If $f:R\to R$ , where $f\left(x\right)=ax+\cos x$  is an invertible function, then

$f:R\to R$ , $f\left(x\right)=ax+\cos x$

$-\infty <x<\infty ,-1\le \cos x\le 1\Rightarrow -\infty <f\left(x\right)<\infty$

$\therefore$  $f$  is onto

For $f\left(x\right)$  to be one-one, $f\left(x\right)$  should be increasing or decreasing

$f\text{'}\left(x\right)=a-\sin x$

$f\text{'}\left(x\right)\ge 0\Rightarrow a\ge \sin x,\forall x,\Rightarrow a\ge 1$

$f\text{'}\left(x\right)\le 0\Rightarrow a\le \sin x,\forall x\Rightarrow a\le -1$

Hence, for $f$  to be invertible

i.e., one-one, onto $a\in (-\infty ,-1]\cup [1,\infty )$ .