Maxima and MinimaHard

Question

$\ Let\ \alpha$ and $\beta$ respectively be the maximum and the minimum values of the function

$$\begin{matrix} f(\theta) = 4 & \left( \sin^{4}\left( \frac{7\pi}{2} - \theta \right) + \sin^{4}(11\pi + \theta) \right) \\ & \ - 2\left( \sin^{6}\left( \frac{3\pi}{2} - \theta \right) + \sin^{6}(9\pi - \theta) \right),\theta \in \mathbf{R} \end{matrix}$$

Then $\alpha + 2\beta$ is equal to :

Options

A.4
B.5
C.3
D.6

Solution

$f(\theta) = 4\left( \sin^{4}\left( \frac{7\pi}{2} - \theta \right) + \sin^{4}(11\pi + \theta) \right) - 2\left( \sin^{6}\left( \frac{3\pi}{2} - \theta \right) + \sin^{6}(9\pi - \theta) \right)$

$${f(\theta) = 4\left( \cos^{4}(\theta) + \sin^{4}(\theta) \right) - 2\left( \cos^{6}\theta + \sin^{6}\theta \right) }{f(\theta) = 4\left( 1 - 2\sin^{2}\theta\cos^{2}\theta \right) - 2\left( 1 - 3\sin^{2}\theta\cos^{2}\theta \right) }{f(\theta) = 2 - 2\sin^{2}\theta\cos^{2}\theta }{f(\theta) = 2 - \frac{\sin^{2}(2\theta)}{2} }{\alpha = f(\theta)_{\text{max~}} = 2 }{\beta = f(\theta)_{\text{min~}} = \frac{3}{2} }{\Rightarrow \alpha + 2\beta = 5 } $$

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