यदि `2f(x-1) - f((1-x)/x)=x` तो `f(x)=?`

Explanation

`2f(x-1) - f((1-x)/x)=x`

`2f(x-1) - f(1/x - 1)=x``...(i)`

यदि `x to 1/x`
अतः `2f(1/x-1) - f(x - 1)=1/x``...(ii)`

समीकरण `2xx(i) + (ii)` से

`4f(x-1)- 2f(1/x-1) ``+ 2f(1/x-1) - f(x-1)``=2x + 1/x`

`3f(x-1)=(2x^2 + 1)/x`

`f(x-1)=(2x^2 + 1)/(3x)`

अब यदि `x-1 to X` अतः `x to X+1`

`f(X)=(2(X+1)^2 + 1)/(3(X+1))`

अतः `f(x)=(2(x+1)^2 + 1)/(3(x+1))`