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JEE Advanced 2013Paper 2
Exam
JEE Advanced
Year
2013
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2013 · Paper 2 · Mathematics · Application of Derivatives

Paragraph: Let f:[0,1] arrow ℝ (the set of all real numbers) be a...

Paragraph: Let f:[0,1]→Rf:[0,1] \rightarrow \mathbb{R} (the set of all real numbers) be a function. Suppose the function ff is twice differentiable, f(0)=f(1)=0f(0)=f(1)=0 and satisfies f′′(x)−2f′(x)+f(x)≥ex,x∈[0,1].f^{\prime \prime}(x)-2 f^{\prime}(x)+f(x) \geq e^{x}, x \in[0,1] .
Question: Which of the following is true for f(x)f(x) ?

Options

  1. A0<f(x)<∞0 < f \left(\text{x}\right) < \infty
  2. B−12<f(x)<12- \dfrac{1}{2} < f \left(\text{x}\right) < \dfrac{1}{2}
  3. C−14<f(x)<1- \dfrac{1}{4} < f \left(\text{x}\right) < 1
  4. D−∞<f(x)<0- \infty < f \left(\text{x}\right) < 0

Correct answer

D. −∞<f(x)<0- \infty < f \left(\text{x}\right) < 0

Step-by-step solution

f″(x)−2f′(x)+f(x)≥ex{\text{f}}^{″} \left(\text{x}\right) - 2 {\text{f}}^{′} \left(\text{x}\right) + \text{f} \left(\text{x}\right) \ge {\text{e}}^{\text{x}}
f″(x)⋅e−x−f′(x)e−x−f′(x)e−x+f(x)e−x≥1{\text{f}}^{″} \left(\text{x}\right) \cdot {\text{e}}^{- \text{x}} - {\text{f}}^{′} \left(\text{x}\right) {\text{e}}^{- \text{x}} - {\text{f}}^{′} \left(\text{x}\right) {\text{e}}^{- \text{x}} + \text{f} \left(\text{x}\right) {\text{e}}^{- \text{x}} \ge 1
ddx(f′(x)e−x)−ddx(f(x)⋅e−x)≥1\dfrac{\text{d}}{\text{dx}} \left({\text{f}}^{′} \left(\text{x}\right) {\text{e}}^{- \text{x}}\right) - \dfrac{\text{d}}{\text{dx}} \left(\text{f} \left(\text{x}\right) \cdot {\text{e}}^{- \text{x}}\right) \ge 1
ddx(f′(x)e−x−f(x)e−x)≥1\dfrac{\text{d}}{\text{dx}} \left({\text{f}}^{′} \left(\text{x}\right) {\text{e}}^{- \text{x}} - \text{f} \left(\text{x}\right) {\text{e}}^{- \text{x}}\right) \ge 1
⇒d2dx2(e−xf(x))≥1⇒ \dfrac{{\text{d}}^{2}}{{\text{dx}}^{2}} \left({\text{e}}^{- \text{x}} \text{f} \left(\text{x}\right)\right) \ge 1 ∀x∈[0,1]\forall \text{x} \in \left[0 \text{,} 1\right]
Let ϕ(x)=e−xf(x)ϕ \left(\text{x}\right) = {\text{e}}^{- \text{x}} \text{f} \left(\text{x}\right)
⇒ϕ(x)⇒ ϕ \left(\text{x}\right) is concave upward
f(0)=f(1)=0f ( 0 ) = f ( 1 ) = 0
⇒ϕ(0)=0=ϕ(1)⇒ ϕ \left(0\right) = 0 = ϕ \left(1\right)
⇒f(x)<0⇒ \text{f} \left(\text{x}\right) < 0
⇒ϕ(x)<0⇒ ϕ \left(\text{x}\right) < 0
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