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The set of values for which x3 + 1 > x2 + x is(a) x > 0 (b) x < 0 (c) x > – 1 (d) – 1 < x < 1 |
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Answer» (c) x > – 1 x3 + 1 > x2 + \(x\) ⇒ x3 + 1 – x2 – \(x\) > 0 ⇒ x3 – x2 + 1 – \(x\) > 0 ⇒ x2 (x – 1) – (x – 1) > 0 ⇒ (x2 – 1) (x – 1) > 0 ⇒ (x + 1) (x – 1)2 > 0 As (x – 1)2 is +ve so the given expression is > 0, when x + 1 > 0 ⇒ x > – 1. |
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