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NCERT Exemplar · Q20

Q.If ∣x+2∣≤9|x + 2| \le 9, then
(A) x∈(−7,11)x \in (-7, 11)
(B) x∈[−11,7]x \in [-11, 7]
(C) x∈(−∞,−7)∪(11,∞)x \in (-\infty, -7) \cup (11, \infty)
(D) x∈(−∞,−7)∪[11,∞)x \in (-\infty, -7) \cup [11, \infty)

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The inequality ∣x+2∣≤9|x + 2| \le 9 means the distance of xx from −2-2 is at most 99, so xx lies between −11-11 and 77 inclusive. The correct answer is x∈[−11,7]x \in [-11, 7], which is option (B).

The core idea here is that an absolute value inequality of the form ∣x−a∣≤b|x - a| \le b describes all points xx whose distance from aa on the number line is at most bb. This is a "closed interval" centered at aa, stretching bb units to the left and bb units to the right.

For ∣x+2∣≤9|x + 2| \le 9, rewrite it as ∣x−(−2)∣≤9|x - (-2)| \le 9. So the center is −2-2, and the radius is 99. The solution is every xx from −2−9-2 - 9 to −2+9-2 + 9, inclusive.

Let's work through it step by step.

  1. Rewrite the inequality in standard form.

    ∣x+2∣≤9|x + 2| \le 9 is the same as ∣x−(−2)∣≤9|x - (-2)| \le 9. This makes it clear: we want all xx whose distance from −2-2 is 99 or less.

  2. Translate the absolute value into a compound inequality.

    For any real cc and d>0d > 0, ∣x−c∣≤d|x - c| \le d means −d≤x−c≤d-d \le x - c \le d.

    Here c=−2c = -2 and d=9d = 9, so:

−9≤x+2≤9-9 \le x + 2 \le 9

  1. Isolate xx in the middle. Subtract 22 from all three parts:

−9−2≤x≤9−2-9 - 2 \le x \le 9 - 2

−11≤x≤7-11 \le x \le 7

  1. Write the solution in interval notation. …

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