If a function $f$ is defined by:
$\begin{cases} f(x) = x-1, & \text{when } -\infty < x < 1 \\ f(x) = 0, & \text{when } x=1 \\ f(x) = x^3-1, & \text{when } 1 < x < \infty \end{cases}$
then at $x=1$,$f$ is:

  • A
    continuous and differentiable
  • B
    continuous but not differentiable
  • C
    discontinuous and differentiable
  • D
    discontinuous and not differentiable

Explore More

Similar Questions

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ equals

$A$ function $f(x) = \begin{cases} 1 + x, & x \le 2 \\ 5 - x, & x > 2 \end{cases}$ is

Let $f: R \rightarrow R$ be defined as $f(x)=\begin{cases} \frac{a-b \cos 2 x}{x^2} & ; x<0 \\ x^2+c x+2 & ; 0 \leq x \leq 1 \\ 2 x+1 & ; x>1 \end{cases}$. If $f$ is continuous everywhere in $R$ and $m$ is the number of points where $f$ is $NOT$ differentiable,then $m+a+b+c$ equals:

If $f:R \to R$ and $f(x)$ is a polynomial function of degree $10$ such that $f(x)=0$ has all real and distinct roots,then the equation $(f'(x))^2 - f(x)f''(x) = 0$ has:

Let $f, g: R \rightarrow R$ be two real-valued functions defined as $f(x)=\begin{cases} -|x+3| & , x < 0 \\ e^{x} & , x \geq 0 \end{cases}$ and $g(x)=\begin{cases} x^{2}+k_{1} x & , x < 0 \\ 4 x+k_{2} & , x \geq 0 \end{cases}$,where $k_{1}$ and $k_{2}$ are real constants. If $(g \circ f)$ is differentiable at $x=0$,then $(g \circ f)(-4)+(g \circ f)(4)$ 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