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(D) Given the quadratic equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$.Comparing this with $ax^2 + bx + c = 0$,we have $a = 2$,$b = -2(m^2 + 1)$,an

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$m^2$

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(D) Given the quadratic equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$.Comparing this with $ax^2 + bx + c = 0$,we have $a = 2$,$b = -2(m^2 + 1)$,and $c = m^4 + m^2 + 1$.Using the properties of roots:Sum of roots: $\alpha + \beta = -b/a = \frac{2(m^2 + 1)}{2} = m^2 + 1$ ... $(i)$Product of roots: $\alpha \beta = c/a = \frac{m^4 + m^2 + 1}{2}$ ... $(ii)$We know that $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.Substituting the values from $(i)$ and $(ii)$:$\alpha^2 + \beta^2 = (m^2 + 1)^2 - 2 \left( \frac{m^4 + m^2 + 1}{2} \right)$$\alpha^2 + \beta^2 = (m^4 + 2m^2 + 1) - (m^4 + m^2 + 1)$$\alpha^2 + \beta^2 = m^4 - m^4 + 2m^2 - m^2 + 1 - 1 = m^2$.

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$,then $\alpha^2 + \beta^2$ equals:

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