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
$$f(x) = \sqrt{13}\sin(x + \varphi) + 1,$$
$$f(x-c) = \sqrt{13}\sin(x + \varphi - c) + 1,$$
where $0 < \varphi < \frac{\pi}{2}$ and $\tan \varphi = \frac{2}{3}$. Then $af(x) + bf(x-c) = 1$ becomes
$$\sqrt{13}a\sin(x + \varphi) + \sqrt{13}b\sin(x + \varphi - c) + a + b = 1.$$
That is,
$$\sqrt{13}a\sin(x + \varphi) + \sqrt{13}b\sin(x + \varphi)\cos c - \sqrt{13}b\sin c\cos(x + \varphi) + (a + b - 1) = 0.$$
Therefore
$$\sqrt{13}(a + b\cos c)\sin(x + \varphi) - \sqrt{13}b\sin c\cos(x + \varphi) + (a + b - 1) = 0.$$
Since the equality above holds for any $x \in \mathbb{R}$, we must have
$$\begin{cases} a + b\cos c = 0, \\ b\sin c = 0, \end{cases} \qquad \textcircled{1}$$
$$\begin{cases} b\sin c = 0, \\ a + b - 1 = 0. \end{cases} \qquad \textcircled{2}$$
$$\begin{cases} a + b - 1 = 0. \end{cases} \qquad \textcircled{3}$$
If $b=0$, then $a=0$ from ①, and this contradicts ③. So $b \neq 0$, 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, $\frac{b\cos c}{a} = -1$.