Question

If function $f\left(x\right)=\{\begin{array}{cc}2x-4,& x\le 2\\ x^2-4,& x>2\end{array}$, then

• A. $f\left(x\right)$ is continuous and differentiable at $x=2$
• B. $f\left(x\right)$ is continuous but not differentiable at $x=2$
• C. $f^{\prime }\left(x\right)$ is continuous and differentiable at $x=2$
• D. $f^{\prime }\left(x\right)$ is continuous but not differentiable at $x=2$

Answer 1

The correct option is B $f\left(x\right)$ is continuous but not differentiable at $x=2$

Given : $f\left(x\right)=\{\begin{array}{cc}2x-4,& x\le 2\\ x^2-4,& x>2\end{array}$
Now,
$f\left(2^{+}\right)=f\left(2^{-}\right)=f\left(2\right)=0$
Therefore, it is continuous at $x=2$

Checking differentiability
$f^{\prime }\left(x\right)=\{\begin{array}{cc}2,& x<2\\ 2x,& x>2\end{array}\Rightarrow f^{\prime }\left(2^{+}\right)=4,f^{\prime }\left(2^{-}\right)=2\therefore L.H.D.\ne R.H.D.$
So, $f\left(x\right)$ is not differentiable at $x=2$.
$\therefore f\left(x\right)$ is continuous but not differentiable at $x=2$

As $f^{\prime }\left(2^{+}\right)\ne f^{\prime }\left(2^{-}\right)$, so $f^{\prime }\left(x\right)$ is not continuous at $x=2$ and also non differentiable.