If a function $f(x) = \begin{cases} \frac{\tan (\alpha + 1)x + \tan 2x}{x}, & \text{if } x > 0 \\ \beta, & \text{at } x = 0 \\ \frac{\sin 3x - \tan 3x}{x^{3}}, & \text{if } x < 0 \end{cases}$ is continuous at $x = 0$,then $|\alpha| + |\beta| =$

The values of $a$ and $b$,so that the function $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous for $0 \leq x \leq \pi$,are respectively given by

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:

If $f(x) = \operatorname{sgn} \left( 3\cos x - \frac{a}{3} \right)$ is continuous for all $x$,then the least positive integral value of $a$ is - (where $\operatorname{sgn}(x)$ denotes the signum function of $x$)

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

If the function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}$ is right continuous at $x = 2$, then $k =$