Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
If x=2sinθ-sin2θ and y=2cosθ-cos2θθ0,2π, then d2ydx2 at θ=π is:
MathematicsDifferentiationJEE MainJEE Main 2020 (09 Jan Shift 2)
Options:
  • A 34
  • B -38
  • C 32
  • D -34
Solution:
2896 Upvotes Verified Answer
The correct answer is: -38

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

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.