If $f(x)$ is a differentiable function such that $f: R \to R$ and $f\left( \frac{1}{n} \right) = 0$ for all $n \ge 1, n \in I$,then:

The set of all those points,where the function $f(x) = \frac{x}{1 + |x|}$ is differentiable,is

$A$ function $f$ is defined as follows:
$f(x) = \begin{cases} \sin x & \text{if } x \le c \\ ax + b & \text{if } x > c \end{cases}$
where $c$ is a known quantity. If $f$ is derivable at $x = c$,then the values of $a$ and $b$ are . . . . . . and . . . . . . respectively.

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 $[x]$ denote the greatest integer less than or equal to $x$. Then the number of points where the function $y = [x] + |1 - x|$ for $-1 \leq x \leq 3$ is not differentiable,is

The set of all values of $x$ for which $f(x) = ||x| - 1|$ is differentiable is