Let the product of ω₁=(8+i) sin θ+(7+4 i) cos θ and ω₂=(1+8 i) sin θ+(4+7 i) cos θ be 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 :

Solution

Maths sir · 4 steps

Name the two real expressions A and B. Then the product collapses to i(A²+B²) and only a sine is left.

1. Spot the pattern. Let A = 8 sin θ + 7 cos θ and B = sin θ + 4 cos θ. Collecting real and imaginary parts gives ω₁ = A + iB and ω₂ = B + iA.
2. Multiply. (A+iB)(B+iA) = AB + iA² + iB² - AB = i(A²+B²). So α = 0 and β = A² + B².

   α+β = (8 sin θ+7 cos θ)²+(sin θ+4 cos θ)² = 65 + 120 sin θ cos θ = 65 + 60 sin 2θ

3. Expand carefully. The squares give 65 sin² θ + 65 cos² θ = 65. The cross terms give 112 sin θ cos θ + 8 sin θ cos θ = 120 sin θ cos θ = 60 sin 2θ.
4. So. sin 2θ runs from -1 to 1, so p = 65 + 60 = 125 and q = 65 - 60 = 5. Then p + q = 130, option (a).

The idea to keep. When two complex numbers look like A+iB and B+iA, their product is i(A²+B²). Spot the swapped parts before you expand anything.

Answer from the NTA final answer key. Solution by PrepWiser.