The value of $k$ which makes $f(x) = \begin{cases} \sin \frac{1}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ continuous at $x = 0$ is

If $f(x) = \begin{cases} \frac{3 \sin(\pi x)}{5x} & x \neq 0 \\ 2K & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $K$ 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 $f(x) = x^3$, $x \in [-1, 1]$. Then which of the following are correct?

For the function $f(x) = \frac{1}{x + 2^{\frac{1}{x - 2}}}$,$x \neq 2$,which of the following holds?

Let $f(x) = \min \{1, 1 + x \sin x \}$ for $0 \leq x \leq 2\pi$. If $m$ is the number of points where $f$ is not differentiable and $n$ is the number of points where $f$ is not continuous,then the ordered pair $(m, n)$ is equal to