Let `f` be a positive function. If `I_1 = int_(1-k)^k x f[x(1-x)]\ dx` and `I_2 = int_(1-k)^k f[x(1-x)]\ dx,` where `2k-1 gt 0.` Then `I_1/I_2` is

Options

A. `2`
B. `k`
C. `1/2`
D. `1`

Answer

C

Explanation

Given `f` is a positive function and
`I_(1)=int_(1-k)^(k)xf[x(1-x)]dx`
`I_(2)=int_(1-k)^(k)f[(1-x)]dx`
Now `I_(1)=int_(1-k)^(k)xf[(1--x)]dx`………………1
`=int_(1-k)^(k)(1-x)f[(1-x)x]dx`………………2
[ Using the property `int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx`]
Adding 1 and 2 we get
`2I_(1)=int_(1-k)^(k)f[x(1-x)]dx=I_(2)` or `(I_(1))/(I_(2))=1/2`