The set of values of $p$ for which the roots of the equation $3 x^{2} + 2 x + p (p - 1) = 0$ are of opposite sign is

(- \infty, 0)

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(0, 1)

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(1, \infty)

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(0, \infty)

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Solution

The correct option is C $(0, 1)$
Given equation
$3 x^{2} + 2 x + p (p - 1) = 0$
Since, roots are of opposite sign.
$\Rightarrow α β < 0$
$\frac{p (p - 1)}{3} < 0$
$\Rightarrow p (p - 1) < 0$
$0 < p < 1$
$\Rightarrow p \in (0, 1)$ .

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