If $f(x) = \begin{cases} x + 1, & x < 2 \\ 2x - 1, & x \ge 2 \end{cases}$,then $f'(2)$ equals

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

Let $f : R \to R$ be a function such that $|f(x)| \leq x^2$,for all $x \in R$. Then,at $x = 0$,$f$ is

Suppose $f(x) = \begin{cases} [\cos \pi x], & x \leq 1 \\ 2\{x\} - 1, & x > 1 \end{cases}$,where $[\cdot]$ and $\{\cdot\}$ denote the greatest integer function and the fractional part of $x$ respectively,then at $x = 1$:

Let $f: R \rightarrow R$ be a function defined by $f(x) = \max \{x, x^2\}$. Let $S$ denote the set of all points in $R$ where $f$ is not differentiable. Then $S$ is:

Let $f(x) = \begin{cases} \max \{|x|, x^2\}, & |x| \le 2 \\ 8 - 2|x|, & 2 < |x| \le 4 \end{cases}$. Let $S$ be the set of points in the interval $(-4, 4)$ at which $f$ is not differentiable. Then $S$