Correct option (A), (B), (C), (D)
f(−1) f(0) < 0
⇒ (2a − b − c) (c + a − 2b) < 0
⇒ (a − b + a − c) (c + a − 2b) < 0
⇒ c + a < 2b
One root is 1 and other is c + a - 2b/a + b - 2c
Therefore, both roots are rational.
Now, discriminant of ax2 + 2bx + c = 0 is 4b2 − 4ac.
Using (c + a) < 2b, we have D > 0.
Also, a, b and c are +ve. Therefore, both the roots are real and −ve.