Question

Let $y=f(x)={\sin }^3\left(\frac{\pi }{3}\left(\cos \left(\frac{\pi }{3\sqrt{2}}{\left(-4x^3+5x^2+1\right)}^{\frac{3}{2}}\right)\right)\right)$. Then, at $x=1$

• A. $2y^{\prime }+\sqrt{3}{\pi }^{2}y=0$
• B. $\sqrt{2}{y}^{\prime }-3{\pi }^{2}y=0$
• C. $2y^{\prime }+3{\pi }^{2}y=0$
• D. $y^{\prime }+3{\pi }^{2}y=0$

Answer 1

Answer: (C)

$y={\sin }^3(\pi /3\mathrm{cosg}(x))$
$g(x)=\frac{\pi }{3\sqrt{2}}{\left(-4x^3+5x^2+1\right)}^{3/2}$
$g(1)=2\pi /3$
$y^{\prime }=3{\sin }^2\left(\frac{\pi }{3}\cos g(x)\right)\times \cos \left(\frac{\pi }{3}\cos g(x)\right)$$\times \frac{\pi }{3}(-\sin g(x))g^{\prime }(x)$
$y^{\prime }(1)=3{\sin }^2\left(-\frac{\pi }{6}\right)\cdot \cos \left(\frac{\pi }{6}\right)\cdot \frac{\pi }{3}\left(-\sin \frac{2\pi }{3}\right)g^{\prime }(1)$
$g^{\prime }(x)=\frac{\pi }{3\sqrt{2}}{\left(-4x^3+5x^2+1\right)}^{1/2}\left(-12x^2+10x\right)$
$g^{\prime }(1)=\frac{\pi }{2\sqrt{2}}(\sqrt{2})(-2)=-\pi$
$y^{\prime }(1)=\frac{/3}{4}\cdot \frac{\sqrt{3}}{2}\cdot \frac{\pi }{/3}\left(\frac{-\sqrt{3}}{2}\right)(-\pi )=\frac{3{\pi }^2}{16}$
$y(1)={\sin }^3(\pi /3\cos 2\pi /3)=-\frac{1}{8}$
$2y^{\prime }(1)+3{\pi }^{2}y(1)=0$