If the function $f(x) = \begin{cases} x + a^2\sqrt{2} \sin x, & 0 \le x < \pi/4 \\ x \cot x + b, & \pi/4 \le x < \pi/2 \\ b \sin 2x - a \cos 2x, & \pi/2 \le x \le \pi \end{cases}$ is continuous in the interval $[0, \pi]$,then the values of $(a, b)$ are:

Let $f(x) = \begin{cases} x^2 - a & x < 3 \\ b\sqrt{x - 2} + a & 3 \leqslant x < 6 \\ 2x + b & x \geqslant 6 \end{cases}$. If $f(x)$ is continuous $\forall x \in R$,then find the value of $\frac{f(1) - f(3)}{4}$.

Statement-$1$: The equation $x \log x = 2 - x$ is satisfied by at least one value of $x$ lying between $1$ and $2$.
Statement-$2$: The function $f(x) = x \log x$ is an increasing function in $[1, 2]$ and $g(x) = 2 - x$ is a decreasing function in $[1, 2]$,and the graphs represented by these functions intersect at a point in $[1, 2]$.

Let $f(x) = \begin{cases} (1+ax)^{1/x} & , x < 0 \\ 1+b & , x = 0 \\ \frac{(x+4)^{1/2}-2}{(x+c)^{1/3}-2} & , x > 0 \end{cases}$ be continuous at $x=0$. Then $e^2bc$ is equal to

If $f(x) = \left[\tan \left(\frac{\pi}{4} + x\right)\right]^{\frac{1}{x}}$ for $x \neq 0$ and $f(x) = k$ for $x = 0$ is continuous at $x = 0$,then $k = \dots$

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: