Let a function $g:[0,4] \rightarrow R$ be defined as
$g(x) = \begin{cases} \max_{0 \leq t \leq x} \{t^3 - 6t^2 + 9t - 3\} & , 0 \leq x \leq 3 \\ 4 - x & , 3 < x \leq 4 \end{cases}$
Then the number of points in the interval $(0,4)$ where $g(x)$ is $NOT$ differentiable is $.....$

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

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 $a - b =$

Let the functions $f, g$ and $h$ be defined as follows:
$f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$g(x) = \begin{cases} x^2 \sin \left( \frac{1}{x} \right) & \text{for } -1 \le x \le 1, x \ne 0 \\ 0 & \text{for } x = 0 \end{cases}$
$h(x) = |x|^3$ for $-1 \le x \le 1$.
Which of these functions are differentiable at $x = 0$?

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:

Consider the function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=e^{-\left|\log _e x\right|}$. If $m$ and $n$ are respectively the number of points at which $f$ is not continuous and $f$ is not differentiable,then $m+n$ is