JEE Main 2025 Apr 7 Shift 2, Mathematics Q17: Some Important Properties Of Functions

The expression changes form at $x=2$ and $x=3$, so the equation is examined on three intervals.

Case1: $x<2$: $|x-2|=2-x,|x-3|=3-x$, giving $x(2-x)+3(3-x)+1=0\Rightarrow x^2+x-10=0\Rightarrow x=\frac{-1\pm \sqrt{41}}{2}$. Only $x=\frac{-1-\sqrt{41}}{2}\approx -3.70$ satisfies $x<2$; this gives 1 valid root.

Case2: $2\leq x<3$: $|x-2|=x-2,|x-3|=3-x$, giving $x(x-2)+3(3-x)+1=0\Rightarrow x^2-5x+10=0$. The discriminant is $25-40=-15<0$, so there are no real roots here.

Case3: $x\ge 3$: $|x-2|=x-2,|x-3|=x-3$, giving $x(x-2)+3(x-3)+1=0\Rightarrow x^2+x-8=0\Rightarrow x=\frac{-1\pm \sqrt{33}}{2}\approx 2.37$ or $-3.37$. Neither value satisfies $x\ge 3$, so both are rejected.

$\therefore$ There is exactly 1 real root, from the first case.