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JEE Main & Advanced JEE Main Paper (Held On 11 April 2015)

  • question_answer
    Let k and K be the minimum and the maximum values of the function\[f(x)=\frac{{{(1+x)}^{0.6}}}{1+{{x}^{0.6}}}\]in \[[0,1]\]respectively, then the ordered pair (k, K) is equal to : [JEE Main Online Paper (Held On 11 April 2015)]  

    A) \[(1,{{2}^{0.6}})\]

    B) \[({{2}^{-0.4}}1,{{2}^{0.6}})\]

    C) \[({{2}^{-0.6}},1)\]

    D) \[({{2}^{-0.4}},1)\]

    Correct Answer: D

    Solution :

      \[f(x)=\frac{{{\left( 1+x \right)}^{{}^{3}/{}_{5}}}}{1+{{x}^{{}^{3}/{}_{5}}}}\] \[f'(x)=0\Rightarrow x=1\] \[\therefore \] \[f(0)=1,\,\,f(1)=\frac{{{2}^{0.6}}}{2}\] \[={{2}^{-0.4}}\] \[\therefore \]\[f(x)\in \left( {{2}^{-0.4}},1 \right)\]


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