If the function $f(x)=3 \cos |x|-6 a x+b$ increases for all $x \in R$, then the range of values of a is given by

$(-\infty,-1 / 2)$

We have,

$f(x)=3 \cos |x|-6 a x+b$

$\Rightarrow f(x)=3 \cos x-6 a x+b$           [∵ cos |x| = cos x for all x]

$\Rightarrow f'(x) =-3 \sin x-6 a$

For f(x) to be increasing on R, we must have

$f'(x)>0$ for all $x \in R$

$\Rightarrow -3 \sin x-6 a>0$ for all $x \in R$

$\Rightarrow \sin x+2 a<0$ for all $x \in R$

$\Rightarrow \sin x<-2 a$ for all $x \in R$

$\Rightarrow 1>-2 a$              [∵ Max. value of sin x is 1]

$\Rightarrow a<-\frac{1}{2}$

$\Rightarrow a \in(-\infty,-1 / 2)$