Let A= \ θ in[0,2 π]: 1+10 Re (2 cos θ+i sin θ/cos θ-3 i sin θ )=0 \ . Then sum θ in A θ 2 is equal to

Mathematics · Complex Numbers

JEE Main 2025 — 8 April, Evening Shift — Question 36

Let $A=\left\{\theta \in[0,2 \pi]: 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0\right\}$ . Then $\sum_{\theta \in A} \theta^{2}$ is equal to

Option A:
$\frac{27}{4} \pi^{2}$
Option B:
$8 \pi^{2}$
Option C:
$6 \pi^{2}$
Option D:
$\frac{21}{4} \pi^{2}$
Correct

Answer: D

Step-by-step solution

$\operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)$

\begin{aligned} & \operatorname{Re}\left(\frac{(2 \cos \theta+i \sin \theta)(\cos \theta+3 i \sin \theta)}{\cos ^{2} \theta+9 \sin ^{2} \theta}\right) \\& =\frac{2 \cos ^{2} \theta-3 \sin ^{2} \theta}{1+8 \sin ^{2} \theta} \\& \text { Now, } 1+10\left(\frac{2 \cos ^{2} \theta-3 \sin ^{2} \theta}{1+8 \sin ^{2} \theta}\right)=0 \\& \quad=1+8 \sin ^{2} \theta+20 \cos ^{2} \theta-30 \sin ^{2} \theta=0 \\& \quad=1-22 \sin ^{2} \theta+20 \cos ^{2} \theta=0 \\& \quad=1+20(\cos 2 \theta)-2 \sin ^{2} \theta=0 \\& \quad=20 \cos 2 \theta+\cos 2 \theta=0 \\& \quad=21 \cos 2 \theta=0 \end{aligned}

2 \theta=(2 n+1) \frac{\pi}{2}, n \in I

2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2}

\Rightarrow \theta=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}

\sum_{\theta \in 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}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Mathematics
Chapter: Complex Numbers
Topic: Properties of Complex Numbers