If a real valued function $f(x) = \begin{cases} \log(1+[x]), & x \geq 0 \\ \sin^{-1}[x], & -1 \leq x < 0 \\ k([x]+|x|), & x < -1 \end{cases}$ is continuous at $x = -1$,then $k =$

  • A
    $-\pi / 2$
  • B
    $-\pi$
  • C
    $\pi$
  • D
    $\pi / 2$

Explore More

Similar Questions

If $f(x) = \begin{cases} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x} & \text{for } -1 \le x < 0 \\ 2x^2 + 3x - 2 & \text{for } 0 \le x \le 1 \end{cases}$ is continuous at $x = 0$,then $k = $

If $f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ 3, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $k$ is equal to . . . . . . .

If $f(x) = \frac{\log_{\sin |x|} \cos^3 x}{\log_{\sin |3x|} \cos^3 (x/2)}$ for $|x| < \frac{\pi}{3}, x \neq 0$ and $f(0) = 4$,then the number of points of discontinuity of $f$ in $\left( -\frac{\pi}{3}, \frac{\pi}{3} \right)$ is:

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} a^2 \cos^2 x + b^2 \sin^2 x, & x \leq 0 \\ e^{ax+b}, & x > 0 \end{cases}$ and is a continuous function,then:

Prove that every rational function is continuous at every point in its domain.

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