If f : R → R defined by f(x)=3 x+2 a cos x-5 is invertible, then 'a'…

For f(x) to be invertible, it should be a bijection.

Clearly, f is onto for all $a \in R$, because range (f) = R.

For f(x) to be one-one, we must have,

$f'(x) \geq 0$ or, $f'(x) \leq 0$ for all $x \in R$

$\Rightarrow 3-2 a \sin x \geq 0$ or, $3-2 a \sin x \leq 0$ for all $x \in R$

$\Rightarrow 2 a \sin x \leq 3$ for all $x \in R$ or, $2 a \sin x \geq 3$ for all $x \in R$

$\Rightarrow |2 a| \leq 3$ or, $-|2 a| \geq 3$              $\left[\begin{array}{r}∵ Max~2 a \sin x=|2 a| \& \\ Min~2 a \sin x=-|2 a|\end{array}\right]$

$\Rightarrow |2 a| \leq 3 \Rightarrow|a| \leq \frac{3}{2} \Rightarrow a \in[-3 / 2,3 / 2]$