Let $f(x) = \frac{1-\tan x}{4x-\pi}$ for $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{1}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is

If a function $f(x) = \begin{cases} ax+b, & x \leq -1 \\ 2x^2+2bx-\frac{a}{2}, & -1 < x < 1 \\ 7, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$, then $(a, b) =$

Consider $f(x) = [x]|x^3 - 2x^2 - x + 2|$ in $[-\frac{3}{2}, \frac{9}{2}]$. The number of points where $f(x)$ is discontinuous is (where $[.]$ denotes the greatest integer function).

For real $x$ with $-10 \leq x \leq 10$,define $f(x) = \int_{-10}^x 2^{[t]} dt$,where for a real number $r$,we denote by $[r]$ the greatest integer less than or equal to $r$. The number of points of discontinuity of $f$ in the interval $(-10, 10)$ is

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 $K = ?$

For the function $f(x) = \frac{1}{x + 2^{\frac{1}{x - 2}}}$,$x \neq 2$,which of the following holds?