Question
Let f be a positive function. Let
I1=∫k1−k xf{x(1−x)}dx, I2=∫k1−kf{x(1−x)}dx
when 2k−1>0. Then I1I2 is [IIT 1997 Cancelled]
I1=∫k1−k xf{x(1−x)}dx, I2=∫k1−kf{x(1−x)}dx
when 2k−1>0. Then I1I2 is [IIT 1997 Cancelled]
Answer: Option C
:
C
I1=∫k1−kxf{x(1−x)}dx
=∫k1−k(1−k+k−x)f[(1−k+k−x){1−(1−k+k−x)}]dx
=∫k1−k(1−x)f{x(1−x)}dx
=∫k1−kf{x(1−x)}dx−∫k1−kxf{x(1−x)}dx=I2−I1
∴2I1=I2⇒I1I2=12
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:
C
I1=∫k1−kxf{x(1−x)}dx
=∫k1−k(1−k+k−x)f[(1−k+k−x){1−(1−k+k−x)}]dx
=∫k1−k(1−x)f{x(1−x)}dx
=∫k1−kf{x(1−x)}dx−∫k1−kxf{x(1−x)}dx=I2−I1
∴2I1=I2⇒I1I2=12
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