12th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-5

  • question_answer
    \[\int{{{7}^{{{7}^{{{7}^{x}}}}}}}{{.7}^{{{7}^{^{x}}}}}{{.7}^{^{^{x}}}}.dx\]is equal to

    A)  \[\frac{{{7}^{{{7}^{{{7}^{x}}}}}}}{(\log 7)}+c\]               

    B)  \[\frac{{{7}^{{{7}^{{{7}^{x}}}}}}}{{{(\log 7)}^{2}}}+c\]    

    C)  \[\frac{{{7}^{{{7}^{{{7}^{x}}}}}}}{{{(\log 7)}^{3}}}+c\]           

    D)  None of these

    Correct Answer: A

    Solution :

    [a] \[I=\int{{{7}^{{{7}^{{{7}^{x}}}}}}{{.7}^{{{7}^{x}}}}{{.7}^{x}}.dx}\]       Let \[z={{7}^{{{7}^{{{7}^{x}}}}}}\] \[\Rightarrow \frac{dz}{dx}={{7}^{{{7}^{{{7}^{x}}}}}}.\frac{d}{dx}\left( {{7}^{{{7}^{x}}}} \right).\log 7\] \[[\sin ce,\frac{d}{dx}({{a}^{x}})={{a}^{x}}.\log a]\] \[\Rightarrow dz={{7}^{{{7}^{{{7}^{x}}}}}}{{.7}^{{{7}^{x}}}}{{.7}^{x}}{{(\log 7)}^{3}}dx\] Now, \[I=\int{\frac{dz}{{{(\log 7)}^{3}}}}=\frac{z}{{{(\log 7)}^{3}}}+c=\frac{{{7}^{{{7}^{{{7}^{x}}}}}}}{{{(\log 7)}^{3}}}+c\]


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