Question

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

• A. $f\left(x\right)$ is monotonically increasing on $[1,\infty )$
• B. $f\left(x\right)$ is monotonically decreasing on $(0,1)$
• C. $f\left(x\right)+f\left(\frac{1}{x}\right)=0,$ 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)

$f\left(2^x\right)$ is an odd function of $x$ on $R$

$f\left(x\right)=\underset{1/x}{\overset{x}{\int }}{e}^{-(t+\frac{1}{t})}\frac{dt}{t}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{e^{-(x+\frac{1}{x})}}{x}+\frac{x}{x^2}\times e^{-(x+\frac{1}{x})}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2e^{-(x+\frac{1}{x})}}{x}$

$\Rightarrow f^{\prime }\left(x\right)>0,\forall x\in (0,\infty )$

Now,
$f\left(x\right)+f\left(\frac{1}{x}\right)=\underset{1/x}{\overset{x}{\int }}{e}^{-(t+\frac{1}{t})}\frac{dt}{t}+\underset{x}{\overset{1/x}{\int }}{e}^{-(t+\frac{1}{t})}\frac{dt}{t}=\underset{1/x}{\overset{x}{\int }}{e}^{-(t+\frac{1}{t})}\frac{dt}{t}-\underset{1/x}{\overset{x}{\int }}{e}^{-(t+\frac{1}{t})}\frac{dt}{t}=0$

Putting $x=2^x$ in
$f\left(x\right)+f\left(\frac{1}{x}\right)=0$ We have,
$f\left(2^x\right)+f\left(\frac{1}{2^x}\right)=0$
$\Rightarrow f\left(2^x\right)=-f\left(2^{-x}\right)\to$ odd function