Which one of the following functions is discontinuous at $x=1$?

If $f(x) = \cos \left[ \frac{\pi}{x} \right] \cos \left( \frac{\pi}{2} (x - 1) \right)$,then $f(x)$ is continuous at: (where $[x]$ is the greatest integer function of $x$)

The values of $a$ and $b$ such that the function $f(x) = \begin{cases} -2 \sin x, & -\pi \leq x \leq -\frac{\pi}{2} \\ a \sin x + b, & -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & \frac{\pi}{2} \leq x \leq \pi \end{cases}$ is continuous in $[-\pi, \pi]$,are

Let $f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases}$. If $f(x)$ is continuous for all $x$,then $k =$

For $a, b > 0$,let $f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0 \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{a} x \sqrt{x}}, & x > 0 \end{cases}$ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to

Is the function defined by $f(x) = x^{2} - \sin x + 5$ continuous at $x = \pi$?