BCECE Engineering BCECE Engineering Solved Paper-2009

  • question_answer
    Let \[g(x)\]be the inverse of function \[f(x)\] and \[f(x)=\frac{1}{1+{{x}^{3}}},\] then \[g(x)\]is equal to

    A) \[\frac{1}{1+{{\{g(x)\}}^{3}}}\]  

    B)         \[\frac{1}{1+{{\{f(x)\}}^{3}}}\]

    C)  \[1+{{\{g(x)\}}^{3}}\]

    D)  \[1+{{\{f(x)\}}^{3}}\]

    Correct Answer: C

    Solution :

    Let \[f(x)=y\] \[\Rightarrow \]\[x={{f}^{-1}}(y)\] \[\Rightarrow \]\[g(y)=x\] Now, g \[(f(x))=\frac{1}{f(x)}\forall x\]                                 \[=1+{{x}^{3}}\]                 \[\Rightarrow \]\[g(y)=1+{{\{g(y)\}}^{3}}\] \[\Rightarrow \]\[g(x)=1+{{\{g(x)\}}^{3}}\]


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