Let f and g be continuous function on [0,a] such that f(x)=f(a−x) and g(x)+g(a−x)=4, then a∫0f(x)g(x)dx is equal to :
A
a∫0f(x)dx
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B
2a∫0f(x)dx
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C
4a∫0f(x)dx
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D
−3a∫0f(x)dx
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Solution
The correct option is B2a∫0f(x)dx Given f(x)=f(a−x) g(x)+g(a−x)=4⇒g(a−x)=4−g(x) I=a∫0f(x).g(x)dx =a∫0f(a−x).g(a−x)dx =a∫0f(x).(4−g(x))dx ⇒I=a∫04f(x)dx−a∫0f(x).g(x)dx ⇒2I=a∫04f(x)dx ⇒I=2a∫0f(x)dx