The quadratic equations $2 x^{2} - (a^{3} + 8 a - 1) x + a^{2} - 4 a = 0$ possesses roots of opposite sign. Then

a \leq 0

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0 < a < 4

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4 \leq a < 8

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a \geq 8

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Solution

The correct option is B $0 < a < 4$
Given, $2 x^{2} - (a^{3} + 8 a - 1) x + a^{2} - 4 a = 0$
Since, roots are opposite sign,
$∴$ Product of the roots $< 0$
$\Rightarrow a^{2} - 4 a < 0$
$\Rightarrow a (a - 4) < 0$
$\Rightarrow 0 < a < 4$

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