Question

Let $f\left(x\right)=\{\begin{array}{cc}\left\{x^2\right\},& -1\le x<1\\ |1-2x|,& 1\le x<2\\ (1-x^2)\text{sgn}(x^2-3x-4),& 2\le x\le 4\end{array}$

where $\left\{k\right\}$ and $\text{sgn}\left(k\right)$ denote fractional part function and signum function of $k$ respectively. If $m$ denotes the number of points of discontinuity of $f\left(x\right)$ in $[-1,4]$ and $n$ denotes the number of points of non-differentiability of $f\left(x\right)$ in $(-1,4),$ then $(m+n)$ is equal to

• A. $2$
• B. $4$
• C. $5$
• D. $3$

Answer 1

The correct option is D $3$

$f\left(x\right)=\{\begin{array}{cc}0,& x=-1\\ x^2,& -1<x<1\\ 2x-1,& 1\le x<2\\ x^2-1& 2\le x<4\\ 0,& x=4\end{array}$

Clearly, $f\left(x\right)$ is discontinuous at $x=-1$
and $x=4$ in $[-1,4]$

$f^{\prime }\left(x\right)=\{\begin{array}{cc}2x,& -1<x<1\\ 2,& 1\le x<2\\ 2x,& 2\le x<4\end{array}$
$f\left(x\right)$ is not differentiable at $x=2$ in $(-1,4).$
$\therefore m=2$ and $n=1$
Hence, $m+n=3$