IMC 1995 · Day 1 · P4

From the definition we have $$F'(x) = \frac{x-1}{\ln x}, \quad x > 1.$$ Therefore $F'(x) > 0$ for $x \in (1, \infty)$. Thus $F$ is strictly increasing and hence one-to-one. Since $$F(x) \ge (x^2 - x) \min \left\{ \frac{1}{\ln t} : x \le t \le x^2 \right\} = \frac{x^2 - x}{\ln x^2} \to \infty$$ as $x \to \infty$, it follows that the range of $F$ is $(F(1+), \infty)$. In order to determine $F(1+)$ we substitute $t = e^v$ in the definition of $F$ and we get $$F(x) = \int_{\ln x}^{2 \ln x} \frac{e^v}{v}\,dv.$$ Hence $$F(x) < e^{2 \ln x} \int_{\ln x}^{2 \ln x} \frac{1}{v}\,dv = x^2 \ln 2$$ and similarly $F(x) > x \ln 2$. Thus $F(1+) = \ln 2$.