A) \[a=\frac{3}{2},\,\,\,b=\frac{3}{2}\]
B) \[a=\frac{3}{4},\,\,\,b=-\frac{3}{4}\]
C) \[a=\frac{3}{4},\,\,\,b=\frac{3}{2}\]
D) \[a=b\]
Correct Answer: C
Solution :
Integrate it by parts taking \[\log \left( 1+\frac{x}{2} \right)\]as first function \[=\left[ \log \left( 1+\frac{x}{2} \right)\frac{{{x}^{2}}}{2} \right]_{0}^{2}-\int_{0}^{1}{\frac{1}{1+\frac{x}{2}}\frac{1}{2}\frac{{{x}^{2}}}{2}}dx\] \[=\frac{1}{2}\log \frac{3}{2}-\frac{1}{2}\int_{0}^{1}{\frac{{{x}^{2}}}{x+2}dx}\] \[=\frac{1}{2}\log \frac{3}{2}-\frac{1}{2}\left[ \frac{1}{2}-2+4\log 3-4\log 2 \right]=\frac{3}{4}+\frac{3}{2}\log \frac{2}{3}\] On comparing with the given value \[a=\frac{3}{4},b=\frac{3}{2}\].
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