Question

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

• A. \(\frac{7}{3}\)
• B. \(\frac{9}{2}\)
• C. \(\frac{9}{4}\)
• D. \(\frac{7}{4}\)

Hint:

For functional equations, substituting specific values of x can simplify the problem and lead to a system of equations. Carefully combine and solve the equations step-by-step.

Answer 1

The Correct Option is C

The given functional equation is:
\[f(x) + f\left(\frac{1}{1-x}\right) = 1 + x.\]
Step 1: Substitute Specific Values of \(x\)
For \(x = 2\):
\[f(2) + f(-1) = 3. \tag{1}\]
For \(x = -1\):
\[f(-1) + f\left(\frac{1}{2}\right) = 0. \tag{2}\]
For \(x = \frac{1}{2}\):
\[f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2}. \tag{3}\]
Step 2: Solve the System of Equations
Add equations (1), (2), and (3):
\[\big(f(2) + f(-1)\big) + \big(f(-1) + f\left(\frac{1}{2}\right)\big) + \big(f\left(\frac{1}{2}\right) + f(2)\big) = 3 + 0 + \frac{3}{2}.\]
Simplify:
\[2f(2) + 2f(-1) + 2f\left(\frac{1}{2}\right) = \frac{9}{2}.\]
Divide through by 2:
\[f(2) + f(-1) + f\left(\frac{1}{2}\right) = \frac{9}{4}. \tag{4}\]
Substitute \(f(-1) + f\left(\frac{1}{2}\right) = 0\) (from equation (2)) into equation (4):
\[f(2) = \frac{9}{4}.\]
Final Result: \(f(2) = \frac{9}{4}\).