(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].