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

Let $\alpha, \beta \in R$ be such that the function $f(x) = \begin{cases} 2 \alpha (x^2 - 2) + 2 \beta x, & x < 1 \\ (\alpha + 3) x + (\alpha - \beta), & x \ge 1 \end{cases}$ is differentiable at all $x \in R$. Then $34(\alpha + \beta)$ is equal to

The number of points,at which the function $f(x) = |2x+1| - 3|x+2| + |x^2+x-2|$,$x \in R$ is not differentiable,is ............

If $f$ is a real-valued differentiable function satisfying $|f(x) - f(y)| \le (x - y)^2$ for all $x, y \in R$ and $f(0) = 0$,then $f(1)$ is equal to:

If $f(x) = \begin{cases} ax^2 + bx - \frac{13}{8}, & x \leq 1 \\ 3x - 3, & 1 < x \leq 2 \\ bx^3 + 1, & x > 2 \end{cases}$ is differentiable $\forall x \in R$,then $a - b =$

If $f(x) = \begin{cases} x^2 & \text{if } x \leqslant x_0 \\ ax + b & \text{if } x > x_0 \end{cases}$ is derivable for all $x \in \mathbb{R}$,then the values of $a$ and $b$ are respectively: