Question

For all ‘x’, x² + 2ax + 10 – 3a > 0, then the interval in which ‘a’ lies is

• A. a< - 5
• B. – 1 < a < 2
• C. a> 5
• D. 2 < a < 5

Answer 1

Answer: (B)

– 1 < a < 2

Answer 2

x² + 2ax + 20 – 3x > 0 \overset{̶}{V}x

⇒ D < 0 ⇒ 4a² – 4(10 – 3a) < 0 ⇒ a² + 3a – 10 < 0

⇒ (a + 5) (a – 2) < 0 ⇒ a ε (- 5, 2)