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Complex Numbers question

2025 · 4 Apr · Shift 2 · Q37
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Complex Numbers question

2025 · 4 Apr · Shift 2 · Q37

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let the product of ω1=(8+i)sin⁡θ+(7+4i)cos⁡θ\omega_1=(8+i) \sin \theta+(7+4 i) \cos \thetaω1​=(8+i)sinθ+(7+4i)cosθ and ω2=(1+8i)sin⁡θ+(4+7i)cos⁡θ\omega_2=(1+8 i) \sin \theta+(4+7 i) \cos \thetaω2​=(1+8i)sinθ+(4+7i)cosθ be α+iβ\alpha+i \betaα+iβ, i=−1i=\sqrt{-1}i=−1​. Let p and q be the maximum and the minimum values of α+β\alpha+\betaα+β respectively. Then p+q\mathrm{p}+\mathrm{q}p+q is equal to :
  1. A
    130
  2. B
    150
  3. C
    160
  4. D
    140
View written solutionFree

Correct answer: A

  1. Write the given complex numbers

We have

ω1=(8+i)sin⁡θ+(7+4i)cos⁡θ\omega_1=(8+i)\sin\theta+(7+4i)\cos\thetaω1​=(8+i)sinθ+(7+4i)cosθ

and

ω2=(1+8i)sin⁡θ+(4+7i)cos⁡θ.\omega_2=(1+8i)\sin\theta+(4+7i)\cos\theta.ω2​=(1+8i)sinθ+(4+7i)cosθ.

Let

ω1=a1+ib1,ω2=a2+ib2.\omega_1=a_1+ib_1,\qquad \omega_2=a_2+ib_2.ω1​=a1​+ib1​,ω2​=a2​+ib2​.

Then

a1=8sin⁡θ+7cos⁡θ,b1=sin⁡θ+4cos⁡θ,a_1=8\sin\theta+7\cos\theta, \qquad b_1=\sin\theta+4\cos\theta,a1​=8sinθ+7cosθ,b1​=sinθ+4cosθ, a2=sin⁡θ+4cos⁡θ,b2=8sin⁡θ+7cos⁡θ.a_2=\sin\theta+4\cos\theta, \qquad b_2=8\sin\theta+7\cos\theta.a2​=sinθ+4cosθ,b2​=8sinθ+7cosθ.

So observe that

a2=b1,b2=a1.a_2=b_1,\qquad b_2=a_1.a2​=b1​,b2​=a1​.
  1. Compute ω1ω2\omega_1\omega_2ω1​ω2​

Using

(a1+ib1)(a2+ib2)=(a1a2−b1b2)+i(a1b2+a2b1),(a_1+ib_1)(a_2+ib_2)=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1),(a1​+ib1​)(a2​+ib2​)=(a1​a2​−b1​b2​)+i(a1​b2​+a2​b1​),

we get

α=a1a2−b1b2,β=a1b2+a2b1.\alpha=a_1a_2-b_1b_2, \qquad \beta=a_1b_2+a_2b_1.α=a1​a2​−b1​b2​,β=a1​b2​+a2​b1​.

Since a2=b1a_2=b_1a2​=b1​ and b2=a1b_2=a_1b2​=a1​,

α=a1b1−b1a1=0.\alpha=a_1b_1-b_1a_1=0.α=a1​b1​−b1​a1​=0.

Also,

β=a12+b12.\beta=a_1^2+b_1^2.β=a12​+b12​.

Hence

α+β=a12+b12.\alpha+\beta=a_1^2+b_1^2.α+β=a12​+b12​.

Therefore

α+β=(8sin⁡θ+7cos⁡θ)2+(sin⁡θ+4cos⁡θ)2.\alpha+\beta=(8\sin\theta+7\cos\theta)^2+(\sin\theta+4\cos\theta)^2.α+β=(8sinθ+7cosθ)2+(sinθ+4cosθ)2.
  1. Expand the expression
(8sin⁡θ+7cos⁡θ)2=64sin⁡2θ+49cos⁡2θ+112sin⁡θcos⁡θ,(8\sin\theta+7\cos\theta)^2=64\sin^2\theta+49\cos^2\theta+112\sin\theta\cos\theta,(8sinθ+7cosθ)2=64sin2θ+49cos2θ+112sinθcosθ, (sin⁡θ+4cos⁡θ)2=sin⁡2θ+16cos⁡2θ+8sin⁡θcos⁡θ.(\sin\theta+4\cos\theta)^2=\sin^2\theta+16\cos^2\theta+8\sin\theta\cos\theta.(sinθ+4cosθ)2=sin2θ+16cos2θ+8sinθcosθ.

Adding,

α+β=65sin⁡2θ+65cos⁡2θ+120sin⁡θcos⁡θ.\alpha+\beta=65\sin^2\theta+65\cos^2\theta+120\sin\theta\cos\theta.α+β=65sin2θ+65cos2θ+120sinθcosθ.

Since

sin⁡2θ+cos⁡2θ=1,\sin^2\theta+\cos^2\theta=1,sin2θ+cos2θ=1,

we get

α+β=65+120sin⁡θcos⁡θ.\alpha+\beta=65+120\sin\theta\cos\theta.α+β=65+120sinθcosθ.

Using

2sin⁡θcos⁡θ=sin⁡2θ,2\sin\theta\cos\theta=\sin 2\theta,2sinθcosθ=sin2θ,

so

sin⁡θcos⁡θ=sin⁡2θ2,\sin\theta\cos\theta=\frac{\sin 2\theta}{2},sinθcosθ=2sin2θ​,

thus

α+β=65+60sin⁡2θ.\alpha+\beta=65+60\sin 2\theta.α+β=65+60sin2θ.
  1. Find maximum and minimum values

Since

−1≤sin⁡2θ≤1,-1\le \sin 2\theta\le 1,−1≤sin2θ≤1,

we have

max⁡(α+β)=65+60=125,\max(\alpha+\beta)=65+60=125,max(α+β)=65+60=125, min⁡(α+β)=65−60=5.\min(\alpha+\beta)=65-60=5.min(α+β)=65−60=5.

So

p=125,q=5.p=125,\qquad q=5.p=125,q=5.

Hence

p+q=125+5=130.p+q=125+5=130.p+q=125+5=130.
  1. Compare with stored answer

Derived answer is 130, which matches option A.

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