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JEE Main 20245 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Application of Derivatives

Let a rectangle A B C D of sides 2 and 4 be inscribed in another...

Let a rectangle ABCDA B C D of sides 2 and 4 be inscribed in another rectangle PQRSP Q R S such that the vertices of the rectangle ABCDA B C D lie on the sides of the rectangle PQRSP Q R S. Let aa and bb be the sides of the rectangle PQRSP Q R S when its area is maximum. Then (a+b)2(a+b)^2 is equal to :

Options

  1. A72
  2. B60
  3. C64
  4. D80

Correct answer

A. 72

Step-by-step solution

 Area =(4cos⁡θ+2sin⁡θ)(2cos⁡θ+4sin⁡θ)=8cos⁡2θ+16sin⁡θcos⁡θ+4sin⁡θcos⁡θ+8sin⁡2θ=8+20sin⁡θcos⁡θ=8+10sin⁡2θ Max Area =8+10=18(sin⁡2θ=1)θ=45∘(a+b)2=(4cos⁡θ+2sin⁡θ+2cos⁡θ+4sin⁡θ)2=(6cos⁡θ+6sin⁡θ)2=36(sin⁡θ+cos⁡θ)2=36(2)2\begin{aligned} & \text{ Area }=(4 \cos \theta+2 \sin \theta)(2 \cos \theta+4 \sin \theta) \\ & =8 \cos ^2 \theta+16 \sin \theta \cos \theta+4 \sin \theta \cos \theta+8 \sin ^2 \theta \\ & =8+20 \sin \theta \cos \theta \\ & =8+10 \sin 2 \theta \\ & \text{ Max Area }=8+10=18(\sin 2 \theta=1) \theta=45^{\circ} \\ & (a+b)^2=(4 \cos \theta+2 \sin \theta+2 \cos \theta+4 \sin \theta)^2 \\ & =(6 \cos \theta+6 \sin \theta)^2 \\ & =36(\sin \theta+\cos \theta)^2 \\ & =36(\sqrt{2})^2\end{aligned}=72$
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