Mathematics - Functions Question with Solution | TestHub

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