The function $f(x)=\sqrt{\frac{3 x^2-5 x-2}{2 x^2-7 x+5}}$ has discontinuous points at $x=$

The value of $k$ which makes $f(x) = \begin{cases} \sin \frac{1}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ continuous at $x = 0$ is

The function $f$ is defined by $f(x) = \begin{cases} 2x - 1, & \text{if } x > 2 \\ k, & \text{if } x = 2 \\ x^2 - 1, & \text{if } x < 2 \end{cases}$. If $f$ is continuous at $x = 2$,then the value of $k$ is equal to:

If the function $f(x) = \begin{cases} [\tan(\frac{\pi}{4} + x)]^{\frac{1}{x}}, & x \neq 0 \\ K, & x = 0 \end{cases}$ is continuous at $x = 0$,then $K = ?$

Let $f(x) = \min \{1, 1 + x \sin x \}$ for $0 \leq x \leq 2\pi$. If $m$ is the number of points where $f$ is not differentiable and $n$ is the number of points where $f$ is not continuous,then the ordered pair $(m, n)$ is equal to

If $f(x) = [x] - [\frac{x}{4}]$,$x \in R$,where $[x]$ denotes the greatest integer function,then