If $[.]$ denotes the greatest integer function,then $f(x) = [x]^2 - [x^2]$ is discontinuous at

If $f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x-2)^2}, x \neq 2 \\ k, x = 2 \end{cases}$ is continuous at $x = 2$, then $k = \rule{1cm}{0.15mm}$

If $f(x)$,defined below,is continuous at $x = 4$,then find the values of $a$ and $b$ given that $f(x)$ is continuous on the interval $[0, 8]$.
$f(x) = \begin{cases} x^2 + ax + b, & 0 \leq x < 2 \\ 3x + 2, & 2 \leq x \leq 4 \\ 2ax + 5b, & 4 < x \leq 8 \end{cases}$

If $f(x) = \begin{cases} x+a, & x \leq 0 \\ |x-4|, & x > 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ (x-4)^2+b, & x \geq 0 \end{cases}$ are continuous on $\mathbb{R}$,then $(g \circ f)(2) + (f \circ g)(-2)$ is equal to.

The values of $a$ and $b$,so that the function $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous for $0 \leq x \leq \pi$,are respectively given by

Let $f$ be defined by $f(x) = \begin{cases} \frac{\tan x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$.
Statement-$1$: $x = 0$ is a point of local minima for $f$.
Statement-$2$: $f'(0) = 0$.