If $f(x) = \begin{cases} x + 2, & -1 < x < 3 \\ 5, & x = 3 \\ 8 - x, & x > 3 \end{cases}$,then at $x = 3$,$f'(x) = $

Which one of the following functions is continuous everywhere in its domain but has at least one point where it is not differentiable?

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} x^{5} \sin \left(\frac{1}{x}\right) + 5x^{2} & , x < 0 \\ 0 & , x = 0 \\ x^{5} \cos \left(\frac{1}{x}\right) + \lambda x^{2} & , x > 0 \end{cases}$. The value of $\lambda$ for which $f''(0)$ exists is:

If $f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geq 1 \\ ax^2 + b, & -1 < x < 1 \end{cases}$ is differentiable $\forall x \in \mathbb{R}$,then one of the values of $a$ and $b$ is-

If $x + |y| = 2y,$ then $y$ as a function of $x,$ at $x = 0$ is

Let $R$ denote the set of all real numbers. Define the function $f: R \rightarrow R$ by $f(x) = \begin{cases} 2-2x^2-x^2 \sin \frac{1}{x} & \text{if } x \neq 0 \\ 2 & \text{if } x=0 \end{cases}$. Then which one of the following statements is True?