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

Given \(\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t\),
\[
0 \leq x \leq 1
\]
Applying Leibnitz theorem, we get
\[
\begin{array}{rlrl}
& & \sqrt{1-\left(f^{\prime}(x)\right)^2} & =f(x) \\
\Rightarrow & & 1-\left(f^{\prime}(x)\right)^2 & =f^2(x) \\
\Rightarrow & & \left(f^{\prime}(x)\right)^2 & =1-f^2(x) \\
\Rightarrow & & f^{\prime}(x) & =\pm \sqrt{1-f^2(x)} \\
\Rightarrow & & \quad \frac{d y}{d x} & =\pm \sqrt{1-y^2} \\
& \text { where } y=f(x) \Rightarrow & \frac{d y}{\sqrt{1-y^2}}=\pm d x
\end{array}
\]
On integrating both sides, we get
\[
\begin{aligned}
& \sin ^{-1}(y)=\pm x+C \\
& \because \quad f(0)=0 \Rightarrow C=0 \Rightarrow y=\pm \sin x \\
& y=\sin x=f(x) \text { given } f(x) \geq 0 \text { for } \\
& x \in[0,1]
\end{aligned}
\]
It is known that \(\sin x < x, \forall x \in R^{+}\)
\(\therefore \quad \sin \left(\frac{1}{2}\right) < \frac{1}{2} \Rightarrow f\left(\frac{1}{2}\right) < \frac{1}{2}\)
and \(\sin \left(\frac{1}{3}\right) < \frac{1}{3} \Rightarrow f\left(\frac{1}{3}\right) < \frac{1}{3}\)