For $a \neq 0$ and $b \neq 0$,if the real valued function $f(x) = \frac{\sqrt[5]{a(625+x)} - 5}{\sqrt[4]{625+bx} - 5}$ is continuous at $x = 0$,then $f(0) =$

If $f(x) = [x] - [\frac{x}{4}]$,$x \in R$,where $[x]$ denotes the greatest integer function,then

If the function given by $f(x) = \begin{cases} -2 \sin x & -\pi \leq x < -\pi/2 \\ a \sin x + b & -\pi/2 \leq x \leq \pi/2 \\ \cos x & \pi/2 < x \leq \pi \end{cases}$ is continuous in $[-\pi, \pi]$,then the value of $(3a + 2b)^3$ is

Define $f: R \to R$ by $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & x > 0 \end{cases}$ Then the value of $a$ so that $f$ is continuous at $x = 0$ is:

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-$2$: $f'(0) = 0$.

For the function $f(x) = \frac{\log_e(1 + x) - \log_e(1 - x)}{x}$ to be continuous at $x = 0$,the value of $f(0)$ should be