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Question

Suppose f, g are continuous and differentiable on [0,b], are non-negative on [0,b] and f is non constant with f(0) = 0, then the minimum value of a0g(x)f(x)dx+b0g(x)f(x)dx on aϵ(0,b] is

A
f(a).g(a)
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B
f(b).g(b)
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C
f(a).g(b)
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D
f(b).g(a)
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Solution

The correct option is C f(a).g(b)
Integrating by parts the integral
a0g(x)f(x)dx+b0g(x)f(x)dx=g(x)f(x)|a0a0g(x)f(x)dx+b0g(x)f(x)dx=f(a).g(a)+bag(x)f(x)dxf(a).g(a)+bag(x)f(x)dx=f(a).g(b)

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