Sequence And Series Question 560
Question: The value of $ {\log_{e}}( 1+ax^{2}+a^{2}+\frac{a}{x^{2}} ) $ is
Options:
A) $ a,( x^{2}-\frac{1}{x^{2}} )-\frac{a^{2}}{2},( x^{4}-\frac{1}{x^{4}} )+\frac{a^{3}}{3},( x^{6}-\frac{1}{x^{6}} )-….. $
B) $ a,( x^{2}+\frac{1}{x^{2}} )-\frac{a^{2}}{2},( x^{4}+\frac{1}{x^{4}} )+\frac{a^{3}}{3},( x^{6}+\frac{1}{x^{6}} )-….. $
C) $ a,( x^{2}+\frac{1}{x^{2}} )+\frac{a^{2}}{2},( x^{4}+\frac{1}{x^{4}} )+\frac{a^{3}}{3},( x^{6}+\frac{1}{x^{6}} )+….. $
D) $ a,( x^{2}-\frac{1}{x^{2}} )+\frac{a^{2}}{2},( x^{4}-\frac{1}{x^{4}} )+\frac{a^{3}}{3},( x^{6}-\frac{1}{x^{6}} )+….. $
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Answer:
Correct Answer: B
Solution:
$ {\log_{e}}[ 1+ax^{2}+a^{2}+\frac{a}{x^{2}} ] $ $ ={\log_{e}}(1+ax^{2}),( 1+\frac{a}{x^{2}} ) $ $ ={\log_{e}}(1+ax^{2})+{\log_{e}}( 1+\frac{a}{x^{2}} ) $ = $ [ ax^{2}-\frac{1}{2}a^{2}x^{4}+\frac{1}{3}a^{3}x^{6}-……….. ] $ $ +[ \frac{a}{x^{2}}-\frac{1}{2}a^{2}( \frac{1}{x^{4}} )+\frac{1}{3}a^{3}( \frac{1}{x^{6}} )-….. ] $ $ =a( x^{2}+\frac{1}{x^{2}} )-\frac{1}{2}.a^{2}( x^{4}+\frac{1}{x^{4}} )+\frac{1}{3}a^{3}( x^{6}+\frac{1}{x^{6}} )-… $
BETA
