Q27 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
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 :
- A4
- B6
- Correct answerC5
- D3
Solution
$f(\theta) = 4\left\{\sin^4\left(\dfrac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right\} - 2\left\{\sin^2\left(\dfrac{3\pi}{2} - \theta\right) + \sin^2(9\pi - \theta)\right\}$
$f(\theta) = 4(\cos^4(\theta) + \sin^4(\theta)) - 2(\cos^6\theta + \sin^6\theta)$
$f(\theta) = 4(1 - 2\sin^2\theta\cos^2\theta) - 2(1 - 3\sin^2\theta\cos^2\theta)$
$f(\theta) = 2 - 2\sin^2\theta\cos^2\theta$
$f(\theta) = 2 - \dfrac{\sin^2(2\theta)}{2}$
$\alpha = f(\theta)_{max} = 2$
$\beta = f(\theta)_{min} = \dfrac{3}{2}$
$\Rightarrow \alpha + 2\beta = 5$
$f(\theta) = 4(\cos^4(\theta) + \sin^4(\theta)) - 2(\cos^6\theta + \sin^6\theta)$
$f(\theta) = 4(1 - 2\sin^2\theta\cos^2\theta) - 2(1 - 3\sin^2\theta\cos^2\theta)$
$f(\theta) = 2 - 2\sin^2\theta\cos^2\theta$
$f(\theta) = 2 - \dfrac{\sin^2(2\theta)}{2}$
$\alpha = f(\theta)_{max} = 2$
$\beta = f(\theta)_{min} = \dfrac{3}{2}$
$\Rightarrow \alpha + 2\beta = 5$
Hence, the correct answer is option: 'C'.