The values of $a$ and $b$ for which the map $f: R \to R,$ given by $f \left(x\right) =ax+ b (a, b \in R),$ is a bijection with $fof$ as identify function, are 

We have $f : R\to R$ defined by $f\left(x\right)=ax+b.$ It is given that $f$ is a bijection. Therefore, it is either strictly increasing or strictly decreasing. This means that either $f^{\mathrm{\prime }}\left(x\right)>0$ or, $f^{\mathrm{\prime }}\left(x\right)<0$ for all $x\in R$

We have, $f^{\mathrm{\prime }}\left(x\right)=a$

$\therefore f^{\mathrm{\prime }}\left(x\right)>0$ or, $f^{\mathrm{\prime }}\left(x\right)<0\Rightarrow a\ne 0.$

If $a\ne 0,$ then the graph of $f \left(x\right)$ is a straight line. So, range of $f=R.$ Thus, $f\left(x\right)$ is a bijection, if $a\ne 0.$

It is given that

     $f\circ f=I$

$\Rightarrow f$ is inverse of itself.

$\Rightarrow f\left(x\right)=f^{-1}\left(x\right)$ for all $x\in R$

$\Rightarrow ax+b=\frac{x-b}{a}$for all $x\in R$     $\left[\because f^{-1}\left(x\right)=\frac{x-b}{a}\right]$

$\Rightarrow x\left(a^2-1\right)+b(a+1)=0$ for all $x\in R$

$\Rightarrow a^2-1=0$ and $b(a+1)=0$

$\Rightarrow a=\pm 1$ and $b(a+1)=0$

When $a=1$, we find that $b(a+1)=0\Rightarrow b=0$

When $a=-1,$ we find that $b(a+1)=0$ is true for all $b\in R.$

Thus, we have

         $(a=1$ and $b=0)$ or, $(a=-1$ and $b\in R).$