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(C) Given $f^{\prime}(x) - 2f(x) > 0$. Multiply by the integrating factor $e^{-2x}$: $e^{-2x} f^{\prime}(x) - 2e^{-2x} f(x) > 0$. This is equivalent t

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$A, C$

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(C) Given $f^{\prime}(x) - 2f(x) > 0$. Multiply by the integrating factor $e^{-2x}$: $e^{-2x} f^{\prime}(x) - 2e^{-2x} f(x) > 0$. This is equivalent to $\frac{d}{dx}(f(x) e^{-2x}) > 0$. Let $g(x) = f(x) e^{-2x}$. Since $g^{\prime}(x) > 0$,$g(x)$ is a strictly increasing function. For $x > 0$,$g(x) > g(0)$. Since $g(0) = f(0) e^0 = 1 \cdot 1 = 1$,we have $f(x) e^{-2x} > 1$,which implies $f(x) > e^{2x}$ for all $x > 0$. Since $f(x) > e^{2x} > 0$ for $x > 0$,and $f^{\prime}(x) > 2f(x)$,we have $f^{\prime}(x) > 2e^{2x} > 0$. Since $f^{\prime}(x) > 0$,$f(x)$ is an increasing function in $(0, \infty)$. Thus,$A$ and $C$ are correct.

If $f: R \rightarrow R$ is a differentiable function such that $f^{\prime}(x) > 2f(x)$ for all $x \in R$,and $f(0) = 1$,then:

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