A) \[-\frac{2}{5}\log 2+\frac{1}{2}\]
B) \[-\frac{2}{5}\log 2-\frac{1}{2}\]
C) \[\frac{2}{5}\log 2+\frac{1}{2}\]
D) None of these
Correct Answer: A
Solution :
We have,\[2f(x)+3f\left( \frac{1}{x} \right)=\frac{1}{x}-2\] ... (i) Replacing\[x\]by\[\left( \frac{1}{x} \right)\], we get \[2f\left( \frac{1}{x} \right)+3f(x)=x-2\] ? (ii) On solving Eqs. (i) and (ii), we get \[f(x)=\frac{-2}{5x}+\frac{3x}{5}-\frac{2}{5}\] Now, \[\int_{1}^{2}{f(x)}\,\,dx=\int_{1}^{2}{\left( \frac{-2}{5x}+\frac{3x}{5}-\frac{2}{5} \right)dx}\] \[={{\left[ \frac{-2}{5}\log x+\frac{3}{5}\frac{{{x}^{2}}}{2}-\frac{2x}{5} \right]}^{2}}\] \[=\left( -\frac{2}{5}\log 2+\frac{6}{5}-\frac{4}{5} \right)-\left( \frac{3}{10}-\frac{2}{5} \right)\] \[=-\frac{2}{5}\log 2+\frac{2}{5}+\frac{1}{10}\] \[=-\frac{2}{5}\log 2+\frac{1}{2}\]You need to login to perform this action.
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