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Question

Let f be a positive function. Let
I1=k1k xf{x(1x)}dx,  I2=k1kf{x(1x)}dx
when 2k1>0. Then I1I2 is               [IIT 1997 Cancelled]



A

2

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B

k

loader
C

1/2

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D

1

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Solution

The correct option is B

1/2


I1=k1k xf{x(1x)}dx
=k1k(1k+kx)f[(1k+kx){1(1k+kx)}]dx
=k1k(1x)f{x(1x)}dx
=k1kf{x(1x)}dxk1kxf{x(1x)}dx=I2I1
2I1=I2I1I2=12

Mathematics

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