A) 0.486
B) 0.243
C) 1.8
D) 0
Correct Answer: D
Solution :
\[\int\limits_{-0.9}^{0.9}{\left\{ [{{x}^{2}}]+\log \left( \frac{2-x}{2+x} \right) \right\}dx}\] \[=\int\limits_{-0.9}^{0.9}{\left\{ [{{x}^{2}}]dx+\int\limits_{-0.9}^{0.9}{\log }\left( \frac{2-x}{2+x} \right) \right\}dx}\] \[=0+\int\limits_{-0.9}^{0.9}{\log \left( \frac{2-x}{2+x} \right)dx}\] Put\[x=-x\Rightarrow f(x)=\log \frac{2-x}{2+x}\] and\[f(-x)=\log \frac{2+x}{2-x}\] \[=-\log \frac{(2+x)}{2-x}=-f(x)\] So, it is an odd function, hence Required integral 1 = 0.
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