Consider the function `f(x)` satisfying the identity `f(x) +f((x-1)/(x))=1+x AA x in R -{0,1}, and g(x)=2f(x)-x+1.`
The domain of `y=sqrt(g(x))` is

Options

A. 2
B. 1
C. 3
D. 0

Answer

D

Explanation

`f(x)+f((x-1)/(x))=1+x " (1)" `
In (1), replace x by `(x-1)/(x)`. Then
`f((x-1)/(x))+f(((x-1)/(x)-1)/((x-1)/(x)))=1+(x-1)/(x)`
`or f((x-1)/(x))+f((1)/(1-x))=1+(x-1)/(x) " (2)" `
Now, from `(1) -(2)`, we have
`f(x)-f((1)/(1-x))=x-(x-1)/(x) " (3)" `
In (3), replace x by `(1)/(x-1)`. Then
`f((1)/(1-x))-f((x-1)/(x))=(1)/(1-x)-((1)/(1-x)-1)/((1)/(1-x))`
` or f((1)/(1-x))-f((x-1)/(x))=(1)/(1-x)-x " (4)" `
Now, from `(1)+(3)+(4)`, we have
`2f(x)=1+x+x-(x-1)/(x)+(1)/(1-x)-x`
` or f(x)=(x^(3)-x^(2)-1)/(2x(x-1))`
`g(x)=1, or (x^(2)-x-1)/(x(x-1)) =1 or -x-1=-x,` which has no solution.