$A$ function $y=f(x)$ with $f(-1)=-249$ has no maximum and has only one minimum at $x=5$ with $f(5)=75$. Which one of the following is true?

  • A
    At some point in $(-1,5)$, $f(x)$ is discontinuous
  • B
    The minimum value cannot be $75$ since $f(-1) < f(5)=75$
  • C
    $f(x)$ is discontinuous at every point of $\mathbb{R}$
  • D
    $f(x)$ is continuous on $\mathbb{R}$

Explore More

Similar Questions

If $f(x) = \frac{\sin(\pi \cos^2 x)}{3x^2}$ for $x \neq 0$ is continuous at $x = 0$,then $f(0) = $

If $f(x) = |x|/x$ for $x \neq 0$ and $1$ for $x = 0$,then the function is

If the function $f(x) = \begin{cases} 1 + \sin \frac{\pi x}{2}, & \text{for } -\infty < x \le 1 \\ ax + b, & \text{for } 1 < x < 3 \\ 6 \tan \frac{x\pi}{12}, & \text{for } 3 \le x < 6 \end{cases}$ is continuous in the interval $(-\infty, 6)$,then the values of $a$ and $b$ are respectively

If the function $f(x) = \frac{1 - \cos 4x}{8x^2}$ for $x \ne 0$ and $f(x) = k$ for $x = 0$ is a continuous function at $x = 0$,then the value of $k$ is:

If the function defined by $f(x) = \begin{cases} (x^2 + e^{\frac{1}{2-x}})^{-1}, & x \neq 2 \\ k, & x = 2 \end{cases}$ is right continuous at $x = 2$, then $k =$

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