JEE Main 31-Aug-2021 Shift 2 Solved Paper

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).