Discussion Forum

Let   $f:(0,\infty)\rightarrow R$   be given by $f(x)= \int_{1/x}^{x} e^{-(t+\frac{1}{t})}\frac{dt}{t}$  Then

Answer:

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(2^x)= $\int_{2^{-x}}^{2^{x}} \frac{e^{-(t+\frac{1}{t})}}{t}dt$

 g(-x)=f(2^-x)

                        =$\int_{2^{x}}^{2^{-x}} \frac{e^{-(t+\frac{1}{t})}}{t}dt$

                    = -g(x)

  $\therefore$     f(2^x)  is an odd function.