2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    If\[f(x)=x-{{x}^{2}}+{{x}^{3}}-{{x}^{4}}+....\]to\[\infty \]for\[|x|<1,\]then\[{{f}^{-1}}(x)\]is equal to:

    A) \[x\]                                     

    B)  \[\left( \frac{x}{1-x} \right)\]    

    C)         \[\frac{1-x}{x}\]              

    D)         \[\frac{1}{x}\]

    E)  \[\left( \frac{x}{1+x} \right)\]

    Correct Answer: B

    Solution :

    \[f(x)=x-{{x}^{2}}+{{x}^{3}}-{{x}^{4}}+....\infty ,|x|<1\] \[f(x)=\frac{x}{1-(-x)}=\frac{x}{1+x}\] \[\Rightarrow \]               \[f(x)=y=\frac{x}{1+x}\] \[\Rightarrow \]               \[y(1+x)=x\] \[\Rightarrow \]               \[x=\frac{y}{1-y}\] \[\therefore \]  \[{{f}^{-1}}(x)=\frac{x}{1-x}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner