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NCERT Exemplar · Q4

Q.Solve for xx: ∣x−1∣≤5|x-1| \le 5, ∣x∣≥2|x| \ge 2.

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Solve each absolute-value inequality separately, then find where both conditions hold simultaneously. The solution is x∈[−4,−2]∪[2,6]x \in [-4, -2] \cup [2, 6].

Absolute-value inequalities encode distance conditions on the number line. The inequality ∣x−1∣≤5|x - 1| \le 5 asks for all points within distance 55 from 11, while ∣x∣≥2|x| \ge 2 demands points at least distance 22 from the origin. We need the intersection of these two regions.

Solving the first inequality: ∣x−1∣≤5|x - 1| \le 5

The absolute value ∣x−1∣|x - 1| measures the distance from xx to 11. Saying this distance is at most 55 means:

−5≤x−1≤5-5 \le x - 1 \le 5

Adding 11 throughout:

−4≤x≤6-4 \le x \le 6

So the first condition gives us x∈[−4,6]x \in [-4, 6].

Solving the second inequality: ∣x∣≥2|x| \ge 2

The absolute value ∣x∣|x| measures distance from the origin. We want points at least 22 units away, which means either x≥2x \ge 2 or x≤−2x \le -2.

Tip

For ∣x∣≥a|x| \ge a (where a>0a > 0), the solution is always x≤−ax \le -a or x≥ax \ge a — the exterior of the interval (−a,a)(-a, a).

So the second condition gives us x∈(−∞,−2]∪[2,∞)x \in (-\infty, -2] \cup [2, \infty).

Finding the intersection

We need both conditions to hold. Let's visualize:

  • First condition: [−4,6][-4, 6] — a single interval
  • Second condition: (−∞,−2]∪[2,∞)(-\infty, -2] \cup [2, \infty) — two rays

The intersection consists of:

  1. Left piece: The overlap of [−4,6][-4, 6] with (−∞,−2](-\infty, -2] is [−4,−2][-4, -2]. …

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