JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Basic Mathematics

Put $t=2^x>0$, turning the equation into $5+|t-1|=t^2-2t$. For $t\ge 1$ it reads $t^2-3t-4=0$, i.e. $(t-4)(t+1)=0$, giving the admissible $t=4$. For $0<t<1$ it reads $t^2-t-6=0$, i.e. $(t-3)(t+2)=0$, whose roots $3$ and $-2$ are both outside $(0,1)$. Only $t=4$ survives, and since $t=2^x$ is one-to-one it yields a single real $x$; the equation has one real root.