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limn→∞(∑k=1nak−1∫(k−1)akaf(x)f(x)+f((2k−1)a−x) dx)=75\lim_{n \to \infty} \left(\sum_{k=1}^n a^{k-1} \int_{(k-1)a}^{ka} {\dfrac{f(x)}{f(x) + f \left( \left(2k-1\right)a-x \right)}} \ dx \right) = \dfrac{7}{5} n→∞lim(k=1∑nak−1∫(k−1)akaf(x)+f((2k−1)a−x)f(x) dx)=57
Given that function f(x)>0f(x) > 0f(x)>0, ∀ x∈R\forall ~ x \in \mathbb{R}∀ x∈R is bounded and satisfies the condition above. If a<1a < 1a<1, find aaa.
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