If $f(x) = \begin{cases} e^x + a & \text{for } x < 0 \\ x - 3 & \text{for } x \geqslant 0 \end{cases}$ is differentiable at $x = 0$,then $a$ equals:

The function $g(x) = \begin{cases} x + b, & x < 0 \\ \cos x, & x \geqslant 0 \end{cases}$ can be made differentiable at $x = 0$.

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 $f(x) = \begin{cases} \frac{x-1}{2x^2-7x+5}, & \text{for } x \neq 1 \\ -\frac{1}{3}, & \text{for } x=1 \end{cases}$,then $f^{\prime}(1)$ is equal to:

$f(x) = |\log_e |x||$ is differentiable at

Let $S$ be the set of points where the function $f(x) = |2 - |x - 3||, x \in R,$ is not differentiable. Then $\sum_{x \in S} f(f(x))$ is equal to