JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Solution: $x=1$ par, $(-4x^3 + 5x^2 + 1) = 2$ .

Andar wali term $\frac{\pi}{3\sqrt{2}} (2)^{3/2} = \frac{\pi}{3\sqrt{2}} (2\sqrt{2}) = \frac{2\pi}{3}$ ho jayegi.

Then $y = \sin^3(\frac{\pi}{3} \cos \frac{2\pi}{3}) = \sin^3(\frac{\pi}{3} \cdot -\frac{1}{2}) = \sin^3(-\frac{\pi}{6}) = -\frac{1}{8}$ .

Chain rule se differentiate karke $x=1$ rakhne par:

y' = 3 \sin^2(-\frac{\pi}{6}) \cos(-\frac{\pi}{6}) \cdot \frac{\pi}{3} (-\sin \frac{2\pi}{3}) \cdot \frac{\pi}{3\sqrt{2}} \cdot \frac{3}{2} (2)^{1/2} (-2)

Solve karne par equation $2y' + \sqrt{3}\pi^2 y = 0$ satisfy hoti hai.

Sahi Option: (4)

Choose the correct answer:

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