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Solve : 

(i) |2x - 3| ≤ \(\frac14\)

(ii) |x – 4| > 7

(iii) 1 < | x – 3 | < 5

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(i) We know that |x – a| < r        ⇒ (a – r) < x < (a + r)

∴  |2x - 3| ≤ \(\frac14\)  ⇒  \(\big(3-\frac14\big)\) < 2\(x\) < \(\big(3+\frac14\big)\) ⇒ \(\frac{11}{4}\)< 2\(x\) < \(\frac{13}{4}\)

⇒  \(\frac{11}{8}\)< \(x\) < \(\frac{13}{8}\)      ⇒ \(x\) ∈ \(\big[​​\frac{11}{8},\frac{13}{8}\big]\)

(ii) Since |x – a| > r ⇒ x < a – r or x > a + r

|x – 4| > 7  ⇒ x < 4 – 7 or x > 4 + 7  ⇒ x < –3 or x > 11 

\(x\)∈ (–∞, –3] or \(x\)∈ [11, ∞) ⇒ \(x\)∈ (–∞, –3] ∪ [11, ∞). 

(iii) Since a < |x – c| < b ⇒ x∈ [–b + c, –a + c] ∪ [a + c, b + c] 

∴ 1 < | x – 3 | < 5  ⇒ x∈ [–5 + 3, –1 + 3] ∪ [1 + 3, 5 + 3] 

\(x\) ∈ [–2, 2] ∪ [4, 8].

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