Let f : R - {0, 1} rightarrow R be a function | vedclass

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(B) Given the functional equation: $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$  $(1)$Substitute $x = 2$ in $(1)$:$f(2) + f\left(\frac{1}{1-2}\right)

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$\frac{9}{4}$

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(B) Given the functional equation: $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$  $(1)$Substitute $x = 2$ in $(1)$:$f(2) + f\left(\frac{1}{1-2}\right) = 1 + 2$$f(2) + f(-1) = 3$  $(2)$Substitute $x = -1$ in $(1)$:$f(-1) + f\left(\frac{1}{1-(-1)}\right) = 1 + (-1)$$f(-1) + f\left(\frac{1}{2}\right) = 0$  $(3)$Substitute $x = \frac{1}{2}$ in $(1)$:$f\left(\frac{1}{2}\right) + f\left(\frac{1}{1-1/2}\right) = 1 + \frac{1}{2}$$f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2}$  $(4)$Now,perform $(2) + (4) - (3)$:$(f(2) + f(-1)) + (f\left(\frac{1}{2}\right) + f(2)) - (f(-1) + f\left(\frac{1}{2}\right)) = 3 + \frac{3}{2} - 0$$2f(2) = \frac{9}{2}$$f(2) = \frac{9}{4}$

Let $f : R - \{0, 1\} \rightarrow R$ be a function such that $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$. Then $f(2)$ is equal to:

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