JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Basic Mathematics

Split at the critical points $x=2$ and $x=3$. For $x<2$ the equation becomes $x(2-x)+3(3-x)+1=0$, i.e. $x^2+x-10=0$, whose roots are $\frac{-1\pm \sqrt{41}}{2}$; only $\frac{-1-\sqrt{41}}{2}$ satisfies $x<2$. For $2\le x<3$ it becomes $x^2-5x+10=0$ with negative discriminant, giving no real root. For $x\ge 3$ it becomes $x^2+x-8=0$ with roots $\frac{-1\pm \sqrt{33}}{2}$, and since $\frac{-1+\sqrt{33}}{2}\approx 2.37<3$, neither qualifies. Hence there is exactly $1$ real root.