Let f be a positive function. Let I(1)=i...

Let `f` be a positive function. Let `I_(1)=int_(1-k)^(k)x f[x(1-x)]dx` , `I_(2)=int_(1-k)^(k)f[x(1-x)]dx`, where `2k-1gt0`. Then `(I_(1))/(I_(2))` is

A

2

B

k

C

`1//2`

D

`1`

Text Solution

Verified by Experts

The correct Answer is:

C

Using `int _(0)^(a) f(x) dx = int _(0) ^(a) f (a - x ) dx`, we have
`I_1 = int_(1-k ) ^(k) (1 - x ) f {(1- x ) (1 - (1 - x ))}dx `
`rArr I_1 = int _(1-k) ^(k) (1- x ) f {x (1 - x )} dx `
`rArr I_1 = int_(1-k) ^(k) f {x( 1- x )} dx - int_(1-k ) ^(k) x f {x (1 - x) } dx `
`rArr I_1 = I_2 - I_1 rArr 2 I_1 = I_2 rArr (I_1)/(I_2) = (1)/(2)`

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