If $f(x) = \begin{cases} 6 \beta - 3 \alpha x, & \text{if } -4 \leq x < -2 \\ 4x + 1, & \text{if } -2 \leq x \leq 2 \end{cases}$ is continuous on $[-4, 2]$,then $\alpha + \beta = $

  • A
    $\frac{-7}{6}$
  • B
    $\frac{4}{7}$
  • C
    $\frac{-4}{7}$
  • D
    $\frac{7}{6}$

Explore More

Similar Questions

Discuss the continuity of the function $f$ given by $f(x) = |x|$ at $x = 0$.

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

If $f(x) = \begin{cases} |x - 3|, & x \geqslant 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & x < 1 \end{cases}$,then $f(x)$ is:

If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by $f(x)=\begin{cases} \frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k, & x=\frac{\pi}{4} \end{cases}$ is continuous,then $k$ is equal to

Let $S_n = 1 + 3x + 9x^2 + 27x^3 + \ldots$ ($n$ terms) and $-\frac{1}{3} < x < \frac{1}{3}$. If $\lim_{n \rightarrow \infty} S_n = f(x)$,then $f(x)$ is discontinuous at the point $x =$

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