Explanation

(c)
Given, $\int_{0}^{x}\sqrt{1 - \left( f^{'}(t) \right)^{2}}\ dt = \int_{0}^{x}{f(t)}\ dt,\ 0 \leq x \leq 1\$
Applying Leibnitz theorem, we get
$\sqrt{1 - \left( f^{'}(x) \right)^{2}} = f(x)$
$\Rightarrow 1 - \left( f^{'}(x) \right)^{2} = f^{2}(x)$
$\Rightarrow \ \left( f^{'}(x) \right)^{2} = 1 - f^{2}(x)$
$\Rightarrow f^{'}(x) = \pm \sqrt{1 - f^{2}(x)}$
$\Rightarrow \ \frac{dy}{dx} = \pm \sqrt{1 - y^{2}},\ where\ y = f(x)$
$\Rightarrow \ \frac{dy}{\sqrt{1 - y^{2}}} = \pm dx$
On integrating both sides, we get
$\sin^{- 1}{(y) = \pm x + C}$
$\because f(0) = 0 \Rightarrow C = 0$
$\therefore y = \pm \sin x$
$y = \sin x = f(x)\$given$\ f(x) \geq 0\$for$\ x \in \lbrack 0,1\rbrack$
It is known that $\sin{x < x,}\forall\ x \in R^{+}$
$\therefore\sin{\left( \frac{1}{2} \right) < \frac{1}{2} \Rightarrow}f\left( \frac{1}{2} \right) < \frac{1}{2}\ and\ \sin{\left( \frac{1}{3} \right) < \frac{1}{3}}$
$\Rightarrow \ f\left( \frac{1}{3} \right) < \frac{1}{3}$