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

If the function $f(x) = \begin{cases} 1 + \sin \frac{\pi x}{2}, & \text{for } -\infty < x \le 1 \\ ax + b, & \text{for } 1 < x < 3 \\ 6 \tan \frac{x\pi}{12}, & \text{for } 3 \le x < 6 \end{cases}$ is continuous in the interval $(-\infty, 6)$,then the values of $a$ and $b$ are respectively

$f(x)= \begin{cases}(1+3x)^{\frac{4}{x}}, & \text{if } x \neq 0 \\ a, & \text{if } x=0 \end{cases}$
If $f$ is continuous at $x=0$,then $\log a=$

If $f(x) = \frac{e^{x^2} - \cos x}{x^2}$ for $x \neq 0$ is continuous at $x = 0$,then $f(0) = $.

The number of points of discontinuity of the function $f(x) = [\frac{x^2}{2}] - [\sqrt{x}]$ for $x \in [0, 4]$,where $[\cdot]$ denotes the greatest integer function,is . . . . . . .

If $f(x)$ is continuous at $x = 0$, where $f(x) = \frac{8^x - 2^x}{k^x - 1}$ for $x \neq 0$ and $f(0) = 2$, then the value of $k$ is ...