Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given,

$f:\left(-2, 2\right)\to ℝ$ be defined by $f\left(x\right)=\left\{\begin{array}{ccc}x\left[x\right]& ,& -2<x<0\\ \left(x-1\right)\left[x\right]& ,& 0\le x<2\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{l}\begin{array}{c}-2x, -2<x<-1\\ -x, -1\le x<0\end{array}\\ \begin{array}{c}0, 0<x<1\\ x-1, 1\le x<2\end{array}\end{array}\right.$

Now plotting the diagram of the above function we get,

Now from above diagram we can say that, $y=f\left(x\right)$ is same as $y=\left|f\left(x\right)\right|$

Hence, the function is not continuous at one point and non differentiable at three points, so $m=1, n=3$

Hence, $m+n=4$