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

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,
$f(g(5))+g(f(5))=−4+16=12$