A) \[0\]
B) \[2\int_{-3}^{3}{f\,(x)dx}\]
C) \[2\int_{0}^{3}{f\,(x)\,\,g\,\,(x)\,\,dx}\]
D) \[2\int_{0}^{3}{[f(x)-g(x)]\,dx}\]
Correct Answer: A
Solution :
\[\int_{-3}^{3}{[f(x)+f(-x)].[g(x)-g(-x)]\,dx}\] \[\because \] \[\int_{-a}^{a}{f(x)\,\,dx}\] \[=\left\{ \begin{matrix} 2\int_{0}^{a}{f(x)\,dx,\,\,if\,f(-x)=f(x),\,i.e.,\,\,even} \\ 0,\,\,\,if(-x)=-f(x),\,i.e.,\,odd \\ \end{matrix} \right.\] Let \[h(x)=\{f(x)+f(-x)\}.\,\{g(x)-g(-x)\}\] \[h(-x)=\{f(-x)+f(x)\}.\,\{g\,(-x)-g(x)\}\] \[=-\{f(x)+f(-x)\}.\{g\,(x)-g(-x)\}\] \[h(-x)=-h\,(x),\,\,i.e.,\,\] odd function \[\therefore \]\[\int_{-3}^{3}{[f(x)+f(-x)].[g(x)g(-x)].dx=0}\]
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