The values of $p$ for which the equation $x^{2} - 6 | x | + 5 - | p | = 0$ has exactly four real roots, is

(0, 2)

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(- 5, 5)

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(2, 10)

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(- 5, 10)

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Solution

The correct option is B $(- 5, 5)$
For four real roots, $t^{2} - 6 t + 5 - | p | = 0 (| x | = t)$ must have two distinct positive roots.
Let $f (t) = t^{2} - 6 t + 5 - | p |$
$\Rightarrow Δ > 0$ and $f (0) > 0$
$\Rightarrow p \in R$ and $5 - | p | > 0$
$\Rightarrow p \in (- 5, 5)$

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