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: (D)

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

Answer 2

Given that

$f\left(x\right)=\int\limits_{1/x}^{x} e^{-(t+\frac{1}{t})} d t$

⇒$\frac{d}{dx}f(x) = \frac{e^{-x+\frac{1}{x}}}{x}\frac{dx}{dx}-\frac{e^{-(\frac1 x+x)}}{1/x}x \frac{d \frac{1}{x}}{dx}$

= $\frac{e^{-(x+\frac{1}{x})}}{x} + xe^{-(x+\frac{1}{x})}x\frac{-1}{x^2}$

= $e^{-(x+\frac{1}{x})} + \frac{1}{x}e^{-(x+\frac{1}{x})}$

= $2e^{-(x+\frac{1}{x})}= 0$

Hence, f(x)+$$f(\frac{1}{x}) =

\int\limits_{1/x}^{x} \frac{e^{-(t+\frac{1}{t}})} {t}dt+\int\limits_{x}^{1/x} \frac{e^{-(t+\frac{1}{t}})}{t} d t$$$$\int\limits_{1/x}^{1/x} \frac{e^{-(t+\frac{1}{t})}}{t} d t=0

Now,

$f(2^x)+f(\frac{1}{2^x})=f(2^x)+f(2^x)=0$

Therefore, $f(2^x)$ is an odd function.

Note, Let 2x= μ

log2μ =x

∴ f(2x) = h(x) an odd function

Then h(-x)