Solution set of the inequation x2−2x≥15 is
(−∞,3]∪[5,∞)
(−∞,−5]∪[3,∞)
(−∞,−3]∪[5,∞)
None of these
The given inequation is x2−2x≥15
⇒ x2−2x−15≥0
⇔ (x−5)(x+3)≥0
⇔ (x−(−3))(x−5)≥0
⇔ either x≤−3 or x≥5
⇔ x∈ (−∞,3] ∪ [5,∞)
(∵ For a<b, (x−a)(x−b)≥0 iff either x≤a or x≥b)