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  • If |x + 2| ≤ 9, then A.x ∈ (–7, 11) B. x ∈ [–11, 7] C. x ∈ (–∞, –7) ∪ (11, ∞) D. x ∈ (–∞, –7) ∪ [11, ∞)

IF \left | x+2 \right |\leq 9 then 

A. x\epsilon \left ( -7,11 \right )

B. x\epsilon \left ( -11,7 \right )

C. x\epsilon \left ( -\infty ,-7 \right ) \upsilon \left ( 11,\infty \right )

D. x\epsilon \left ( -\infty ,-7 \right ) \upsilon \left [ 11,\infty \right )

Answers (1)

Given:  \left | x-2 \right |\leq 9Thus, there will be two cases,

(x+2) ≤ 9 → x ≤ 7

Thus, x \epsilon (-∞,7)       …….. (i)

& -(x+2) ≤ 9 → -x-2 ≤ 9 → -x ≤ 11 → x ≥ -11

Thus, x \epsilon [-11, ∞]    …….. (ii)

Therefore, x \epsilon [-11,7]      ………. [From (i) & (ii)]

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