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

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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

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Given:  $ f(x) + f(x - 1) = 1 $ $ f(x^2 - x) = 5 $ $ g(x) = x^2 $ Step 1: Substituting $ x = 1 $ in equations (1) and (2): $ f(0) = 5 $ from equation (2), since $ x^2 - x = 0 $ $ f(1) + f(0) = 1 \Rightarrow f(1) = 1 - 5 = -4 $ Step 2: Substituting $ x = 2 $ in equation (1): $ f(2) + f(1) = 1 \Rightarrow f(2) = 1 - (-4) = 5 $ Observation: $ f(n) = 5 $ if $ n $ is even $ f(n) = -4 $ if $ n $ is odd Final Calculation: $ f(g(5)) + g(f(5)) = f(25) + g(-4) $ $ f(25) = -4 $ since 25 is odd $ g(-4) = (-4)^2 = 16 $ $ f(g(5)) + g(f(5)) = -4 + 16 = 12 $

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

Correct Answer: 12

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