Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $(\sin x \cos y)(f(2 x+2 y)-f(2 x-2 y))=(\cos x$ $\sin \mathrm{y})(f(2 \mathrm{x}+2 \mathrm{y})+f(2 \mathrm{x}-2 \mathrm{y}))$, for all $\mathrm{x}, \mathrm{y} \in \mathbf{R}$.
If $f^{\prime}(0)=\frac{1}{2}$, then the value of $24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)$ is:

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Aicharya · Accepted Answer

$\begin{aligned} & (\sin x \cos y)(f(2 x+2 y)-f(2 x-2 y))=(\cos x \sin y) \\ & (f(2 x+2 y)+f(2 x-2 y)) \\ & f(2 x+2 y)(\sin (x-y))=f(2 x-2 y) \sin (x+y) \\ & \frac{f(2 x+2 y)}{\sin (x+y)}=\frac{f(2 x-2 y)}{\sin (x-y)} \\ & \text { Put } 2 x+2 y=m, 2 x-2 y=n \\ & \frac{f(m)}{\sin \left(\frac{m}{2}\right)}=\frac{f(n)}{\sin \left(\frac{n}{2}\right)}=\mathrm{K} \\ & \Rightarrow f(m)=K \sin \left(\frac{m}{2}\right) \\ & \therefore f(x)=K \sin \left(\frac{x}{2}\right) \\ & f^{\prime}(x)=\frac{K}{2} \cos \left(\frac{x}{2}\right) \\ & P_{\text {Put }} x=0 ; \frac{1}{2}=\frac{K}{2} \Rightarrow K=1 \\ & f^{\prime}(x)=\frac{1}{2} \cos \frac{x}{2}\end{aligned}$
$\begin{aligned} & f^{\prime \prime}(x)=-\frac{1}{4} \sin \frac{x}{2} \\ & 4 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=\left(-\frac{1}{4} \sin \left(\frac{5 \pi}{6}\right)\right) 24 \\ & =\frac{-24}{8}=-3\end{aligned}$

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a twice differentiable function such that $(\sin x \cos y)(f(2…

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