Let y = f(x) = ^3 ( 3 ( 3 2 ( -4x^3 + 5x^2 + 1)^ 32)) Then, at x = 1,

Let y = f(x) = ^3 ( 3 ( 3 2 ( -4x^3 + 5x^2 + 1)^ 32)) Then, at x… — Solution | JEE Mathematics

MCQMediumJEE 2023Applications of Derivatives (Monotonicity, Extrema)

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

A
$2y' + 3y = 0$
B
$2y' + 3y' = 0$
C
$y' - 3y' = 0$
D
$y' + 3y' = 0$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

y = \sin^3 \left( \frac{\pi}{3} \cos g(x) \right)

where

g(x) = \frac{3\sqrt{2}}{\pi}\left(-4x^3 + 5x^2 + 1\right)^{3/2}

Find: The correct relation at $x = 1$ .

From the solution working,

g(1) = \frac{2\pi}{3}

and

g'(x) = \frac{3\sqrt{2}}{\pi}\left(-4x^3 + 5x^2 + 1\right)^{1/2}(-12x^2 + 10x)

g'(1) = -\pi

Differentiate using the chain rule:

y' = 3\sin^2\left(\frac{\pi}{3}\cos g(x)\right)\cos\left(\frac{\pi}{3}\cos g(x)\right) \cdot \frac{\pi}{3}(-\sin g(x))g'(x)

At $x = 1$ ,

y(1) = \sin^3\left(\frac{\pi}{3}\cos \frac{2\pi}{3}\right) = -\frac{1}{8}

Using the values shown in the solution,

y'(1) = \frac{3\pi^2}{16}

Hence,

2y'(1) + 3\pi^2 y(1) = 0

Therefore, the relation obtained from the solution is $2y' + 3\pi^2 y = 0$ at $x=1$ . However, the solution's marks Option C as the correct option even though the listed options do not match this derived relation. The answer is taken from the solution authority: C.

Value-based verification

Using the values extracted from the solution,

y(1) = -\frac{1}{8}, \qquad y'(1) = \frac{3\pi^2}{16}

Now check the derived linear combination:

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

So the working is internally consistent, but the printed options appear corrupted.

Common mistakes

A common mistake is to differentiate $\sin^3(u)$ as $3\sin^2(u)\cos(u)$ and then stop there. This is incomplete because $u = \frac{\pi}{3}\cos g(x)$ is itself a function of $x$ . Continue with the chain rule through both outer layers.
Students may evaluate $g(1)$ incorrectly by mishandling the power $\left(-4x^3+5x^2+1\right)^{3/2}$ . First substitute $x=1$ carefully, then apply the exponent. Do not distribute the exponent termwise.
Another mistake is to miss the negative sign while differentiating $\cos g(x)$ . Since $\frac{d}{dx}(\cos g(x)) = -\sin g(x)\,g'(x)$ , omitting the minus sign changes the final relation.

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Answer

Step-by-step solution

Common mistakes

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