Exams Joint Entrance Examination Mathematics Complex Numbers Question 21 Question 21 of 455 Medium MCQ
Let the product of ω 1 = ( 8 + i ) sin θ + ( 7 + 4 i ) cos θ {\omega }_{1}=(8+i)\sin \theta +(7+4i)\cos \theta ω 1 = ( 8 + i ) sin θ + ( 7 + 4 i ) cos θ and ω 2 = ( 1 + 8 i ) sin θ + ( 4 + 7 i ) cos θ {\omega }_{2}=(1+8i)\sin \theta +(4+7i)\cos \theta ω 2 = ( 1 + 8 i ) sin θ + ( 4 + 7 i ) cos θ be α + i β \alpha +i\beta α + i β , i = − 1 i=\sqrt{-1} i = − 1 . Let p and q be the maximum and the minimum values of α + β \alpha +\beta α + β respectively. Then p + q p+q p + q is equal to :
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ω 1 = ( 8 sin θ + 7 cos θ ) + i ( sin θ + 4 cos θ ) ω 2 = ( sin θ + 4 cos θ ) + i ( 8 sin θ + 7 cos θ ) α = ( 8 sin θ + 7 cos θ ) + ( sin θ + 4 cos θ ) − ( sin θ + 4 cos θ ) + ( 8 sin θ + 7 cos θ ) = 0 β = ( 8 sin θ + 7 cos θ ) 2 + ( sin θ + 4 cos θ ) 2 \begin{gathered} {\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} \end{gathered} ω 1 = ( 8 sin θ + 7 cos θ ) + i ( sin θ + 4 cos θ ) ω 2 = ( sin θ + 4 cos θ ) + i ( 8 sin θ + 7 cos θ ) α = ( 8 sin θ + 7 cos θ ) + ( sin θ + 4 cos θ ) − ( sin θ + 4 cos θ ) + ( 8 sin θ + 7 cos θ ) = 0 β = ( 8 sin θ + 7 cos θ ) 2 + ( sin θ + 4 cos θ ) 2
= 65 sin 2 θ + 65 cos 2 θ + 56 sin 2 θ + 4 sin 2 θ = 65 + 60 sin 2 θ ( α + β ) m a x = 125 = p ( α + β ) m i n = 5 = q p + q = 130 \begin{gathered} =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{gathered} = 65 sin 2 θ + 65 cos 2 θ + 56 sin 2 θ + 4 sin 2 θ = 65 + 60 sin 2 θ ( α + β ) ma x = 125 = p ( α + β ) min = 5 = q p + q = 130 Answer the question to unlock