For all twice differentiable functions $f: R \rightarrow R$, with $f(0)=f(1)=f^{\prime}(0)=0$,

(B) $f^{\prime \prime}(x)=0$, for some $x \in(0,1)$

(A) $f^{\prime \prime}(x) \neq 0$, at every point $x \in(0,1)$

(D) $f^{\prime \prime}(x)=0$, at every point $x \in(0,1)$

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Application of Derivatives

For all twice differentiable functions

f : R \to R

, with

f (0) = f (1) = f^{'} (0) = 0

Answer: (B)

Because

f (0) = f (1)

and

f

is differentiable, Rolle's theorem gives a point

c \in (0, 1)

with

f^{'} (c) = 0

. We already have

f^{'} (0) = 0

, so

f^{'}

vanishes at both

0

and

c

. Applying Rolle's theorem to

f^{'}

[0, c]

yields a point

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

where

f^{''} (c_{1}) = 0

. Hence

f^{''} (x) = 0

for some

x \in (0, 1)

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option