Question

Directions for questions 49 and 50: Read the information given below and answer the question.
Let \(f\) be a function defined on the set of integers. Assume that \(f\) satisfies the following properties:
1. \(f(0) \neq 0\);
2. \(f(1) = 3\); and
3. \(f(x)f(y) = f(x+y) + f(x-y)\) for all integers \(x\) and \(y\).

What is \(f(3)\)?

• A. 7
• B. 18
• C. 123
• D. 322

Hint:

Use x = y = 0 to pin down f(0), then set y = 1 to build the recurrence f(n+1) = 3f(n) - f(n-1).

Answer 1

The Correct Option is B

Step 1: Find f(0).
Put \(x = y = 0\) in the given rule \(f(x)f(y) = f(x+y) + f(x-y)\).
This gives \(f(0)^2 = f(0) + f(0) = 2f(0)\).
So \(f(0)^2 - 2f(0) = 0\), which means \(f(0)\big(f(0) - 2\big) = 0\).
So \(f(0) = 0\) or \(f(0) = 2\). We are told \(f(0) \neq 0\), so \(f(0) = 2\).

Step 2: Build a recurrence for f(n).
Put \(y = 1\) and \(x = n\) in the rule:
\[ f(n)f(1) = f(n+1) + f(n-1) \]
Since \(f(1) = 3\), this becomes
\[ f(n+1) = 3f(n) - f(n-1) \]
This lets us build every value of \(f\) from the two starting values \(f(0)=2\) and \(f(1)=3\).

Step 3: Compute f(2) and f(3).
\(f(2) = 3f(1) - f(0) = 3(3) - 2 = 7\).
\(f(3) = 3f(2) - f(1) = 3(7) - 3 = 18\).

Step 4: Check the options.
\(f(3) = 18\), which is option 2. Option 1 (7) is actually \(f(2)\), a common slip if you stop one step early. Options 3 and 4 (123, 322) are later terms of the same sequence (\(f(5)\) and \(f(6)\)), not \(f(3)\).

Final Answer:
\(f(3) = 18\).\[ \boxed{f(3) = 18} \]