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

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

(A) $2y^{\prime}+\sqrt{3}\pi^{2}y=0$

(C) $\sqrt{2}y^{\prime}-3\pi^{2}y=0$

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Differentiation

Let

y = f (x) = sin^{3} (\frac{π}{3} (cos (\frac{π}{3 2} (- 4 x^{3} + 5 x^{2} + 1)^{\frac{3}{2}})))

.
Then, at

x = 1

Answer: (B)

Let

g (x) = \frac{π}{3 2} (- 4 x^{3} + 5 x^{2} + 1)^{3/2}

. At

x = 1

the inner bracket is

2

, so

g (1) = \frac{π}{3 2} (2)^{3/2} = \frac{2 π}{3}

, giving

y (1) = sin^{3} (\frac{π}{3} cos \frac{2 π}{3}) = sin^{3} (- \frac{π}{6}) = - \frac{1}{8}

. Differentiating,

g^{'} (1) = \frac{π}{3 2} \cdot \frac{3}{2} (2)^{1/2} (- 2) = - π

, and by the chain rule

y^{'} (1) = 3 sin^{2} (- \frac{π}{6}) cos (- \frac{π}{6}) \cdot \frac{π}{3} (- sin \frac{2 π}{3}) g^{'} (1) = \frac{3 π ^{2}}{16}

. Then

2 y^{'} (1) + 3 π^{2} y (1) = \frac{3 π ^{2}}{8} - \frac{3 π ^{2}}{8} = 0

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option