JEE Main 2025 (Online) 2nd April Morning ShiftQuestion 1
Let $f:\mathbf{R}\to\mathbf{R}$ be a twice differentiable function such that $(\sin x\cos y)\left(f(2x+2y)-f(2x-2y)\right)=(\cos x\sin y)\left(f(2x+2y)+f(2x-2y)\right)$, for all $x,y\in\mathbf{R}$. If $f'(0)=\frac{1}{2}$, then the value of $24f''\left(\frac{5\pi}{3}\right)$ is :
Differential Equation
Solution
Concept. Divide by $\cos x\cos y$; the equation says $\frac{f(2p)}{\sin p}$ is the same for every $p$. Formulas used. (1) $\tan x\pm\tan y=\dfrac{\sin(x\pm y)}{\cos x\cos y}$Step 1 - Rearrange. Dividing by $\cos x\cos y$: $$f(2x+2y)(\tan x-\tan y)=f(2x-2y)(\tan x+\tan y).$$ By (1), $f(2p)\sin q=f(2q)\sin p$ with $p=x+y$, $q=x-y$. Step 2 - The form of $f$. $p$ and $q$ are independent, so $\dfrac{f(2p)}{\sin p}=k$ (a constant). Hence $f(z)=k\sin\frac z2$. Step 3 - Find $k$. $f'(z)=\frac k2\cos\frac z2$, so $f'(0)=\frac k2=\frac12$ and $k=1$. Step 4 - Value. $f''(z)=-\frac14\sin\frac z2$, so $f''\left(\frac{5\pi}{3}\right)=-\frac14\sin\frac{5\pi}{6}=-\frac18$ and $24f''=-3$. Answer: (B)