Define $f(x) = \begin{cases} \frac{1-\sin x}{(\pi-2x)^2} & \text{, if } x \neq \frac{\pi}{2} \\ k & \text{, if } x = \frac{\pi}{2} \end{cases}$. If $f(x)$ is continuous at $x = \frac{\pi}{2}$,then $k =$

  • A
    $-\frac{1}{8}$
  • B
    $\frac{1}{8}$
  • C
    $\frac{\pi}{8}$
  • D
    $\frac{\pi}{2}$

Explore More

Similar Questions

Examine whether $f(x) = \sin |x|$ is a continuous function.

If $f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & x < 0 \\ a, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4}, & x > 0 \end{cases}$ is continuous at $x = 0$,then the value of $a$ will be

The function $f : R \rightarrow R$ defined by $f(x) = \lim_{n \rightarrow \infty} \frac{\cos(2 \pi x) - x^{2n} \sin(x-1)}{1 + x^{2n+1} - x^{2n}}$ is continuous for all $x$ in.

Let $[t]$ denote the greatest integer less than or equal to $t$. Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a function defined by $f(x) = [\frac{x}{2} + 3] - [\sqrt{x}]$. Let $S$ be the set of all points in the interval $[0, 8]$ at which $f$ is not continuous. Then $\sum_{a \in S} a$ is equal to:

Statement-$1$: The equation $x \log x = 2 - x$ is satisfied by at least one value of $x$ lying between $1$ and $2$.
Statement-$2$: The function $f(x) = x \log x$ is an increasing function in $[1, 2]$ and $g(x) = 2 - x$ is a decreasing function in $[1, 2]$,and the graphs represented by these functions intersect at a point in $[1, 2]$.

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