JEE Main 202331 Jan, Morning shiftMathematicsMediumLimit, Continuity and Differentiability420× since 2002Let y=f(x)=sin3(π3(cos(π32(−4x3+5x2+1)32)))y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right)y=f(x)=sin3(3π(cos(32π(−4x3+5x2+1)23))). Then, at x = 1,AOption A. 2y′+3π2y=02 y^{\prime}+\sqrt{3} \pi^{2} y=02y′+3π2y=0BOption B. y′+3π2y=0y^{\prime}+3 \pi^{2} y=0y′+3π2y=0COption C. 2y′−3π2y=0\sqrt{2} y^{\prime}-3 \pi^{2} y=02y′−3π2y=0DOption D. 2y′+3π2y=02 y^{\prime}+3 \pi^{2} y=02y′+3π2y=0CheckSkip · show solution