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Question

The value of 21[f{g(x)}]1.f{g(x)}.g(x)dx, where g(1)=g(2), is equal to?

A
1
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B
2
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C
0
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D
none of these
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Solution

The correct option is B 0
Let I=21[f{g(x)}]1.f{g(x)}.g(x)dx

Substitute [f{g(x)}]=tf{g(x)}.g(x)dx=dt

I=f(g(2))f(g(1))tdt=[t22]f(g(2))f(g(1))=0g(1)=g(2)

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