Calculus – Functional Equation – JEE Main 2 Apr 2025 Shift 1

Solution:

Rearrange the given equation to group the function terms:
$$f(2x+2y)(\sin x \cos y – \cos x \sin y) = f(2x-2y)(\cos x \sin y + \sin x \cos y)$$

Using the compound angle formulas $\sin(A-B)$ and $\sin(A+B)$:
$$f(2x+2y)\sin(x-y) = f(2x-2y)\sin(x+y)$$

Rewrite as a ratio:
$$\frac{f(2x+2y)}{\sin(x+y)} = \frac{f(2x-2y)}{\sin(x-y)}$$

Let $2x+2y = m$ and $2x-2y = n$. Then $x+y = \frac{m}{2}$ and $x-y = \frac{n}{2}$.
Substituting these into the equation implies the ratio is constant for any independent variables:
$$\frac{f(m)}{\sin(m/2)} = \frac{f(n)}{\sin(n/2)} = K \quad (\text{constant})$$

Thus, the function is:
$$f(x) = K \sin\left(\frac{x}{2}\right)$$

Differentiate $f(x)$ to find $K$:
$$f'(x) = \frac{K}{2} \cos\left(\frac{x}{2}\right)$$

Given $f'(0) = \frac{1}{2}$, substitute $x=0$:
$$\frac{1}{2} = \frac{K}{2} \cos(0) \Rightarrow \frac{K}{2} = \frac{1}{2} \Rightarrow K = 1$$

So, $f(x) = \sin\left(\frac{x}{2}\right)$. Now, find the second derivative:
$$f'(x) = \frac{1}{2} \cos\left(\frac{x}{2}\right) \Rightarrow f”(x) = -\frac{1}{4} \sin\left(\frac{x}{2}\right)$$

Calculate the required value at $x = \frac{5\pi}{3}$:
$$24 f”\left(\frac{5\pi}{3}\right) = 24 \left( -\frac{1}{4} \sin\left(\frac{5\pi}{6}\right) \right)$$

$$= -6 \sin\left(\frac{5\pi}{6}\right) = -6 \left(\frac{1}{2}\right)$$

$$= -3$$

Ans. (2)