(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].**