Let A= \ θ ∈[0,2 π]: 1+10 Re((2 θ+i θ)/( θ-3 i θ))=0\ . Then ∑_ θ ∈ A θ^2 is equal to

Correct answer: A. (21)/(4) π^2 aligned & 1+10 Re((2 θ+i θ)/( θ-3 i θ))=0 & z+ z =2 Re(z) & 2 θ+i θ θ-3 i θ + 2 θ-i θ θ+3 i θ =2 ×((-1)/(10)) & ((2 ^2 θ-3 ^2 θ)+(2 ^2 θ)-(3 ^2 θ))/( ^2 θ+9 ^2 θ)=(-2)/(10) & (2 ^2 θ-3 ^2 θ)/( ^2 θ+9 ^2 θ)=(-1)/(10) & 20 ^2 θ-30 ^2 θ=- ^2 θ-9 ^2 θ & 21 ^2 θ-21 ^2 θ=0 & 2 θ=0 & 2 θ=(π)/(2), (3 π)/(2), (5 π)/(2), (7 π)/(2) & ∑ θ^2=(π^2)/(16)+(9 π^2)/(16)+(25 π^2)/(16)+(49 π^2)/(16)=(84 π^2)/(16)=(21 π^2)/(4) aligned

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Complex Number

Let A= \ θ ∈[0,2 π]: 1+10 Re((2 θ+i θ)/( θ-3 i θ))=0\ . Then ∑_ θ ∈...

Let

\mathrm{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

Options

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

Correct answer

A. $\frac{21}{4} \pi^2$

Step-by-step solution

\begin{aligned} & 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0 \\ & \therefore \mathrm{z}+\overline{\mathrm{z}}=2 \operatorname{Re}(\mathrm{z}) \\ & \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}=2 \times\left(\frac{-1}{10}\right) \\ & \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{-2}{10} \\ & \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 \\ & 21 \cos ^2 \theta-21 \sin ^2 \theta=0 \\ & \Rightarrow \cos 2 \theta=0 \\ & 2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \\ & \Rightarrow \sum \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}\end{aligned}

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