JEE Main 2025 Jan 24 Shift 2, Mathematics Q18: Some Important Properties Of Functions

The LHS factorises as $(x+1)(x+2)$, which is non-negative for $x\le -2$ or $x\ge -1$ and negative on $(-2,-1)$. The RHS, $\min \{|x-3|,|x+2|\}$, is non-negative everywhere. So any solution must satisfy $x\le -2$ or $x\ge -1$.

Compare $|x-3|$ and $|x+2|$: their crossover occurs at the midpoint $x=1/2$. For $x<1/2$, $|x+2|<|x-3|$, so $\min =|x+2|$. For $x>1/2$, $\min =|x-3|$.

Region $x\le -2$: $\min =|x+2|=-(x+2)$. Equation: $(x+1)(x+2)=-(x+2)$, i.e. $(x+2)(x+1+1)=0\Rightarrow (x+2)(x+2)=0\Rightarrow x=-2$. Valid (one solution).

Region $-1\le x<1/2$: $\min =|x+2|=x+2$. Equation: $(x+1)(x+2)=x+2\Rightarrow (x+2)(x+1-1)=0\Rightarrow x(x+2)=0\Rightarrow x=0$ (the root $x=-2$ is outside this region). One solution.

Region $x\ge 1/2$: $\min =|x-3|$. For $1/2\le x\le 3$, equation $(x+1)(x+2)=3-x\Rightarrow x^2+4x-1=0\Rightarrow x=-2\pm \sqrt{5}$. The positive root $-2+\sqrt{5}\approx 0.24$ does not satisfy $x\ge 1/2$, so no solution here. For $x>3$, $(x+1)(x+2)=x-3$ has no real root (LHS grows much faster).

Total: $x=-2$ and $x=0$. Two real solutions.

Correct option: (1)