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Let $$f:R\rightarrow R$$ and $$g:R\rightarrow R$$  be continuous functions, then the value of $$\displaystyle\int_{\displaystyle-\frac{\pi}{2}}^{\displaystyle\frac{\pi}{2}}{(f(x)+f(-x))(g(x)-g(-x))dx}$$, is equal to


A
1
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B
0
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C
1
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D
none of these
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Solution

The correct option is C $$0$$
Let $${ I }=\int _{ -\cfrac { \pi }{ 2 } }^{ \cfrac { \pi }{ 2 } }{ \left( f\left( x \right) +f\left( -x \right) \right) \left( g\left( x \right) -g\left( -x \right) \right) dx } $$ ...(1)
Using property $$\int _{ a }^{ b }{ f\left( x \right) dx } =\int _{ a }^{ b }{ f\left( a+b-x \right) dx } $$
$${ I }=\int _{ -\cfrac { \pi }{ 2 } }^{ \cfrac { \pi }{ 2 } }{ \left( f\left( -x \right) +f\left( x \right) \right) \left( g\left( -x \right) -g\left( x \right) \right) dx } $$ ...(2)
Adding (1) and (2), we get
$$2I=\int _{ -\cfrac { \pi }{ 2 } }^{ \cfrac { \pi }{ 2 } }{ \left( f\left( -x \right) +f\left( x \right) \right) \left( \left( g\left( x \right) -g\left( -x \right) \right) +\left( g\left( -x \right) -g\left( x \right) \right) \right) dx } \\ =0\Rightarrow I=0$$

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