Let f x = x - 1 x 2 - 2 x - 3 + x - 3 , x ∈ ℝ . If m and M are respectively the number of points…

Given,

$f\left(x\right)=\left|\left(x-1\right)\left(x+1\right)\left(x-3\right)\right|+\left(x-3\right)$

$f\left(x\right)=\left\{\begin{array}{cc}\left(x-3\right)\left(x^2\right),& 3\le x\le 4\\ \left(x-3\right)\left(2-x^2\right),& 1\le x<3\\ \left(x-3\right)\left(x^2\right),& 0<x<1\end{array}\right.$

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{cc}3x^2-6x,& 3<x<4\\ -3x^2+6x+2,& 1<x<3\\ 3x^2-6x,& 0<x<1\end{array}\right.$

$f^{\text{'}}\left(3^{+}\right)>0f^{\text{'}}\left(3^{-}\right)<0\to$ Minimum

$f^{\text{'}}\left(1^{+}\right)>0f^{\text{'}}\left(1^{-}\right)<0\to$ Minimum

For $x\in \left(1,3\right),f^{\text{'}}\left(x\right)=0$ at one point $\to$ Maximum

For $x\in \left(3,4\right), f^{\text{'}}\left(x\right)\ne 0$

For $x\in \left(0,1\right), f^{\text{'}}\left(x\right)+0$

So points of minima are $2$, so $m=2$

and point of maxima is $1$, so $M=1$

So, $M+n=3$