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

Let $f(x) = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {2\sin x} \right)}^{2n}}}}{{{3^n} - {{\left( {2\cos x} \right)}^{2n}}}}; n \in Z$,$x \ne m\pi \pm \frac{\pi }{6}; m \in Z$ and $f\left( {m\pi \pm \frac{\pi }{6}} \right) = 0$. Then which of the following is true?

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

Let $S_n = 1 + 3x + 9x^2 + 27x^3 + \ldots$ ($n$ terms) and $-\frac{1}{3} < x < \frac{1}{3}$. If $\lim_{n \rightarrow \infty} S_n = f(x)$,then $f(x)$ is discontinuous at the point $x =$

Find all points of discontinuity of $f$,where $f$ is defined by $f(x) = \begin{cases} x + 1, & \text{if } x \ge 1 \\ x^2 + 1, & \text{if } x < 1 \end{cases}$. Is $f$ a continuous function?

Let $[x]$ denote the greatest integer function and $f(x) = \max\{1+x+[x], 2+x, x+2[x]\}$ for $0 \leq x \leq 2$. Let $m$ be the number of points in $[0, 2]$ where $f$ is not continuous,and $n$ be the number of points in $(0, 2)$ where $f$ is not differentiable. Then $(m+n)^2+2$ is equal to: