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

Answer 1

Answer: (A), (C)

$f\left(x\right)$ is increasing in $(0,\infty )$

$f^{\prime }\left(x\right)-2f\left(x\right)>0$
$e^{-2x}\cdot f^{\prime }\left(x\right)-2e^{-2x}\cdot f\left(x\right)>0$
$\frac{d}{dx}\left(e^{-2x}f\right(x\left)\right)>0\Rightarrow e^{-2x}\cdot f\left(x\right)$ is increasing function.
$e^{-2x}\cdot f\left(x\right)>1$ for all $x\in (0,\infty \text{)}$
$f\left(x\right)>e^{2x}$
$\because f^{\prime }\left(x\right)>2f\left(x\right)>e^{2x}>0$
$\therefore f\left(x\right)$ is increasing
Also as, $f^{\prime }\left(x\right)=\frac{f\left(x\right)-f\left(0\right)}{x-0}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{f\left(x\right)-1}{x}$
i.e., $f^{\prime }\left(x\right)>e^{2x}\forall x\in (0,\text{1)}$
$<e^{2x}\forall x\in (1,\infty \text{)}$