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. not continuous at $x=1$
• B. not differentiable at $x=1$
• C. continuous and differentiable at $x=1$
• D. continuous at $x=1$ but not differentiable at $x=1$

Answer 1

The correct option is C continuous and differentiable at $x=1$

$f\left(x\right)=\{\begin{array}{cc}|x-3|& ,x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}& ,x<1\end{array}$
We need to check for the continuity and differentiability at $x=1$.
Now, $f\left(1\right)=2$
LHL $=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=2$

RHL $=|1-3|=2$

$\therefore$ LHL$=$RHL$=f\left(1\right)$
Hence, the function is continuous at $x=1$.
Let us check for the differentiability at $x=1$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}-1& ,1<x<3\\ 1& ,x>3\\ \frac{2x}{4}-\frac{3}{2}& ,x<1\end{array}$
$RH{D}_{atx=1}=-1$
$LH{D}_{atx=1}=\frac{2x}{4}-\frac{3}{2}=\frac{2}{4}-\frac{3}{2}=-1$
Hence, the function is differentiable at $x=1$.
Hence, option C is correct.