Properties of Differentiable Function Satisfying Inequality Condition

To determine the correct statements, let us define an auxiliary function $g: \mathbb{R} \rightarrow \mathbb{R}$ given by: $g(x) = e^{-2x} f(x)$

Differentiating $g(x)$ with respect to $x$ using the product rule gives: $g'(x) = -2 e^{-2x} f(x) + e^{-2x} f'(x) = e^{-2x} \left( f'(x) - 2f(x) \right)$

We are given that $f'(x) > 2f(x)$ for all $x \in \mathbb{R}$, which implies: $f'(x) - 2f(x) > 0 \quad \text{for all } x \in \mathbb{R}$

Since the exponential function $e^{-2x} > 0$ for all $x \in \mathbb{R}$, we have: $g'(x) > 0 \quad \text{for all } x \in \mathbb{R}$

This implies that $g(x)$ is a strictly increasing function on $\mathbb{R}$.

Now, for any $x \in (0, \infty)$, since $x > 0$ and $g(x)$ is strictly increasing: $g(x) > g(0)$

Using $f(0) = 1$, we can evaluate $g(0)$: $g(0) = e^{0} f(0) = 1 \cdot 1 = 1$

Therefore: $e^{-2x} f(x) > 1 \implies f(x) > e^{2x} \quad \text{for all } x \in (0, \infty)$

This confirms that Option (C) is correct.

Next, for $x \in (0, \infty)$, since $e^{2x} > 1 > 0$, it follows from $f(x) > e^{2x}$ that: $f(x) > 0 \quad \text{for all } x \in (0, \infty)$

Using the given condition $f'(x) > 2f(x)$: $f'(x) > 2f(x) > 2e^{2x} > 0 \quad \text{for all } x \in (0, \infty)$

Since $f'(x) > 0$ for all $x \in (0, \infty)$, the function $f(x)$ is strictly increasing in $(0, \infty)$. This confirms that Option (A) is correct and Option (B) is incorrect.

Finally, since $f'(x) > 2e^{2x} > e^{2x}$ for $x \in (0, \infty)$, Option (D) is incorrect.

Correct Options: (A) and (C)