Let $f:[0,1] \rightarrow \mathbb{R}$ (the set of all real numbers) be a function. Suppose the function $f$ is twice differentiable, $f(0)=f(1)=0$ and satisfies $f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^{x}, x \in[0,1]$. If the function $\mathrm{e}^{-x} f(x)$ assumes its minimum in the interval $[0,1]$ at $x=\frac{1}{4}$, which of the following is true ?

(C) $f^{\prime}(x)<f(x), \quad 0<x<\frac{1}{4}$

(A) $f^{\prime}(x)<f(x), \frac{1}{4}<x<\frac{3}{4}$

(B) $f^{\prime}(x)>f(x), \quad 0<x<\frac{1}{4}$

(D) $f^{\prime}(x)<f(x), \frac{3}{4}<x<1$

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Application of Derivatives

Let

f : [0, 1] \to R

(the set of all real numbers) be a function. Suppose the function

f

is twice differentiable,

f (0) = f (1) = 0

and satisfies

f^{''} (x) - 2 f^{'} (x) + f (x) \geq e^{x}, x \in [0, 1]

.

If the function

e^{- x} f (x)

assumes its minimum in the interval

[0, 1]

x = \frac{1}{4}

, which of the following is true ?

Answer: (C)

Let

ϕ (x) = e^{- x} f (x)

. Then

ϕ^{'} (x) = e^{- x} (f^{'} (x) - f (x))

.

Since

ϕ

has a minimum at

x = 1/4

, we have

ϕ^{'} (1/4) = 0

, which means

f^{'} (1/4) = f (1/4)

.

From the previous question,

ϕ^{''} (x) = e^{- x} (f^{''} (x) - 2 f^{'} (x) + f (x)) > 0

for all

x \in [0, 1]

.

This means

ϕ^{'} (x)

is strictly increasing on

[0, 1]

.

Since

ϕ^{'} (1/4) = 0

For $x < 1/4$ : $ϕ^{'} (x) < 0$ , so $f^{'} (x) - f (x) < 0$ , i.e., $f^{'} (x) < f (x)$
For $x > 1/4$ : $ϕ^{'} (x) > 0$ , so $f^{'} (x) - f (x) > 0$ , i.e., $f^{'} (x) > f (x)$

Therefore,

f^{'} (x) < f (x)

for

0 < x < 1/4

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Single correct option