JEE Challenger
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MathMedAdv 2014 • Paper 1

Monotonicity and Properties of Exponential Integral Function

Consider the function f:(0,∞)→Rf: (0, \infty) \rightarrow \mathbb{R} defined by f(x)=∫1/xxe−(t+1t)dttf(x) = \int_{1/x}^{x} e^{-\left(t + \frac{1}{t}\right)} \frac{dt}{t} Which of the following statements are correct?

Options

A

f(x)f(x) is monotonically increasing on [1,∞)[1, \infty)

Correct
B

f(x)f(x) is monotonically decreasing on (0,1)(0, 1)

C

f(x)+f(1x)=0f(x) + f\left(\frac{1}{x}\right) = 0, for all x∈(0,∞)x \in (0, \infty)

Correct
D

f(2x)f(2^x) is an odd function of xx on R\mathbb{R}

Correct

Topics & Concepts

IntegralsDefinite IntegrationProperties of Definite IntegralsMonotonicity

Step-by-Step Solution

To determine the correct statements about the given function f:(0,∞)→Rf:(0, \infty) \rightarrow \mathbb{R} defined by f(x)=∫1/xxe−(t+1t)dttf(x) = \int_{1/x}^{x} e^{-\left(t + \frac{1}{t}\right)} \frac{dt}{t} we analyze its derivative and functional properties step-by-step.

Step 1: Derivative and Monotonicity Analysis

Using the Leibniz rule for differentiation under the integral sign: f′(x)=e−(x+1x)⋅1x⋅ddx(x)−e−(1x+11/x)⋅11/x⋅ddx(1x)f'(x) = e^{-\left(x + \frac{1}{x}\right)} \cdot \frac{1}{x} \cdot \frac{d}{dx}(x) - e^{-\left(\frac{1}{x} + \frac{1}{1/x}\right)} \cdot \frac{1}{1/x} \cdot \frac{d}{dx}\left(\frac{1}{x}\right)

Simplifying each term: f′(x)=e−(x+1x)1x−e−(1x+x)⋅x(−1x2)f'(x) = e^{-\left(x + \frac{1}{x}\right)} \frac{1}{x} - e^{-\left(\frac{1}{x} + x\right)} \cdot x \left(-\frac{1}{x^2}\right) f′(x)=1xe−(x+1x)+1xe−(x+1x)=2xe−(x+1x)f'(x) = \frac{1}{x} e^{-\left(x + \frac{1}{x}\right)} + \frac{1}{x} e^{-\left(x + \frac{1}{x}\right)} = \frac{2}{x} e^{-\left(x + \frac{1}{x}\right)}

For all x∈(0,∞)x \in (0, \infty):

  • x>0x > 0
  • e−(x+1x)>0e^{-\left(x + \frac{1}{x}\right)} > 0

Thus, f′(x)>0f'(x) > 0 for all x∈(0,∞)x \in (0, \infty).

  1. Since f′(x)>0f'(x) > 0 for all x∈[1,∞)x \in [1, \infty), f(x)f(x) is monotonically increasing on [1,∞)[1, \infty). Therefore, Statement (A) is correct.
  2. Since f′(x)>0f'(x) > 0 for all x∈(0,1)x \in (0, 1), f(x)f(x) is monotonically increasing on (0,1)(0, 1), not decreasing. Therefore, Statement (B) is incorrect.

Step 2: Relation between f(x)f(x) and f(1x)f\left(\frac{1}{x}\right)

Evaluating f(1x)f\left(\frac{1}{x}\right): f(1x)=∫x1/xe−(t+1t)dttf\left(\frac{1}{x}\right) = \int_{x}^{1/x} e^{-\left(t + \frac{1}{t}\right)} \frac{dt}{t}

Using the integral property ∫abg(t)dt=−∫bag(t)dt\int_{a}^{b} g(t) dt = -\int_{b}^{a} g(t) dt: f(1x)=−∫1/xxe−(t+1t)dtt=−f(x)f\left(\frac{1}{x}\right) = -\int_{1/x}^{x} e^{-\left(t + \frac{1}{t}\right)} \frac{dt}{t} = -f(x)

Rearranging gives: f(x)+f(1x)=0for all x∈(0,∞)f(x) + f\left(\frac{1}{x}\right) = 0 \quad \text{for all } x \in (0, \infty)

Therefore, Statement (C) is correct.

Step 3: Symmetry of f(2x)f(2^x)

Let g(x)=f(2x)g(x) = f(2^x). To check if g(x)g(x) is an odd function, we compute g(−x)g(-x): g(−x)=f(2−x)=f(12x)g(-x) = f(2^{-x}) = f\left(\frac{1}{2^x}\right)

Using the property proven in Step 2, f(1u)=−f(u)f\left(\frac{1}{u}\right) = -f(u) for u=2xu = 2^x: g(−x)=−f(2x)=−g(x)g(-x) = -f(2^x) = -g(x)

Since g(−x)=−g(x)g(-x) = -g(x) for all x∈Rx \in \mathbb{R}, f(2x)f(2^x) is an odd function of xx on R\mathbb{R}. Therefore, Statement (D) is correct.

Conclusion

The correct statements are (A), (C), and (D).

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