Let f : \left[\right. \frac{1}{2} , \textrm{ } 1 \left]\right. \rightarrow R (the set of all real numbers) be a positive, non‑constant and differentiable function such that f '(x) < 2f(x) and f \left(\right. \frac{1}{2} \left.\right) = 1. Then the value of \int_{1 / 2}^{1} f \left(\right. x \left.\right) \textrm{ } dx lies in the interval

Engineering

Mathematics

Properties of Definite Integral

Question

Let f : $[\frac{1}{2}, 1] \to R$ (the set of all real numbers) be a positive, non‑constant and differentiable function such that f '(x) < 2f(x) and $f (\frac{1}{2})$ = 1. Then the value of $\int_{1 / 2}^{1} f (x) dx$ lies in the interval

$(0, \frac{e - 1}{2})$

(2e – 1, 2e)

$(\frac{e - 1}{2}, e - 1)$

(e – 1, 2e – 1)

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Given, f '(x) < 2f(x)

$∴$ f '(x) – 2f(x) < 0 $\forall x \in [\frac{1}{2}, 1]$

$∴ \frac{d}{dx} (f (x) \cdot e^{- 2 x}) < 0$

Hence, f(x) e^–2xis a decreasing function in $[\frac{1}{2}, 1]$

$f (x) e^{- 2 x} \leq f (\frac{1}{2}) . e^{- 1}$

$f (x) e^{- 2 x} \leq 1 \times \frac{1}{e}$

0 $\leq$ f(x) $\leq$ e^{2x –1}

Hence, $0 < \int_{1 / 2}^{1} f (x) dx \leq \int_{1 / 2}^{1} e^{2 x - 1} dx$

$0 < I \leq \frac{e - 1}{2}$

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