Let $f:(0, \infty) \in R$ be given $f(x)=\int\limits_{1 / x}^x e^{-\left(t+\frac{1}{t}\right)} \frac{1}{t} d t$, 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 & explanation
Correct answer: option 3
We have,
$f(x) =\int\limits_{1 / x}^x e^{-\left(t+\frac{1}{t}\right)} \frac{1}{t} d t$
$\Rightarrow f'(x) =\frac{1}{x} e^{-\left(x+\frac{1}{x}\right)}+\frac{x}{x^2} e^{-\left(x+\frac{1}{x}\right)}$
$\Rightarrow f'(x)=\frac{2}{x} e^{-\left(x+\frac{1}{x}\right)}>0$ for all $x \in[1, \infty)$
∴ $f(x)$ is monotonically increasing on $[1, \infty)$.
So, option (a) is correct.
Again, $f(x)=\int\limits_{1 / x}^x e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$
$\Rightarrow f\left(\frac{1}{x}\right)=\int\limits_x^{1 / x} e^{-\left(t+\frac{1}{t}\right)} \frac{d t}{t}$
$\Rightarrow f\left(\frac{1}{x}\right)=\int\limits_{1 / u}^u u e^{-\left(\frac{1}{u}+u\right)}\left(\frac{-1}{u^2}\right) d u$, where $t=\frac{1}{u}$
$\Rightarrow f\left(\frac{1}{x}\right)=-\int\limits_{1 / u}^u e^{-\left(u+\frac{1}{u}\right)} \frac{1}{u} d u=-\int\limits_{1 / x}^x e^{-\left(t+\frac{1}{t}\right)} \frac{1}{t} d t=-f(x)$
∴ $f(x)+f\left(\frac{1}{x}\right)=0$. So, option (c) is correct.
Finally, $f(x)=\int\limits_{1 / x}^x e^{-\left(t+\frac{1}{t}\right)} \frac{1}{t} d t$
$\Rightarrow g(x)=f\left(2^x\right)=\int\limits_{2^{-x}}^{2^x} e^{-\left(t+\frac{1}{t}\right)} \frac{1}{t} d t$
$\Rightarrow g(-x)=\int\limits_{2^x}^{2^{-x}} e^{-\left(t+\frac{1}{t}\right)} \frac{1}{t} d t=-\int\limits_{2^{-x}}^{2^x} e^{-\left(u+\frac{1}{u}\right)} \frac{1}{u} d u=-g(x)$
∴ $g(x)=f\left(2^x\right)$ is an odd function of $x$ on $R$.
So, option (d) is correct.