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Question

Let f be a given function such that f(x)>0, ∀x \in D_f. Let \)
I1k1kxf(x(1x))dx
I2=k1kf(x(1x))dx, where 2k1>0
then I1I2 is equal to

A
1
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B
k
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C
12
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D
2
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Solution

The correct option is C 12
I1=k1kxf(x(1x))dx
=k1k(1x)f((1x)x)dx
(baf(x)dx=baf(a+bx)dx)
=k1kf(x(1x))dxk1kxf((1x)x)dx
I1=I2I1
2I1I2
I1I2=12

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