Determine all ordered pairs of real numbers $(a, b)$ such that the line $y = ax + b$ intersects the curve $y = \ln(1 + x^2)$ in exactly one point.

I really liked this problem, and thought it makes for a good “research” problem for calculus classes. For notation purposes, let $g(x) = \ln(1 + x^2)$. The following is also very sketch-like, and conversational. We first note that logarithmic curve is even, and we can simply consider $a \ge 0$, since $y = -ax + b$ will have the same number of intersection points. Simply looking at the graph, it’s clear that $(a, b) = (0, 0)$ is such a solution, and for fixed $a = 0$, that no other $y$-intercept will work.

Intuitively, for large enough slope, a line will also only intersect $g(x)$ at one point, no matter the intercept since it “outgrows” the log. We can formalize this by noting that $g'(x) = \frac{2x}{1 + x^2}$ with

- $\lim_{x\to \pm \infty} g'(x) = 0$,
- $g'(0) = 0, g'(1) = 1$
- $g'(x) \le 1$ by trivial inequality.

Assume not, then there exists two or more zeros for the function $g(x) – (ax + b)$. By the mean value theorem, then there would exist a point $c$ such that $g'(c) – a = 0$, which cannot happen since $|g'(c)| \le |a|$.

Now, we need to examine the region where $1 > a > 0$. For $b > 0$, if the $y$-intercept is large enough then the logarithmic growth will not catch up with the linear growth. To find this “large enough” intercept, we set simply find the line where it intersects $g(x)$; this will occur where the derivatives equal.

Solving $g'(x) = a$, we have $\frac{2x}{1 + x^2} = a \implies x^*_{\pm} = \frac{1 \pm \sqrt{1 – a^2}}{a}$. The intersection $b$ will be $ax^*_{+} + b = \ln(1 + (x^*_p)^2) \implies b = \ln(1 + (x^*_p)^2) – ax^*_p$, hence any $b > \ln(1 + ((x^*_p)^2) – ax^*_p$ suffices.

On the other hand, the root $x^*_{-}$ corresponds to the cases where $b < 0$ where the line “ducks” underneath first before intersecting the log curve, and hence any $b < \ln(1 + (x^*_m)^2) – ax^*_m$ will work.

Finally,the case of $a = 1$ is left as an exercise because this is too long.