Let $g: R \rightarrow R$ a differentiable function with $g(0)=0, g'(0)=0$ and $g'(1) \neq 0$. Let $f(x)=\left\{\begin{array}{r}\frac{x}{|x|} g(x), x \neq 0 \\ 0, x=0\end{array}\right.$ and $h(x)=e^{|x|}$ for all $x \in R$. Let foh $(x)$ denote $f(h(x))$ and (hof) $(x)$ denoteh $(f(x))$.Then which of the following is (are) true?

(a) f is differentiable at x = 0

(b) h is differentiable at x = 0

(c) foh is differentiable at x = 0

(d) hof is differentiable at x = 0

(a), (d)

We have,

$h(x)=e^{|x|}=\left\{\begin{array}{l} e^{-x}, x<0 \\ e^x, x \geq 0 \end{array}\right.$

∴ (LHD of h(x) at x = 0) = $\lim\limits_{x \rightarrow 0^{-}} \frac{h(x)-h(0)}{x-0}$

$=-\lim\limits_{x \rightarrow 0^{-}} \frac{e^{-x}-1}{-x}$

$=-\lim\limits_{x \rightarrow 0^{-}} \frac{e^{-x}-1}{-x}=-1$

and, (RHD of h(x) at x = 0) = $\lim\limits_{x \rightarrow 0^{+}} \frac{h(x)-h(0)}{x-0}$

$=\lim\limits_{x \rightarrow 0^{+}} \frac{e^x-1}{x}=1$

∴ (LHD at f(x) at x = 0) $\neq$ (RHD of h(x) at x = 0)

So, h(x) is not differentiable at x = 0 and hence option (b) is not true.

We have,

$f(x)=\left\{\begin{array}{c} \frac{x}{|x|} g(x), x \neq 0 \\ 0, x=0 \end{array} \text { or, } f(x)=\left\{\begin{array}{r} g(x), x>0 \\ 0, x=0 \\ -g(x), x<0 \end{array}\right.\right.$

∴  (LHD of f(x) at x = 0) = $\lim\limits_{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$

$=\lim\limits_{x \rightarrow 0^{-}} \frac{-g(x)}{x}$              [∵ f(0) = 0]

$=-\lim\limits_{x \rightarrow 0^{-}} \frac{g(x)-g(0)}{x-0}$              [∵ g(0) = 0]

$=-g'(0)=0$

∴  (LHD of f(x) at x = 0) = $\lim\limits_{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$=\lim\limits_{x \rightarrow 0^{+}} \frac{f(x)}{x}$

$=\lim\limits_{x \rightarrow 0^{+}} \frac{g(x)-g(0)}{x-0}=g'(0)=0$

∴ (LHD of f(x) at x = 0) = (RHD of f(x) at x = 0)

So, f(x) is differentiable at x = 0.

So, option (a) is true

We have,

$f(x)=\left\{\begin{array}{r}\frac{x}{|x|} g(x), x \neq 0 \\ 0, x=0\end{array}\right.$

$\Rightarrow \quad f(x)=\left\{\begin{array}{r}-g(x), x<0 \\ 0, x=0 \\ g(x), x>0\end{array}\right.$

and, $h(x)=e^{|x|}>0$  for all  $x \in R$

∴  $f(h(x))=g\left(e^{|x|}\right)$  for all  $x \in R$

$\Rightarrow f(h(x))=\left\{\begin{array}{c}g\left(e^{-x}\right), x<0 \\ g(1), x=0 \\ g\left(e^x\right), x>0\end{array}\right.$

(LHD of foh(x) at x = 0) = $\lim\limits_{x \rightarrow 0^{-}} \frac{f(h(x))-f(h(0))}{x-0}$

$=\lim\limits_{x \rightarrow 0^{-}} \frac{g\left(e^{-x}\right)-g(1)}{x}$

$=\lim\limits_{x \rightarrow 0^{-}} \frac{g\left(e^{-x}\right)-g(1)}{e^{-x}-1} \times \frac{e^{-x}-1}{x}$

$=\lim\limits_{x \rightarrow 0^{-}} \frac{g\left(e^{-x}\right)-g(1)}{e^{-x}-1} \times \lim\limits_{x \rightarrow 0} \frac{e^{-x}-1}{x}$

$=g'(1) \times(-1)=-g'(1)$

(RHD of foh(x) at x = 0) = $\lim\limits_{x \rightarrow 0^{+}} \frac{f(h(x))-f(h(0))}{x-0}$

$=\lim\limits_{x \rightarrow 0^{+}} \frac{g\left(e^x\right)-g(1)}{e^x-1} \times \frac{e^x-1}{x}$

$=g'(1) \times 1=g'(1)$

(LHD of foh(x) at x = 0) ≠ (RHD of foh(x) at x = 0)

So, foh is not differentiable at x = 0. Hence, option (c) is not true.

Again,

$h(x)=e^{|x|}$  and  $f(x)=\left\{\begin{array}{r} \frac{x}{|x|} g(x), x \neq 0 \\ 0, x=0 \end{array}\right. $

$\Rightarrow h o f(x)=h(f(x))=\left\{\begin{array}{l} e^{|f(x)|}, x \neq 0 \\ e^0=1, x=0 \end{array}\right. $

$\Rightarrow h o f(x)=\left\{\begin{array}{r} e^{|f(x)|}, x \neq 0 \\ 1, x=0 \end{array}\right.$

∴  (LHD of hof at x = 0) 

$=\lim\limits_{x \rightarrow 0^{-}} \frac{h o f(x)-h o f(0)}{x-0}$

$=\lim\limits_{x \rightarrow 0^{-}} \frac{e^{|g(x)|}-1}{x}$

$=\lim\limits_{x \rightarrow 0^{-}} \frac{e^{|g(x)|}-1}{|g(x)|} \times \frac{|g(x)|}{x}$

$=\lim\limits_{x \rightarrow 0^{-}} \frac{e^{|g(x)|}-1}{|g(x)|} \times\left|\frac{g(x)-g(0)}{x-0}\right| \times \frac{|x-0|}{x}$

$=1 \times g'(0) \times(-1)=0$

(RHD of hof at x = 0)

$=\lim\limits_{x \rightarrow 0^{+}} \frac{h o f(x)-h o f(0)}{x-0}$

$=\lim\limits_{x \rightarrow 0^{+}} \frac{h(f(x))-h(f(0))}{x}$

$=\lim\limits_{x \rightarrow 0^{+}} \frac{e^{|g(x)|}-1}{|g(x)|} \times \frac{|g(x)|}{x}$

$=\lim\limits_{x \rightarrow 0^{+}} \frac{e^{|g(x)|}-1}{|g(x)|} \times\left|\frac{g(x)-g(0)}{x-0}\right| \times \frac{|x|}{x}$

$=1 \times\left|g'(0)\right| \times 1=0$

So, hof is differentiable at x = 0. Consequently, option (d) is true.