Let $f:(0, \infty) \rightarrow \mathbb{R}$ be given by \[ f(x)=\int_{\frac{1}{x}}^{x} e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t} \] Then

(A) $f(x)$ is monotonically increasing on $[1, \infty)$; (C) $f(x)+f\left(\frac{1}{x}\right)=0$, for all $x \in(0, \infty)$; (D) $f\left(2^{x}\right)$ is an odd function of $x$ on $\mathbb{R}$

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

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Definite Integrals

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Let

f : (0, \infty) \to R

be given by

f (x) = \int_{\frac{1}{x}}^{x} e^{- (t + \frac{1}{t})} \frac{d t}{t}

Then

(A) $f (x)$ is monotonically increasing on $[1, \infty)$ ✓ Correct
(B) $f (x)$ is monotonically decreasing on $(0, 1)$
(C) $f (x) + f (\frac{1}{x}) = 0$ , for all $x \in (0, \infty)$ ✓ Correct
(D) $f (2^{x})$ is an odd function of $x$ on $R$ ✓ Correct

Answer: (A), (C), (D)

Solution

Using Leibniz rule,

f^{'} (x) = e^{- (x + \frac{1}{x})} \cdot \frac{1}{x} - e^{- (\frac{1}{x} + x)} \cdot \frac{1}{1/ x} \cdot (- \frac{1}{x ^{2}}) = \frac{2 e ^{- (x + \frac{1}{x})}}{x} > 0

for all

x > 0

. So

f (x)

is monotonically increasing on

[1, \infty)

, making (A) correct. For (C),

f (\frac{1}{x}) = \int_{x}^{1/ x} e^{- (t + \frac{1}{t})} \frac{d t}{t} = - \int_{1/ x}^{x} e^{- (t + \frac{1}{t})} \frac{d t}{t} = - f (x)

, so (C) is correct. Substituting

x = 2^{y}

, we get

f (2^{y}) + f (2^{- y}) = 0

, which means

f (2^{y})

is an odd function of

y

, so (D) is correct.

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