x2+b2=1−2bx ...... (i) x2+b2+2bx=1 (x+b)2=1 x+b=±1 ∴x=1−b,−1−b x2+a2=1−2ax ...... (ii) x2+a2+2ax=1 (x+a)2=1 ∴x=1−a,−1−a It is given that the equations have only one root in common, then there are 2 possibilities. (i) 1−b=−1−a a−b=−2 a−b+2=0 or b−a=2 (ii) −1−b=1−a a−b=2 ∴∣a−b∣=2 If we take 1−b=1−a or −1−a=−b−1 we get a=b but then both roots will be common, which is not possible. Hence, options A,B and C are correct.