A) 1
B) -1
C) 1 or -1
D) 0
Correct Answer: A
Solution :
(a): \[\left( {{\log }_{\frac{1}{2}}}2 \right)\left( {{\log }_{\frac{1}{3}}}3 \right)\left( {{\log }_{\frac{1}{4}}}4 \right)..........\left( {{\log }_{\frac{1}{99}}}99 \right)\] \[\left( \frac{\log 2}{{{\log }_{\frac{1}{2}}}2} \right)\left( \frac{\log 3}{{{\log }_{\frac{1}{3}}}3} \right)\left( \frac{\log 4}{{{\log }_{\frac{1}{4}}}4} \right)..........\left( \frac{\log 99}{{{\log }_{\frac{1}{99}}}99} \right)\] \[\left( \therefore {{\log }_{b}}a=\frac{\log a}{\log b} \right)\] \[=\left( \frac{\log 2}{-\log 2} \right)\left( \frac{\log 3}{-\log 3} \right)\left( \frac{\log 4}{-\log 4} \right)..........\left( \frac{\log 99}{-\log 99} \right)\] \[=\left( -1 \right)\times (-1)\times \left( -1 \right)\times ,....\times \left( -1 \right)\] (\[\therefore \]number of terms is even) = 1
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