Area =(4cosθ+2sinθ)(2cosθ+4sinθ)=8cos2θ+16sinθcosθ+4sinθcosθ+8sin2θ=8+20sinθcosθ=8+10sin2θ Max Area =8+10=18(sin2θ=1)θ=45∘(a+b)2=(4cosθ+2sinθ+2cosθ+4sinθ)2=(6cosθ+6sinθ)2=36(sinθ+cosθ)2=36(2)2 =72
JEE Main 2024 — Mathematics Calculus
Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a+b)2 is equal to :
Held on 5 Apr 2024 · Verified 6 Jul 2026.
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