Reason : Let q1(x) and q2(x) be the quotients when p(x) is divided by (x–1)and(x–3) respectively. Also asume that q3(x) is quotient when p(x) is divided by (x–1)(x–3). Now A.G.C I, p(x)=q1(x)×(x–1)+3...(1) and A.G.C II, p(x)=q2(x)×(x–3)+5 ...(2) Also A.G.C III, p(x)=q3(x)(x–1)(x–3)+r(x) ...(3) Now at x = 1, from eqn no. (1), p(1) = 3...(4) at x = 3, from eqn no. (2), p(3) = 5...(5) Now put x = 1 in equation no. (3) we get p(1)=r(1)⇒r(1)=3 [using equation (4)]...(6) Similarly from (3) & (5) we get r(3) = 5...(7) from equation no.(3) it is clear that divisor (x–1)(x–3) is of degree two so its remainder r(x) will be of degree one or zero so Let r(x)=Ax+B put r = 1 & r = 3 is equation (8) we get r(1)=A+B⇒A+B=3 [using (6)]...(8) similarly r(3)=3A+B⇒3A+B=5 [using 7] ...(9) on solving equation no.(8) & (9) we getA = 1 & B = 2⇒ r(x)=1x+2 ...(10)Now put x = –2 in equation no.(10) we get r(–2)=–2+2=0 Hence option no.(3) is correct.