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…
- $140$
- $130$
- $160$
- $150$
Solution
& \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)
Asked in: JEE Main 2025 (04 Apr Shift 2)