MathematicsMedium63×since 2002Q3673Let 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,A2y′+3π2y=02 y^{\prime}+\sqrt{3} \pi^{2} y=02y′+3π2y=0By′+3π2y=0y^{\prime}+3 \pi^{2} y=0y′+3π2y=0C2y′−3π2y=0\sqrt{2} y^{\prime}-3 \pi^{2} y=02y′−3π2y=0D2y′+3π2y=02 y^{\prime}+3 \pi^{2} y=02y′+3π2y=0Check answerSkip