Answer:
Option A,C,D
Explanation:
Plan (i) $f(x)= \int_{\phi(x)}^{\psi(x)} f(t)dt$ then
$\frac{d}{dx}[I(x)]=f\left\{\psi(x)\right\}\left\{\frac{d}{dx}\psi(x)\right\}$ - $f\left\{\phi(x)\right\}\left\{\frac{d}{dx}\phi(x)\right\}$
(ii) if f'(x) > o, $\forall x \in[a,b]$ then f(x) is monotonically increasing on [a,b]
(iii) $\int_{a}^{b} f(x) dx=-\int_{b}^{a} f(x) dx$
(iv) if f(-x) =-f(x) , $\forall x \in R$ , then f(x) is an odd function of x on R
Given,
$f(x)= \int_{1/x}^{x} \frac{e^{-(t+\frac{1}{t})}}{t} dt$
$f'(x)=1.\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}-\left(\frac{-1}{x^{2}}\right)\frac{e^{-(\frac{1}{x}+x)}}{1/x}$
$=\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}+\frac{e^{-\left(x+\frac{1}{x}\right)}}{x}$
$=\frac{2e^{-\left(x+\frac{1}{x}\right)}}{x}$
As, f'(x)>0, $\forall x \in(0,\infty)$
$\therefore$ f(x) is monotonically increasing on (0, $\infty$)
$\Rightarrow$ options (a) is correct and (b) is wrong.
Now, $f(x)+f(\frac{1}{x})=\int_{1/x}^{x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t}dt+\int_{x}^{1/x} \frac{e^{-\left(t+\frac{1}{t}\right)}}{t}dt$
= 0, $\forall x\in(0,\infty)$
Now, let
g(x) =f(2x)= $\int_{2^{-x}}^{2^{x}} \frac{e^{-(t+\frac{1}{t})}}{t}dt$
g(-x)=f(2-x)
=$\int_{2^{x}}^{2^{-x}} \frac{e^{-(t+\frac{1}{t})}}{t}dt$
= -g(x)
$\therefore$ f(2x) is an odd function.
