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Question

Let f:RR and g:RR be continuous functions, then the value of π2π2(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=π2π2(f(x)+f(x))(g(x)g(x))dx ...(1)
Using property baf(x)dx=baf(a+bx)dx
I=π2π2(f(x)+f(x))(g(x)g(x))dx ...(2)
Adding (1) and (2), we get
2I=π2π2(f(x)+f(x))((g(x)g(x))+(g(x)g(x)))dx=0I=0

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