Let f ( ) = 3 ( sin 4 ( 3 2 - ) + sin 4 ( 3 ) ) - 2 ( 1 - sin 2 2 ) and S = { [ 0 , ] : f ' () = - 3 2 } . If 4 =S, then f ( ) is equal to [2023]

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Solution

Q.
Let $f (θ) = 3 (\sin^{4} (\frac{3 π}{2} - θ) + \sin^{4} (3 π + θ)) - 2 (1 - \sin^{2} 2 θ)$ and $S = {θ \in [0, π] : f' (θ) = - \frac{\sqrt{3}}{2}} .$ If $4 β = \sum_{θ \in S} θ,$ then $f (β)$ is equal to [2023]

1 $\frac{11}{8}$

2 $\frac{3}{2}$

3 $\frac{9}{8}$

4 $\frac{5}{4}$

Ans.
(4)

$Given f (θ) = 3 (\sin^{4} (\frac{3 π}{2} - θ) + \sin^{4} (3 π + θ))$ $- 2 (1 - \sin^{2} 2 θ)$

$= 3 (\cos^{4} θ + \sin^{4} θ)$ $- 2 \cos^{2} 2 θ$

$= 3 (1 - \frac{1}{2} \sin^{2} 2 θ) - 2 \cos^{2} 2 θ = 3 - \frac{3}{2} \sin^{2} 2 θ - 2 \cos^{2} 2 θ$

$= 3 - \frac{3}{2} (1 - \cos^{2} 2 θ) - 2 \cos^{2} 2 θ = \frac{3}{2} - \frac{1}{2} \cos^{2} 2 θ$

$= \frac{3}{2} - \frac{1}{2} (\frac{1 + \cos 4 θ}{2}) = \frac{5}{4} - \frac{\cos 4 θ}{4}$

$Now, f' (θ) = \sin 4 θ = \frac{- \sqrt{3}}{2}$

$\Rightarrow 4 θ = n π + {(- 1)}^{n} \frac{- π}{3} \Rightarrow θ = \frac{n π}{4} + {(- 1)}^{n} \frac{- π}{12}$

$\Rightarrow θ = (\frac{π}{4} + \frac{π}{12}), (\frac{π}{2} - \frac{π}{12}), (\frac{3 π}{4} + \frac{π}{12}), (π - \frac{π}{12})$

$\Rightarrow 4 β = \frac{π}{4} + \frac{π}{2} + \frac{3 π}{4} = \frac{5 π}{2}$

$\Rightarrow β = \frac{5 π}{8} \Rightarrow f (β) = \frac{5}{4} - \frac{\cos \frac{5 π}{2}}{4} = \frac{5}{4}$

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