Let f(x) = |(x - 1)( x ^ 2 - 2x - 3)| + x - 3, x ∈ ℝ . If m and M...

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{cases}\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{cases}\right)</p><p>{f}^{'}\left(x\right) = \left\{$\begin{cases}3{x}^{2} - 6x, & 3 < x < 4 \\ - 3{x}^{2} + 6x + 2, & 1 < x < 3 \\ 3{x}^{2} - 6x, & 0 < x < 1\end{cases}$\right)

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

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

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

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

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

So points of minima are $2$, so $m = 2$

and point of maxima is $1$, so $M = 1$

So, $M + n = 3$