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MathematicsSingle correct+2Question 43 of 60

JEE Advanced 2013 · Paper 1, Mathematics question 43

Let f:[12,1]→Rf:\left[\frac12,1\right]\to\mathbb R (the set of all real numbers) be a positive, non-constant and differentiable function such that f′(x)<2f(x)f'(x)<2f(x) and f(12)=1f\left(\frac12\right)=1. Then the value of ∫1/21f(x) dx\displaystyle\int_{1/2}^1f(x)\,dx lies in the interval

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Answer and worked solution

Official answer: D

Integrating factor. f′−2f<0f'-2f<0 means (e−2xf)′<0\left(e^{-2x}f\right)'<0, so e−2xf(x)<e−1f(12)=e−1e^{-2x}f(x)<e^{-1}f\left(\frac12\right)=e^{-1} for x>12x>\frac12, i.e. f(x)<e2x−1f(x)<e^{2x-1}.

Bounds. 0<∫1/21f dx<∫1/21e2x−1dx=e−120<\displaystyle\int_{1/2}^1f\,dx<\int_{1/2}^1e^{2x-1}dx=\frac{e-1}2 (ff is positive).

(D)

Topics:Properties of definite integralsMonotonicity
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