मान लीजिए y=f(x)=sin ^3left(frac{pi}{3}cos le | vedclass

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(B) दिया गया है $y = \sin^3\left(\frac{\pi}{3} \cos(g(x))\right)$ जहाँ $g(x) = \frac{\pi}{3\sqrt{2}}(-4x^3 + 5x^2 + 1)^{3/2}$.$x=1$ पर,$g(1) = \frac{\

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$2 y^{\prime}+3 \pi^2 y=0$

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(B) दिया गया है $y = \sin^3\left(\frac{\pi}{3} \cos(g(x))\right)$ जहाँ $g(x) = \frac{\pi}{3\sqrt{2}}(-4x^3 + 5x^2 + 1)^{3/2}$.$x=1$ पर,$g(1) = \frac{\pi}{3\sqrt{2}}(-4+5+1)^{3/2} = \frac{\pi}{3\sqrt{2}}(2)^{3/2} = \frac{\pi}{3\sqrt{2}}(2\sqrt{2}) = \frac{2\pi}{3}$.$y(1) = \sin^3\left(\frac{\pi}{3} \cos\left(\frac{2\pi}{3}\right)\right) = \sin^3\left(\frac{\pi}{3} \cdot \left(-\frac{1}{2}\right)\right) = \sin^3\left(-\frac{\pi}{6}\right) = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8}$.अब,$y' = 3\sin^2\left(\frac{\pi}{3}\cos(g(x))\right) \cdot \cos\left(\frac{\pi}{3}\cos(g(x))\right) \cdot \left(-\frac{\pi}{3}\sin(g(x))\right) \cdot g'(x)$.$g'(x) = \frac{\pi}{3\sqrt{2}} \cdot \frac{3}{2}(-4x^3 + 5x^2 + 1)^{1/2} \cdot (-12x^2 + 10x) = \frac{\pi}{2\sqrt{2}} \sqrt{-4x^3 + 5x^2 + 1} (-12x^2 + 10x)$.$g'(1) = \frac{\pi}{2\sqrt{2}} \cdot \sqrt{2} \cdot (-2) = -\pi$.$y'$ में $x=1$ रखने पर:$y'(1) = 3\sin^2(-\pi/6) \cdot \cos(-\pi/6) \cdot \left(-\frac{\pi}{3}\sin(2\pi/3)\right) \cdot (-\pi) = 3 \cdot \frac{1}{4} \cdot \frac{\sqrt{3}}{2} \cdot \left(-\frac{\pi}{3} \cdot \frac{\sqrt{3}}{2}\right) \cdot (-\pi) = \frac{3\pi^2}{16}$.विकल्प $B$ की जाँच करने पर: $2y'(1) + 3\pi^2 y(1) = 2\left(\frac{3\pi^2}{16}\right) + 3\pi^2\left(-\frac{1}{8}\right) = \frac{3\pi^2}{8} - \frac{3\pi^2}{8} = 0$.

मान लीजिए $y=f(x)=\sin ^3\left(\frac{\pi}{3}\cos \left(\frac{\pi}{3 \sqrt{2}}\left(-4 x^3+5 x^2+1\right)^{\frac{3}{2}}\right)\right)$. तो,$x =1$ पर,

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