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: Get f(0) from the given properties.
Setting \(x=y=0\) in \(f(x)f(y)=f(x+y)+f(x-y)\) gives \(f(0)^2 = 2f(0)\), so \(f(0)=0\) or \(f(0)=2\).
The condition \(f(0) \neq 0\) rules out the first case, so \(f(0)=2\).

Step 2: Find f(2) by direct substitution.
Put \(x=y=1\):
\[ f(1)f(1) = f(2) + f(0) \]
\[ 3 \times 3 = f(2) + 2 \]
\[ f(2) = 9 - 2 = 7 \]

Step 3: Find f(3) by direct substitution.
Put \(x=2, y=1\):
\[ f(2)f(1) = f(3) + f(1) \]
\[ 7 \times 3 = f(3) + 3 \]
\[ f(3) = 21 - 3 = 18 \]

Step 4: Match with the options.
This direct substitution, done without setting up any general recurrence, gives the same value $f(3)=18$, confirming option 2 is correct.
Notice that this substitution route only ever uses values we already know, $f(0)$, $f(1)$ and the newly found $f(2)$, plugged straight back into the same rule with a fresh pair of $x$ and $y$. There is no need to write down a formula like $f(n+1)=3f(n)-f(n-1)$ at all; each answer just falls out of the property directly.
Also check that the working is self-consistent: since $f(2)=7$, redo the check with $x=1,y=1$ once more as a sanity test, giving $f(1)f(1)=9=f(2)+f(0)=7+2=9$, which matches, so the value of $f(2)$ used to reach $f(3)$ is trustworthy.
So the final value is confirmed from two independent substitutions, and it agrees with the option list.
\[ \boxed{f(3) = 18} \]