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 B $(0, 1)$
Roots are of opposite sign means that the product of roots should be negative.
Hence, $\frac{p (p - 1)}{3} < 0$
$\Rightarrow p (p - 1) < 0 p^{2} - p < 0.$
This happens for 0 < p < 1.

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