JEE MAIN - Mathematics Hindi (2023 - 31st January Morning Shift - No. 15)

माना $$y=f(x)=\sin ^{3}\left(\frac{\pi}{3}\left(\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^{3}+5 x^{2}+1\right)^{\frac{3}{2}}\right)\right)\right)$$ है । तो $$x=1$$ पर
$$2y'+\sqrt3\pi^2y=0$$
$$y'+3\pi^2y=0$$
$$\sqrt2y'-3\pi^2y=0$$
$$2y'+3\pi^2y=0$$

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