Let f( heta ) = 3\left( {{{\sin }^4}\left( {{{3\pi } \over 2} - heta } ight) + {{\sin }^4}(3\pi + heta )} ight) - 2(1 - {\sin ^2}2 heta ) and S = \left\{ { heta \in [0,\pi ]:f'( heta ) = - {{\sqrt 3 } \over 2}} ight\}. If 4\beta = \sum\limits_{ heta \in S} heta , then f(\beta ) is e

Question

Let f(	heta ) = 3\left( {{{\sin }^4}\left( {{{3\pi } \over 2} - 	heta } ight) + {{\sin }^4}(3\pi + 	heta )} ight) - 2(1 - {\sin ^2}2	heta ) and S = \left\{ {	heta \in [0,\pi ]:f'(	heta ) = - {{\sqrt 3 } \over 2}} ight\}. If 4\beta = \sum\limits_{	heta \in S} 	heta , then f(\beta ) is equal to

Accepted Answer

Answer: C. \frac{5}{4} f(	heta)=3\left(\sin ^{4}\left(\frac{3 \pi}{2}-	hetaight)+\sin ^{4}(3 x+	heta)ight)-2\left(1-\sin ^{2} 2 	hetaight) S=\left\{	heta \in[0, \pi]: f^{\prime}(	heta)=-\frac{\sqrt{3}}{2}ight\} \Rightarrow \mathrm{f}(	heta)=3\left(\cos ^{4} 	heta+\sin ^{4} 	hetaight)-2 \cos ^{2} 2 	heta \Rightarrow \mathrm{f}(	heta)=3\left(1-\frac{1}{2} \sin ^{2} 2 	hetaight)-2 \cos ^{2} 2 	heta \Rightarrow \mathrm{f}(	heta)=3-\frac{3}{2} \sin ^{2} 2 	heta-2 \cos ^{2} 	heta =\frac{3}{2}-\frac{1}{2} \cos ^{2} 2 	heta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 	heta}{2}ight) f(	heta)=\frac{5}{4}-\frac{\cos 4 	heta}{4} f^{\prime}(	heta)=\sin 4 	heta \Rightarrow f^{\prime}(	heta)=\sin 4 	heta=-\frac{\sqrt{3}}{2} \Rightarrow 4 	heta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{3} \Rightarrow 	heta=\frac{\mathrm{n} \pi}{4}+(-1)^{\mathrm{n}} \frac{\pi}{12} \begin{aligned} & \Rightarrow 	heta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}ight),\left(\frac{\pi}{2}+\frac{\pi}{12}ight),\left(\frac{3 \pi}{4}-\frac{\pi}{12}ight) \\ & \Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2} \\ & \Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f(\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4} \end{aligned}

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