If the roots of the quadratic equation $(4 p - p^{2} - 5) x^{2} - (2 p - 1) x + 3 p = 0$ lie on either side of unity, then the number of integral values of $p$ is

2

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3

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4

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1

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Solution

The correct option is A $2$
Given equation
$(4 p - p^{2} - 5) x^{2} - (2 p - 1) x + 3 p = 0$
We know that
$4 p - p^{2} - 5 = - (p^{2} - 4 p + 4) - 1 = - 1 - (p - 2)^{2} ∴ 4 p - p^{2} - 5 < 0 \forall p \in R$

According to the question,
$1$ must lie in between the roots.
So,
$f (1) > 0$
$\Rightarrow 4 p - p^{2} - 5 - 2 p + 1 + 3 p > 0$
$\Rightarrow p^{2} - 5 p + 4 < 0$
$\Rightarrow (p - 1) (p - 4) < 0$
$\Rightarrow p \in (1, 4)$
Intergral value of
$∴ p = 2, 3$

Hence, the number of integral value of $p$ is $2$ .

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