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NCERT Exemplar · Q19

Q.If ∣x−1∣>5|x - 1| > 5, then
(A) x∈(−4,6)x \in (-4, 6)
(B) x∈[−4,6]x \in [-4, 6]
(C) x∈[−∞,−4)∪(6,∞)x \in [-\infty, -4) \cup (6, \infty)
(D) x∈[−∞,−4)∪[6,∞)x \in [-\infty, -4) \cup [6, \infty)

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An absolute-value inequality ∣x−1∣>5|x - 1| > 5 means the distance from xx to 11 exceeds 55, so xx lies more than 55 units away from 11 on either side: x<−4x < -4 or x>6x > 6.

The absolute value ∣x−1∣|x - 1| measures the distance between xx and 11 on the number line. When we write ∣x−1∣>5|x - 1| > 5, we're asking: for which values of xx is this distance strictly greater than 55?

Picture the number line with 11 at the center. Points exactly 55 units away are 1−5=−41 - 5 = -4 (to the left) and 1+5=61 + 5 = 6 (to the right). The inequality ∣x−1∣>5|x - 1| > 5 demands that xx be farther than these boundary points—so xx must lie either to the left of −4-4 or to the right of 66.

Now let's translate this into algebra.

Step-by-step solution

  1. Rewrite the absolute-value inequality. The definition of absolute value tells us that ∣A∣>B|A| > B (for B>0B > 0) splits into two cases:

A>BorA<−B.A > B \quad \text{or} \quad A < -B.

Here A=x−1A = x - 1 and B=5B = 5, so

x−1>5orx−1<−5.x - 1 > 5 \quad \text{or} \quad x - 1 < -5.

  1. Solve each inequality separately.
    • From x−1>5x - 1 > 5, add 11 to both sides:

x>6.x > 6.

  • From x−1<−5x - 1 < -5, add 11 to both sides:

x<−4.x < -4.

  1. Combine the solution sets. The word "or" means we take the union of the two intervals. In interval notation: x∈(−∞,−4)∪(6,∞).x \in (-\infty, -4) \cup (6, \infty). …

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