Question

The function

$f\left(x\right)=\{\begin{array}{cc}|x-3|,& x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& x<1\end{array}$,

is

• A. continuous at $x=1$
• B. differentiable at $x=1$
• C. discontinuous at $x=1$
• D. differentiable at $x=3$

Answer 1

Answer: (A), (B)

The correct options are
A

continuous at $x=1$

B

differentiable at $x=1$

Here, $f\left(x\right)=\{\begin{array}{cc}|x-3|,& x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& x<1\end{array}$
$\therefore$ RHL at $x=1,\underset{h\to 0}{lim}|1+h-3|=2$

LHL at $x=1,\underset{h\to 0}{lim}\frac{(1-h{)}^2}{4}-\frac{3(1-h)}{2}+\frac{13}{4}=2\therefore f\left(x\right)\text{is continuous at}x=1\text{Again,}f\left(x\right)=\{\begin{array}{cc}-(x-3),& 1\le x<3\\ (x-3),& x\ge 3\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& x<1\end{array}\therefore f^{\prime }\left(x\right)=\{\begin{array}{cc}-1,& 1\le x<3\\ 1,& \ge 3\\ \frac{x}{2}-\frac{3}{2},& x<1\end{array}$
$\therefore \begin{array}{c}\text{RHD at}x=1\Rightarrow -1\\ \text{LHD at}x=1\Rightarrow -1\end{array}]$ differentiable at $x=1$.

Again,$\begin{array}{c}\text{RHD at}x=3\Rightarrow 1\\ \text{LHD at}x=3\Rightarrow -1\end{array}]$ not differentiable at $x=3$.