Question

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

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

Answer 1

Answer: (C)

$ y =\sin ^3(\pi / 3 \operatorname{cosg}(x))$
$ g(x)=\frac{\pi}{3 \sqrt{2}}\left(-4 x^3+5 x^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(-4 x^3+5 x^2+1\right)^{1 / 2}\left(-12 x^2+10 x\right)$
$ g^{\prime}(1)=\frac{\pi}{2 \sqrt{2}}(\sqrt{2})(-2)=-\pi $
$y^{\prime}(1)=\frac{\not 3}{4} \cdot \frac{\sqrt{3}}{2} \cdot \frac{\pi}{\not 3 }\left(\frac{-\sqrt{3}}{2}\right)(-\pi)=\frac{3 \pi^2}{16}$
$ y(1)=\sin ^3(\pi / 3 \cos 2 \pi / 3)=-\frac{1}{8} $
$ 2 y^{\prime}(1)+3 \pi^2 y(1)=0$