JEE Main 2024 — 8 April, Shift 1 — Question 26

$f(\theta)=\frac{\sin ^{4} \theta+3 \cos ^{2} \theta}{\sin ^{4} \theta+\cos ^{2} \theta}$

$f(\theta)=1+\frac{2 \cos ^{2} \theta}{\sin ^{4} \theta+\cos ^{2} \theta}$

$f(\theta)=\frac{2 \cos ^{2} \theta}{\cos ^{4} \theta-\cos ^{2} \theta+1}+1$

$f(\theta)=\frac{2}{\cos ^{2} \theta+\sec ^{2} \theta-1}+1$

$\left.\mathrm{f}(\theta)\right|_{\min .}=1$

$\mathrm{f}(\theta)_{\text {max. }}=3$

$\mathrm{S}=\frac{64}{1-1 / 3}=96$