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 \[m+n=1,\,\,\,mn=-1\] \[\therefore \,\,\frac{N}{D}=\frac{{{3}^{m+n}}}{{{3}^{mn}}}=\frac{{{3}^{1}}}{{{3}^{-1}}}={{3}^{2}}=9\].You need to login to perform this action.
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