Application of Derivatives 2 Question 18
####18. If $f:(0, \infty) \rightarrow R$ be given by
(2014 Adv.)
$ f(x)=\int_{1 / x}^{x} e^{-t+\frac{1}{t}} \frac{d t}{t} $
Then,
(a) $f(x)$ is monotonically increasing on $[1, \infty)$
(b) $f(x)$ is monotonically decreasing on $[0,1)$
(c) $f(x)+f \frac{1}{x}=0, \forall x \in(0, \infty)$
(d) $f\left(2^{x}\right)$ is an odd function of $x$ on $R$
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Answer:
Correct Answer: 18. (c)
Solution:
- Given, $f(x)=\int_{\frac{1}{x}}^{x} \frac{e^{-t+\frac{1}{t}}}{t} d t$
$ \begin{aligned} f^{\prime}(x) & =1 \cdot \frac{e^{-x+\frac{1}{x}}}{x}-\frac{-1}{x^{2}} \frac{e^{-\frac{1}{x}+x}}{1 / x} \\ & =\frac{e^{-x+\frac{1}{x}}}{x}+\frac{e^{-x+\frac{1}{x}}}{x}=\frac{2 e^{-x+\frac{1}{x}}}{x} \end{aligned} $
As $\quad f^{\prime}(x)>0, \forall x \in(0, \infty)$
$\therefore f(x)$ is monotonically increasing on $(0, \infty)$.
$\Rightarrow$ Option (a) is correct and option (b) is wrong.
Now, $f(x)+f \frac{1}{x}=\int_{1 / x}^{x} \frac{e^{-t+\frac{1}{t}}}{t} d t+\int_{x}^{1 / x} \frac{e^{-t+\frac{1}{t}}}{t} d t$ $=0, \forall x \in(0, \infty)$
Now, let $\quad g(x)=f\left(2^{x}\right)=\int_{2^{-x}}^{2^{x}} \frac{e^{-t+\frac{1}{t}}}{t} d t$
$ g(-x)=f\left(2^{-x}\right)=\int_{2^{2}}^{2^{-x}} \frac{e^{-t+\frac{1}{t}}}{t} d t=-g(x) $
$\therefore f\left(2^{x}\right)$ is an odd function.
