Let and be the roots of the equation x2 + 2ax + (3a + 10) = 0 such that < 1 < . Then the set of all possible values of a is :
- (- , -115 ) (5, )
- (- , -115 )
- (- , -3)
- (- , -2) (5, )
Answer
(2) (- , -115 )
Solution
Let f(x) = x2 + 2ax + (3a + 10) (upward parabola)
1 lies between the roots ⇔ f(1) < 0
1 + 2a + 3a + 10 < 0
5a + 11 < 0 ⇒ a < - 115
(f(1) < 0 already ensures real, distinct roots)
a (- , -115 )
Answer: (2)