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(D) Let the roots of the quadratic equation $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$ be $\alpha$ and $\beta$.The sum of the squares of the roots is

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$2$

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(D) Let the roots of the quadratic equation $x^2 + (3 - \lambda)x + (2 - \lambda) = 0$ be $\alpha$ and $\beta$.The sum of the squares of the roots is given by $S = \alpha^2 + \beta^2$.Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$,we have:$\alpha + \beta = -(3 - \lambda) = \lambda - 3$$\alpha\beta = 2 - \lambda$Substituting these into the expression for $S$:$S = (\lambda - 3)^2 - 2(2 - \lambda)$$S = \lambda^2 - 6\lambda + 9 - 4 + 2\lambda$$S = \lambda^2 - 4\lambda + 5$To find the least value,we complete the square:$S = (\lambda^2 - 4\lambda + 4) + 1 = (\lambda - 2)^2 + 1$The expression $S$ attains its minimum value when $(\lambda - 2)^2 = 0$,which occurs at $\lambda = 2$.

The value of $\lambda$ such that the sum of the squares of the roots of the quadratic equation $x^2 + (3 - \lambda)x + 2 = \lambda$ has the least value is:

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