⇒Derivative

Correct Option is (A)

$\mathrm{f}(x)$ can be written as

$\begin{array}{rl}& \mathrm{f}(x)=\{\begin{array}{c}\frac{x}{1-x},x\le 0\\ \frac{x}{1+x},x>0\end{array}\\ \therefore & {\mathrm{f}}^{\prime }(x)=\{\begin{array}{ll}\frac{(1-x)+x}{(1+x{)}^2},& x\le 0\\ \frac{(1+x)-x}{(1+x{)}^2},& x>0\end{array}\\ \therefore & {\mathrm{f}}^{\prime }(x)=\frac{1}{(1+x{)}^2}\mathrm{\forall }x\in (-\mathrm{\infty },\mathrm{\infty })\\ \therefore & \text{ Derivative of }\mathrm{f}(x)\text{ exists }\mathrm{\forall }x\in (-\mathrm{\infty },\mathrm{\infty })\end{array}$

Correct Option is (C)

$\begin{array}{rl}& y=\frac{\log \tan x}{\log \sin x}\\ & \frac{dy}{dx}=\frac{(\log \sin x)\left(\frac{1}{\tan x}\right)\cdot {\sec }^{2}x-(\log \tan x)\left(\frac{1}{\sin x}\right)(\cos x)}{(\log \sin x{)}^2}\\ & \text{ At }x=\frac{\pi }{4}\\ & {\left(\frac{dy}{dx}\right)}_{x=\frac{\pi }{4}}=\frac{\log \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{1}\right)(\sqrt{2}{)}^2-(\log 1)\left(\frac{\sqrt{2}}{1}\right)\left(\frac{1}{\sqrt{2}}\right)}{{[\log \left(\frac{1}{\sqrt{2}}\right)]}^2}\\ & =\frac{-2\times \frac{1}{2}(\log 2)-0}{\frac{1}{4}(\log 2{)}^2}\dots [\because \log 1=0]\\ & =\frac{-4}{\log 2}\end{array}$

Correct Option is (B)

Given: ${\mathrm{f}}^{\prime }(x)=\mathrm{f}(x)$ for all $x\in \mathrm{R}$

$\Rightarrow \frac{{\mathrm{f}}^{\prime }(x)}{\mathrm{f}(x)}=1$

Integrating on both sides, we get

$\begin{array}{rl}& \log |\mathrm{f}(x)|=x+\mathrm{c}\\ & \Rightarrow \mathrm{f}(x)={\mathrm{e}}^{x+\mathrm{c}}\\ & \Rightarrow \mathrm{f}(x)={\mathrm{e}}^x\cdot {\mathrm{e}}^{\mathrm{c}}\\ & \Rightarrow \mathrm{f}(x)={\mathrm{e}}^x\cdot {\mathrm{c}}_1\dots \text{ (i) }[{\mathrm{e}}^{\mathrm{c}}={\mathrm{c}}_1]\end{array}$

As $\mathrm{f}(1)=2$

$\begin{array}{rl}& \therefore c_1\cdot e=2\\ & \Rightarrow c_1=\frac{2}{e}\end{array}$

Equation (i) becomes

$f(x)={\mathrm{e}}^x\cdot \frac{2}{\mathrm{e}}$

Now, $\mathrm{h}(x)=\mathrm{f}(\mathrm{f}(x))$

$\begin{array}{rl}\therefore & h^{\prime }(x)=f^{\prime }(f(x))\times f^{\prime }(x)\\ \therefore & h^{\prime }(1)=f^{\prime }(f(1))\times f^{\prime }(1)\\ & \Rightarrow h^{\prime }(1)=f^{\prime }(2)\times f^{\prime }(1)\\ & \Rightarrow h^{\prime }(1)=e^2\times \frac{2}{e}\times 2\\ & \Rightarrow h^{\prime }(1)=4e\end{array}$

Correct Option is (B)

$\begin{array}{rl}\mathrm{h}(x)& =\mathrm{f}(\mathrm{g}(x))\\ & =\mathrm{f}({\sin }^{-1}x)\\ \therefore \mathrm{h}(x)& ={\mathrm{e}}^{{\sin }^{-1}x}\end{array}$

Differentiating w.r.t. $x$, we get

$\begin{array}{rl}& \begin{array}{rl}{\mathrm{h}}^{\prime }(x)& ={\mathrm{e}}^{{\sin }^{-1}x}\cdot \frac{\mathrm{d}}{\mathrm{d}x}({\sin }^{-1}x)\\ & ={\mathrm{e}}^{{\sin }^{-1}x}\cdot \frac{1}{\sqrt{1-x^2}}\end{array}\\ & \text{ Now, }\frac{{\mathrm{h}}^{\prime }(x)}{\mathrm{h}(x)}=\frac{{\mathrm{e}}^{{\sin }^{-1}x}\cdot \frac{1}{\sqrt{1-x^2}}}{{\mathrm{e}}^{{\sin }^{-1}x}}=\frac{1}{\sqrt{1-x^2}}\end{array}$

Correct Option is (D)

$\begin{array}{rl}& \text{ Let }y=\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x){)}^2\\ \therefore & \frac{dy}{dx}=f^{\prime }(f(f(x)))\cdot {\mathrm{f}}^{\prime }(\mathrm{f}(x))\cdot {\mathrm{f}}^{\prime }(x)+2\mathrm{f}(x){\mathrm{f}}^{\prime }(x)\\ & {\frac{dy}{dx}|}_{x=1}=f^{\prime }(f(f(1)))\cdot f^{\prime }(f(1))\cdot f^{\prime }(1)+2f(1)f^{\prime }(1)\\ & =3\cdot 3\cdot 3+2\cdot 1\cdot 3\\ & =33\end{array}$