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

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$1$

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(C) Let the roots of the equation $x^2-(a-2)x-(a+1)=0$ be $\alpha$ and $\beta$. From the relation between roots and coefficients,we have $\alpha+\beta = a-2$ and $\alpha\beta = -(a+1)$. The sum of the squares of the roots is given by $S = \alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta$. Substituting the values,$S = (a-2)^2 - 2(-(a+1)) = (a-2)^2 + 2(a+1)$. Expanding this,$S = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6$. To find the minimum value,we complete the square: $S = (a^2 - 2a + 1) + 5 = (a-1)^2 + 5$. Since $(a-1)^2 \ge 0$,the minimum value of $S$ occurs when $(a-1)^2 = 0$,which implies $a = 1$.

If the sum of the squares of the roots of the equation $x^2-(a-2)x-(a+1)=0$ is least for an appropriate value of the variable parameter $a$,then the value of $a$ will be

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