If $f(x) = \begin{cases} x, & \text{if } x \text{ is irrational} \\ 0, & \text{if } x \text{ is rational} \end{cases}$,then $f$ is

Let $f(x) = \begin{cases} x, & x \in \mathbb{Q} \\ 1 - x, & x \notin \mathbb{Q} \end{cases}$. Then at $x = \frac{1}{2}$,$f(x)$ is:

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

For the function $f(x) = \frac{\log_e(1 + x) - \log_e(1 - x)}{x}$ to be continuous at $x = 0$,the value of $f(0)$ should be

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

If $f(x) = \begin{cases} \frac{x - 1}{2}, & 0 \leqslant x < 1 \\ 1/2, & 1 \leqslant x < 2 \end{cases}$ and $g(x) = (2x + 1)(x - k) + 3$ for $0 \leqslant x < \infty$,then $g(f(x))$ will be continuous at $x = 1$ if $k$ is equal to: