Question

If x is real and x² + 3x + 2 > 0, x² – 3x – 4 < 0, then which of the following is correct?

(a) – 1 < x < 4 

(b) 2 < x < 4 

(c) – 1 < x < 1 

(d) – 1 < x < 1 or 2 < x < 4

Answer 1

(d) – 1 < x < 1 or 2 < x < 4

x² – 3x + 2 > 0 

⇒ (x – 1) (x – 2) > 0

⇒ \(x\)∈ (– ∞, 1) ∪ (2, ∞)               …(i)

and x² – 3\(x\) – 4 < 0 ⇒ (\(x\) – 4) (\(x\) + 1) < 0

⇒ – 1 < x < 4             …(ii) 

∴ Combining the solution sets in (i) and (ii), we have 

– 1 < x < 1 or 2 < x < 4