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

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Complex Numbers

Let the product of

ω_{1} = (8 + i) sin θ + (7 + 4 i) cos θ

and

ω_{2} = (1 + 8 i) sin θ + (4 + 7 i) cos θ

α + i β

i = - 1

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

α + β

respectively. Then

p + q

is equal to :-

Answer: (B)

Separating real and imaginary parts,

ω_{1} = (8 sin θ + 7 cos θ) + i (sin θ + 4 cos θ)

and

ω_{2} = (sin θ + 4 cos θ) + i (8 sin θ + 7 cos θ)

. Writing

a = 8 sin θ + 7 cos θ

and

b = sin θ + 4 cos θ

, these are

ω_{1} = a + ib

and

ω_{2} = b + ia

, so

ω_{1} ω_{2} = (a + ib) (b + ia) = i (a^{2} + b^{2})

. Thus

α = 0

and

α + β = a^{2} + b^{2} = 65 sin^{2} θ + 120 sin θ cos θ + 65 cos^{2} θ = 65 + 60 sin 2 θ

. Its maximum is

125

and its minimum is

5

, so

p + q = 130

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option