Let f be a non-negative function defined on t | vedclass

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(C) Given the equation $\int_0^x \sqrt{1-(f'(t))^2} dt = \int_0^x f(t) dt$ for $0 \leq x \leq 1$.Applying the Leibniz integral rule by differentiating

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$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$

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(C) Given the equation $\int_0^x \sqrt{1-(f'(t))^2} dt = \int_0^x f(t) dt$ for $0 \leq x \leq 1$.Applying the Leibniz integral rule by differentiating both sides with respect to $x$,we get:$\sqrt{1-(f'(x))^2} = f(x)$Squaring both sides:$1-(f'(x))^2 = f^2(x)$$(f'(x))^2 = 1 - f^2(x)$$f'(x) = \pm \sqrt{1 - f^2(x)}$Let $y = f(x)$,then $\frac{dy}{dx} = \pm \sqrt{1 - y^2}$.Separating variables:$\frac{dy}{\sqrt{1 - y^2}} = \pm dx$Integrating both sides:$\sin^{-1}(y) = \pm x + C$Since $f(0) = 0$,we have $\sin^{-1}(0) = 0 + C$,so $C = 0$.Thus,$y = \pm \sin(x)$. Since $f$ is a non-negative function,$f(x) = \sin(x)$.We know that for $x > 0$,$\sin(x) Therefore,$\sin\left(\frac{1}{2}\right) Hence,$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$.

Let $f$ be a non-negative function defined on the interval $[0,1]$. If $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} dt = \int_0^x f(t) dt$ for $0 \leq x \leq 1$ and $f(0)=0$,then:

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