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JEE Advanced 2013Paper 1
Exam
JEE Advanced
Year
2013
Paper
Paper 1
Subject
Mathematics
Chapter
Differential Equations

JEE Advanced 2013 · Paper 1 · Mathematics · Differential Equations

Let f :[(1)/(2)1] ⟶ R (the set of all real numbers) be a positive,...

Let f:[121]⟶Rf \text{:}\left[\dfrac{1}{2}1\right] ⟶ R (the set of all real numbers) be a positive, non-constant and differentiable function such that f′(x)<2f(x){f}^{'}\left(x\right) < 2f \left(x\right) and f(12)=1f \left(\dfrac{1}{2}\right) = 1. Then the value of ∫1/21f(x)dx\mathop{∫}_{1/2}^{1} f \left(x\right) \text{dx} lies in the interval:

Options

  1. A(2e−12e)\left(2 e - 1 2 e \right)
  2. B\left(e &ApplyFunction; - 1 2 e &ApplyFunction; - 1\right)
  3. C\left(\dfrac{e &ApplyFunction; - 1}{2} e &ApplyFunction; - 1\right)
  4. D(0,e−12)\left(0 \text{,} \dfrac{e - 1}{2}\right)

Correct answer

D. (0,e−12)\left(0 \text{,} \dfrac{e - 1}{2}\right)

Step-by-step solution

f′(x)<2f(x){f}^{'}\left(x\right) < 2f\left(x\right)
⇒f′(x)f(x)<2\text{⇒}\dfrac{{f}^{'}\left(x\right)}{f\left(x\right)} < 2

⇒∫12xf′(x)f(x)dx=∫12x2dx⇒ {∫}_{\dfrac{1}{2}}^{x}\dfrac{f ' \left(x\right)}{f\left(x\right)}dx = {∫}_{\dfrac{1}{2}}^{x}2dx

⇒lnf(x)<2x−1⇒ lnf\left(x\right) < 2x - 1

⇒f(x)<e2x−1⇒ f\left(x\right) < {e}^{2x - 1}
⇒0<∫121f(x)dx<∫121e(2x−1)dx⇒ 0 < {∫}_{\dfrac{1}{2}}^{1}f\left(x\right)dx < {∫}_{\dfrac{1}{2}}^{1}{e}^{\left(2x - 1\right)}dx

⇒0<∫121f(x)dx<[e2x−12]121⇒ 0 < {∫}_{\dfrac{1}{2}}^{1}f\left(x\right)dx < {\left[\dfrac{{e}^{2x - 1}}{2}\right]}_{\dfrac{1}{2}}^{1}
⇒0<∫121f(x)dx<e−12⇒ 0 < {∫}_{\dfrac{1}{2}}^{1}f\left(x\right)dx < \dfrac{e - 1}{2}

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