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

We have,

   $\begin{array}{l} f\left(x\right)=3\cos \left|x\right|-6ax+b\\ \Rightarrow f\left(x\right)=3\cos x-6ax+b[\because \cos |x|=\cos x\text{ for all }x]\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=-3\sin x-6a\end{array}$

For $f \left(x\right)$ to be increasing on $R,$ we must have

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

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

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

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

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

$\begin{array}{l}\Rightarrow a<-\frac{1}{2}\\ \Rightarrow a\in (-\mathrm{\infty },-1/2).\end{array}$