A) \[f({{a}^{2}})\]
B) \[f\left( \frac{1}{a} \right)\]
C) \[f(-a)\]
D) \[f\left| \frac{-a}{a-1} \right|\]
Correct Answer: A
Solution :
[a] Given \[f(x)=\frac{x}{x-1};\] Then, \[f(a)=\frac{a}{a=1}\] and \[f(a+1)=\frac{a+1}{a}\] So, \[\frac{f(a)}{f(a+1)}=\frac{a}{a-1}\times \frac{a}{a+1}=\frac{{{a}^{2}}}{{{a}^{2}}-1}=f({{a}^{2}})\]
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