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Let f be a non-negative function defined on the interval $$[0,1]$$. If$$\int_{0}^{x}\sqrt{1 - (f'(t))^2} dt = \int_{0}^{x} f(t)dt, 0 \leq x \leq 1$$, and f(0) = 0, then
$$f\left(\frac{1}{2}\right) < \frac{1}{2}$$ and $$f\left(\frac{1}{3}\right) > \frac{1}{3}$$
$$f\left(\frac{1}{2}\right) > \frac{1}{2}$$ and $$f\left(\frac{1}{3}\right) > \frac{1}{3}$$
$$f\left(\frac{1}{2}\right) < \frac{1}{2}$$ and $$f\left(\frac{1}{3}\right) < \frac{1}{3}$$
$$f\left(\frac{1}{2}\right) > \frac{1}{2}$$ and $$f\left(\frac{1}{3}\right) < \frac{1}{3}$$
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