Question

For the function $f\left(x\right)=\{\begin{array}{cc}{x}^2-5,& x\le 3\\ \sqrt{x+13},& x>3\end{array}$

Consider the following statements:

$1.$The function is discontinuous at $x=3.$

$2.$The function is not differentiable at $x=0.$
Then which of the above statement(s) is correct:

• A. $1$ only
• B. $2$ only
• C. Both $1$ and $2$
• D. Neither $1$ nor $2$

Answer 1

The correct option is D Neither $1$ nor $2$

Given : $f\left(x\right)=\{\begin{array}{cc}{x}^2-5,& x\le 3\\ \sqrt{x+13},& x>3\end{array}$

Continuity at $x=3$
$\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3^{-}}{lim}(x^2-5)=9-5=4\underset{x\to 3^{+}}{lim}f\left(x\right)=\underset{x\to 3^{+}}{lim}\sqrt{x+13}=\sqrt{16}=4L.H.L.=R.H.L.=f\left(3\right)$

$\therefore$ It is continuous at $x=3$

$f\left(x\right)=x^2-5,x\le 3\Rightarrow f^{\prime }\left(x\right)=2x,x<3$

So the function $f\left(x\right)$ is differentiable at $x=0$.

So, both statements are incorrect.