JEE Main 2024DifferentiationDifferentiationHardQuestionLet 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,OptionsA2y′+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 AnswerHide SolutionSolutionf(x)=sin3(π3cos(π32(−4x3+5x2+1)3/2))f(x)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right)f(x)=sin3(3πcos(32π(−4x3+5x2+1)3/2)) f′(x)=3sin2(π3cos(π32(−4x3+5x2+1)3/2))cos(π3cos(π32(−4x3+5x2+1)3/2))π3(−sin(π32(−4x3+5x2+1)3/2))π3232(−4x3+5x3+1)1/2(−12x2+10x)\begin{aligned} & f^{\prime}(x)=3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3}\left(-\sin \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3 \sqrt{2}} \frac{3}{2}\left(-4 x^{3}+5 x^{3}+1\right)^{1 / 2}\left(-12 x^{2}+10 x\right) \end{aligned}f′(x)=3sin2(3πcos(32π(−4x3+5x2+1)3/2))cos(3πcos(32π(−4x3+5x2+1)3/2))3π(−sin(32π(−4x3+5x2+1)3/2))32π23(−4x3+5x3+1)1/2(−12x2+10x) f′(1)=3π216f^{\prime}(1)=\frac{3 \pi^{2}}{16}f′(1)=163π2 f(1)=sin3(π3cos(π3222))=sin3(−π6)=−18∴2f′(1)+3π2f(1)=0\begin{aligned} & f(1)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2 \sqrt{2}\right)\right) \\\\ &=\sin ^{3}\left(-\frac{\pi}{6}\right)=\frac{-1}{8} \\\\ & \therefore 2 f^{\prime}(1)+3 \pi^{2} f(1)=0 \end{aligned}f(1)=sin3(3πcos(32π22))=sin3(−6π)=8−1∴2f′(1)+3π2f(1)=0