If $f(x) = [x] - [\frac{x}{4}]$,$x \in R$,where $[x]$ denotes the greatest integer function,then

  • A
    Both $\lim_{x \to 4^-} f(x)$ and $\lim_{x \to 4^+} f(x)$ exist but are not equal
  • B
    $\lim_{x \to 4^-} f(x)$ exists but $\lim_{x \to 4^+} f(x)$ does not exist
  • C
    $\lim_{x \to 4^+} f(x)$ exists but $\lim_{x \to 4^-} f(x)$ does not exist
  • D
    $f$ is continuous at $x = 4$

Explore More

Similar Questions

If $f(x) = \begin{cases} x^2 - 1, & \text{if } x \ge 2 \\ x + 1, & \text{if } x < 2 \end{cases}$, then $\lim_{x \to 1} f(x) + \lim_{x \to 2} f(x) =$

If the function $f(x)=\begin{cases} \frac{1-\cos x}{x^{2}}, & \text{for } x \neq 0 \\ k, & \text{for } x=0 \end{cases}$ is continuous at $x=0$,then the value of $k$ is

Let $f(x) = \frac{g(x)}{h(x)}$,where $g$ and $h$ are continuous functions on the open interval $(a, b)$. Which of the following statements is true for $a < x < b$?

Discuss the continuity of the following functions:
a) $f(x) = \sin x + \cos x$
b) $f(x) = \sin x - \cos x$
c) $f(x) = \sin x \times \cos x$

The values of $a$ and $b$,so that the function $f(x) = \begin{cases} x+a \sqrt{2} \sin x, & 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b, & \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x, & \frac{\pi}{2} < x \leq \pi \end{cases}$ is continuous for $0 \leq x \leq \pi$,are respectively given by

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