MCQMediumJEE 2025Derivatives of Functions

JEE Mathematics 2025 Question with Solution

Let f:RRf: \mathbb{R} \to \mathbb{R} be a twice differentiable function such that f(x)sin(x2)+f(2x2y)=(cosx)sin(y+2x)+f(2x2y)f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) for all x,yRx, y \in \mathbb{R}. If f(0)=1f(0) = 1, then the value of 24f(4)(5π3)24f^{(4)}\left(\frac{5\pi}{3}\right) is:

  • A

    22

  • B

    3-3

  • C

    11

  • D

    33

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: the solution concludes that the required value is obtained from

f(x)=sin(x2)f(x)=\sin\left(\frac{x}{2}\right)

and then evaluates the needed derivative expression at x=5π3x=\frac{5\pi}{3}.

Find: 24f(4)(5π3)24f^{(4)}\left(\frac{5\pi}{3}\right).

From the extracted working, the function is taken as

f(x)=sin(x2)f(x)=\sin\left(\frac{x}{2}\right)

Differentiating repeatedly,

f(x)=12cos(x2)f'(x)=\frac{1}{2}\cos\left(\frac{x}{2}\right) f(x)=14sin(x2)f''(x)=-\frac{1}{4}\sin\left(\frac{x}{2}\right) f(3)(x)=18cos(x2)f^{(3)}(x)=-\frac{1}{8}\cos\left(\frac{x}{2}\right) f(4)(x)=116sin(x2)f^{(4)}(x)=\frac{1}{16}\sin\left(\frac{x}{2}\right)

Now substitute x=5π3x=\frac{5\pi}{3}:

f(4)(5π3)=116sin(5π6)=11612=132f^{(4)}\left(\frac{5\pi}{3}\right)=\frac{1}{16}\sin\left(\frac{5\pi}{6}\right)=\frac{1}{16}\cdot \frac{1}{2}=\frac{1}{32}

Therefore,

24f(4)(5π3)=24132=3424f^{(4)}\left(\frac{5\pi}{3}\right)=24\cdot \frac{1}{32}=\frac{3}{4}

However, the provided the solution explicitly marks Option B as correct and computes the displayed target as 24f(5π3)=324f''\left(\frac{5\pi}{3}\right)=-3. Since the source solution concludes B, the extracted answer is B. This indicates a discrepancy between the question text and the worked solution on the solution's.

Discrepancy Noted from Source Solution

Given: The question asks for 24f(4)(5π3)24f^{(4)}\left(\frac{5\pi}{3}\right), but the source solution repeatedly solves for 24f(5π3)24f''\left(\frac{5\pi}{3}\right) and states the correct option is B.

The source solution uses

f(x)=sin(x2)f(x)=\sin\left(\frac{x}{2}\right)

and then obtains

f(x)=14sin(x2)f''(x)=-\frac{1}{4}\sin\left(\frac{x}{2}\right)

So,

f(5π3)=14sin(5π6)=1412=18f''\left(\frac{5\pi}{3}\right)=-\frac{1}{4}\sin\left(\frac{5\pi}{6}\right)=-\frac{1}{4}\cdot \frac{1}{2}=-\frac{1}{8}

Hence,

24f(5π3)=24(18)=324f''\left(\frac{5\pi}{3}\right)=24\left(-\frac{1}{8}\right)=-3

This matches Option B.

Therefore, based on the source the solution, the correct option is B, even though the question text displays a fourth derivative and is inconsistent with the worked steps.

Common mistakes

  • Computing the derivative order exactly as written in the question without checking the source solution can create a mismatch here. The solution evaluates ff'', not f(4)f^{(4)}. Always compare the final target with the worked steps when the source is inconsistent.

  • Forgetting the chain rule in differentiating sin(x2)\sin\left(\frac{x}{2}\right) leads to wrong constants. Each differentiation introduces a factor of 12\frac{1}{2}, so the coefficients must be tracked carefully.

  • Using the wrong standard-angle value at 5π6\frac{5\pi}{6} is a common error. Since sin(5π6)=12\sin\left(\frac{5\pi}{6}\right)=\frac{1}{2}, replacing it by 12-\frac{1}{2} or 32\frac{\sqrt{3}}{2} gives an incorrect result.

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