If the function $f(x) = \frac{\log(1 + ax) - \log(1 - bx)}{x}$,$x \neq 0$ is continuous at $x = 0$,then $f(0) = $ . . . . . .

Let $f: R \rightarrow R$ be defined by $f(x)=\begin{cases} \alpha+\frac{\sin [x]}{x}, & \text{if } x>0 \\ 2, & \text{if } x=0 \\ \beta+\left[\frac{\sin x-x}{x^3}\right], & \text{if } x < 0 \end{cases}$ where $[x]$ denotes the greatest integer function. If $f$ is continuous at $x=0$,then $\beta-\alpha$ is equal to

If $f(x) = \cos \left[ \frac{\pi}{x} \right] \cos \left( \frac{\pi}{2} (x - 1) \right)$,then $f(x)$ is continuous at: (where $[x]$ is the greatest integer function of $x$)

The number of points of discontinuity of $f(x)$ where $f(x) = | | |x + [x]| - 3[x] | - 5[x] |$ on $[-2, 2]$ is (where $[ \cdot ]$ denotes the greatest integer function).

Let $f$ and $g$ be real-valued functions. If $\lim _{x \rightarrow 0} \frac{2 f(x)-g(x)}{[f(x)+7]^{2 / 3}}=\frac{7}{4}$, $\lim _{x \rightarrow 0} f(x)=1$ and $\lim _{x \rightarrow 0} g(x)=\alpha$, then $h(x)= \begin{cases} \sin (\alpha x), & 0 \leq x \leq \frac{\pi}{10} \\ \cos (2 \alpha x), & \frac{\pi}{10} < x \leq \frac{\pi}{5} \end{cases}$ is:

If $f(x)$ is continuous on its domain $[-2,2]$,where $f(x) = \begin{cases} \frac{\sin ax}{x} + 3, & -2 \leq x < 0 \\ 2x + 7, & 0 \leq x \leq 1 \\ \sqrt{x^2+8} - b, & 1 < x \leq 2 \end{cases}$ then the value of $2a + 3b$ is