JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 15)
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,
$$2 y^{\prime}+\sqrt{3} \pi^{2} y=0$$
$$y^{\prime}+3 \pi^{2} y=0$$
$$\sqrt{2} y^{\prime}-3 \pi^{2} y=0$$
$$2 y^{\prime}+3 \pi^{2} y=0$$
Explanation
$f(x)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right)$
$$ \begin{aligned} & f^{\prime}(x)=3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3}\left(-\sin \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3 \sqrt{2}} \frac{3}{2}\left(-4 x^{3}+5 x^{3}+1\right)^{1 / 2}\left(-12 x^{2}+10 x\right) \end{aligned} $$
$f^{\prime}(1)=\frac{3 \pi^{2}}{16}$
$$ \begin{aligned} & f(1)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2 \sqrt{2}\right)\right) \\\\ &=\sin ^{3}\left(-\frac{\pi}{6}\right)=\frac{-1}{8} \\\\ & \therefore 2 f^{\prime}(1)+3 \pi^{2} f(1)=0 \end{aligned} $$
$$ \begin{aligned} & f^{\prime}(x)=3 \sin ^{2}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \cos \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3}\left(-\sin \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{3 / 2}\right)\right) \\\\ & \frac{\pi}{3 \sqrt{2}} \frac{3}{2}\left(-4 x^{3}+5 x^{3}+1\right)^{1 / 2}\left(-12 x^{2}+10 x\right) \end{aligned} $$
$f^{\prime}(1)=\frac{3 \pi^{2}}{16}$
$$ \begin{aligned} & f(1)=\sin ^{3}\left(\frac{\pi}{3} \cos \left(\frac{\pi}{3 \sqrt{2}} 2 \sqrt{2}\right)\right) \\\\ &=\sin ^{3}\left(-\frac{\pi}{6}\right)=\frac{-1}{8} \\\\ & \therefore 2 f^{\prime}(1)+3 \pi^{2} f(1)=0 \end{aligned} $$
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