Practicing Success
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 $[1, \infty)$ |
(a), (b), (c) (a), (b), (d) (a), (c), (d) (b), (c), (d) |
(a), (c), (d) |
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. |