$x^{2} - 6 x + 5 \leq 0, x^{2} - 2 x > 0$ where $x$ is an integer. Find all the possible values of $x$ .

Only $1, 5$

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Only $1, 3, 5$

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Only $3, 4, 5$

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Only $4, 5$

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Solution

The correct option is C
Only $3, 4, 5$

$x^{2} - 6 x + 5 \leq 0$
$\Rightarrow (x - 1) (x - 5) \leq 0$
$\Rightarrow 1 \leq x \leq 5 \dots \dots (1)$

$x^{2} - 2 x > 0$
$\Rightarrow x (x - 2) > 0$
$\Rightarrow x < 0$ or $x > 2 \dots \dots (2)$

From $(1) & (2),$
$2 < x \leq 5$
But $x$ is integer so
$\Rightarrow x = 3, 4, 5$
Hence, The correct answer is option C.

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