If $(p^{2} - 4 p + 5, 2) = (2 p - 3, | p - 2 |)$ , then the number of values of $p$ is

0

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1

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2

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3

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Solution

The correct option is B $1$
Here, $p^{2} - 4 p + 5 = 2 p - 3$
$\Rightarrow p^{2} - 6 p + 8 = 0$
$\Rightarrow p^{2} - 4 p - 2 p + 8 = 0$
$\Rightarrow (p - 4) (p - 2) = 0$
$\Rightarrow p = 2, 4 . . . (1)$
Also,
$2 = | p - 2 |$
$\Rightarrow p = 0, 4 . . . (2)$

Value of $p$ common from $(1)$ and $(2)$ is $p = 4$
$∴$ There is only one value of $p$

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