The points at which the function $f(x) = \frac{x + 1}{x^2 + x - 12}$ is discontinuous,are

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f(x)=x-[x]$,$g(x)=1-x+[x]$,and $h(x)=\min \{f(x), g(x)\}$ for $x \in [-2, 2]$. Then $h$ is :

If $f(x) = \begin{cases} e^{1/x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,then:

If the function $f$ is continuous at $x = \pi$, where $f(x) = \frac{1 - \cos[7(x - \pi)]}{5(x - \pi)^2}$ for $x \neq \pi$, then $f(\pi) =$

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 ...

Let $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} \right)} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f(x)$ is: