Let f be a non-negative function defined on the interval [0,1]. If

$\int\limits_0^x \sqrt{1-\left\{f^{\prime}(t)\right\}^2} d t=\int\limits_0^x f(t) d t, 0 \leq x \leq 1$ and $f(0)=0$, then

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

We have,

$\int\limits_0^x \sqrt{1-\left\{f'(t)\right\}^2} d t=\int\limits_0^x f(t) d t$

Differentiating w.r. to $x$, we get

$\sqrt{1-\left\{f'(x)\right\}^2}=f(x)$

$\Rightarrow \left\{f'(x)\right\}^2=1-\{f(x)\}^2$

$\Rightarrow f'(x)= \pm \sqrt{1-\{f(x)\}^2}$

$\Rightarrow \frac{f'(x)}{\sqrt{1-\{f(x)\}^2}}= \pm 1$

$\Rightarrow \int\limits \frac{f'(x)}{\sqrt{1-\{f(x)\}^2}} d x= \pm \int\limits 1 . d$

$\Rightarrow \sin ^{-1}\{f(x)\}= \pm x+C$

It is given that $f(0)=0$. Therefore, $C=0$.

∴  $\sin ^{-1}\{f(x)\}= \pm x$

But, $f(x)>0$ for all $x \in[0,1]$.

∴  $\sin ^{-1}\{f(x)\}=x \Rightarrow f(x)=\sin x$

We know that

$\sin x<x$ for all $x>0$

$\Rightarrow f(x)<x$ for all $x>0$

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