Question

Consider 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 $x=0$.
Which of the statements is$/$ are correct?

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

Answer 1

The correct option is A $1$ only

$1.\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3^{-}}{lim}(x^2-5)=9-5=4$

$\underset{x\to 3^{+}}{lim}f\left(x\right)=\underset{x\to 3^{+}}{lim}\sqrt{x+13}=\sqrt{16}=4$

LHL$=$RHL

$\Rightarrow \underset{x\to 3}{lim}f\left(x\right)=4$

$\Rightarrow$ It is continuous

$2.f\left(x\right)=x^2-5$

$f^{\prime }\left(x\right)=2x$ for $x\le 3$

$f^{\prime }\left(x\right)$ is continuous for $x\le 3$

At $x=0f^{\prime }\left(x\right)$ is continuous

$\Rightarrow f\left(x\right)$ is differentiable at $x=0$