Question

Let $f:(0, \infty) \rightarrow R$ be given by $f(x)=\int\limits_{1 / x}^{x} e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$, then

• A. $f(x)$ is monotonically increasing on $[1, \infty$ )
• B. $f(x)$ is monotonically decreasing on $(0,1)$
• 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 $R$.

Answer 1

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

$f '( x )=\frac{2 e ^{-\left( x +\frac{1}{ x }\right)}}{ x }$
Which is increasing in $[1, \infty)$
Also, $f(x)+f\left(\frac{1}{x}\right)=0$
$g(x)=f\left(2^{x}\right)=\int\limits_{2^{-x}}^{2^{x}} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t $
$g(-x)=\int\limits_{2^{x}}^{2^{-x}} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t} d t=-g(x)$
Hence, an odd function