The function $f(x) = x^2 \sin \frac{1}{x}$ for $x \ne 0$ and $f(0) = 0$ at $x = 0$ is:

Let $f(x) = \begin{cases} x^2 \left| \cos \frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x=0 \end{cases}$,$x \in \mathbb{R}$,then $f$ is

The set of points where $f(x) = \frac{x}{4+|x|}$ is differentiable is

Let $g: R \rightarrow R$ be a differentiable function with $g(0)=0, g^{\prime}(0)=0$ and $g^{\prime}(1) \neq 0$. Let $f(x)=\begin{cases} \frac{x}{|x|} g(x), & x \neq 0 \\ 0, & x=0 \end{cases}$ and $h(x)=e^{|x|}$ for all $x \in R$. Let $(f \circ h)(x)$ denote $f(h(x))$ and $(h \circ f)(x)$ denote $h(f(x))$. Then which of the following is (are) true?
$(A)$ $f$ is differentiable at $x=0$
$(B)$ $h$ is differentiable at $x=0$
$(C)$ $f \circ h$ is differentiable at $x=0$
$(D)$ $h \circ f$ is differentiable at $x=0$

Let $f$ be defined on $D = R - \{-1, 1\}$ by $f(x) = \frac{|x|}{1 - |x|}$,then

The function given by $y = ||x| - 1|$ is differentiable for all real numbers except the points