JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Application of Derivatives

Let

f

be any continuous function on

[0, 2]

and twice differentiable on

(0, 2)

. If

f (0) = 0, f (1) = 1

and

f (2) = 2

, then

Answer: (D)

Solution

The Mean Value Theorem on

[0, 1]

gives

c_{1} \in (0, 1)

with

f^{'} (c_{1}) = \frac{f ( 1 ) - f ( 0 )}{1} = 1

, and on

[1, 2]

gives

c_{2} \in (1, 2)

with

f^{'} (c_{2}) = \frac{f ( 2 ) - f ( 1 )}{1} = 1

. Applying Rolle's theorem to

f^{'}

[c_{1}, c_{2}]

, where

f^{'} (c_{1}) = f^{'} (c_{2}) = 1

, yields

c \in (c_{1}, c_{2}) \subset (0, 2)

with

f^{''} (c) = 0

. Hence

f^{''} (x) = 0

for some

x \in (0, 2)

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option