If $f\left(x\right)=\frac{1+x^2}{1+x}$ for $x\ge 0$ the least value of $f\left(x\right)$ is $\beta$ at $x=\alpha$, then

The correct option is D ${\alpha }^2+\beta =1$
$f\left(x\right)=\frac{1+x^2}{x+1}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(x+1)2x-(x^2+1)}{(x+1{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{x^2+2x-1}{(x+1{)}^2}$
From $f^{\prime }\left(x\right)=0$
$\Rightarrow x^2+2x-1=0$
$\Rightarrow (x+1{)}^2-2=0$
$\Rightarrow [x-(\sqrt{2}-1\left)\right][x-(-\sqrt{2}-1\left)\right]=0$
$\Rightarrow x=\sqrt{2}-1,-\sqrt{2}-1$
But given that $x\ge 0$
Now check function value at $x=1,\sqrt{2}-1$ and when $x\to \infty$
$f\left(0\right)=1$
$f(\sqrt{2}-1)=\frac{3-2\sqrt{2}+1}{\sqrt{2}}=2\sqrt{2}-2$
and $\underset{x\to \infty }{lim}f\left(x\right)$ also tends to $\infty$
Hence, least value of $f\left(x\right)$ is at $\sqrt{2}-1$ and least value is $2\sqrt{2}-2$
$\therefore \alpha =\sqrt{2}-1={\alpha }^2=3-2\sqrt{2}$
and $\beta =2\sqrt{2}-2=2(\sqrt{2}-1)=2\alpha$
$\therefore \beta =2\alpha$

$\therefore {\alpha }^2+\beta =3-2\sqrt{2}+2\sqrt{2}-2=1$
Hence, ${\alpha }^2+\beta =1$