QJEEM21Mafe5c: JEE Main 2021 Mathematics - Class 12 JEEM21Mafe5c Solution: Let f be any continuous function on [0, 2] and twice…

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

$$\therefore$$ h(x) satisfies all the condition of Rolle's theorem.

$$\therefore$$ there exist C₁ $$\in$$(0, 1) such that h'(c₁) = 0

$$\Rightarrow$$ f'(₁) $$-$$ 1 = 0 $$\Rightarrow$$ f'(c₁) = 1

also there exist c₂ $$\in$$(1, 2) such that h'(c₂) = 0

$$\Rightarrow$$ f'(c₂) = 1

Now, using Rolle's theorem on [c₁, c₂] for f'(x)

We have f''(c) = 0, c$$\in$$(c₁, c₂)

Hence, f''(x) = 0 for some x$$\in$$(0, 2).