The correct answer is D. We are given the functional equation: $f(x) + f\left(\frac{1}{1-x}\right) = x$ for $x \neq 0, 1$. Our goal is to find the value of $f(2)$. This type of problem often involves a cyclical substitution of the function's argument. Step 1: Substitute $x=2$ into the given functional equation. $f(2) + f\left(\frac{1}{1-2}\right) = 2$ $f(2) + f(-1) = 2$ (Equation 1) Step 2: Observe the argument of the second term, which is $g(x) = \frac{1}{1-x}$. Let's see what happens if we apply this transformation repeatedly. First transformation: $x \to g(x) = \frac{1}{1-x}$ If we substitute $x$ with $g(x)$ in the original equation, we get: $f(g(x)) + f(g(g(x))) = g(x)$ Let's evaluate $g(g(x))$: $g(g(x)) = \frac{1}{1-g(x)} = \frac{1}{1-\frac{1}{1-x}} = \frac{1}{\frac{(1-x)-1}{1-x}} = \frac{1-x}{-x} = 1-\frac{1}{x}$. So, substituting $x$ with $\frac{1}{1-x}$ in the original equation yields: $f\left(\frac{1}{1-x}\right) + f\left(1-\frac{1}{x}\right) = \frac{1}{1-x}$ Step 3: Now, we apply this equation for a value derived from $x=2$. The first derived value was $-1$. So, let $x=-1$ in the equation from Step 2. $f\left(\frac{1}{1-(-1)}\right) + f\left(1-\frac{1}{-1}\right) = \frac{1}{1-(-1)}$ $f\left(\frac{1}{2}\right) + f(1+1) = \frac{1}{2}$ $f\left(\frac{1}{2}\right) + f(2) = \frac{1}{2}$ (Equation 2) Step 4: We have now generated $f(1/2)$. Let's apply the transformation one more time. The next argument in the sequence is $g(g(x)) = 1 - \frac{1}{x}$. If we substitute $x$ with $g(g(x))$ in the original equation, we get: $f(g(g(x))) + f(g(g(g(x)))) = g(g(x))$ Let's evaluate $g(g(g(x)))$: $g(g(g(x))) = \frac{1}{1-g(g(x))} = \frac{1}{1-(1-\frac{1}{x})} = \frac{1}{\frac{1}{x}} = x$. So, substituting $x$ with $1-\frac{1}{x}$ in the original equation yields: $f\left(1-\frac{1}{x}\right) + f(x) = 1-\frac{1}{x}$ Step 5: Apply this equation for a value derived from $x=2$. The second derived value was $1/2$. So, let $x=1/2$ in the equation from Step 4. $f\left(1-\frac{1}{1/2}\right) + f\left(\frac{1}{2}\right) = 1-\frac{1}{1/2}$ $f(1-2) + f\left(\frac{1}{2}\right) = 1-2$ $f(-1) + f\left(\frac{1}{2}\right) = -1$ (Equation 3) Step 6: We now have a system of three linear equations involving $f(2)$, $f(-1)$, and $f(1/2)$: (1) $f(2) + f(-1) = 2$ (2) $f(1/2) + f(2) = 1/2$ (3) $f(-1) + f(1/2) = -1$ To solve for $f(2)$, we can add all three equations: $(f(2) + f(-1)) + (f(1/2) + f(2)) + (f(-1) + f(1/2)) = 2 + 1/2 + (-1)$ $2f(2) + 2f(-1) + 2f(1/2) = 3/2$ $2(f(2) + f(-1) + f(1/2)) = 3/2$ $f(2) + f(-1) + f(1/2) = 3/4$ (Equation 4) Step 7: Subtract Equation 3 from Equation 4 to isolate $f(2)$. $(f(2) + f(-1) + f(1/2)) - (f(-1) + f(1/2)) = 3/4 - (-1)$ $f(2) = 3/4 + 1$ $f(2) = 7/4$ Thus, the value of $f(2)$ is $\frac{7}{4}$. The final answer is $\boxed{\frac{7}{4}}$.