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Question

Let f be a positive function
Let I1=k1kx f{x(1x)}dx and 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 C 12
I1=k1k(1k+kx)f{(1k+kx)(1(1k+kx))}dx
(baf(x)dx=baf(a+bx)dx)
=k1k(1x)f(x(1x))dx
=k1kf{x(1x)}dxk1kxf{x(1x)}dx
I1=I2I1
I1I2=12

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