Question

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

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

Answer 1

Answer: (A), (B), (D)

continuous at $x=1$
non-differentiable at $x=3$
differentiable at $x=1$

Here $f\left(x\right)=\{\begin{array}{c}|x-3|x\ge 1\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}x<1\end{array}$
$\therefore R.H.L$ at $x=1=\underset{h\to 0}{lim}|1+h-3|=2$
And $L.H.L$ at $x=1=\underset{h\to 0}{lim}\frac{{(1-4)}^2}{4}-\frac{3(1-h)}{2}+\frac{13}{4}$
$=\frac{1}{4}-\frac{3}{2}+\frac{13}{4}=\frac{14}{4}-\frac{3}{2}=2$
$\therefore f\left(x\right)$ is continuous at $x=1$
Again $f\left(x\right)=\{\begin{array}{c}-(x-3)1\le x<3\\ (x-3)x\ge 3\\ \frac{x^2}{4}-\frac{3x}{2}+\frac{13}{4}x<1\end{array}\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{c}-11\le x<3\\ 1x\ge 3\\ \frac{x}{2}-\frac{3}{2}x<1\end{array}$
$R.H.D$ at $x=1\Rightarrow -1$
And $L.H.D$ at $x=1\Rightarrow \frac{1}{2}-\frac{3}{2}=-1$
Hence it is differentiable at $x=1$
Again $R.H.D$ at $x=3\Rightarrow 1$
And $L.H.D$ at $x=3\Rightarrow -1$
Hence not differentiable at $x=3$