JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Application of Derivatives

Writing $g(x)=|x+2|-2|x|$ piecewise gives $g=x-2$ for $x<-2$, $g=3x+2$ for $-2\le x<0$, and $g=2-x$ for $x\ge 0$; so $g$ rises to a peak $2$ at $x=0$ and vanishes at $x=-\frac{2}{3}$ and $x=2$. Then $f=|g|$ touches the axis at $x=-\frac{2}{3}$ and $x=2$, giving two local minima, and has a peak at $x=0$, one local maximum. At $x=-2$ the graph only changes slope while still decreasing, so it is not an extremum. With two minima and one maximum, $m+n=3$.