Let $\mathrm{f}$ be a non-negative function in $[0,1]$ and twice differentiable in $(0,1) .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} d t=\int_{0}^{x} f(t) d t$
$0 \leq x \leq 1$ and $f(0)=0$, then $\lim _{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} f(t) d t:$
Correct Option: , 4
$\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \mathrm{dt}=\int_{0}^{x} f(t)$ dt $\quad 0 \leq x \leq 1$
differentiating both the sides
$\sqrt{1-\left(f^{\prime}(x)\right)^{2}}=f(x)$
$\Rightarrow 1-\left(f^{\prime}(x)\right)^{2}=f^{2}(x)$
$\frac{f^{\prime}(x)}{\sqrt{1-f^{2}(x)}}=1$
$\sin ^{-1} f(x)=x+C$
$\because f(0)=0 \Rightarrow \mathrm{C}=0 \Rightarrow f(\mathrm{x})=\sin \mathrm{x}$
Now $\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sin t d t}{x^{2}}\left(\frac{0}{0}\right)=\frac{1}{2}$
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