Question

If $f:R\to R$ is a differentiable function such that $f^{\prime }\left(x\right)>2f\left(x\right)$ for all $x\in R$, and $f\left(0\right)=1,$ then

• A. $f\left(x\right)$ is increasing in$(0,\infty )$
• B. $f\left(x\right)$ is decreasing in$(0,\infty )$
• C. $f\left(x\right)>e^{2x}$ in $(0,\infty )$
• D. $f^{\prime }\left(x\right)<e^{2x}$ in $(0,\infty )$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(x\right)$ is increasing in$(0,\infty )$
C $f\left(x\right)>e^{2x}$ in $(0,\infty )$

$f^{\prime }\left(x\right)-2f\left(x\right)>0\forall x\in R\cdots \left(1\right)$
$\text{Let}y=f\left(x\right)$
$\text{Then}\frac{dy}{dx}-2y>0$
Integrating function $\text{I.F.}=e^{\int -2dx}=e^{-2x}$

Multiplying by $e^{-2x},$ we get
$e^{-2x}\frac{dy}{dx}-2y{e}^{-2x}>0$
$\Rightarrow \frac{d}{dx}(y.e^{-2x})=\frac{d}{dx}\left(f\right(x).e^{-2x})>0$

Suppose $g\left(x\right)=f\left(x\right).e^{-2x}$
Then $g^{\prime }\left(x\right)>0\forall x\in R\left[\text{From (1)}\right]$
$\therefore g\left(x\right)=f\left(x\right).e^{-2x}$ is strictly increasing for all $x\in R$
Also, $g\left(0\right)=1$
$\Rightarrow g\left(x\right)>g\left(0\right)=1\text{for all}x>0$
$\Rightarrow f\left(x\right).e^{-2x}>1\forall x\in (0,\infty )$
Hence $f\left(x\right)>e^{2x}\forall x\in (0,\infty )$
$\text{As}{f}^{\prime }\left(x\right)>2f\left(x\right)\Rightarrow f^{\prime }\left(x\right)>2e^{2x}>2\forall x\in (0,\infty )$
$\therefore f\left(x\right)$ is strictly increasing in $(0,\infty )$