The condition for which the quadratic equation $(p^{2} - 3 p + 2) x^{2} + (p^{2} - 4) x + (p - 1) = 0$ will have a graph which is an upward opening parabola is

p \in (1, 2)

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p \in R

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p \in (- \infty, - 1) \cup (2, \infty)

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p \in (- \infty, 1) \cup (2, \infty)

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Solution

The correct option is D $p \in (- \infty, 1) \cup (2, \infty)$
For $f (x) = a x^{2} + b x + c$ , the graph is an upward opening parabola when $a > 0$ .

Here we have $f (x) = (p^{2} - 3 p + 2) x^{2} + (p^{2} - 4) x + (p - 1)$
So for an upward opening parabolic graph,
$p^{2} - 3 p + 2 > 0$
$\Rightarrow (p - 1) (p - 2) > 0$
So using wavy curve method, we get

$p \in (- \infty, 1) \cup (2, \infty)$

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