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 \(f(7)\)?

• A. 123
• B. 322
• C. 843
• D. 1126

Hint:

Reuse the recurrence f(n+1) = 3f(n) - f(n-1) built from f(0) = 2 and f(1) = 3, and iterate it up to n = 7.

Answer 1

The Correct Option is C

Step 1: Recall the recurrence.
As in the setup for this pair of questions, put \(x=n, y=1\) in \(f(x)f(y)=f(x+y)+f(x-y)\) to get
\[ f(n+1) = 3f(n) - f(n-1) \]
with \(f(0)=2\) (forced since \(f(0)\neq 0\)) and \(f(1)=3\).

Step 2: Build up the values step by step.
\(f(2) = 3(3) - 2 = 7\)
\(f(3) = 3(7) - 3 = 18\)
\(f(4) = 3(18) - 7 = 47\)
\(f(5) = 3(47) - 18 = 123\)
\(f(6) = 3(123) - 47 = 322\)
\(f(7) = 3(322) - 123 = 843\)

Step 3: Check the options.
\(f(7)=843\) is option 3. Options 1 and 2 (123, 322) are the earlier terms \(f(5)\) and \(f(6)\), so picking those means stopping too early. Option 4 (1126) does not match the recurrence at all.

Final Answer:
\[ \boxed{f(7) = 843} \]