ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 4.1 complex nubers

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

  1. A \(\frac{21}{4} \pi^2\)
  2. B \(8 \pi^2\)
  3. C \(\frac{27}{4} \pi^2\)
  4. D \(6 \pi^2\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{21}{4} \pi^2\)

Step-by-step Solution

Detailed explanation

\(1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0 \) \( \therefore \mathrm{z}+\overline{\mathrm{z}}=2 \operatorname{Re}(\mathrm{z}) \)…
From JEE Mains
Explore more questions on app