Q92 / 155 · WBJEE 2022 · Mathematics
Let $f$ be a non-negative function defined in $[0, \pi / 2], f^{\prime}$ exists and be continuous for all $x$ and $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t$ and $f(0)=0$. Then
- A$f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) > \frac{1}{3}$
- B$f\left(\frac{1}{2}\right) > \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$
- Correct answerC$f\left(\frac{4}{3}\right) < \frac{4}{3}$ and $f\left(\frac{2}{3}\right) < \frac{2}{3}$
- D$\mathrm{f}\left(\frac{4}{3}\right) > \frac{4}{3}$ and $\mathrm{f}\left(\frac{2}{3}\right) > \frac{2}{3}$
Solution
$\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t$
Using Leibnitz Rule,
$\begin{aligned}
&\sqrt{1-\left(f^{\prime}(x)\right)^2}=f(x) \\
&\therefore f^{\prime}(x)=\pm \sqrt{1-f^2(x)} \Rightarrow \int \frac{d f(x)}{\sqrt{1-f^2(x)}}=\pm \int d x \\
&\sin ^{-1}(f(x))=\pm x+c \\
&\therefore f(x)=\sin (\pm x+C) \\
&\because f(0)=0 \Rightarrow C=0 \\
&\therefore f(x)=\sin x \text { or }-\sin x
\end{aligned}$
But $f$ is non-negative on $[0, \pi / 2]$
$\begin{aligned}
&\therefore \mathrm{f}(\mathrm{x})=\operatorname{Sin} \mathrm{x} \\
&\because \mathrm{f}(\mathrm{x}) < \mathrm{x} \text { for all } \mathrm{x} > 0 \\
&\therefore \mathrm{f}\left(\frac{4}{3}\right) < \frac{4}{3} \text { and } \mathrm{f}\left(\frac{2}{3}\right) < \frac{2}{3}
\end{aligned}$
Hence, the correct answer is option: 'C'.