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

Let f: (0, ∞) → R be given by f(x) = ∫_(1/x)ˣ e^(-(t + 1/t)) dt/t. Then $f(x) = \int_{1/x}^{x} e^{-\left(t + \frac{1}{t}\right)} \frac{dt}{t}, \quad x > 0$ $\text{Check the four properties}$ $f'(x) = \varphi(x) - \varphi\!\left(\tfrac{1}{x}\right)\cdot\left(-\tfrac{1}{x^2}\right), \quad \varphi(t) = \frac{e^{-(t+1/t)}}{t}$ $f'(x) = \frac{e^{-(x+1/x)}}{x} + \frac{e^{-(x+1/x)}}{x} = \frac{2\,e^{-(x+1/x)}}{x}$ $f'(x) = \frac{2\,e^{-(x+1/x)}}{x} > 0 \quad \text{for all } x > 0$ $\Rightarrow f \text{ is increasing on } (0,\infty)$ $(1): \text{ increasing on } [1,\infty) \ \Rightarrow \ \text{TRUE}$ $(2): \text{ decreasing on } (0,1) \ \Rightarrow \ \text{FALSE}$ $f\!\left(\frac{1}{x}\right) = \int_{x}^{1/x}\frac{e^{-(t+1/t)}}{t}\,dt = -\int_{1/x}^{x}\frac{e^{-(t+1/t)}}{t}\,dt = -f(x)$ $\Rightarrow f(x) + f\!\left(\frac{1}{x}\right) = 0 \ \Rightarrow \ (3) \text{ TRUE}$ $F(x) = f(2^x): \quad F(-x) = f(2^{-x}) = f\!\left(\frac{1}{2^x}\right)$ $= -f(2^x) = -F(x) \Rightarrow F \text{ is odd} \ \Rightarrow \ (4) \text{ TRUE}$ $\text{Correct statements: } (1), (3), (4)$