If $|x+2| \leq 9$, then

$x \in(-\infty,-7) \cup(11, \infty)$

$x \in(-\infty,-7) \cup[11, \infty)$

JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

|x+2| \leq 9

, then

View answer and explanation

Correct option: B

x \in[-11,7]

Option B (x ∈ [-11,7]) is correct because the absolute value inequality |x+2| ≤ 9 implies that -9 ≤ x+2 ≤ 9.

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