A differentiable function $f:\mathbb R \to \mathbb R$ such that $f(0)=0$ and $f^{\prime}(x)+2 f(x)>0$ for all $x \in \mathbb{R}$

From the inequality, we know that $f'(0)>0$ hence $f$ is strictly increasing at $0$. In particular, it must change sign and statements B and C are false. Also since it is increasing and $f(0)=0$, statement D is also false.

Suppose $f$ vanishes at $x_0 \ne 0$, if it does not change its sign, then $f'(x_0)+2f(x_0)=0$, hence a contradiction. Thus, if $f$ vanishes at such $x_0 \ne 0$, it must change sign.

Since $f(0)=0$ and strictly increasing, if $f$ changes sign there is a first positive $x_0$ where it happens and $f'(x_0)<0$. So $f'(x_0)+2f(x_0)<0$, contradiction. From the same argument, let $x_0$ the greatest negative root of $f$, then $f'(x_0)<0$ and $f'(x_0)+2f(x_0)<0$, also a contradiction.

As a conclusion, $f$ must vanish at $0$ only and is strictly increasing at this point. So statement A is true.

Consider the function $e^{2x}f(x)$; we have

$(e^{2x}f(x))' = 2e^{2x}f(x) + e^{2x}f'(x) = e^{2x}(f'(x) + 2f(x)) > 0, \tag 1$

since

$f'(x) + 2f(x) > 0 \tag 2$

by hypothesis; furthermore,

$e^{2 \cdot 0}f(0) = e^0 f(0) = f(0) = 0, \tag 3$

also by hypothesis. Thus we see from (1) and (3) that $e^{2x}f(x)$ is strictly monotonically increasing, and takes the value $0$ at $x = 0$. Thus

$x > 0 \Longrightarrow e^{2x}f(x) > 0 \tag 4$

and

$x < 0 \Longrightarrow e^{2x}f(x) < 0. \tag 5$

Since

$\forall x \in \Bbb R, e^{2x} > 0, \tag 6$

it follows from (4) and (5) that

$x > 0 \Longrightarrow f(x) > 0 \tag 7$

and

$x < 0 \Longrightarrow f(x) < 0, \tag 8$

i.e., (A) is the requisite result.

Note Added in Edit, Friday 7 January 2022 12:09 AM PST: It is easy to see that essentially the same argument applies if the "$2$" in $f'(x) + 2f(x) > 0$ is replaced by any real $k > 0$, so that $f'(x) + 2f(x) > 0$ becomes $f'(x) + kf(x) > 0$ End of Note.