Let f:ℝ-0,1→ ℝ be a function such that f(x)+f(1/(1-x))=1+x. Then f(2) is equal to

Solution

Maths sir · 3 steps

Keep applying x → 1/(1-x) to 2: it cycles back in three steps, giving three equations.

1. Follow the chain. Start at x = 2: 1/(1-2) = -1, then 1/(1-(-1)) = 1/2, then 1/(1-1/2) = 2. The map returns to 2 after three steps, so only f(2), f(-1) and f(1/2) appear.

   Put x =	Equation
   2	f(2) + f(-1) = 3
   -1	f(-1) + f(1/2) = 0
   1/2	f(1/2) + f(2) = 3/2

2. Add all three. Each value appears twice, so 2[f(2) + f(-1) + f(1/2)] = 3 + 0 + 3/2 = 9/2. This gives f(2) + f(-1) + f(1/2) = 9/4.

   f(2) = 9/4 - [f(-1) + f(1/2)] = 9/4 - 0

3. So. f(2) = 9/4, which is option (a).

The idea to keep. In f(x) + f(g(x)) equations, keep applying g until the input repeats. 1/(1-x) always cycles in three steps, so add the three equations and subtract a pair.

Answer from the NTA final answer key. Solution by PrepWiser.