Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined as $f(x) = \begin{cases} 5, & \text{if } x \le 1 \\ a + bx, & \text{if } 1 < x < 3 \\ b + 5x, & \text{if } 3 \le x < 5 \\ 30, & \text{if } x \ge 5 \end{cases}$. Then $f$ is

  • A
    continuous if $a = 5$ and $b = 5$
  • B
    continuous if $a = 5$ and $b = 10$
  • C
    continuous if $a = 0$ and $b = 5$
  • D
    not continuous for any values of $a$ and $b$

Explore More

Similar Questions

If a function $f(x) = \begin{cases} ax+b, & x \leq -1 \\ 2x^2+2bx-\frac{a}{2}, & -1 < x < 1 \\ 7, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$, then $(a, b) =$

If $f$ is a continuous real-valued function defined on a closed interval $[a, b]$,then the range of the function is . . . . . .

Let $f$ be defined by $f(x) = \begin{cases} \frac{\tan x}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$.
Statement-$1$: $x = 0$ is a point of local minima for $f$.
Statement-$2$: $f'(0) = 0$.

Difficult
View Solution

If $f: R \rightarrow R$ defined as $f(x) = \frac{x^3+2x^2+x+2}{x^2+x-2}$ (when $x \neq -2$) is continuous at $x = -2$, then $f(-2)$ is equal to

If a real-valued function $f(x) = \begin{cases} \frac{2x^2+(k+2)x+9}{3x^2-7x-6} & , \text{for } x \neq 3 \\ l & , \text{for } x=3 \end{cases}$ is continuous at $x=3$ and $l$ is a finite value,then $l-k=$

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