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 number of roots of the equation `g(x)=1` is

Options

A. `(-oo,5]`
B. `[1,oo)`
C. `(-oo,1) cup [5, oo)`
D. None of these

Answer

C

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))`
`y=g(x)=(x^(2)-x-1)/(x(x-1)) or (y-1)x^(2)+(1-y)x+1=0`
Now, x is real, Therefore, ` D ge 0 or (1-y)^(2)-4(y-1) ge 0`
`or (y-1) (y-5) ge 0`
`or y in (-oo,1) cup [5,oo)`