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

  • A
    $-\frac{1}{2}$
  • B
    $\frac{1}{2}$
  • C
    $1$
  • D
    $-1$

Explore More

Similar Questions

Let $f, g: R \to R$ be two functions defined by $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right), & x \ne 0 \\ 0, & x = 0 \end{cases}$ and $g(x) = x f(x)$.
Statement $I$: $f$ is a continuous function at $x = 0$.
Statement $II$: $g$ is a differentiable function at $x = 0$.

For what value of $k$ is the function $f(x) = \begin{cases} \frac{\text{log}(1+2x) \sin x^{\circ}}{x^2}, & x \neq 0 \\ k, & x = 0 \end{cases}$ continuous at $x = 0$?

The number of points at which the function $f(x) = \frac{\sqrt{11+|x|-6\sqrt{2+|x|}}}{6-2\sqrt{2+|x|}}$ is discontinuous in $(-\infty, \infty)$ is

Define $f: R \rightarrow R$ by $f(x) = [x] + \sqrt{x - [x]}$ for $x \in R$,where $[x]$ denotes the greatest integer function. Then the set of points at which $f$ is continuous is

If $f(x) = \begin{cases} \frac{\log_{e} x}{x-1} & x \neq 1 \\ k & x=1 \end{cases}$ is continuous at $x=1$,then the value of $k$ is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo