Question

There are two positive integers – $x$ and $y$. A function of $x$ and $y$ is defined such that:
$f(0,y) = y + 1$,
$f(x + 1, 0) = f(x, 1)$,
$f(x + 1, y + 1) = f(x, f(x + 1, y))$
What is the value of $f(1, 2)$?

• A. 2
• B. 4
• C. 3
• D. Cannot be determined

Hint:

When working with recursive functional equations, break them down step-by-step and substitute from the base case upward.

Answer 1

The Correct Option is B

We use the recursive definition step-by-step: Step 1: Start with $f(1,2)$ From the 3rd rule: \[ f(1,2) = f(0, f(1,1)) \] Step 2: Evaluate $f(1,1)$ Again from the 3rd rule: \[ f(1,1) = f(0, f(1,0)) \] Step 3: Evaluate $f(1,0)$ From the 2nd rule: \[ f(1,0) = f(0,1) \] Step 4: Evaluate $f(0,1)$ From the 1st rule: \[ f(0,1) = 1 + 1 = 2 \] Step 5: Substitute back \[ f(1,0) = 2 \] \[ f(1,1) = f(0, 2) = 2 + 1 = 3 \] \[ f(1,2) = f(0, 3) = 3 + 1 = 4 \] \[ \boxed{4} \]