Question

Let $f$ be a non-negative function defined on the interval $[0,1]$.If ${\int }_0^x\sqrt{1-\left(f^{\prime }\right(t){)}^2}dt={\int }_0^{x}f\left(t\right)dt$ for $0\le x\le 1$ and $f\left(0\right)=0$ then which of the following is true?

• 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 $f\left(\frac{1}{3}\right)>\frac{1}{3}$
• C. $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$
• D. $f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

Answer 1

The correct option is C $f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$

Given ${\int }_0^x{\sqrt{1-f^{\prime }\left(t\right)}}^{2}dt={\int }_0^{x}f\left(t\right)dt,0\le x\le 1$
Differentiating both sides w.r.t x by using leinitz rule, we get
${\int }_0^x\sqrt{1-{\left(f^{\prime }\left(t\right)\right)}^2}=f\left(x\right)\Rightarrow f^{\prime }\left(x\right)=\pm \sqrt{1-{\left(f\left(x\right)\right)}^2}$
$\Rightarrow \int \frac{f^{\prime }\left(x\right)}{\sqrt{1-{\left(f\left(x\right)\right)}^2}}dx=\pm \int dx\Rightarrow {\sin }^{-1}\left(f\left(x\right)\right)=\pm x+c$
Put $x=0\Rightarrow {\sin }^{-1}\left(f\left(0\right)\right)=c$
$\Rightarrow c={\sin }^{-1}\left(0\right)=0(\because f\left(0\right)=0)\therefore f\left(x\right)=\pm \sin x$
But $f\left(x\right)\ge 0,\forall x\in [0,1]$
$\therefore f\left(x\right)=\sin x$
$\therefore \sin \left(\frac{1}{2}\right)<\frac{1}{2}$ and $\sin \left(\frac{1}{3}\right)<\frac{1}{3}$
$\Rightarrow f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$