If $f : \mathbb R \to \mathbb R$ is a differentiable function such that $f '(x) > 2f(x)$ for all $x\in \mathbb R$, and $f(0) = 1$, then which of the following are correct?
(A) $f '(x) < e^{2x}$ in $(0, \infty)$
(B) $f(x)$ is increasing in $(0, \infty)$
(C) $f(x)$ is decreasing in $(0, \infty)$
(D) $f(x) > e^{2x}$ in $(0, \infty)$
My attempt is as follows:-
Case $1$: $f(x)>0$ $\forall$ $x\in \mathbb R$
$$\dfrac{f'(x)}{f(x)}>2$$
Integrating both sides
$$\int_{0}^{x}\dfrac{f'(x)}{f(x)}dx>\int_{0}^{x}2dx$$ $$\ln\left|f(x)\right|-\ln\left|f(0)\right|>2x$$
As $f(x)>0$, $$\ln(f(x))>2x$$
$$f(x)>e^{2x}$$
So $(D)$ option looks correct
As $f(x)>0$ $\forall$ $x\in \mathbb R$, so $f'(x)>0$ because $f'(x)>2f(x)$. Hence we can say function is increasing $\forall$ $x\in \mathbb R$.
Hence $(B)$ option looks correct, but I am getting these answers only under the assumption $f(x)>0$ $\forall$ $x\in \mathbb R$, is it fine?
Case $2$: $f(x)<0$ $\forall$ $x\in \mathbb R$
$$\dfrac{f'(x)}{f(x)}<2$$
Integrating both sides
$$\int_{0}^{x}\dfrac{f'(x)}{f(x)}dx>\int_{0}^{x}2dx$$ $$f(x)<e^{2x}$$
But from here we are not able to match any options.
Given that $$f'>2f \implies e^{-2x} f'- 2f e^{-2x} >0 \implies \frac{d (f e^{-2x})}{dx} >0, \forall x \in R.$$ So $g(x)=f(x)e^{-2x}$ is an inrceasing function. Then $$x >0 \implies g(x) >g(0) \implies e^{-2x} f(x) >1, x \in (0,\infty).$$ So (D) option is correct.