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(C) Given $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} dt = \int_0^x f(t) dt$. Applying the Leibniz rule by differentiating both sides with respect

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$f(4/3) < 4/3$ and $f(2/3) < 2/3$

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(C) Given $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} dt = \int_0^x f(t) dt$. Applying the Leibniz rule by differentiating both sides with respect to $x$, we get $\sqrt{1-\left(f^{\prime}(x)\right)^2} = f(x)$. Squaring both sides, $1 - (f^{\prime}(x))^2 = f^2(x)$, which implies $(f^{\prime}(x))^2 = 1 - f^2(x)$. Thus, $f^{\prime}(x) = \pm \sqrt{1 - f^2(x)}$. Separating variables, $\int \frac{df}{\sqrt{1-f^2}} = \pm \int dx$, which gives $\sin^{-1}(f(x)) = \pm x + C$. Since $f(0) = 0$, we have $\sin^{-1}(0) = 0 + C$, so $C = 0$. Thus, $f(x) = \sin(x)$ or $f(x) = -\sin(x)$. Since $f$ is non-negative on $[0, \pi/2]$, we must have $f(x) = \sin(x)$. We know that for $x > 0$, $\sin(x) Therefore, $f(4/3) = \sin(4/3) < 4/3$ and $f(2/3) = \sin(2/3) < 2/3$.

Let $f$ be a non-negative function defined in $[0, \pi / 2]$, $f^{\prime}$ exists and is continuous for all $x$, and $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} dt = \int_0^x f(t) dt$ with $f(0) = 0$. Then

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