China Mathematical Competition

Let $c=\pi$. Then $f(x)+f(x-c)=2$ for any $x\in \mathbb{R}$. Now let $a=b=\frac{1}{2}$, and $c=\pi$. We have $$af(x)+bf(x-c)=1$$ for any $x\in \mathbb{R}$. Consequently, $\frac{b\cos c}{a}=-1$. So Answer is (C).

More generally, we have $$f(x)=\sqrt{13}\sin (x+\phi )+1,$$ $$f(x-c)=\sqrt{13}\sin (x+\phi -c)+1,$$ where $0<\phi <\frac{\pi }{2}$ and $\tan \phi =\frac{2}{3}$. Then $af(x)+bf(x-c)=1$ becomes $$\sqrt{13}a\sin (x+\phi )+\sqrt{13}b\sin (x+\phi -c)+a+b=1.$$ That is, $$\sqrt{13}a\sin (x+\phi )+\sqrt{13}b\sin (x+\phi )\cos c-\sqrt{13}b\sin c\cos (x+\phi )+(a+b-1)=0.$$ Therefore $$\sqrt{13}(a+b\cos c)\sin (x+\phi )-\sqrt{13}b\sin c\cos (x+\phi )+(a+b-1)=0.$$ Since the equality above holds for any $x\in \mathbb{R}$, we must have $$\{\begin{array}{l}a+b\cos c=0,\\ b\sin c=0,\end{array}\stackrel{◯}{1}$$ $$\{\begin{array}{l}b\sin c=0,\\ a+b-1=0.\end{array}\stackrel{◯}{2}$$ $$\{\begin{array}{l}a+b-1=0.\end{array}\stackrel{◯}{3}$$ If $b=0$, then $a=0$ from ①, and this contradicts ③. So $b\ne 0$, and $\sin c=0$ from ②. Therefore $c=2k\pi +\pi$ or $c=2k\pi$ ($k\in \mathbb{Z}$). If $c=2k\pi$, then $\cos c=1$, and it leads to a contradiction between ① and ③. So $c=2k\pi +\pi$ ($k\in \mathbb{Z}$) and $\cos c=-1$. From ① and ③, we get $a=b=\frac{1}{2}$. Consequently, $\frac{b\cos c}{a}=-1$.