1. [3] Let $f$ be a differentiable real-valued function defined on the positive real numbers. The tangent lines to the graph of $f$ always meet the $y$-axis 1 unit lower than where they meet the function. If $f(1)=0$, what is $f(2)$ ?
Answer:
$\ln 2$

Solution: The tangent line to $f$ at $x$ meets the $y$-axis at $f(x)-1$ for any $x$, so the slope of the tangent line is $f^{\prime}(x)=\frac{1}{x}$, and so $f(x)=\ln (x)+C$ for some $a$. Since $f(1)=0$, we have $C=0$, and so $f(x)=\ln (x)$. Thus $f(2)=\ln (2)$.