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 $[t]$ 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 :

Consider the function $f(x) = \begin{cases} \frac{x+5}{x-2}, & \text{if } x \neq 2 \\ 1, & \text{if } x=2 \end{cases}$. Then,$f(f(x))$ is discontinuous

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 $f(x) = \frac{1+\cos \pi x}{\pi(1-x)^2}$ for $x \neq 1$ is continuous at $x=1$,then $f(1)$ is equal to

Let $f: R \rightarrow R$ be the function defined by $f(x) = \begin{cases} 5, & \text{if } x \leq 1 \\ a+bx, & \text{if } 1 < x < 3 \\ b+5x, & \text{if } 3 \leq x < 5 \\ 30, & \text{if } x \geq 5 \end{cases}$. Then $f$ is: