JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Application of Derivatives

First, factor: $x^2-2x-3=(x-3)(x+1)$, so $f(x)=|(x-1)(x-3)(x+1)|+x-3$.

The expression $(x-1)(x-3)(x+1)$ changes sign at $x=1$ and $x=3$ (in the interval $(0,4)$).

For $x\in (0,1)$: $(x-1)<0$, $(x-3)<0$, $(x+1)>0$ → product is positive.
For $x\in (1,3)$: $(x-1)>0$, $(x-3)<0$, $(x+1)>0$ → product is negative.
For $x\in (3,4)$: all factors positive → product is positive.

So:


Differentiating:


$f$ is non-differentiable at $x=1$ and $x=3$. In $(1,3)$, $f^{\prime }(x)=0$ gives $x=1+\sqrt{5/3}$.

Analyzing the sign changes of $f^{\prime }$ and behavior at non-differentiable points, we find 2 local minima and 1 local maximum in $(0,4)$, so $m+M=3$.