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A function $f(x)$ is increasing on R if its derivative $f'(x) \ge 0$ for all $x \in R$.
Given $f(x) = \sin x - ax + b$, we need to find its derivative $f'(x)$. $$f'(x) = \frac{d}{dx}(\sin x - ax + b) = \cos x - a$$
For $f(x)$ to be increasing on R, we must have $f'(x) \ge 0$ for all $x \in R$. $$\cos x - a \ge 0$$ $$a \le \cos x$$
We know that the range of $\cos x$ is $[-1, 1]$. Therefore, the maximum value of $\cos x$ is 1 and the minimum value is -1. $$-1 \le \cos x \le 1$$
Since $a \le \cos x$ for all $x \in R$, $a$ must be less than or equal to the minimum value of $\cos x$. However, we need $a \le \cos x$ to hold true for all $x$. Therefore, $a$ must be less than or equal to the *minimum* value of $\cos x$, which is -1. $$a \le -1$$
Final Answer: $a \le -1$