Let the product of ω_1=(8+i) θ+(7+4 i) θ and ω_2=(1+8 i) θ+(4+7 i) θ be α+i β , i= -1 . Let p and q be the maximum and the minimum values of α+β respectively.

Correct answer: B. 130 aligned & ω_1=(8 θ+7 θ)+i( θ+4 θ) & ω_2=( θ+4 θ)+i(8 θ+7 θ) & ω_1 ω_2=8 ^2 θ+7 θ θ+32 θ θ+ & 28 ^2 θ-8 ^2 θ-32 θ θ-7 θ θ & -28 ^2 θ+i( ^2 θ+4 θ θ+4 θ θ. & +16 ^2 θ+64 ^2 θ+56 θ θ+56 θ & . θ+49 ^2 θ) & ω_1 ω_2=0+i(65 ^2 θ+120 θ θ+65 ^2 θ) & α+β=65+60 2 q & α+.β|_ =125 & α+.β|_ =5 aligned Ans. =125+5=130 option (2)

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Complex Number

Let the product of ω_1=(8+i) θ+(7+4 i) θ and ω_2=(1+8 i) θ+(4+7 i)...

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

\alpha+i \beta

\mathrm{i}=\sqrt{-1}

. Let p and q be the maximum and the minimum values of

\alpha+\beta

respectively.

Options

A $140$
B $130$
C $160$
D $150$

Correct answer

B. $130$

Step-by-step solution

\begin{aligned}& \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) \\ & \omega_1 \omega_2=8 \sin ^2 \theta+7 \sin \theta \cos \theta+32 \sin \theta \cos \theta+ \\ & 28 \cos ^2 \theta-8 \sin ^2 \theta-32 \sin \theta \cos \theta-7 \sin \theta \cos \theta \\ & -28 \cos ^2 \theta+i\left(\sin ^2 \theta+4 \sin \theta \cos \theta+4 \sin \theta \cos \theta\right. \\ & +16 \cos ^2 \theta+64 \sin ^2 \theta+56 \sin \theta \cos \theta+56 \sin \theta \\ & \left.\cos \theta+49 \cos ^2 \theta\right) \\ & \omega_1 \omega_2=0+i\left(65 \sin ^2 \theta+120 \sin \theta \cos \theta+65 \cos ^2 \theta\right) \\ & \alpha+\beta=65+60 \sin 2 q \\ & \alpha+\left.\beta\right|_{\max }=125 \\ & \alpha+\left.\beta\right|_{\min }=5\end{aligned}

\text{ Ans. }=125+5=130$option (2)

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