Question
If $$f\left( x \right) = \frac{x}{{x - 1}},$$ then $$\frac{{f\left( a \right)}}{{f\left( {a + 1} \right)}}$$ is equal to :
A.
$$f\left( {{a^2}} \right)$$
B.
$$f\left( {\frac{1}{a}} \right)$$
C.
$$f\left( { - a} \right)$$
D.
$$f\left[ {\frac{{ - a}}{{a - 1}}} \right]$$
Answer :
$$f\left( {{a^2}} \right)$$
Solution :
$$\eqalign{
& {\text{Given, }}f\left( x \right) = \frac{x}{{x - 1}} \cr
& {\text{Then,}}\,f\left( a \right) = \frac{a}{{a - 1}}\,{\text{and }}f\left( {a + 1} \right) = \frac{{a + 1}}{a} \cr
& {\text{So, }}\frac{{f\left( a \right)}}{{f\left( {a + 1} \right)}} = \frac{a}{{a - 1}} \times \frac{a}{{a + 1}} = \frac{{{a^2}}}{{{a^2} - 1}} = f\left( {{a^2}} \right) \cr} $$