Question

Let $f:(0,\infty )\to R$ be given by $f(x)=\underset{1/x}{\overset{x}{\int }}{e}^{-\left(t+\frac{1}{t}\right)}\frac{dt}{t}$, 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)

$f^{\prime }(x)=\frac{2e^{-\left(x+\frac{1}{x}\right)}}{x}$
Which is increasing in $[1,\infty )$
Also, $f(x)+f\left(\frac{1}{x}\right)=0$
$g(x)=f\left(2^x\right)=\underset{2^{-x}}{\overset{2^x}{\int }}\frac{e^{-\left(t+\frac{1}{t}\right)}}{t}dt$
$g(-x)=\underset{2^x}{\overset{2^{-x}}{\int }}\frac{e^{-\left(t+\frac{1}{t}\right)}}{t}dt=-g(x)$
Hence, an odd function