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  4. Let f(θ)=3( sin 4((3 π/2)-θ)+ sin 4(3 π+θ))-2(1- sin 2 2 θ) and S= θ ∈[0, π]: f prime(θ)=-(√3/2) . If 4 β= displaystyle∑θ ∈ S θ, then f(β) is equal to

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Q. Let $f(\theta)=3\left(\sin ^4\left(\frac{3 \pi}{2}-\theta\right)+\sin ^4(3 \pi+\theta)\right)-2\left(1-\sin ^2 2 \theta\right)$ and $S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\}$. If $4 \beta=\displaystyle\sum_{\theta \in S} \theta$, then $f(\beta)$ is equal to

JEE MainJEE Main 2023Continuity and Differentiability

Solution:

$f (\theta)=3\left(\sin ^4\left(\frac{3 \pi}{2}-\theta\right)+\sin ^4(3 x+\theta)\right)-2\left(1-\sin ^2 2 \theta\right) $
$ S =\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\} $
$ \Rightarrow f(\theta)=3\left(\cos ^4 \theta+\sin ^4 \theta\right)-2 \cos ^2 2 \theta $
$ \Rightarrow f(\theta)=3\left(1-\frac{1}{2} \sin ^2 2 \theta\right)-2 \cos ^2 2 \theta $
$ \Rightarrow f(\theta)=3-\frac{3}{2} \sin ^2 2 \theta-2 \cos ^2 \theta$
$=\frac{3}{2}-\frac{1}{2} \cos ^2 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right)$
$f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4}$
$ f^{\prime}(\theta)=\sin 4 \theta$
$ \Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2} $
$\Rightarrow 4 \theta= n \pi+(-1)^{ n } \frac{\pi}{3}$
$ \Rightarrow \theta=\frac{ n \pi}{4}+(-1)^{ n } \frac{\pi}{12}$
$ \Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right) $
$ \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}$