Let $$f$$ be a non-negative function defined on the interval [0, 1]. If $$\int\limits_0^x {\sqrt {1 - {{\left( {f'\left( t \right)} \right)}^2}dt} = \int\limits_0^x {f\left( t \right)dx,\,\,\,0 \leqslant x \leqslant 1,} } $$         and $$f\left( 0 \right) = 0,$$   then-

Solution :
Given that $$f$$ is a non negative function defined on $$\left[ {0,\,1} \right]$$  and $$\int\limits_0^x {\sqrt {1 - {{\left( {f'\left( t \right)} \right)}^2}} } dt = \int\limits_0^x {f\left( t \right)dt,\,\,\,\,0 \leqslant x \leqslant 1} $$
Differentiating both sides with respect to $$x,$$ we get
$$\eqalign{ & \sqrt {1 - {{\left[ {f'\left( x \right)} \right]}^2}} = f\left( x \right) \cr & \Rightarrow 1 - {\left[ {f'\left( x \right)} \right]^2} = {\left[ {f\left( x \right)} \right]^2} \cr & \Rightarrow {\left[ {f'\left( x \right)} \right]^2} = 1 - {\left[ {f\left( x \right)} \right]^2} \cr & \Rightarrow \frac{d}{{dx}}f\left( x \right) = \pm \sqrt {1 - {{\left[ {f\left( x \right)} \right]}^2}} \cr & \Rightarrow \pm \frac{{d\,f\left( x \right)}}{{\sqrt {1 - {{\left[ {f\left( x \right)} \right]}^2}} }} = dx \cr} $$
Integrating both sides with respect to $$x,$$ we get
$$ \pm \frac{{d\,f\left( x \right)}}{{\sqrt {1 - {{\left[ {f\left( x \right)} \right]}^2}} }} = \int {dx} \,\,\, \Rightarrow \pm {\sin ^{ - 1}}f\left( x \right) = x + C$$
$$\because $$ Given that $$f\left( 0 \right) = 0\,\,\, \Rightarrow C = 0$$
Hence $$f\left( x \right) = \pm \sin \,x$$
But as $$f\left( x \right)$$  is a non negative function on $$\left[ {0,\,1} \right]$$
$$\therefore f\left( x \right) = \sin \,x$$
Now $$\sin \,x < x,\,\forall \,x > 0$$
$$\therefore \,f\left( {\frac{1}{2}} \right) < \frac{1}{2}{\text{ and }}f\left( {\frac{1}{3}} \right) < \frac{1}{3}$$