JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 12)
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 $\mathrm{p}+\mathrm{q}$ is equal to :
130
150
160
140
Explanation
$$\begin{aligned} & \omega_1=(8 \sin \theta+7 \cos \theta)+i(\sin \theta+4 \cos \theta) \\ & \omega_2=(\sin \theta+4 \cos \theta)+i(8 \sin \theta+7 \cos \theta) \\ & \alpha=(8 \sin \theta+7 \cos \theta)+(\sin \theta+4 \cos \theta) \\ & \quad-(\sin \theta+4 \cos \theta)+(8 \sin \theta+7 \cos \theta)=0 \\ & \beta=(8 \sin \theta+7 \cos \theta)^2+(\sin \theta+4 \cos \theta)^2 \end{aligned}$$
$$\begin{aligned} & =65 \sin ^2 \theta+65 \cos ^2 \theta+56 \sin 2 \theta+4 \sin 2 \theta \\ & =65+60 \sin 2 \theta \\ & (\alpha+\beta)_{\max }=125=p \\ & (\alpha+\beta)_{\min }=5=q \\ & p+q=130 \end{aligned}$$
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