Let  $f\left(x\right)$ be invertable function and let  $f^{-1}\left(x\right)$  be its inverse, let equation  $f\left(f^{-1}\left(x\right)\right)=f^{-1}\left(x\right)$ has two real roots $\alpha , \beta$ (within the domain of  $f\left(x\right)$ ) then which of the following is(are) true ?

$$\begin{gathered} Let f\left(x\right)=x^2 \\ If f\left(x\right)=y\Rightarrow x=f^{-1}\left(y\right) \\ {x}^2=y \\ x=\sqrt{y} \\ \therefore f^{-1}\left(y\right)=\sqrt{y} \\ {f}^{-1}\left(x\right)=\sqrt{x} \\ f\left(f^{-1}\left(x\right)\right)=f^{-1}\left(x\right) \\ f\left(\sqrt{x}\right)=\sqrt{x} \\ {\left(\sqrt{x}\right)}^2=\sqrt{x} \\ \sqrt{x}\left(\sqrt{x-1}\right)=0 \\ \sqrt{x}=0,\sqrt{x}=1 \\ x=0\left(or\right)x=1 \end{gathered}$$

$$\begin{gathered} \left(a\right) f\left(x\right)=\frac{1}{x} \\ {x}^2=\frac{1}{x} \\ {x}^3-1=0 \\ (x^2+x+1)=0; x=1; only one root \end{gathered}$$

$$\begin{gathered} \left(b\right) f^{-1}\left(x\right)=x \\ \sqrt{x}=x \\ \sqrt{x}(1-\sqrt{x})=0 \\ \sqrt{x}=0or\sqrt{x}=1 \\ 2 real roots \end{gathered}$$

$$\begin{gathered} \left(c\right) f\left(x\right)=x \\ {x}^2-x=0 \\ x|x-1|=0 \\ x=0or1 \end{gathered}$$
2 real roots

$$\begin{gathered} \left(d\right) (0,0),(\alpha ,f\left(\alpha \right)),(\beta ,f\left(\beta \right))\Rightarrow (0,0),(\alpha ,{\alpha }^2),(\beta ,{\beta }^2) \\ Area=\frac{1}{2}|\alpha {\beta }^2-\beta {\alpha }^2|=\frac{1}{2}\left|\alpha \beta (\beta -\alpha )\right| \end{gathered}$$

$$\begin{gathered} If \alpha =0, \beta =1 \\ Area \Delta AOB=0 \end{gathered}$$