Let $F$ be the set of all continuous, real-valued functions which are solutions to:

$[f(x)]^2 = \int_{0}^{x} [f(t)f'(t)-f(t)-f'(t)-1] dt + 100$

Evaluate: $\frac{1}{|F|}\sum_{f(x) \in F} |f(100)|$

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Details and Assumptions:
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- $|S|$ denotes the cardinality of $S$

The answer is 100.

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Two functions are equal if and only if they share a point and their derivatives are equal.

Applying this principle, we check that the $LHS$ at $x=0$ is the same as the $RHS$ :

$[f(0)]^{2} = \int_{0}^{0} [f(t)f'(t)-f(t)-f'(t)-1] dt + 100 = 100$

So $f(0)=\pm 10$ .

Now, checking that their derivatives are equal:

$2f(x)f'(x) = f(x)f'(x)-f(x)-f'(x)-1 \Rightarrow [f(x)+1][f'(x)+1]=0$

And so $f(x)=-1$ or $f'(x)=-1$ . This gives $f(x)=-1, c_{1}-x$

Using what we know about $f(0)$ , we reject $f(x)=-1$ , and get that $c_{1}=\pm 10$ . We end up with:

$f(x) = 10-x, \quad f(x)=-10-x$

Hence, $|F|=2$ and our sum is $\frac{1}{2}(90+110)=\boxed{100}$

We can also reject all piecewise functions consisting of a combination of these three functions, since combining the constant function with any of the other linear functions will result in a function that is not differentiable at some point. Also, combining the two non-constant functions will cause the resulting function to be discontinuous