Question

The set of values for which x³ + 1 > x² + x is

(a) x > 0 

(b) x < 0 

(c) x > – 1 

(d) – 1 < x < 1

Answer 1

(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.