JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Continuity and Differentiability

(A) For $x\ne 0$, $f(x)=\frac{x}{|x|}g(x)=\pm g(x)$. At $x=0$:

For $h>0$: ${\lim }_{h\to 0^{+}}\frac{g(h)}{h}=g^{\prime }(0)=0$.
For $h<0$: ${\lim }_{h\to 0^{-}}\frac{-g(h)}{h}=-g^{\prime }(0)=0$.
So $f^{\prime }(0)=0$ exists. ✓

(B) $h(x)=e^{|x|}$. At $x=0$:

Not equal, so $h$ is not differentiable at $x=0$. ✗

(C) $(f\circ h)(x)=f(e^{|x|})=\frac{e^{|x|}}{e^{|x|}}g(e^{|x|})=g(e^{|x|})$.
At $x=0$: $(f\circ h)(0)=g(1)$.

From the right: ${\lim }_{x\to 0^{+}}\frac{g(e^x)-g(1)}{x}$. From the left: ${\lim }_{x\to 0^{-}}\frac{g(e^{-x})-g(1)}{x}$.
These limits are not equal in general since $g^{\prime }(1)\ne 0$. ✗

(D) $(h\circ f)(x)=h(f(x))=e^{|f(x)|}$. Since $f(0)=0$ and $f$ is continuous at 0, and $e^{|x|}$ is differentiable at $x=0$ from the right and left with the same derivative (actually, $e^{|x|}$ has derivative 1 from the right and -1 from the left at 0, but composed with $f$ which has derivative 0 at 0)...

Actually, $(h\circ f)(x)=e^{|f(x)|}$. Near $x=0$, $f(x)\approx 0$, so $(h\circ f)(x)\approx e^0=1$. By chain rule considerations and the fact that $f^{\prime }(0)=0$, $(h\circ f)$ is differentiable at $x=0$. ✓