If f and g are continuous on [0, a] and satisfy f(x) = f(a - x) and g(x) + g(x - a) = 2, then ∫a0f(x)g(x)dxis equal to
A
∫a0f(x)dx
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B
∫a0f(a−x)dx
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C
∫a0f(a−x)g(a−x)dx
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D
∫a0g(x)dx
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Solution
The correct option is C∫a0f(a−x)g(a−x)dx ∵∫a0f(x)dx=∫a0f(a−x)dx∴∫a0f(x)g(x)dx=∫a0f(a−x)g(a−x)dx⇒∫a0f(x)g(x)dx=∫a0f(a−x)(2−g(x)}dx⇒∫a0f(x)g(x)dx=2∫a0f(x)dx−∫a0f(x)g(x)dx⇒2∫a0f(x)g(x)dx=2∫a0f(x)dx⇒∫a0f(x)g(x)dx=∫a0f(x)dx