JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Quadratic Equations and Inequalities

For $({\sin }^4\theta +{\cos }^4\theta )x^2+(\sin 2\theta )x+({\sin }^6\theta +{\cos }^6\theta )=0$ to have real roots, discriminant $\ge 0$.

Using ${\sin }^4\theta +{\cos }^4\theta =1-\frac{1}{2}{\sin }^{2}2\theta$ and ${\sin }^6\theta +{\cos }^6\theta =1-\frac{3}{4}{\sin }^{2}2\theta$, and writing $u={\sin }^{2}2\theta \in [0,1]$:



Expanding: $u-4+6u-\frac{3}{2}{u}^2\ge 0\Rightarrow -\frac{3}{2}{u}^2+6u-4\ge 0\Rightarrow 3u^2-12u+8\le 0$.

Roots of $3u^2-12u+8=0$: $u=\frac{12\pm \sqrt{144-96}}{6}=2\pm \frac{2\sqrt{3}}{3}$, i.e. $u\approx 0.845$ or $u\approx 3.155$. The quadratic (opening upward) is $\le 0$ between its roots, so $3u^2-12u+8\le 0$ for $u\in \left[2-\frac{2\sqrt{3}}{3},2+\frac{2\sqrt{3}}{3}\right]$.

Intersecting with the physical range $u\in [0,1]$ (since $u={\sin }^{2}2\theta$) gives $S=\left[2-\frac{2\sqrt{3}}{3},1\right]$.

So $\alpha =2-\frac{2\sqrt{3}}{3}$ (smallest) and $\beta =1$ (largest).