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 · 2025 · 08 Apr 2025 · Shift 2 · Complex Numbers

Let

A = {θ \in [0, 2 π] : 1 + 10 Re (\frac{2 c o s θ + i s i n θ}{c o s θ - 3 i s i n θ}) = 0}

. Then

\sum_{θ \in A} θ^{2}

is equal to

Answer: (A)

Multiplying numerator and denominator by

cos θ + 3 i sin θ

, the real part of

\frac{2 cos θ + i sin θ}{cos θ - 3 i sin θ}

\frac{2 cos ^{2} θ - 3 sin ^{2} θ}{cos ^{2} θ + 9 sin ^{2} θ}

. The condition

1 + 10 Re = 0

becomes

\frac{2 cos ^{2} θ - 3 sin ^{2} θ}{cos ^{2} θ + 9 sin ^{2} θ} = - \frac{1}{10}

. Cross-multiplying gives

20 cos^{2} θ - 30 sin^{2} θ = - cos^{2} θ - 9 sin^{2} θ

, i.e.

21 cos^{2} θ - 21 sin^{2} θ = 0

, so

cos 2 θ = 0

. For

θ \in [0, 2 π]

this gives

θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}

, and

\sum θ^{2} = \frac{π ^{2}}{16} (1 + 9 + 25 + 49) = \frac{84 π ^{2}}{16} = \frac{21 π ^{2}}{4}

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option