If $f(x)$ is continuous at $x = 3$ where $f(x) = \begin{cases} ax + 1, & \text{for } x \leq 3 \\ bx + 3, & \text{for } x > 3 \end{cases}$,then

  • A
    $a + b = - \frac{2}{3}$
  • B
    $a - b = - \frac{2}{3}$
  • C
    $a - b = \frac{2}{3}$
  • D
    $a + b = \frac{2}{3}$

Explore More

Similar Questions

Let $[x]$ denote the greatest integer function and $f(x) = \max\{1+x+[x], 2+x, x+2[x]\}$ for $0 \leq x \leq 2$. Let $m$ be the number of points in $[0, 2]$ where $f$ is not continuous,and $n$ be the number of points in $(0, 2)$ where $f$ is not differentiable. Then $(m+n)^2+2$ is equal to:

If $f(x) = \begin{cases} \frac{3 \sin(\pi x)}{5x} & x \neq 0 \\ 2K & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $K$ is:

If $f(x) = \begin{cases} x+a, & x \leq 0 \\ |x-4|, & x > 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ (x-4)^2+b, & x \geq 0 \end{cases}$ are continuous on $\mathbb{R}$,then $(g \circ f)(2) + (f \circ g)(-2)$ is equal to.

If $f(x)= \begin{cases} \frac{x-[x]}{x-2}, & x>2 \\ b, & x=2 \\ \frac{|x^2-x-2|}{a(2+x-x^2)}, & -1 < x \leq 2 \\ 2a-b, & x \leq -1 \end{cases}$ is continuous on $R$,then $\lim _{x \rightarrow 0} \frac{\sin ^2 ax+x \tan bx}{x^2}=$

If $f(x) = \frac{3^{x+3} - 3^{-x} - 2}{\tan x \cdot \log(1+x)}$ for $x \neq 0$, is continuous at $x = 0$, then the value of $f(0)$ is equal to ...

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