Correct option: C.$f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$
c
Given $I \sqrt{ }\left(1-( f ( t ))^2\right)$ for int $0 \rightarrow x d t=\int f(t) d t$, for int $0 \rightarrow x, 0 \leq x \leq 1$
Apph'ing Leibnitz theorem, we get
$\sqrt{1-\left(f^{\prime}(x)\right)^2}=f(x)$
$\Rightarrow \quad 1-\left(f^{\prime}(x)\right)^2=f^2(x)$
$\Rightarrow \quad\left(f^{\prime}(x)\right)^2=1-f^2(x)$
$\Rightarrow \quad f^{\prime}(x)= \pm \sqrt{1-f^2(x)}$
$\Rightarrow \quad \frac{d y}{d x}= \pm \sqrt{1-y^2} \text {, where } y=f(x)$
$\Rightarrow \quad \frac{d y}{\sqrt{1-y^2}}= \pm d x$
On integrating both sides, we get
$ \sin ^{-1}(y) $$ = \pm x+C$
$\because $$ f(0) $$ =0 \Rightarrow C=0$
$\therefore $$ y $$ = \pm \sin x$
$y=\sin x=f(x) \text { given } f(x) \geq \text { ofor } x \in[0,1]$
It is known that $\sin x<x, \forall x \in R^{+}$
$\therefore \sin \left(\frac{1}{2}\right)<\frac{1}{2} \Rightarrow f\left(\frac{1}{2}\right)<\frac{1}{2} \text { and } \sin \left(\frac{1}{3}\right)<\frac{1}{3}$
$\Rightarrow \quad f\left(\frac{1}{3}\right)<\frac{1}{3}$