JEE Main 2020MathematicsDifferentiationMediumMCQ

JEE Main 2020 — Differentiation Question with Solution

JEE Main 2020 (09 Jan Shift 2)

Question

If x=2sinθ-sin2θ and y=2cosθ-cos2θ, θ∈0,2π, then d2ydx2 at θ=π is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B-38

Step-by-step explanation

dxdθ=2cosθ-2cos2θ

dydθ=-2sinθ+2sin2θ

∴ dydx=sin2θ-sinθcosθ-cos2θ

=2sinθ2.cos3θ22sinθ2.sin3θ2=cot3θ2

d2ydx2=ddθdydxdθdx=-32cosec23θ2.dθdx

⇒d2ydx2=-32cosec23θ22cosθ-cos2θ

⇒d2ydx2θ=π=34-1-1=38

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About this question

This is a previous-year question from JEE Main 2020, covering the Differentiation chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.