JEE Main 20209 Jan, Evening shiftMathematicsHardLimit, Continuity and Differentiability420× since 2002If x=2sinθ−sin2θx = 2\sin \theta - \sin 2\thetax=2sinθ−sin2θ and y=2cosθ−cos2θy = 2\cos \theta - \cos 2\thetay=2cosθ−cos2θ, θ∈[0,2π]\theta \in \left[ {0,2\pi } \right]θ∈[0,2π], then d2ydx2{{{d^2}y} \over {d{x^2}}}dx2d2y at θ\thetaθ = π\piπ is :AOption A. 38{3 \over 8}83BOption B. 32{3 \over 2}23COption C. 34{3 \over 4}43DOption D. -34{3 \over 4}43CheckSkip · show solution
