Discussion
If \(f(x) = \frac{x}{x - 1}\) then \(\frac{f(a)}{f(a + 1)}\) is equal to
Answer: A
The ratio equals \(f(a^{2})\).
Substitute into \(f(x) = \frac{x}{x-1}\):
\[ \begin{aligned} f(a) &= \frac{a}{a-1}, \quad f(a+1) = \frac{a+1}{a} \\ \frac{f(a)}{f(a+1)} &= \frac{a}{a-1} \cdot \frac{a}{a+1} \\ &= \frac{a^{2}}{a^{2} - 1} \end{aligned} \]
That is exactly \(f\) evaluated at \(a^{2}\), since \(f(a^{2}) = \frac{a^{2}}{a^{2}-1}\).
The other choices do not match: \(f(-a) = \frac{a}{a+1}\), \(f(-a^{2}) = \frac{a^{2}}{a^{2}+1}\), \(f(1/a) = \frac{1}{1-a}\) and \(f(1/a^{2}) = \frac{1}{1-a^{2}}\).
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