f(0)=0,f(1)=1 and f(2)=2 Let h(x)=f(x)−x Clearly h(x) is continuous and twice differentiable on (0,2) Also, h(0)=h(1)=h(2)=0 ∴h(x) satisfies all the condition of Rolle's theorem. ∴ There exist C1∈(0,1) such that h′(c1)=0 ⇒f′(c1)−1=0 ⇒f′(c1)=1 also there exist c2∈(1,2) such that h′(c2)=0 f′(c2)=1 Now, using Rolle's theorem on [c1,c2] for f′(x) We have f′′(c)=0,c∈(c1,c2) Hence, f′′(x)=0 for some x∈(0,2).