JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Basic Mathematics

Note $\min \{|x-3|,|x+2|\}$ switches at $x=\frac{1}{2}$ (where $|x-3|=|x+2|$), giving


Solve $x^2+3x+2$ equal to each piece, keeping only roots inside the valid range.

For $x<-2$: $x^2+3x+2=-x-2\Rightarrow x^2+4x+4=0\Rightarrow (x+2{)}^2=0\Rightarrow x=-2$ (boundary case, check separately).

For $-2\le x<\frac{1}{2}$: $x^2+3x+2=x+2\Rightarrow x^2+2x=0\Rightarrow x(x+2)=0\Rightarrow x=0$ or $x=-2$, both lie in this range.

For $\frac{1}{2}\le x<3$: $x^2+3x+2=3-x\Rightarrow x^2+4x-1=0\Rightarrow x=-2\pm \sqrt{5}$. Neither root ($\approx 0.24$ or $\approx -4.24$) lies in $[\frac{1}{2},3)$.

For $x\ge 3$: $x^2+3x+2=x-3\Rightarrow x^2+2x+5=0$, discriminant $4-20<0$, no real root.

Checking $x=-2$ directly in the original equation: LHS $=4-6+2=0$, and $\min \{|-2-3|,|-2+2|\}=\min \{5,0\}=0$ — it satisfies the equation.

So the valid solutions are $x=-2$ and $x=0$, giving