In order that the function $f(x) = (x + 1)^{1/x}$ is continuous at $x = 0$,$f(0)$ must be defined as

If $f(x) = \begin{cases} \frac{1-\sin^3 x}{3 \cos^2 x}, & x < \frac{\pi}{2} \\ \alpha, & x = \frac{\pi}{2} \\ \frac{\beta(1-\sin x)}{(\pi-2 x)^2}, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $\alpha \beta =$

If $f(x) = \begin{cases} kx + 1, & x \leq \frac{\pi}{2} \\ \sin x, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $k = $ . . . . . . .

If the function $f(x) = \begin{cases} x + a \sqrt{2} \sin x & \text{if } 0 \leq x \leq \frac{\pi}{4} \\ 2x \cot x + b & \text{if } \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous in $[0, \pi]$,then $a - b = $

Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & x>0 \\ 2, & x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & x < 0 \end{cases}$. If $f$ is continuous at $x=0$,find the value of $\alpha + \beta$.

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: