A function f(x) is given such that it satisfies all the real values of x, and a relation is given as,
$$f\left(x\right)\ +\ 2f\left(\dfrac{1}{1-x}\right)\ =\ x$$
What is the value of $$f\left(2\right)$$?
Solution
We are given the equation,
$$f\left(x\right)\ +\ 2f\left(\dfrac{1}{1-x}\right)\ =\ x$$
Substituting the value of x = 2 in the equation, we get,
$$f\left(2\right)\ +\ 2f\left(-1\right)\ =\ 2$$ -----(1)
Substituting x = -1, we get
$$f\left(-1\right)\ +\ 2f\left(\dfrac{1}{2}\right)\ = -1$$ -----(2)
Substituting x = $$\dfrac{1}{2}$$ we get,
$$f\left(\dfrac{1}{2}\right)\ +\ 2f\left(2\right)\ =\ \dfrac{1}{2}$$ -----(3)
By applying (1) - 2*(2) + 4*(3), we get,
$$f\left(2\right)\ +\ 2f\left(-1\right)\ -\ 2f\left(-1\right)\ -\ 4f\left(\dfrac{1}{2}\right)\ +\ 4f\left(\dfrac{1}{2}\right)\ +\ 8f\left(2\right)\ =\ 2\ -\ 2\times\ \left(-1\right)\ +\ \ 4\times\ \dfrac{1}{2}$$
$$f\left(2\right)+\ 8f\left(2\right)\ =\ 2\ +\ 2\ +\ 2$$
$$9f\left(2\right)\ =\ 6$$
$$f\left(2\right)\ =\ \dfrac{2}{3}$$
The correct answer is option A.
