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
Let $f$ be a continuous,periodic even function defined on $\mathbb{R}$ such that $f(0) = 1$,$f(2) = -1$ and the period of $f$ is $4$. The minimum number of roots of the equation $f(x) = 0$ in the interval $[-10, 10]$ will be:
Show that the function defined by $f(x)=\cos \left(x^{2}\right)$ is a continuous function.
Easy
View SolutionIf $f(x) = \begin{cases} x^2 + \alpha, & x \ge 0 \\ 2\sqrt{x^2 + 1} + \beta, & x < 0 \end{cases}$ is continuous at $x = 0$ and $f(\frac{1}{2}) = 2$,then $\alpha^2 + \beta^2$ is
If $[x]$ is the greatest integer function and $f(x) = \begin{cases} 2[x] - \frac{x}{|x|}, & x \neq 0 \\ 1, & x = 0 \end{cases}$ is a real-valued function, then $f$ is
Discuss the continuity of the function $f,$ where $f$ is defined by $f(x) = \begin{cases} 3, & \text{if } 0 \le x \le 1 \\ 4, & \text{if } 1 < x < 3 \\ 5, & \text{if } 3 \le x \le 10 \end{cases}$ at $x=3.$
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo