If $f:R\to R$ is a differentiable function such that ${f}'\left( x \right)>2f\left( x \right)$ for all $x\in R$ 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 $\left( 0,\infty \right)$

Hint: We solve this question by first considering the given inequality. Then we assume that $y=f\left( x \right)$ and solve the differential equation using the formula for integrating factor, ${{e}^{\int{P\left( x \right)dx}}}$. Then we substitute the value $x=0$ and find the inequality of $f\left( x \right)$. Then we differentiate $f\left( x \right)$ again to find whether $f\left( x \right)$ is an increasing function or decreasing function.

Complete step by step answer:
We are given that $f:R\to R$ is a differentiable function and ${f}'\left( x \right)>2f\left( x \right)$.
We are also given that $f\left( 0 \right)=1$.
Now, let us consider the given condition, ${f}'\left( x \right)>2f\left( x \right)$.
We can write it as,
$\begin{align}
  & \Rightarrow {f}'\left( x \right)>2f\left( x \right) \\
 & \Rightarrow {f}'\left( x \right)-2f\left( x \right)>0 \\
\end{align}$
Now let us assume that $y=f\left( x \right)$. Then our above inequality becomes,
$\Rightarrow \dfrac{dy}{dx}-2y>0$
We can see that it is in the form of $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$. We can solve such differential equations by multiplying it with an integrating factor ${{e}^{\int{P\left( x \right)dx}}}$.

Here $P=-2$. So, integrating factor is,
$\Rightarrow {{e}^{\int{-2dx}}}={{e}^{-2x}}$
Multiplying our inequality with it we get,
\[\begin{align}
  & \Rightarrow {{e}^{-2x}}\dfrac{dy}{dx}-{{e}^{-2x}}2y>0\left( {{e}^{-2x}} \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( {{e}^{-2x}}y \right)>0 \\
\end{align}\]
Substituting the value of y we get,
\[\Rightarrow \dfrac{d}{dx}\left( {{e}^{-2x}}f\left( x \right) \right)>0\]
Now let us assume that $g\left( x \right)={{e}^{-2x}}f\left( x \right)$. Then we have,
$\Rightarrow {g}'\left( x \right)>0$
Now let us consider the property that a function $f\left( x \right)$ is said to be an increasing function if ${f}'\left( x \right)>0$.
As we have from above that ${g}'\left( x \right)>0$, we can say that $g\left( x \right)$ is an increasing function.
So, we can say that, for $x\in \left( 0,\infty \right)$
$\begin{align}
  & \Rightarrow g\left( x \right)>g\left( 0 \right) \\
 & \Rightarrow {{e}^{-2x}}f\left( x \right)>{{e}^{-2\left( 0 \right)}}f\left( 0 \right) \\
 & \Rightarrow {{e}^{-2x}}f\left( x \right)>{{e}^{0}}\left( 1 \right) \\
 & \Rightarrow {{e}^{-2x}}f\left( x \right)>1 \\
 & \Rightarrow f\left( x \right)>{{e}^{2x}} \\
\end{align}$
So, we get that $f\left( x \right)>{{e}^{2x}}$ for $x\in \left( 0,\infty \right)$.
Now let us differentiate the above inequality, $f\left( x \right)>{{e}^{2x}}$.
Now let consider the formula for differentiation,
$\dfrac{d}{dx}\left( {{e}^{ax}} \right)=a{{e}^{ax}}$
Using it we get,
$\Rightarrow {f}'\left( x \right)>2{{e}^{2x}}>0$
As ${f}'\left( x \right)>0$, we can say that $f\left( x \right)$ is an increasing function.

So, the correct answer is “Option C and D”.

Note: There is a possibility of one making a mistake by stopping immediately after we found that Option C, $f\left( x \right)>{{e}^{2x}}$ for $x\in \left( 0,\infty \right)$ is correct. But we need to make sure that the options are correct or not in this type of questions, because even if we get that $f\left( x \right)>{{e}^{2x}}$ for $x\in \left( 0,\infty \right)$, there might be a possibility of $f\left( x \right)$ being an increasing or decreasing function.