[UPLIFTJEE]

Let $\mathrm{A}=\left\{\theta \in[0,2 \pi]: 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}\right)=0\right\}$. Then $\sum_{\theta \in \mathrm{A}} \theta^{2}$ is equal to [JEE Main 08 April 2025 S II]
(A) $\frac{21}{4} \pi^{2}$
(B) $8 \pi^{2}$
(C) $\frac{27}{4} \pi^{2}$
(D) $6 \pi^{2}$

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[UPLIFTJEE]

Solution

$1+10 \operatorname{Re}\left(\frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}\right)=0(\therefore 2 \operatorname{Re}(\mathrm{z})=\mathrm{z}+\overline{\mathrm{z}})$

\[ \begin{aligned} & \Rightarrow 1+5\left(\frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}+\frac{2 \cos \theta-\mathrm{i} \sin \theta}{\cos \theta+3 \mathrm{i} \sin \theta}\right)=0 \\ & \Rightarrow \frac{\left(2 \cos ^{2} \theta-3 \sin ^{2} \theta\right)+\left(2 \cos ^{2} \theta\right)-\left(3 \sin ^{2} \theta\right)}{\cos ^{2} \theta+9 \sin ^{2} \theta}=-\frac{1}{5} \\ & \Rightarrow \frac{2 \cos ^{2} \theta-3 \sin ^{2} \theta}{\cos ^{2} \theta+9 \sin ^{2} \theta}=-\frac{1}{10} \Rightarrow 20 \cos ^{2} \theta-30 \sin ^{2} \theta=-\cos ^{2} \theta-9 \sin ^{2} \theta \\ & \Rightarrow 21 \cos ^{2} \theta-21 \sin ^{2} \theta=0 \Rightarrow \cos 2 \theta=0 \text { As } \theta \in[0,2 \pi] \\ & \Rightarrow 2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \end{aligned} \]

Now $\sum_{\theta \in \mathrm{A}} \theta^{2}=\frac{\pi^{2}}{16}+\frac{9 \pi^{2}}{16}+\frac{25 \pi^{2}}{16}+\frac{49 \pi^{2}}{16}=\frac{84 \pi^{2}}{16}=\frac{21 \pi^{2}}{4}$

[UPLIFTJEE]

Answer

A

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