JEE Main 2023 — Mathematics Calculus

The given equation can be written as

$$y={\mathrm{sin}}^{3}(\frac{\pi }{3}\mathrm{cos}g(x))$$y=sin3(3π​cosg(x))

where,

$$g(x)=\frac{\pi }{3\sqrt{2}}{(-4{x}^{3}+5{x}^{2}+1)}^{3/2}$$g(x)=32​π​(−4x3+5x2+1)3/2

$$\Rightarrow {g}^{'}(x)=\frac{\pi }{2\sqrt{2}}{(-4{x}^{3}+5{x}^{2}+1)}^{1/2}(-12{x}^{2}+10x)$$⇒g′(x)=22​π​(−4x3+5x2+1)1/2(−12x2+10x)

$$\Rightarrow {g}^{'}(1)=\frac{\pi \sqrt{2}}{2\sqrt{2}}(-2)=-\pi$$⇒g′(1)=22​π2​​(−2)=−π

And,

$$g(1)=\frac{2\pi }{3}=(\pi -\frac{\pi }{3})$$g(1)=32π​=(π−3π​)

Now,

$${y}^{'}=3{\mathrm{sin}}^{2}(\frac{\pi }{3}\mathrm{cos}g(x))\times \mathrm{cos}(\frac{\pi }{3}\mathrm{cos}g(x))\times \frac{\pi }{3}(-\mathrm{sin}g(x)){g}^{'}(x)$$y′=3sin2(3π​cosg(x))×cos(3π​cosg(x))×3π​(−sing(x))g′(x)

$${y}^{'}(1)=3{\mathrm{sin}}^{2}(-\frac{\pi }{6})\cdot \mathrm{cos}(\frac{\pi }{6})\cdot \frac{\pi }{3}(-\mathrm{sin}\frac{2\pi }{3}){g}^{'}(1)$$y′(1)=3sin2(−6π​)⋅cos(6π​)⋅3π​(−sin32π​)g′(1)

$${y}^{'}(1)=\frac{3}{4}\cdot \frac{\sqrt{3}}{2}\cdot \frac{\pi }{3}(\frac{-\sqrt{3}}{2})(-\pi )=\frac{3{\pi }^{2}}{16}$$y′(1)=43​⋅23​​⋅3π​(2−3​​)(−π)=163π2​

$$y(1)={\mathrm{sin}}^{3}(\frac{\pi }{3}\mathrm{cos}\frac{2\pi }{3})=-\frac{1}{8}$$y(1)=sin3(3π​cos32π​)=−81​

Therefore,

$$2{y}^{'}(1)+3{\pi }^{2}y(1)=0$$2y′(1)+3π2y(1)=0