If the function defined by $f(x) = \frac{\log (1+x)^{1+x}}{x^2} - \frac{1}{x}, x \neq 0$ is continuous at $x=0$,then $6 f(0)$ is equal to

Let $f(x) = \begin{cases} \frac{x^3 + x^2 - 16x + 20}{(x - 2)^2}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases}$. If $f(x)$ is continuous for all $x$,then $k =$

Find the values of $k$ so that the function $f$ is continuous at the indicated point.
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$

If the function $f(x)$ is continuous in $0 \leq x \leq \pi$,then the value of $2a+3b$ is where $f(x) = \begin{cases} x+a \sqrt{2} \sin x & \text{if } 0 \leq x < \frac{\pi}{4} \\ 2x \cot x + b & \text{if } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}$

The function $f(x) = \frac{x^2 - 1}{x^3 - 1}$ is not defined at $x = 1$. If the function is continuous at $x = 1$,then the value of $f(1)$ will be:

The function $f(x) = \begin{cases} sgn([x]) & x \notin I \\ [sgn(x)] & x \in I \end{cases}$ is (where $sgn()$ denotes the signum function and $[.]$ denotes the greatest integer function):