If  $f:R-\{-1,K\}\to R-\{\alpha ,\beta \}$  is a bijective function defined by  $f\left(x\right)=\frac{(2x-1)(2x^2-4px+p^3)}{(x+1)(x^2-p^{2}x+p^2)}$, where  $P\ge 0$, then which of the following is/are TRUE?

If  $x+1$s a factor of  $2x^2-4px+p^3\Rightarrow p^{3}4p+2=0$
Then  $2x-1$ is a factor of  $x^2-p^{2}x+p^2\Rightarrow 2p^2+1=0=0P\notin R$
So this case is not true
Now,  $\frac{2x^2-4px+p^3}{x^2-p^{2}x+p^2}$ will be some constant for bijective of  $f\left(x\right)$. So
 $\frac{2}{1}=\frac{-4p}{-p^2}=\frac{p^3}{p^2}\Rightarrow p=2\Rightarrow f\left(x\right)=\frac{4x-2}{x+1},x\ne 2$
Domain of  $f\left(x\right)$  is  $R-\{-1,2\}$, Range of  $f\left(x\right)$  is  $R-\{2,4\}$
$K=2,\alpha +\beta =6$