Correct Answer - 4
Given `f(x)=x^(3)-(3x^(2))/2+x+1/4=1/4(4x^(3)(4x^(3)-6x^(2)+4x+1)`
`=1/4(4x^(3)-6x^(2)+4x-1+2)`
`=1/4[x^(4)-(1-x)^(4)]+2/4`
`:.f(1-x)=1/4[(1-x)^(4)-x^(4)]+2/4`
`:.f(x)+f(1-x)=2/4+2/4=1`
Replacing `x` by `f(x)`,...................1 we have
`f[f(x)]+f[1-f(x)]=1`...............2
Now, `I=int_(1//4)^(3//4)f(f(x))dx`.............3
Also `I=int_(1//4)^(3//4)f(f(1-x))dx=int_(1//4)^(3//4) f(1-f(x))dx` [using 1 ] ...........4
Adding 3 and 4 we get
`2I=int_(1//4)^(3//4)[f(f(x))+f(1-f(x))]dx=int_(1//4)^(3//4)dx=1/2`
or `I=1/4`
`:.I^(-1)=4`
