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A function $f(x)$ is strictly decreasing if its derivative $f'(x) < 0$ for all $x$ in its domain.
Given $f(x) = a(x - \cos x)$, we need to find its derivative $f'(x)$. $$f'(x) = a(1 + \sin x)$$
For $f(x)$ to be strictly decreasing, $f'(x) < 0$ for all $x \in R$. $$a(1 + \sin x) < 0$$ Since $-1 \le \sin x \le 1$, we have $0 \le 1 + \sin x \le 2$. Thus, $1 + \sin x$ is always non-negative. For $a(1 + \sin x) < 0$, we must have $a < 0$ and $1 + \sin x > 0$. However, $1 + \sin x$ can be equal to 0 when $x = (2n+1)\pi - \frac{\pi}{2}$, where $n$ is an integer. So, for $f'(x)$ to be strictly less than 0, $a$ must be strictly less than 0.
Since $a < 0$, $a$ belongs to the interval $(-\infty, 0)$.
Final Answer: (-∞,0)