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(D) Given $f^{\prime}(x) < 2f(x)$,we have $f^{\prime}(x) - 2f(x) < 0$. Multiply by the integrating factor $e^{-2x}$: $e^{-2x} f^{\prime}(x) - 2e^{-2x}

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$(0, (e - 1)/2)$

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(D) Given $f^{\prime}(x) Multiply by the integrating factor $e^{-2x}$: $e^{-2x} f^{\prime}(x) - 2e^{-2x} f(x) $\frac{d}{dx} (e^{-2x} f(x)) This implies that $g(x) = e^{-2x} f(x)$ is a strictly decreasing function on $[1/2, 1]$. Since $x \ge 1/2$,we have $g(x) $e^{-2x} f(x) Thus,$f(x) Integrating both sides from $1/2$ to $1$: $\int_{1/2}^1 f(x) dx Since $f(x) > 0$,the integral is greater than $0$. Therefore,$\int_{1/2}^1 f(x) dx \in (0, (e - 1)/2)$.

Let $f : [1/2, 1] \rightarrow \mathbb{R}$ be a positive,non-constant and differentiable function such that $f^{\prime}(x) < 2f(x)$ and $f(1/2) = 1$. Then the value of $\int_{1/2}^1 f(x) dx$ lies in the interval

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