According to question. f′′(x)=g′′(x) Integrating w.r.t. x, we get f′(x)=g′(x)+C1 Put x=1⇒f′(1)=g′(1)+C1 ⇒‌4=6+C1 ∴‌C1=−2 ∴‌f′(x)=g′(x)−2 Again, integrating w.r.t.x, we get f(x)=g(x)−2x+C2 ⇒‌3=9−4+C2⇒C2=−2 ∴‌f(x)=g(x)−2x−2 Put x=1 , we get f(1)−g(1)=−2(1)−2=−4