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:

If $f(x) = \begin{cases} \frac{\tan(2p-7)x + \tan 3x}{x}, & x < 0 \\ p-q, & x=0 \\ q\left(\frac{\sqrt{x^2+x}-\sqrt{x}}{x^{3/2}}\right), & x > 0 \end{cases}$. If $f(x)$ is continuous at $x=0$,then $\frac{q}{p} = $

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 a function $f(x)$ defined by $f(x) = \begin{cases} ax^2 + bx + c, & x \leq -1 \\ 2x^2 + 4x + 1, & -1 < x < 1 \\ cx^2 + bx + a, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$,and $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$,then find $\lim_{x \rightarrow -2} f(x)$.

If $f(x) = \left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}$ is continuous at $x=0$,then $f(0)$ 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