Let the product of ω1=(8+i)sinθ+(7+4i)cosθ and ω2=(1+8i)sinθ+(4+7i)cosθ be α + iβ,

Question

Let the product of \(\omega_{1}=(8+i) \sin \theta+(7+4 i) \cos \theta\) and \(\omega_{2}=(1+8 i) \sin \theta+(4+7 \mathrm{i}) \cos \theta\) be \(\alpha+\mathrm{i} \beta,\ \mathrm{i}=\sqrt{-1}.\) Let p and q be the maximum and the minimum values of \(\alpha+\beta\) respectively.

(1) 140

(2) 130

(3) 160

(4) 150  

Answer 1

Correct option is: (2) 130 

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

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

\(\omega_{1} \omega_{2}=8 \sin ^{2} \theta+7 \sin \theta \cos \theta+32 \sin \theta \cos \theta+\)

\(28 \cos ^{2} \theta-8 \sin ^{2} \theta-32 \sin \theta \cos \theta-7 \sin \theta \cos \theta\)

\(-28 \cos ^{2} \theta+\mathrm{i}\left(\sin ^{2} \theta+4 \sin \theta \cos \theta+4 \sin \theta \cos \theta\right.\)

\(+16 \cos ^{2} \theta+64 \sin ^{2} \theta+56 \sin \theta \cos \theta+56 \sin \theta\)

\(\left.\cos \theta+49 \cos ^{2} \theta\right)\)

\(\omega_{1} \omega_{2}=0+i\left(65 \sin ^{2} \theta+120 \sin \theta \cos \theta+65 \cos ^{2} \theta\right)\)

\(\alpha+\beta=65+60 \sin 2 q\)

\(\alpha+\left.\beta\right|_{\max }=125\)

\(\alpha+\left.\beta\right|_{\text {min }}=5\)  

= 125 + 5 = 130