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Real numbers $$x_1, x_2, \ldots, x_{2007}$$ are chosen such that $$(x_1, x_2), (x_2, x_3), \ldots, (x_{2006}, x_{2007})$$ are all points on the graph of $$y = \frac{1}{x - 1}$$.
The condition is $$x_{i+1} = \frac{1}{x_i - 1}$$. a fails, since $$x_1 = 2$$ gives $$x_2 = 1$$ and then $$x_3$$ is undefined. All terms are equal only at the fixed points $$x = \frac{1 \pm \sqrt{5}}{2}$$ of $$x^2 - x - 1 = 0$$, so b fails and there are only two such choices, not infinitely many (c fails). Since $$\frac{1}{x_i - 1}$$ is never 0, no $$x_i$$ with $$i \ge 2$$ is zero, so d fails, and the answer is "None of these".
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