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. Continuous at $x=3$
• C. Differentiable at $x=1$
• D. Differentiable at $x=3$

Answer 1

Answer: (A), (B), (C)

Differentiable at $x=1$

$f\left(x\right)=\{\begin{array}{cc}x-3,& x\ge 3\\ 3-x,& 1\le x<3\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4},& x<1\end{array}$
Clearly, $f\left(1^{+}\right)=f\left(1^{-}\right)=f\left(1\right)=2$
and similarly, $f\left(3^{+}\right)=f\left(3^{-}\right)=f\left(3\right)=0$
Hence, $f\left(x\right)$ is continuous at $x=3$ and $x=1$

Now for differentiablity,
$f^{\prime }\left(x\right)=\{\begin{array}{cc}1,& x>3\\ -1,& 1<x<3\\ \frac{x}{2}-\frac{3}{2},& x<1\end{array}$

$\therefore f^{\prime }\left(1^{+}\right)=-1$ and $f^{\prime }\left(1^{-}\right)=-1$
Clearly, $L.H.D.=R.H.D.$, so $f$ is differentiable at $x=1$
$\Rightarrow f^{\prime }\left(3^{+}\right)=1$ and $f^{\prime }\left(3^{-}\right)=-1$
Clearly, $f\left(x\right)$ is not differentiable at $x=3.$