Question

If $f:\mathbb{R}\to \mathbb{R}$ is a differentiable function such that ${f}'\left( x \right)>2f\left( x \right)$ for all $x\in \mathbb{R}$ , and $f\left( 0 \right)=1$, then
(a) $f\left( x \right)$ is decreasing in $\left( 0,\infty \right)$
(b) ${f}'\left( x \right)<{{e}^{2x}}$ in $\left( 0,\infty \right)$
(c) $f\left( x \right)$ is increasing in $\left( 0,\infty \right)$
(d) $f\left( x \right)>{{e}^{2x}}$ in $\left( 0,\infty \right)$
This question can have multiple correct options.

Answer 1

Hint: First we will show that \[{{e}^{-2x}}f\left( x \right)\] is an increasing function from the data given in the question. Then we will try to see which relation among ${f}'\left( x \right)<{{e}^{2x}}$and $f\left( x \right)>{{e}^{2x}}$turns out to be true. Then we will try to find if f(x) is an increasing or decreasing function in $\left( 0,\infty \right)$.

Complete step-by-step answer:
We know, $f:\mathbb{R}\to \mathbb{R}$ is a differentiable function such that ${f}'\left( x \right)>2f\left( x \right)$ for all $x\in \mathbb{R}$ and $f\left( 0 \right)=1$.

We also know that f(x) is an increasing function if $f\left( {{x}_{1}} \right)\le f\left( {{x}_{2}} \right)$ when ${{x}_{1}}\le {{x}_{2}}$.

Let us recall a few differentiation formulas and concepts according to the question:
(i) \[\dfrac{d}{dx}\left[ {{e}^{mx}} \right]=m{{e}^{mx}}\]
(ii) \[\dfrac{d}{dx}\left[ uv \right]=u\dfrac{d}{dx}\left[ v \right]+v\dfrac{d}{dx}\left[ u \right]\] where u and v are functions of x.
(iii) if \[\dfrac{d}{dx}\left[ h\left( x \right) \right]>0\] where h(x) is a function of x, then h(x) is an increasing function.

Now, from the data given in question,
$\begin{align}
  & {f}'\left( x \right)>2f\left( x \right) \\
 & \Rightarrow {f}'\left( x \right)-2f\left( x \right)>0 \\
\end{align}$
Multiplying this equation with ${{e}^{-2x}}$ , we get
\[\begin{align}
  & {{e}^{-2x}}\left[ {f}'\left( x \right)-2f\left( x \right) \right]>{{e}^{-2x}}\left[ 0 \right] \\
 & \Rightarrow {{e}^{-2x}}{f}'\left( x \right)-2{{e}^{-2x}}f\left( x \right)>0 \\
 & \Rightarrow {{e}^{-2x}}\dfrac{d}{dx}\left[ f\left( x \right) \right]+\dfrac{d}{dx}\left[ {{e}^{-2x}} \right]f\left( x \right)>0\text{ from }\left( i \right) \\
 & \Rightarrow \dfrac{d}{dx}\left[ {{e}^{-2x}}f\left( x \right) \right]>0\text{ from }\left( ii \right) \\
\end{align}\]

From (iii), \[{{e}^{-2x}}f\left( x \right)\] is an increasing function.

Let us assume \[g\left( x \right)={{e}^{-2x}}f\left( x \right)\]
Then for $x=0$ ,
\[\begin{align}
  & g\left( 0 \right)={{e}^{-2\cdot 0}}f\left( 0 \right) \\
 & ={{e}^{0}}\cdot 1 \\
 & =1\text{ }\left( \because {{e}^{0}}=1 \right)
\end{align}\]

And for $x>0$
\[g\left( x \right)>g\left( 0 \right)\] since g(x) is an increasing function in $\left( 0,\infty \right)$
\[\begin{align}
  & \Rightarrow {{e}^{-2x}}f\left( x \right)>1 \\
 & \Rightarrow f\left( x \right)>\dfrac{1}{{{e}^{-2x}}} \\
 & \Rightarrow f\left( x \right)>{{e}^{2x}}\text{ }\ldots \left( iv \right) \\
\end{align}\]
Thus, option (d) is correct.

We know from the question that
${f}'\left( x \right)>2f\left( x \right)$
Putting the value of equation (iv) in above expression, we get
${f}'\left( x \right)>2f\left( x \right)>2{{e}^{2x}}\text{ }\ldots (v)$

We know that the exponential function ${{e}^{x}}$ is an increasing function in $\left( 0,\infty \right)$
Thus, above expression becomes
$\begin{align}
  & {f}'\left( x \right)>2f\left( x \right)>2{{e}^{2x}}>0 \\
 & \Rightarrow {f}'\left( x \right)>0 \\
\end{align}$

From (iii), $f\left( x \right)$ is an increasing function in $\left( 0,\infty \right)$
Thus, option (c) is also correct.

So, the correct answers are “Option C and D”.

Note: Since f(x) is an increasing function in $\left( 0,\infty \right)$, it can not be decreasing at the same time in $\left( 0,\infty \right)$. Therefore, option (a) is wrong. And from expression (v), ${f}'\left( x \right)>{{e}^{2x}}$, which contradicts option (b). Thus, option (b) is also wrong.