Question

If $f\left(x\right)=\{\begin{array}{cc}{x}^2\left(\text{sgn}\right[x\left]\right)+\left\{x\right\},& 0\le x<2\\ \sin x+|x-3|,& 2\le x<4\end{array},$
where [.] and {.} represent the greatest integer and the fractional part function, respectively. Then

• A. $f\left(x\right)$ is differentiable at $x=1$
• B. $f\left(x\right)$ is continuous but non-differentiable at $x=1$
• C. $f\left(x\right)$ is non-differentiable at $x=2$
• D. $f\left(x\right)$ is discontinuous at $x=2$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $f\left(x\right)$ is continuous but non-differentiable at $x=1$
C $f\left(x\right)$ is non-differentiable at $x=2$
D $f\left(x\right)$ is discontinuous at $x=2$

$f\left(x\right)=\{\begin{array}{cc}{x}^2\left(\text{sgn}\right[x\left]\right)+\left\{x\right\},& 0\le x<2\\ \sin x+|x-3|,& 2\le x<4\end{array}$

$\text{sgn}\left(x\right)=\{\begin{array}{cc}1,& x>0\\ 0,& x=0\\ -1,& x<0\end{array}$

For continuity at $x=1$
$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}{x}^2\left(\text{sgn}\right[x\left]\right)+\left\{x\right\}=1\text{sgn}\left(1\right)+0=1+0=1$
$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}{x}^2\left(\text{sgn}\right[x\left]\right)+\left\{x\right\}=1\text{sgn}\left(0\right)+1=0+1=1$

Also, $f\left(1\right)=1$
$\therefore \text{L.H.L}=\text{R.H.L}=f\left(1\right)$
Hence, $f\left(x\right)$ is continuous at $x=1$

Now, for differentiability,
$f^{\prime }\left(1^{+}\right)=\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}$
$=\underset{h\to 0}{lim}\frac{(1+h{)}^2(\text{sgn}[1+h])+\{1+h\}-1}{h}$
$=\underset{h\to 0}{lim}\frac{(1+h{)}^2+h-1}{h}$
$=\underset{h\to 0}{lim}\frac{h^2+3h}{h}$
$=\underset{h\to 0}{lim}(h+3)$
$=3$

$f^{\prime }\left(1^{-}\right)=\underset{h\to 0}{lim}\frac{f(1-h)-f\left(1\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{(1-h{)}^2(\text{sgn}[1-h])+\{1-h\}-1}{-h}$
$=\underset{h\to 0}{lim}\frac{(1-h{)}^2\times 0+1-h-1}{-h}$
$=\underset{h\to 0}{lim}\frac{-h}{-h}$
$=1$
$\Rightarrow f^{\prime }\left(1^{+}\right)\ne f^{\prime }\left(1^{-}\right)$
Hence, $f\left(x\right)$ is not differentiable at $x=1.$

Now, at $x=2$
$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}{x}^2\left(\text{sgn}\right[x\left]\right)+\left\{x\right\}=4\text{sgn}\left(1\right)+1=4+1=5$
$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}\sin x+|x-3|=\sin 2+1$

$\therefore \text{L.H.L}\ne \text{R.H.L}$
Hence, $f\left(x\right)$ is discontinuous at $x=2$ and then $f\left(x\right)$ is also non-differentiable at $x=2.$