If \lambda \in R is such that the sum of the cubes of the roots of the equation, x2 + (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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Answer: B. 2\sqrt 5 Let \alpha , \beta are the roots of the equation, 	herefore \alpha + \beta = \lambda - 2 and \alpha \beta = 10 - \lambda {\alpha ^3} + {\beta ^3} = (\alpha + \beta )3 - 3\alpha \beta (\alpha + \beta ) = (\lambda - 2)3 - 3(10 - \lambda )(\lambda - 2) = \lambda ^3 - 3\lambda ^2 - 24\lambda + 52 Let f(\lambda ) = \lambda ^3 - 3\lambda ^2 - 24\lambda + 52 	herefore {{df(\lambda )} \over {d\lambda }} = 3\lambda ^2 - 6\lambda - 24 	herefore at maximum of minimum {{df(\lambda )} \over {d\lambda }} = 0 	herefore \lambda ^2 - 2\lambda - 8 = 0 \Rightarrow (\lambda + 2) (\lambda - 4) = 0 \Rightarrow \lambda = -2, 4 {{{d^2}f(\lambda )} \over {d{\lambda ^2}}} = 2\lambda - 2 When \lambda = -2 {{{d^2}f(\lambda )} \over {d{\lambda ^2}}} = - 6 < 0 	herefore at \lambda = -2, f(\lambda ) has maximum value. When \lambda = 4 {{{d^2}f(\lambda )} \over {d{\lambda ^2}}} = 6 > 0 	herefore at \lambda = 4, f(\lambda ) has minimum value. 	herefore When \lambda = 4 equation is, x2 - 2x + 6 = 0 	herefore (\alpha - \beta )2 = (\alpha + \beta )2 - 4\alpha \beta \Rightarrow x2 - 4 	imes 6 = - 20 \Rightarrow (\alpha - \beta ) = 2\sqrt 5 i \Rightarrow \left| {\alpha - \beta } ight| = 2\sqrt 5 (ans)

If $$\lambda $$ $$ \in $$ R is such that the sum of the cubes of the roots of the equation, x2 + (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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