A) \[\frac{x}{1+x}\]
B) \[\frac{x}{1-x}\]
C) \[\frac{1-x}{x}\]
D) \[\frac{1}{x}\]
Correct Answer: B
Solution :
[b] Given \[f(x)=x-{{x}^{2}}+{{x}^{3}}-{{x}^{4}}+...to\infty \] \[\Rightarrow y=\frac{x}{1+x}\] (Infinite G.P.) \[\Rightarrow y+xy=x\Rightarrow y=x(1-y)\] \[\Rightarrow x=\frac{y}{1-y}\] \[\Rightarrow {{f}^{-1}}(y)=\frac{y}{1-y}\Rightarrow {{f}^{1}}(x)=\frac{x}{1-x}\]You need to login to perform this action.
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