If the function $f(x) = \frac{1-\sin 2x + \cos 2x}{1+\sin 2x + \cos 2x}$ for $x \neq \frac{\pi}{2}$ and $f(x) = k$ for $x = \frac{\pi}{2}$ is continuous at $x = \frac{\pi}{2}$,then $k = $

Let $\alpha \in R$ be such that the function $f(x) = \begin{cases} \frac{\cos^{-1}(1-\{x\}^2) \sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$,where $\{x\} = x - [x]$ and $[x]$ is the greatest integer less than or equal to $x$. Then:

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

Find all the points of discontinuity of the function $f$ defined by
$f(x) = \begin{cases} x + 2, & \text{if } x < 1 \\ 0, & \text{if } x = 1 \\ x - 2, & \text{if } x > 1 \end{cases}$

Let $f(x) = [2x^2 + 1]$ and $g(x) = \begin{cases} 2x - 3, & x < 0 \\ 2x + 3, & x \geq 0 \end{cases}$,where $[t]$ denotes the greatest integer function $\leq t$. Then,in the open interval $(-1, 1)$,the number of points where $f(g(x))$ is discontinuous is equal to:

Show that the function $f$ given by $f(x) = \begin{cases} x^3 + 3, & \text{if } x \neq 0 \\ 1, & \text{if } x = 0 \end{cases}$ is not continuous at $x = 0$.