The sum of all real values of x satisfying th | vedclass

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(C) The equation $f(x)^{g(x)} = 1$ holds if:$1)$ $f(x) = 1$$x^2 - 5x + 5 = 1$ $\Rightarrow x^2 - 5x + 4 = 0$ $\Rightarrow (x-1)(x-4) = 0$ $\Rightarrow

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$3$

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(C) The equation $f(x)^{g(x)} = 1$ holds if:$1)$ $f(x) = 1$$x^2 - 5x + 5 = 1$ $\Rightarrow x^2 - 5x + 4 = 0$ $\Rightarrow (x-1)(x-4) = 0$ $\Rightarrow x = 1, 4$.$2)$ $f(x) = -1$ and $g(x)$ is an even integer.$x^2 - 5x + 5 = -1$ $\Rightarrow x^2 - 5x + 6 = 0$ $\Rightarrow (x-2)(x-3) = 0$ $\Rightarrow x = 2, 3$.For $x = 2$,$g(2) = 2^2 + 4(2) - 60 = 4 + 8 - 60 = -48$ (even),so $x = 2$ is a solution.For $x = 3$,$g(3) = 3^2 + 4(3) - 60 = 9 + 12 - 60 = -39$ (odd),so $x = 3$ is rejected.$3)$ $f(x) = 0$ and $g(x) > 0$ (not applicable here as $f(x)$ is not $0$ for these roots) or $g(x) = 0$ and $f(x) 
eq 0$.$g(x) = x^2 + 4x - 60 = 0$ $\Rightarrow (x+10)(x-6) = 0$ $\Rightarrow x = -10, 6$.For $x = -10$,$f(-10) = (-10)^2 - 5(-10) + 5 = 100 + 50 + 5 = 155 
eq 0$.For $x = 6$,$f(6) = 6^2 - 5(6) + 5 = 36 - 30 + 5 = 11 
eq 0$.Thus,the valid values of $x$ are $1, 4, 2, -10, 6$.The sum is $1 + 4 + 2 - 10 + 6 = 3$.

List-$I$	List-$II$
$A$. All the roots are negative	$I$. $(b - 3)^2 = 36 + P^2$ for $P \in R$
$B$. Two roots are complex	$II$. $-3 < b < 9$
$C$. Two roots are positive	$III$. $b \in (-\infty, -3) \cup (9, \infty)$
$D$. All roots are real and distinct	$IV$. $b = 9$
	$V$. $b = -3$

The sum of all real values of $x$ satisfying the equation $(x^2 - 5x + 5)^{x^2 + 4x - 60} = 1$ is:

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