Let $f(x) = \begin{cases} x^p \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Then $f(x)$ is continuous but not differentiable at $x = 0$ if:

Find the values of $k$ so that the function $f$ is continuous at the indicated point.
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$

If $f(x) = \frac{e^{x^2} - \cos x}{x^2}$ for $x \neq 0$ is continuous at $x = 0$,then $f(0) = $.

If the function $f(x)$ is continuous on its domain $[-2, 2]$,where $f(x) = \begin{cases} \frac{\sin ax}{x} + 3, & -2 \leq x < 0 \\ x + 5, & 0 \leq x \leq 1 \\ \sqrt{x^2 + 8} - b, & 1 < x \leq 2 \end{cases}$,then $7a + b + 1$ is equal to:

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

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{2 \sin x-\sin 2 x}{2 x \cos x}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$, then the value of $a$ so that $f$ is continuous at $x=0$ is