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: Fix f(0) and f(1).
From \(x=y=0\): \(f(0)^2 = 2f(0)\), so \(f(0)=0\) or \(2\); since \(f(0)\neq 0\), \(f(0)=2\). We are given \(f(1)=3\).

Step 2: Reach f(2) and f(4) using x = y (a doubling step).
Put \(x=y=1\): \(f(1)^2 = f(2)+f(0)\), so \(9 = f(2)+2\), giving \(f(2)=7\).
Put \(x=y=2\): \(f(2)^2 = f(4)+f(0)\), so \(49 = f(4)+2\), giving \(f(4)=47\).

Step 3: Get f(3), then combine 4 and 3 to jump straight to 7.
Put \(x=2,y=1\): \(f(2)f(1)=f(3)+f(1)\), so \(21=f(3)+3\), giving \(f(3)=18\).
Now put \(x=4,y=3\) directly in the functional rule:
\[ f(4)f(3) = f(7) + f(1) \]
\[ 47 \times 18 = f(7) + 3 \]
\[ 846 = f(7) + 3 \]
\[ f(7) = 843 \]

Step 4: Conclusion.
Jumping straight from \(f(4)\) and \(f(3)\) to \(f(7)\) skips computing \(f(5)\) and \(f(6)\) one at a time, and still lands on the same value, 843, confirming option 3.
\[ \boxed{f(7) = 843} \]