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(B) Let the roots of the equation $x^2 + (2 - \lambda) x + (10 - \lambda) = 0$ be $\alpha$ and $\beta$.From the properties of roots,$\alpha + \beta =

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$2\sqrt{5}$

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(B) Let the roots of the equation $x^2 + (2 - \lambda) x + (10 - \lambda) = 0$ be $\alpha$ and $\beta$.From the properties of roots,$\alpha + \beta = - (2 - \lambda) = \lambda - 2$ and $\alpha \beta = 10 - \lambda$.The sum of the cubes of the roots is $S = \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.Substituting the values,$S = (\lambda - 2)^3 - 3(10 - \lambda)(\lambda - 2) = (\lambda - 2) [(\lambda - 2)^2 - 3(10 - \lambda)]$.$S = (\lambda - 2) [\lambda^2 - 4\lambda + 4 - 30 + 3\lambda] = (\lambda - 2)(\lambda^2 - \lambda - 26) = \lambda^3 - 3\lambda^2 - 24\lambda + 52$.To find the minimum,we differentiate $S$ with respect to $\lambda$: $\frac{dS}{d\lambda} = 3\lambda^2 - 6\lambda - 24 = 0$.$3(\lambda^2 - 2\lambda - 8) = 0 \implies 3(\lambda - 4)(\lambda + 2) = 0$. Thus,$\lambda = 4$ or $\lambda = -2$.Checking the second derivative: $\frac{d^2S}{d\lambda^2} = 6\lambda - 6$. For $\lambda = 4$,$6(4) - 6 = 18 > 0$ (minimum).The difference of the roots is $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta} = \sqrt{(\lambda - 2)^2 - 4(10 - \lambda)} = \sqrt{\lambda^2 - 4\lambda + 4 - 40 + 4\lambda} = \sqrt{\lambda^2 - 36}$.For $\lambda = 4$,$|\alpha - \beta| = \sqrt{16 - 36} = \sqrt{-20} = i\sqrt{20} = 2i\sqrt{5}$. The magnitude is $|2i\sqrt{5}| = 2\sqrt{5}$.

If $\lambda \in R$ is such that the sum of the cubes of the roots of the equation $x^2 + (2 - \lambda) x + (10 - \lambda) = 0$ is minimum,then the magnitude of the difference of the roots of this equation is

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