A) \[\frac{x}{x-1}\]
B) \[{{\left( \frac{x}{x-1} \right)}^{19}}\]
C) \[\frac{19x}{x-1}\]
D) x
Correct Answer: A
Solution :
[a] \[\because f(x)=\frac{x}{x-1}\] \[\therefore (fof)(x)=f\{f(x)\}=f\left( \frac{x}{x-1} \right)\] \[=\frac{\frac{x}{x-1}}{\frac{x}{x-1}-1}=\frac{\frac{x}{x-1}}{\frac{x-x+1}{x-1}}=\frac{\frac{x}{x-1}}{\frac{1}{x-1}}=x.\] \[\Rightarrow (fofof)(x)=f(fof)(x)=f(x)=\frac{x}{x-1}\] \[\Rightarrow \underbrace{(fofof....of)}_{19\,times}(x)=f(fof)(x)=f(x)=\frac{x}{x-1}\]
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