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(D) Given $f(x) = x + \int_{0}^{1} (x - t) f(t) dt$.Expanding the integral: $f(x) = x + x \int_{0}^{1} f(t) dt - \int_{0}^{1} t f(t) dt$.Let $A = \int

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$(6, 8)$

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(D) Given $f(x) = x + \int_{0}^{1} (x - t) f(t) dt$.Expanding the integral: $f(x) = x + x \int_{0}^{1} f(t) dt - \int_{0}^{1} t f(t) dt$.Let $A = \int_{0}^{1} f(t) dt$ and $B = \int_{0}^{1} t f(t) dt$.Then $f(x) = (1 + A)x - B$. Let $C = 1 + A$,so $f(x) = Cx - B$.Now,$A = \int_{0}^{1} (Ct - B) dt = [C \frac{t^2}{2} - Bt]_{0}^{1} = \frac{C}{2} - B$.Since $C = 1 + A$,we have $A = \frac{1+A}{2} - B \Rightarrow 2A = 1 + A - 2B \Rightarrow A = 1 - 2B$.Also,$B = \int_{0}^{1} t(Ct - B) dt = \int_{0}^{1} (Ct^2 - Bt) dt = [C \frac{t^3}{3} - B \frac{t^2}{2}]_{0}^{1} = \frac{C}{3} - \frac{B}{2}$.Substituting $C = 1 + A = 1 + (1 - 2B) = 2 - 2B$:$B = \frac{2 - 2B}{3} - \frac{B}{2} \Rightarrow B = \frac{2}{3} - \frac{2B}{3} - \frac{B}{2} = \frac{2}{3} - \frac{7B}{6}$.$B + \frac{7B}{6} = \frac{2}{3} \Rightarrow \frac{13B}{6} = \frac{2}{3} \Rightarrow B = \frac{4}{13}$.Then $A = 1 - 2(\frac{4}{13}) = 1 - \frac{8}{13} = \frac{5}{13}$.$C = 1 + A = 1 + \frac{5}{13} = \frac{18}{13}$.Thus,$f(x) = \frac{18}{13}x - \frac{4}{13}$.Checking the point $(6, 8)$: $f(6) = \frac{18}{13}(6) - \frac{4}{13} = \frac{108 - 4}{13} = \frac{104}{13} = 8$.Therefore,the point $(6, 8)$ lies on the curve.

Let $f$ be a real-valued continuous function on $[0, 1]$ and $f(x) = x + \int_{0}^{1} (x - t) f(t) dt$. Then which of the following points $(x, y)$ lies on the curve $y = f(x)$?

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