Let $f:[-2,2] \rightarrow \mathbb{R}$ be a continuous function such that $f(x)$ assumes only irrational values. If $f(\sqrt{2})=\sqrt{2},$ then

If $f(x) = \begin{cases} \frac{\sqrt{1+ax}-\sqrt{1-ax}}{x}, & -1 \leq x < 0 \\ \frac{x^2+2}{x-2}, & 0 \leq x \leq 1 \end{cases}$ is continuous on $[-1,1]$, then $a=$

Let $f(x) = \begin{cases} \frac{(x - 1)(6x - 1)}{2x - 1}, & \text{if } x \neq \frac{1}{2} \\ 0, & \text{if } x = \frac{1}{2} \end{cases}$. Then at $x = \frac{1}{2}$,

Let $f : [-1,3] \to R$ be defined as $f(x) = \begin{cases} |x| + [x], & -1 \leq x < 1 \\ x + |x|, & 1 \leq x < 2 \\ x + |x|, & 2 \leq x \leq 3 \end{cases}$ where $[t]$ denotes the greatest integer less than or equal to $t$. Then,$f$ is discontinuous at:

If the function $f(x)$,defined below,is continuous everywhere,then $k$ equals: $f(x)=\begin{cases} \frac{2^x-1}{\sqrt{1+x}-1}, & x \neq 0 \\ k, & x=0 \end{cases}$

If the function $f(x) = \left(\frac{5x-8}{8-3x}\right)^{\frac{3}{2x-4}}$ for $x \neq 2$ and $f(2) = k$ is continuous at $x = 2$,then $k =$