JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Application of Derivatives

Put $t=e^{2x}>0$; then $f(x)=0$ becomes $t^4-t^3-3t^2-t+1=0$. Dividing by $t^2$ and setting $u=t+\frac{1}{t}$, so that $t^2+\frac{1}{t^2}=u^2-2$, reduces the equation to $u^2-u-5=0$, with roots $u=\frac{1\pm \sqrt{21}}{2}$. For $t>0$ one has $u=t+\frac{1}{t}\ge 2$, so only $u=\frac{1+\sqrt{21}}{2}\approx 2.79$ is admissible. Then $t^2-ut+1=0$ has discriminant $u^2-4>0$ and two positive roots, and each positive $t$ gives one $x$ through $e^{2x}=t$. Hence the curve meets the $x$-axis at two points.