JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 4)
Let $$f(\theta ) = 3\left( {{{\sin }^4}\left( {{{3\pi } \over 2} - \theta } \right) + {{\sin }^4}(3\pi + \theta )} \right) - 2(1 - {\sin ^2}2\theta )$$ and $$S = \left\{ {\theta \in [0,\pi ]:f'(\theta ) = - {{\sqrt 3 } \over 2}} \right\}$$. If $$4\beta = \sum\limits_{\theta \in S} \theta $$, then $$f(\beta )$$ is equal to
$$\frac{9}{8}$$
$$\frac{3}{2}$$
$$\frac{5}{4}$$
$$\frac{11}{8}$$
Explanation
$f(\theta)=3\left(\sin ^{4}\left(\frac{3 \pi}{2}-\theta\right)+\sin ^{4}(3 x+\theta)\right)-2\left(1-\sin ^{2} 2 \theta\right)$
$S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\}$
$\Rightarrow \mathrm{f}(\theta)=3\left(\cos ^{4} \theta+\sin ^{4} \theta\right)-2 \cos ^{2} 2 \theta$
$\Rightarrow \mathrm{f}(\theta)=3\left(1-\frac{1}{2} \sin ^{2} 2 \theta\right)-2 \cos ^{2} 2 \theta$
$\Rightarrow \mathrm{f}(\theta)=3-\frac{3}{2} \sin ^{2} 2 \theta-2 \cos ^{2} \theta$
$=\frac{3}{2}-\frac{1}{2} \cos ^{2} 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right)$
$f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4}$
$f^{\prime}(\theta)=\sin 4 \theta$
$\Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2}$
$\Rightarrow 4 \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{3}$
$\Rightarrow \theta=\frac{\mathrm{n} \pi}{4}+(-1)^{\mathrm{n}} \frac{\pi}{12}$
$$ \begin{aligned} & \Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right) \\\\ & \Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2} \\\\ & \Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f(\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4} \end{aligned} $$
$S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\}$
$\Rightarrow \mathrm{f}(\theta)=3\left(\cos ^{4} \theta+\sin ^{4} \theta\right)-2 \cos ^{2} 2 \theta$
$\Rightarrow \mathrm{f}(\theta)=3\left(1-\frac{1}{2} \sin ^{2} 2 \theta\right)-2 \cos ^{2} 2 \theta$
$\Rightarrow \mathrm{f}(\theta)=3-\frac{3}{2} \sin ^{2} 2 \theta-2 \cos ^{2} \theta$
$=\frac{3}{2}-\frac{1}{2} \cos ^{2} 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right)$
$f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4}$
$f^{\prime}(\theta)=\sin 4 \theta$
$\Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2}$
$\Rightarrow 4 \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{3}$
$\Rightarrow \theta=\frac{\mathrm{n} \pi}{4}+(-1)^{\mathrm{n}} \frac{\pi}{12}$
$$ \begin{aligned} & \Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right) \\\\ & \Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2} \\\\ & \Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f(\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4} \end{aligned} $$
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