JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Quadratic Equations and Inequalities

Put $t=3^x>0$, so the equation becomes $t(t-1)+2=|t-1|+|t-2|$, i.e. $t^2-t+2=|t-1|+|t-2|$. For $t<1$ the right side is $(1-t)+(2-t)=3-2t$, giving $t^2+t-1=0$, which has one positive root below $1$. For $1\le t<2$ it reduces to $t^2-t+1=0$ and for $t\ge 2$ to $t^2-3t+5=0$, each with no real root. So exactly one admissible $t$ survives, and since $3^x=t$ fixes a single $x$, the solution set is a singleton.