Question

If $f\left(x\right)=\{\begin{array}{c}{-x}^2,whenx\le 0\\ 5x-4,when0<x\le 1\\ 4x^2-3x,when1<x<2\\ 3x+4,whenx\ge 2\end{array}$ then

• A. $f\left(x\right)$ is continuous at $x=0$
• B. $f\left(x\right)$ is continuous at $x=2$
• C. $f\left(x\right)$ is discontinuous at $x=1$
• D. None of the above

Answer 1

The correct option is B $f\left(x\right)$ is continuous at $x=2$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=0$

$f\left(0\right)=0,\underset{x\to 0^{+}}{lim}f\left(x\right)=-4$

$\therefore f\left(x\right)$ is discontinuous at $x=0$

and $\underset{x\to 1^{-}}{lim}f\left(x\right)=1,\underset{x\to 1^{+}}{lim}f\left(x\right)=1,f\left(1\right)=1$

$\therefore f\left(x\right)$ is continuous at $x=1$.

Also, $\underset{x\to 2^{-}}{lim}f\left(x\right)=4{\left(2\right)}^2-3\left(2\right)=10$

$f\left(2\right)=10$

and $\underset{x\to 2^{+}}{lim}f\left(x\right)=3\left(2\right)+4=10$

Hence, $f\left(x\right)$ is continuous at $x=2$.