All the values of $x$ for which $x^{2} - 5 x + 6$ is non-negative are

(2, 3)

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[2, 3]

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(- \infty, 2) \cup (3, \infty)

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(- \infty, 2] \cup [3, \infty)

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Solution

The correct option is D $(- \infty, 2] \cup [3, \infty)$
$x^{2} - 5 x + 6$ is non-negative.
$\Rightarrow x^{2} - 5 x + 6 \geq 0 \Rightarrow (x - 2) (x - 3) \geq 0$

$∴ x \in (- \infty, 2] \cup [3, \infty)$

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