JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer If \[m,\,n\] are the roots of the equation \[{{x}^{2}}-x-1=0\],  then the value of    \[\frac{\left( 1+m{{\log }_{e}}3+\frac{{{(m{{\log }_{e}}3)}^{2}}}{2\,!\,}+...\infty  \right)\,\,\left( 1+n{{\log }_{e}}3+\frac{{{(n{{\log }_{e}}3)}^{2}}}{2\,!\,}+..\infty  \right)\,}{\left( 1+mn{{\log }_{e}}3+\frac{{{(mn{{\log }_{e}}3)}^{2}}}{2\,!}+.....\infty  \right)}\]

    A) 9

    B) 3

    C) 0

    D) 1

    Correct Answer: A

    Solution :

    Numerator \[N={{e}^{m\,{{\log }_{e}}3}}\times {{e}^{n{{\log }_{e}}3}}\] \[={{e}^{{{\log }_{e}}{{3}^{m}}}}\times {{e}^{{{\log }_{e}}{{3}^{n}}}}={{3}^{m}}\times {{3}^{n}}={{3}^{m+n}}\] Denominator \[D={{e}^{mn{{\log }_{e}}3}}={{3}^{mn}}\] whereas given \[m+n=1,\,\,\,mn=-1\] \[\therefore \,\,\frac{N}{D}=\frac{{{3}^{m+n}}}{{{3}^{mn}}}=\frac{{{3}^{1}}}{{{3}^{-1}}}={{3}^{2}}=9\].

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