If $f(x) = \begin{cases} 3 + x; & x \geqslant 0 \\ 2 - 3x; & x < 0 \end{cases}$,then $\lim_{x \to 0} f(f(x))$ is equal to -

Let $a, b \in \mathbb{R}$ $(a \neq 0)$. If the function $f$ is defined as $f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \leq x < 1 \\ a, & 1 \leq x < \sqrt{2} \\ \frac{2b^2-4b}{x}, & \sqrt{2} \leq x < \infty \end{cases}$ is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is

The function $f(t) = \frac{1}{t^2 + t - 2}$,where $t = \frac{1}{x - 1}$,is discontinuous at

Consider $f(x) = \begin{cases} \frac{x^2}{|x|}, & x \ne 0 \\ 0, & x = 0 \end{cases}$

If a function $f$ defined by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x>0 \end{cases}$ is continuous at $x=0$,then $a=$

Let $f(x) = \frac{1-\tan x}{4x-\pi}$ for $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{1}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is