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Question

Let f(x) be a positive function. Let
I1=k1kxf{x(1x)}dx,
I2=k1kf{x(1x)}dx,
where 2k1>0, then I1I2 is

A
2
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B
k
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C
12
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D
1
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Solution

The correct option is B 12
I1=k1kxf{x(1x)}dx
=k1k(1x)f{(1x)(1(1x))}dx
(baf(x)dx=baf(a+bx)dx)
=k1k(1x)f{(1x)x}dx
=k1kf{x(1x)}dx
k1kxf{x(1x)}dx
Therefore, I1=I2I12I1=I2
I1I2=12

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