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(B) Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - (a - 2)x - a + 1 = 0$.From the properties of roots,we have $\alpha + \beta = a - 2$ a

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$1$

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(B) Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - (a - 2)x - a + 1 = 0$.From the properties of roots,we have $\alpha + \beta = a - 2$ and $\alpha \beta = -a + 1$.Let $S$ be the sum of the squares of the roots,so $S = \alpha^2 + \beta^2$.Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$,we get:$S = (a - 2)^2 - 2(-a + 1)$$S = a^2 - 4a + 4 + 2a - 2$$S = a^2 - 2a + 2$.To find the value of $a$ for which $S$ is minimum,we differentiate $S$ with respect to $a$:$\frac{dS}{da} = 2a - 2$.Setting $\frac{dS}{da} = 0$,we get $2a - 2 = 0$,which implies $a = 1$.Since $\frac{d^2S}{da^2} = 2 > 0$,the function $S$ has a local minimum at $a = 1$.Thus,the value of $a$ is $1$.

The value of $a$ such that the sum of the squares of the roots of the equation $x^2 - (a - 2)x - a + 1 = 0$ assumes the least value is:

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