The function $f(x) = \frac{\log(1+ax) - \log(1-bx)}{x}$ is not defined at $x=0$. The value which should be assigned to $f$ at $x=0$ so that it is continuous at $x=0$ is

Find the values of $a$ and $b$ such that the function defined by $f(x) = \begin{cases} 5, & \text{if } x \le 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \ge 10 \end{cases}$ is a continuous function.

If the function $f(x) = \begin{cases} -2 \sin x, & x \leq \frac{-\pi}{2} \\ A \sin x+B, & \frac{-\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & x \geq \frac{\pi}{2} \end{cases}$ is continuous everywhere,then the values of $A$ and $B$ are respectively

Let $f(x) = \begin{cases} x^3+8; x < 0 \\ x^2-4; x \ge 0 \end{cases}$ and $g(x) = \begin{cases} (x-8)^{1/3}; x < 0 \\ (x+4)^{1/2}; x \ge 0 \end{cases}$. Then the number of points, where the function $g \circ f$ is discontinuous, is ————

Let $f:R \rightarrow R$ be defined as $f(x) = \begin{cases} 0, & x \text{ is irrational} \\ \sin |x|, & x \text{ is rational} \end{cases}$. Then, which of the following is true?

If $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $k$ is equal to . . . . . . .