Given $f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & \text{if } x < 0 \\ a, & \text{if } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{if } x > 0 \end{cases}$
If $f(x)$ is continuous at $x=0$,then the value of $a$ 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-$2$: $f'(0) = 0$.

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} x-1, & \text{for } x \leq 1 \\ 2-x^2, & \text{for } 1 < x \leq 3 \\ x-10, & \text{for } 3 < x < 5 \\ 2x, & \text{for } x \geq 5 \end{cases}$,then the set of points of discontinuity of $f$ is

Let $[x]$ denote the greatest integer less than or equal to $x$ and $f(x) = [\tan^2 x]$. Which of the following is true?

Let $f:[-1,2] \rightarrow R$ be defined by $f(x)=[x^2-3]$ where $[\cdot]$ denotes the greatest integer function. The number of points of discontinuity for the function $f$ in the interval $(-1,2)$ is:

If $f(x)$ is continuous at $x = 0$, where $f(x) = \frac{8^x - 2^x}{k^x - 1}$ for $x \neq 0$ and $f(0) = 2$, then the value of $k$ is ...