Let f: R R be a twice differentiable function such that ( x y)(f(2 x+2 y)-f(2 x-2 y))=( x …

Question

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

Accepted Answer

$$\begin{aligned}
& \sin (x-y) f(2 x+2 y)=f(2 x-2 y) \sin (x+y) \
& \frac{f(2 x+2 y)}{\sin (x+y)}=\frac{f(2 x-2 y)}{\sin (x-y)}=k(\text { say }) \
& f(2 x+2 y)=k \sin (x+y) \
& f(2 x)=5 \sin x \quad(\because y=0) \
& f(x)=k \sin \frac{x}{2} \
& f^{\prime}(x)=\frac{k}{2} \cos \frac{x}{2} \
& f^{\prime}(0)=\frac{1}{2} \Rightarrow k=1 \
& f(x)=\sin \frac{x}{2} \Rightarrow f^{\prime}(x)=\frac{1}{2} \cos \frac{x}{2} \
& f^{\prime \prime}(x)=-\frac{1}{4} \sin \frac{x}{2} \
& 24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=-3
\end{aligned}$$

Solutions · 1