JEE Main 2026 (Online) 23rd January Morning ShiftQuestion 1
Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta)=4\left(\sin^4\left(\frac{7\pi}{2}-\theta\right)+\sin^4(11\pi+\theta)\right)-2\left(\sin^6\left(\frac{3\pi}{2}-\theta\right)+\sin^6(9\pi-\theta)\right)$, $\theta\in\mathbf{R}$. Then $\alpha+2\beta$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. Reduce each angle; $f$ becomes a linear function of $p=\sin^2\theta\cos^2\theta$. Formulas used. (1) $\sin^4\theta+\cos^4\theta=1-2p$, $\sin^6\theta+\cos^6\theta=1-3p$, with $p=\sin^2\theta\cos^2\theta=\frac14\sin^22\theta\in\left[0,\frac14\right]$Step 1 - Reduce the angles. $$\sin\left(\tfrac{7\pi}{2}-\theta\right)=-\cos\theta,\quad\sin(11\pi+\theta)=-\sin\theta,$$ $$\sin\left(\tfrac{3\pi}{2}-\theta\right)=-\cos\theta,\quad\sin(9\pi-\theta)=\sin\theta.$$ Even powers remove the signs. Step 2 - Simplify. By (1), $$f=4(1-2p)-2(1-3p)=2-2p.$$ Step 3 - Max and min. $p=0$ gives $\alpha=2$; $p=\frac14$ gives $\beta=\frac32$. Step 4 - Value. $\alpha+2\beta=2+3=5$. Answer: (C)