Show that for $a \ge 1,f(x) = \sqrt 3 \sin x - \cos x - 2ax + b$ is decreasing in $R$.

We have, $a \ge 1,$ $f(x) = \sqrt 3 \sin x - \cos x - 2ax + b$
Therefore, ${f^\prime }(x) = \sqrt 3 \cos x - ( - \sin x) - 2a$
$= \sqrt 3 \cos x + \sin x - 2a$

$= 2\left[ {\frac{{\sqrt 3 }}{2} \cdot \cos x + \frac{1}{2} \cdot \sin x} \right] - 2a$

$= 2\left[ {\cos \frac{\pi }{6} \cdot \cos x + \sin \frac{\pi }{6} \cdot \sin x} \right] - 2a$

$= 2\left( {\cos \frac{\pi }{6} - x} \right) - 2a$

$= 2\left[ {\left( {\cos \frac{\pi }{6} - x} \right) - a} \right]$

We know that, $\cos x \in [ - 1,1]$

and $a \ge 1$
So, $2\left[ {\cos \left( {\frac{\pi }{6} - x} \right) - a} \right] \le 0$

Therefore, ${f^\prime }(x) \le 0$

Hence, $f(x)$ is a decreasing function in $R$.