A) \[-1/2\]
B) 0
C) 1
D) \[2\log (1/2)\]
Correct Answer: A
Solution :
\[\int_{-1/2}^{1/2}{\left\{ [x]+\log \left( \frac{1+x}{1-x} \right) \right\}}dx\] \[=\int_{-1/2}^{1/2}{[x]dx+\int_{-1/2}^{1/2}{\log \left( \frac{1+x}{1-x} \right)}}dx\] \[=\int_{-1/2}^{1/2}{[x]dx+0}\] \[\left[ \because \log \left( \frac{1+x}{1-x} \right) \right.\]is an odd function of\[x\] \[\Rightarrow \] \[\left. \int_{-1/2}^{1/2}{\log \left( \frac{1+x}{1-x} \right)dx}=0 \right]\] \[=\int_{-1/2}^{0}{[x]}\,dx+\int_{0}^{1/2}{[x]}\,dx\] \[=[-x]_{-1/2}^{0}+0=0-1/2=-1/2\]
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