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How many integers are there in the solution set of |2x + 6| < \(\frac{19}{2}\) ?(a) None (b) Two (c) Fourteen (d) Nine

Answer»

(d) Nine

| 2x + 6 | < \(\frac{19}{2}\)

⇒ 2x + 6 < \(\frac{19}{2}\) or - (2x + 6) < \(\frac{19}{2}\)

⇒ - \(\frac{19}{2}\) < 2x + 6 < \(\frac{19}{2}\)

⇒ - \(\frac{19}{2}\) - 6 < 2x < \(\frac{19}{2}\) - 6 ⇒ \(-\frac{31}{2}\) < 2x < \(\frac72\)

⇒ \(-\frac{31}{2}\) < x < \(\frac72\) ⇒ - 7.75 < x > 1.75

∴ The integers between –7.75 and 1.75 are –7, –6, –5, –4, –3, –2, –1, 0, 1, i.e., nine in number.



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