**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 ax^{2} + 2bx + c = 0 is 4b^{2} − 4ac.

Using (c + a) < 2b, we have D > 0.

Also, a, b and c are +ve. Therefore, both the roots are real and −ve.