JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Quadratic Equations and Inequalities

With $t=e^x$ every real $x$ gives exactly one $t>0$, and the equation becomes

which factorises as

For $g(t)=t^3+3t-1$, $g^{\prime }(t)=3t^2+3>0$, so $g$ is strictly increasing; with $g(0)=-1$ and $g(1)=3$ it crosses zero exactly once, at a point of $(0,1)$.
For $h(t)=t^3-4t-1$, $h^{\prime }(t)=3t^2-4$ vanishes at $t=\frac{2}{\sqrt{3}}$, the only positive critical point, and $h\left(\frac{2}{\sqrt{3}}\right)<0$. Since $h(0)=-1$ and $h(3)=14$, $h$ falls to a negative minimum and then rises through zero once, so it too has exactly one positive zero.
Each positive value of $t$ returns one $x=\ln t$, so the equation has two real roots.