If $\alpha$ and $\beta$ are roots of $x^2$ - $(k+1)$ $x$ + $\frac{1}{2}$ $(k^2+k+1)$ $=$ 0, then ${\alpha }^2+{\beta }^2$ is equal

The correct option is B $k$
$\alpha$ and $\beta$ are roots of the equation $x^2-(k+1)x+\frac{1}{2}(k^2+k+1)=0$.

$\alpha +\beta =-\frac{-(k+1)}{1}=k+1$ ------ ( 1 )

$\alpha \beta =\frac{k^2+k+1}{2}$ ----- ( 2 )

Now,

$\Rightarrow$ $(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$\Rightarrow$ ${(k+1)}^2={\alpha }^2+{\beta }^2+2\times \frac{k^2+k+1}{2}$

$\Rightarrow$ $k^2+2k+1={\alpha }^2+{\beta }^2+k^2+k+1$

$\Rightarrow$ $k^2+2k+1-k^2-k-1={\alpha }^2+{\beta }^2$

$\therefore$ ${\alpha }^2+{\beta }^2=k$