$\begin{array}{1 1} a=1\;and\;b \in R \\ a=1,b=0\;or\;a=-1 ,b \in R \\ a= \pm 1,b \in R \\ a= \pm 1, b=0 \end{array}$

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Since $f(x)=ax+b$ is linear, it is either increasing or decreasing.

$\Rightarrow\:f'(x)>0\:\:or\:\:f'(x)<0$

$\Rightarrow\:a>0\:\:or\:\:a<0$ and $a\neq 0$ because $f$ is bijection.

Also given that $fof=I_(x)=x$

$\Rightarrow f(x)=f^{-1}(x)$

Let $y=f(x)=ax+b\:\:\Rightarrow\:x=f^{-1}(y)=\large\frac{y-b}{a}$

$\Rightarrow\:f^{-1}(x)=\large\frac{x-b}{a}$

Since $f=f^{-1}$, $ax+b=\large\frac{x-b}{a}$

Equating the coefficients on both the sides we get

$a=\large\frac{1}{a}$ and $b=-\large\frac{b}{a}$

$\Rightarrow\:a^2=1\:\:and\:\:a=-1\:or\:b=0$

$\Rightarrow a=1\:and\:b=0\:\:or\:a=-1\:and\:b\in R$

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