JEE Main 2025 Apr 2 Shift 1, Mathematics Q4: Differentiability Of A Function

Rearrange the functional identity by grouping $f(2x+2y)$ and $f(2x-2y)$:

$$f(2x+2y)(\sin x\cos y-\cos x\sin y)=f(2x-2y)(\cos x\sin y+\sin x\cos y)$$

$$f(2x+2y)\sin (x-y)=f(2x-2y)\sin (x+y)$$

$$\frac{f(2x+2y)}{\sin (x+y)}=\frac{f(2x-2y)}{\sin (x-y)}$$

Substituting $m=2x+2y$ and $n=2x-2y$:

$$\frac{f(m)}{\sin (m/2)}=\frac{f(n)}{\sin (n/2)}=K\text{ (constant)}$$

Hence $f(x)=K\sin (x/2)$.

Finding K: $f^{\prime }(x)=\frac{K}{2}\cos (x/2)$ and $f^{\prime }(0)=\frac{K}{2}=\frac{1}{2}$, so $K=1$.

$$f(x)=\sin (x/2),f^{\prime \prime }(x)=-\frac{1}{4}\sin (x/2)$$

At $x=\frac{5\pi }{3}$:

$$24f^{\prime \prime }\left(\frac{5\pi }{3}\right)=24\times \left(-\frac{1}{4}\right)\sin \left(\frac{5\pi }{6}\right)=-6\times \frac{1}{2}=-3$$

$\therefore$ Option (2) is correct.