}{d x^2}}
$
is a constant independent of aand $b$.

Answer :
It is given that, $(x-a)^2+(y-b)^2=c^2$
Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{d}{d x}\left[(x-a)^2\right]+\frac{d}{d x}\left[(y-b)^2\right]=\frac{d}{d x}\left(c^2\right) \\
& \Rightarrow 2(x-a) \cdot \frac{d}{d x}(x-a)+2(y-b) \cdot \frac{d}{d x}(y-b)=0 \\
& \Rightarrow 2(x-a) \cdot 1+2(y-b) \cdot \frac{d y}{d x}=0 \\
& \Rightarrow \frac{d y}{d x}=\frac{-(x-a)}{y-b} \\
& \therefore \frac{d^2 y}{d x^2}=\frac{d}{d x}\left[\frac{-(x-a)}{y-b}\right]
\end{aligned}
$

$\begin{aligned}
& =-\left[\frac{(y-b) \cdot \frac{d}{d x}(x-a)-(x-a) \cdot \frac{d}{d x}(y-b)}{(y-b)^2}\right] \\
& =-\left[\frac{(y-b)-(x-a) \cdot \frac{d y}{d x}}{(y-b)^2}\right] \\
& =-\left[\frac{(y-b)-(x-a) \cdot\left\{\frac{-(x-a)}{y-b}\right\}}{(y-b)^2}\right] \\
& \text { [Using (1)] } \\
& =-\left[\frac{(y-b)^2+(x-a)^2}{(y-b)^3}\right] \\
& \therefore\left[\frac{1+\left(\frac{d y}{d x}\right)^2}{\frac{d^2 y}{d x^2}}\right]^{\frac{3}{2}}=\frac{\left[1+\frac{(x-a)^2}{(y-b)^2}\right]^{\frac{3}{2}}}{-\left[\frac{(y-b)^2+(x-a)^2}{(y-b)^3}\right]}=\frac{\left[\frac{(y-b)^2+(x-a)^2}{(y-b)^2}\right]^{\frac{3}{2}}}{-\left[\frac{(y-b)^2+(x-a)^2}{(y-b)^3}\right]} \\
&
\end{aligned}$

$
\begin{aligned}
& =\frac{\left[\frac{c^2}{(y-b)^2}\right]^{\frac{3}{2}}}{-\frac{c^2}{(y-b)^3}}=\frac{\frac{c^3}{(y-b)^3}}{-\frac{c^2}{(y-b)^3}} \\
& =-c \text {, which is constant and is independent of } a \text { and } b
\end{aligned}
$

Hence, proved.

Miscellaneous Exercise Question 16:

If $\cos y=x \cos (a+y)$, with $\cos a \neq \pm 1$, prove that $\frac{d y}{d x}=\frac{\cos ^2(a+y)}{\sin a}$

Answer :
It is given that, $\cos y=x \cos (a+y)$
$
\begin{aligned}
& \therefore \frac{d}{d x}[\cos y]=\frac{d}{d x}[x \cos (a+y)] \\
& \Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y) \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}[\cos (a+y)] \\
& \Rightarrow-\sin y \frac{d y}{d x}=\cos (a+y)+x \cdot[-\sin (a+y)] \frac{d y}{d x} \\
& \Rightarrow[x \sin (a+y)-\sin y] \frac{d y}{d x}=\cos (a+y)
\end{aligned}
$

Since $\cos y=x \cos (a+y), x=\frac{\cos y}{\cos (a+y)}$
Then, equation (1) reduces to
$
\begin{aligned}
& {\left[\frac{\cos y}{\cos (a+y)} \cdot \sin (a+y)-\sin y\right] \frac{d y}{d x}=\cos (a+y)} \\
& \Rightarrow[\cos y \cdot \sin (a+y)-\sin y \cdot \cos (a+y)] \cdot \frac{d y}{d x}=\cos ^2(a+y) \\
& \Rightarrow \sin (a+y-y) \frac{d y}{d x}=\cos ^2(a+b) \\
& \Rightarrow \frac{d y}{d x}=\frac{\cos ^2(a+b)}{\sin a}
\end{aligned}
$

Hence, proved.

Miscellaneous Exercise 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:
It is given that, $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$
$
\begin{aligned}
& \therefore \frac{d x}{d t}=a \cdot \frac{d}{d t}(\cos t+t \sin t) \\
& =a\left[-\sin t+\sin t \cdot \frac{d}{d x}(t)+t \cdot \frac{d}{d t}(\sin t)\right] \\
& =a[-\sin t+\sin t+t \cos t]=a t \cos t \\
& \frac{d y}{d t}=a \cdot \frac{d}{d t}(\sin t-t \cos t) \\
& =a\left[\cos t-\left\{\cos t \cdot \frac{d}{d t}(t)+t \cdot \frac{d}{d t}(\cos t)\right\}\right] \\
& =a[\cos t-\{\cos t-t \sin t\}]=a t \sin t \\
& \therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a t \sin t}{a t \cos t}=\tan t \\
&
\end{aligned}
$

Then, $\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}(\tan t)=\sec ^2 t \cdot \frac{d t}{d x}$
$
\begin{aligned}
& =\sec ^2 t \cdot \frac{1}{a t \cos t} \quad\left[\frac{d x}{d t}=a t \cos t \Rightarrow \frac{d t}{d x}=\frac{1}{a t \cos t}\right] \\
& =\frac{\sec ^3 t}{a t}, 0 \end{aligned}
$

Miscellaneous Exercise Question 18:

If $f(x)=|x|^3$, show that $f^{\prime \prime}(x)$ exists for all real $x$, and find it.

Answer :
It is known that, $|x|= \begin{cases}x, & \text { if } x \geq 0 \\ -x, & \text { if } x<0\end{cases}$
Therefore, when $x \geq 0, f(x)=|x|^3=x^3$
In this case, $f^{\prime}(x)=3 x^2$ and hence, $f^{\prime \prime}(x)=6 x$
When $x<0, f(x)=|x|^3=(-x)^3=-x^3$
In this case, $f^{\prime}(x)=-3 x^2$ and hence, $f^{\prime \prime}(x)=-6 x$
Thus, for $f(x)=|x|^3, f^{\prime \prime}(x)$ exists for all real $x a n d$ is given by,
$
f^{\prime \prime}(x)= \begin{cases}6 x, & \text { if } x \geq 0 \\ -6 x, & \text { if } x<0\end{cases}
$

Miscellaneous Exercise 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.

Given
$
\sin (A+B)=\sin A \cos B+\cos A \sin B
$

Consider A \& B are function of $x$

Differentiating both side w.r.t.x.
$
\begin{aligned}
& \frac{d(\sin (A+B))}{d x}=\frac{d(\sin A \cos B+\cos A \sin B)}{d x} \\
& \frac{d(\sin (A+B))}{d x}=\frac{d(\sin A \cdot \cos B)}{d x}+\frac{d(\cos A \cdot \sin B)}{d x} \\
& \cos (A+B) \cdot \frac{d(A+B)}{d x}=\frac{d(\sin A \cdot \cos B)}{d x}+\frac{d(\cos A \cdot \sin B)}{d x}
\end{aligned}
$

$\begin{aligned}
&\text { teachoo.com }\\
&\begin{aligned}
& \text { Using Product rule } \\
& \text { As }(u v)^{\prime}=u^{\prime} v+v^{\prime} u \\
& \cos (A+B) \cdot\left(\frac{d A}{d x}+\frac{d B}{d x}\right) \\
& =\left(\frac{d(\sin A)}{d x} \cdot \cos B+\frac{d(\cos B)}{d x} \sin \mathrm{A}\right)+\left(\frac{d(\cos A)}{d x} \cdot \sin B+\frac{d(\sin B)}{d x} \cdot \cos \mathrm{A}\right) \\
& =\cos A \cdot \frac{d A}{d x} \cdot \cos \mathrm{B}-\sin B \cdot \frac{d B}{d x} \sin A-\sin A \cdot \frac{d A}{d x} \cdot \sin B+\cos B \cdot \frac{d B}{d x} \cdot \cos \mathrm{A} \\
& =\cos A \cdot \frac{d A}{d x} \cdot \cos \mathrm{B}-\sin A \cdot \frac{d A}{d x} \sin \mathrm{B}-\sin B \cdot \frac{d B}{d x} \cdot \sin A+\cos \mathrm{B} \cdot \frac{d B}{d x} \cdot \cos \mathrm{A} \\
& =\frac{d A}{d x}(\cos A \cos B-\sin A \sin B)+\frac{d B}{d x}(-\sin B \sin A+\cos B \cos A)
\end{aligned}
\end{aligned}$
$
=(\cos A \cos B-\sin A \sin B)\left(\frac{d A}{d x}+\frac{d B}{d x}\right)
$

Thus,
$
\begin{aligned}
& \cos (A+B) \cdot\left(\frac{d A}{d x}+\frac{d B}{d x}\right)=(\cos A \cos B-\sin A \sin B)\left(\frac{d A}{d x}+\frac{d B}{d x}\right) \\
& \cos (\boldsymbol{A}+\boldsymbol{B})=\boldsymbol{\operatorname { c o s }} \boldsymbol{A} \boldsymbol{\operatorname { c o s }} \boldsymbol{B}-\sin \boldsymbol{A} \sin \boldsymbol{B}
\end{aligned}
$

Hence proved

Miscellaneous Exercise Question 20

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

Answer:

Let us consider the function $\quad f(x)=|x|+|x-1|$
$f$ is continuous everywhere but it is not differentiable at $x=0$ and $x=1$.

Miscellaneous ExerciseQuestion 21:

If $y=\left|\begin{array}{ccc}f(x) & g(x) & h(x) \\ l & m & n \\ a & b & c\end{array}\right|$, prove that $\frac{d y}{d x}=\left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{array}\right|$

Answer:
$
\begin{aligned}
& y=\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right| \\
& \Rightarrow y=(m c-n b) f(x)-(l c-n a) g(x)+(l b-m a) h(x) \\
& =(m c-n b) f^{\prime}(x)-(l c-n a) g^{\prime}(x)+(l b-m a) h^{\prime}(x) \\
& =\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
l & m & n \\
a & b & c
\end{array}\right| \\
&
\end{aligned}
$

Then, $\frac{d y}{d x}=\frac{d}{d x}[(m c-n b) f(x)]-\frac{d}{d x}[(l c-n a) g(x)]+\frac{d}{d x}[(l b-m a) h(x)]$

Thus, $\frac{d y}{d x}=\left|\begin{array}{ccc}f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\ l & m & n \\ a & b & c\end{array}\right|$

Miscellaneous ExerciseQuestion 22 :
If $y=e^{a \cos ^{-1} x},-1 \leq x \leq 1$, show that $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-a^2 y=0$

Answer :
It is given that, $y=e^{a \cos ^{-1} x}$
Taking logarithm on both the sides, we obtain
$\log y=a \cos ^{-1} x \log e$
$\log y=a \cos ^{-1} x$
Differentiating both sides with respect to $x$, we obtain
$
\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=a \times \frac{-1}{\sqrt{1-x^2}} \\
& \Rightarrow \frac{d y}{d x}=\frac{-a y}{\sqrt{1-x^2}}
\end{aligned}
$

By squaring both the sides, we obtain
$
\begin{aligned}
& \left(\frac{d y}{d x}\right)^2=\frac{a^2 y^2}{1-x^2} \\
& \Rightarrow\left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=a^2 y^2 \\
& \left(1-x^2\right)\left(\frac{d y}{d x}\right)^2=a^2 y^2
\end{aligned}
$

Again differentiating both sides with respect to $x$, we obtain

$
\begin{aligned}
& \left(\frac{d y}{d x}\right)^2 \frac{d}{d x}\left(1-x^2\right)+\left(1-x^2\right) \times \frac{d}{d x}\left[\left(\frac{d y}{d x}\right)^2\right]=a^2 \frac{d}{d x}\left(y^2\right) \\
& \Rightarrow\left(\frac{d y}{d x}\right)^2(-2 x)+\left(1-x^2\right) \times 2 \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}=a^2 \cdot 2 y \cdot \frac{d y}{d x} \\
& \Rightarrow\left(\frac{d y}{d x}\right)^2(-2 x)+\left(1-x^2\right) \times 2 \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}=a^2 \cdot 2 y \cdot \frac{d y}{d x} \\
& \Rightarrow-x \frac{d y}{d x}+\left(1-x^2\right) \frac{d^2 y}{d x^2}=a^2 \cdot y \quad\left[\frac{d y}{d x} \neq 0\right] \\
& \Rightarrow\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-a^2 y=0 \quad
\end{aligned}
$

Hence, proved.