Let f be a non-negative function defined on the interval [0, 1]. If \[\int\limits_0^x {\sqrt {1 - {{\left( {f'\left( t \right)} \right)}^2}} dt = } \int\limits_0^x {f\left( t \right)dt,0 \leqslant x \leqslant 1} \] and f(0) = 0, then-
$
  {\text{A}}{\text{. f}}\left( {\dfrac{1}{2}} \right) < \dfrac{1}{2}{\text{ and f}}\left( {\dfrac{1}{3}} \right) > \dfrac{1}{3} \\
  {\text{B}}{\text{. f}}\left( {\dfrac{1}{2}} \right) > \dfrac{1}{2}{\text{ and f}}\left( {\dfrac{1}{3}} \right) > \dfrac{1}{3} \\
  {\text{C}}{\text{. f}}\left( {\dfrac{1}{2}} \right) < \dfrac{1}{2}{\text{ and f}}\left( {\dfrac{1}{3}} \right) < \dfrac{1}{3} \\
  {\text{D}}{\text{. f}}\left( {\dfrac{1}{2}} \right) > \dfrac{1}{2}{\text{ and f}}\left( {\dfrac{1}{3}} \right) < \dfrac{1}{3} \\
$

Hint: Here in the given question we have to Use Leibnitz’s rule and differentiate the expression. Find f(x) and then take use of standard inequalities, accordingly to reach the answer.

Complete step-by-step answer:
We are given with-
  \[\int\limits_0^x {\sqrt {1 - {{\left( {f'\left( t \right)} \right)}^2}} dt = } \int\limits_0^x {f\left( t \right)dt} \]
According to the Leibnitz rule-
$\dfrac{{d(\int\limits_{a\left( x \right)}^{b\left( x \right)} {f(x)} )}}{{dx}} = f\left( b \right).b'\left( x \right) - f\left( a \right).a'\left( x \right)$
Accordingly,
$
  \dfrac{{d(\int\limits_0^x {\sqrt {1 - {{\left( {f'\left( t \right)} \right)}^2}} dt)} }}{{dx}} = \dfrac{{d(\int\limits_0^x {f\left( t \right)dt} )}}{{dx}} \\
  \sqrt {1 - {{\left( {f'\left( x \right)} \right)}^2}} .\left( 1 \right) - 0 = f\left( x \right).\left( 1 \right) - 0 \\
  \sqrt {1 - {{\left( {f'\left( x \right)} \right)}^2}} = f\left( x \right) \\
  1 - {\left( {f'\left( x \right)} \right)^2} = {(f\left( x \right))^2} \\
  1 - {\left( {f\left( x \right)} \right)^2} = {\left( {f'\left( x \right)} \right)^2} \\
   \pm \sqrt {1 - {{\left( {f\left( x \right)} \right)}^2}} = f'\left( x \right) \\
   \pm \sqrt {1 - {{\left( {f\left( x \right)} \right)}^2}} = \dfrac{{d(f\left( x \right))}}{{dx}} \\
  dx = \pm \dfrac{{d(f\left( x \right))}}{{\sqrt {1 - {{\left( {f\left( x \right)} \right)}^2}} }} \\
$
Consider f(x) as t and then integrate both sides-
$

  dx = \pm \dfrac{{d(f\left( x \right))}}{{\sqrt {1 - {{\left( {f\left( x \right)} \right)}^2}} }} \\
  dx = \pm \dfrac{{d(t)}}{{\sqrt {1 - {{\left( t \right)}^2}} }} \\
  \int {dx} = \pm \int {\dfrac{{d(t)}}{{\sqrt {1 - {t^2}} }}} \\
  x + C = \pm {\sin ^{ - 1}}\left( t \right) \\
  \because t = f\left( x \right) \\
  x + C = \pm {\sin ^{ - 1}}\left( {f\left( x \right)} \right) \\
$
It is given that- f(0) = 0
Substituting x=0,
$
  0 + C = \pm {\sin ^{ - 1}}\left( {f\left( 0 \right)} \right) \\
  C = \pm {\sin ^{ - 1}}\left( 0 \right) \\
  C = 0 \\
  x = \pm {\sin ^{ - 1}}\left( {f\left( x \right)} \right) \\
   \pm \sin \left( x \right) = f\left( x \right) \\
 $
As it has been given that f(x) is a non-negative function in [0, 1]
$\sin \left( x \right) = f\left( x \right)$
From standard equalities we know that,
$\sin x < x,x \in {R^ + }$
$
  \sin \left( {\dfrac{1}{2}} \right) < \dfrac{1}{2} \\
  \sin \left( {\dfrac{1}{3}} \right) < \dfrac{1}{3} \\
  {\text{As both }}\dfrac{1}{2}{\text{ and }}\dfrac{1}{3}{\text{ are lying }}\left[ {0,1} \right] \\
$
The correct option is C.

Note: Whenever we get this type of question the key concept of solving is we have to understand According to the Leibnitz rule-$\dfrac{{d(\int\limits_{a\left( x \right)}^{b\left( x \right)} {f(x)} )}}{{dx}} = f\left( b \right).b'\left( x \right) - f\left( a \right).a'\left( x \right)$ . and we should have also remembered formulae of integration.