Let $\alpha, \beta$ are the roots of the given equation.
Then $\alpha + \beta = a-2$, $\alpha. \beta= 1-a$
$(\alpha + \beta)^2= \alpha^2 + \beta^2 + 2 \alpha \beta$
$(a-2)^2= \alpha^2 + \beta^2 + 2 - 2a$
$\alpha^2 + \beta^2= a^2 -2a +2= x$(let)
We have to minimize $x$, differentiating $x$ wrt to $a$
$\frac{dx}{da} = 2a-2 = 0, a=1$
$\frac{d^2 x} {d a^2} = 2 > 0$(minima)
So for $a=1$, $x$ will be minimum.
Hence option B) is correct