• # 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

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 $m+n=1,\,\,\,mn=-1$ $\therefore \,\,\frac{N}{D}=\frac{{{3}^{m+n}}}{{{3}^{mn}}}=\frac{{{3}^{1}}}{{{3}^{-1}}}={{3}^{2}}=9$.