The values of $A$ and $B$ such that the function $f(x) = \begin{cases} -2\sin x, & x \le -\frac{\pi}{2} \\ A\sin x + B, & -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & x \ge \frac{\pi}{2} \end{cases}$ is continuous everywhere are

If $f(x) = \frac{2 - \sqrt{x + 4}}{\sin 2x}, (x \neq 0),$ is a continuous function at $x = 0$,then $f(0)$ equals:

Let $f$ be a differentiable function on the open interval $(a, b)$. Which of the following statements must be true?
$I$. $f$ is continuous on the closed interval $[a, b]$
$II$. $f$ is bounded on the open interval $(a, b)$
$III$. If $a < a_1 < b_1 < b$,and $f(a_1) < 0 < f(b_1)$,then there is a number $c$ such that $a_1 < c < b_1$ and $f(c) = 0$

If $f(x) = \begin{cases} 2^{1/x}, & x \ne 0 \\ 3, & x = 0 \end{cases}$,then:

The number of points at which the function $f(x) = \frac{\sqrt{11+|x|-6\sqrt{2+|x|}}}{6-2\sqrt{2+|x|}}$ is discontinuous in $(-\infty, \infty)$ is

Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.