If $\alpha \in R - \{-1\}$ and $f(x) = |(|x| + \alpha)(|x| - 1)|$,then the number of points at which $f(x)$ is not differentiable is:

Let $f : (-1, 1) \to \mathbb{R}$ be a function defined by $f(x) = \min\{-|x|, -\sqrt{1 - x^2}\}$. If $K$ is the set of all points at which $f$ is not differentiable,then $K$ has exactly

If the function $f(x) = \begin{cases} \frac{1}{|x|} & , |x| \geq 2 \\ ax^2 + 2b & , |x| < 2 \end{cases}$ is differentiable on $\mathbb{R}$,then $48(a+b)$ is equal to . . . . . . .

Let the function $f: R \rightarrow R$ be defined by $f(x)=x-x^2+(x-1) \sin x$ and let $g: R \rightarrow R$ be an arbitrary function. Let $f g: R \rightarrow R$ be the product function defined by $(f g)(x)=f(x) g(x)$. Then which of the following statements is/are $TRUE$?
$(A)$ If $g$ is continuous at $x=1$,then $f g$ is differentiable at $x=1$
$(B)$ If $fg$ is differentiable at $x=1$,then $g$ is continuous at $x=1$
$(C)$ If $g$ is differentiable at $x=1$,then $f g$ is differentiable at $x=1$
$(D)$ If $fg$ is differentiable at $x=1$,then $g$ is differentiable at $x=1$

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 $f(x) = |x|$. Then which of the following is true?