Question

If $f:R\to R$ is a differentiable function such that $f^{\prime }\left(x\right)>2f\left(x\right)\forall x\in R$ 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. $f\left(x\right)<e^{2x}$ in $(0,\infty )$

Answer 1

The correct option is A $f\left(x\right)$ is increasing in $(0,\infty )$

Given that,

$f^{\prime }\left(x\right)>2f\left(x\right),f\left(0\right)=1$

$\int \frac{df\left(x\right)}{f\left(x\right)}>\int 2dx$

$\ln f\left(x\right)>2x+c$

$f\left(x\right)>k{e}^{2x},k>0$

$f\left(0\right)=1$

$1>k\therefore 0<k<1$

$\therefore f\left(x\right)>k{e}^{2x}>0$

$f^{\prime }\left(x\right)>2f\left(x\right)>0$

$\therefore f\left(x\right)$ is increasing function (since $f^{\prime }\left(x\right)>0$) in $(0,\infty )$