Question

Suppose for all integers x, there are two functions f and g such that \(f(x)+f(x−1)−1=0\) and \(g(x)=x^2\). If \(f(x^2−x)=5\),then the value of the sum \(f(g(5))+g(f(5))\) is

Answer 1

Correct Answer: 12

Given:

1. \(f(x)+f(x−1)=1\)
2. \(f(x^2 - x) = 5\)
3. \(g(x) = x^2\)

By substituting \(x=1\) into equations (1) and (2), we get:
\(f(0)=5\)
\(f(1)+f(0)=1\\ f(1)=1−5=−4\)

Next, substituting \(x=2\) into equation (1):
\(f(2)+f(1)=1 \\f(2)=1+4=5\)

We observe that:

• \(f(n)=5\) if n is even
• \(f(n)=−4\) if n is odd

Finally, we find:

\(f(g(5))+g(f(5))=f(25)+g(−4)\)
Since 25 is odd, \(f(25)=−4\)
And,
\(g(-4) = (-4)^2 = 16\)
So,
\(f(g(5))+g(f(5))=−4+16=12\)