Let the product of ω 1 =(8+i) sin θ+(7+4 i) cos θ and ω 2 =(1+8 i) sin θ+(4+7 i) cos θ be α+i β, i=√(-1) . Let p and q be the maximum and the minimum values of α+β respectively. Then p+ q is equal to:

Mathematics · Complex Numbers

JEE Main 2025 — 4 April, Evening Shift — Question 21

Let the product of $\omega_{1}=(8+i) \sin \theta+(7+4 i) \cos \theta$ and $\omega_{2}=(1+8 i) \sin \theta+(4+7 i) \cos \theta$ be

$\alpha+i \beta, i=\sqrt{-1}$ . Let $p$ and $q$ be the maximum and the minimum values of $\alpha+\beta$ respectively. Then $p+$ $q$ is

equal to:

$\omega_{1}=(8 \sin \theta+7 \cos \theta)+i(\sin \theta+4 \cos \theta)$

$\omega_{2}=(\sin \theta+4 \cos \theta)+i(8 \sin \theta+7 \cos \theta)$

$\alpha=(8 \sin \theta+7 \cos \theta)+(\sin \theta+4 \cos \theta)$

$-(\sin \theta+4 \cos \theta)+(8 \sin \theta+7 \cos \theta)=0$

$\beta=(8 \sin \theta+7 \cos \theta)^{2}+(\sin \theta+4 \cos \theta)^{2}$

$=65 \sin ^{2} \theta+65 \cos ^{2} \theta+56 \sin 2 \theta+4 \sin 2 \theta$

$=65+60 \sin 2 \theta$

$(\alpha+\beta)_{\max }=125=p$

$(\alpha+\beta)_{\min }=5=q$

$p+q=130$

Answer key and solution verified before publishing.

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