f(x+1)=a1+10(x+1)+a2(x+1)2+a3(x+1)3+(x+1)4 Coefficient of x3 in f(x+1)=a3+4 g(x+2)=b1+3(x+2)+b2(x+2)2+b3(x+2)3+(x+2)4 Coefficient of x3 in g(x+2)=b3+8 ⇒ Coefficient of x3 in h(x)=f(x+1)−g(x+2) is a3+4−b3−8=a3−b3−4 But f(x)≠g(x)∀x ⇒f(x)−g(x)≠0 ⇒f(x)−g(x)=0 have no real roots (a1−b1)+7x+(a2−b2)x2+(a3−b3)x3=0 have no real roots ⇒a3−b3=0 ⇒ Coefficient of x3 in h(x)=−4 Option (C) is correct.