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BCECE Engineering BCECE Engineering Solved Paper-2013

  • question_answer
    If \[g(x)=1+\sqrt{x}\]and \[f\{g(x)\}\]\[=3+2\sqrt{x}+x,\] then \[f(x)\]is equal to

    A) \[1+2{{x}^{2}}\]           

    B)        \[~2+{{x}^{2}}\]               

    C)        \[1+x\]                

    D)        \[~2+x\]

    Correct Answer: B

    Solution :

    Given, \[g(x)=1+\sqrt{x}\] and        \[f\{g(x)\}=3+2\sqrt{x}+x\]         ?.(i) \[\Rightarrow \]               \[f(1+\sqrt{x})=3+2\sqrt{x}+x\] Put \[1+\sqrt{x}=y\]\[\Rightarrow \]\[x={{(y-1)}^{2}}\] \[\therefore \] \[f(y)=3+2(y-1)+{{(y-1)}^{2}}=2+{{y}^{2}}\] \[\therefore \]  \[f(x)=2+{{x}^{2}}\]


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