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

Answer 1

Answer: (C), (D)

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

Given : $f:R\to R$ is a differential function such that $f^{\prime }\left(x\right)>2f\left(x\right)$

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

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

Integrating both sides, we get

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

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

$⟹f\left(x\right)+C>e^{2x}$

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

$⟹f\left(0\right)+C>e^0⟹C>0$

$⟹f\left(x\right)>e^{2x}$

$\therefore$ D is correct

Since, $e^{2x}$ increases in the interval $(0,\infty )$

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

Hence, C and D are correct.