The value of $k$ so that the function $f(x) = \begin{cases} k(2x - x^2), & x < 0 \\ \cos x, & x \ge 0 \end{cases}$ is continuous at $x = 0$,is

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 points of discontinuity of $f,$ where $f(x) = \begin{cases} \frac{\sin x}{x}, & \text{if } x < 0 \\ x + 1, & \text{if } x \ge 0 \end{cases}$

If $\begin{aligned} f(x) &= \frac{4 \sin \pi x}{5 x} \text{ for } x \neq 0 \\ &= 2k \text{ for } x = 0 \end{aligned}$ is continuous at $x = 0$,then the value of $k$ is

Let $f(x) = \begin{cases} \frac{\tan^2 \{x\}}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{\{x\} \cot \{x\}} & \text{for } x < 0 \end{cases}$ where $[x]$ is the greatest integer function and $\{x\}$ is the fractional part function of $x$,then:

If $f(x) = \begin{cases} x^2 - 1, & \text{if } x \ge 2 \\ x + 1, & \text{if } x < 2 \end{cases}$, then $\lim_{x \to 1} f(x) + \lim_{x \to 2} f(x) =$