If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function such that $f'(x)>2f(x)$ for all $x \in \mathbb{R}$, and $f(0)=1$, then

(A) $f(x)$ is increasing in $(0, \infty)$; (C) $f(x)>e^{2x}$ in $(0, \infty)$

(B) $f(x)$ is decreasing in $(0, \infty)$

(D) $f'(x)<e^{2x}$ in $(0, \infty)$

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Application of Derivatives

f : R \to R

is a differentiable function such that

f^{'} (x) > 2 f (x)

for all

x \in R

, and

f (0) = 1

, then

Answer: (A), (C)

Given

f^{'} (x) > 2 f (x)

, we can write:

f^{'} (x) - 2 f (x) > 0

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

\frac{d}{d x} (f (x) e^{- 2 x}) > 0

This means

f (x) e^{- 2 x}

is strictly increasing.

For

x > 0

f (x) e^{- 2 x} > f (0) e^{0} = 1

f (x) > e^{2 x}

Since

f (x) > e^{2 x} > 0

for

x > 0

, and

f^{'} (x) > 2 f (x) > 0

, the function

f (x)

is increasing in

(0, \infty)

.

Also,

f^{'} (x) > 2 f (x) > 2 e^{2 x}

, so option (D) is incorrect.

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · One or more correct options