Question

Let f be a non-negative function defined on the interval $[0,1]$. If ${\int }_0^x\sqrt{1-\left(f^{\prime }\right(t){)}^2}dt={\int }_0^{x}f\left(t\right)dt,0\le x\le 1$, and $f\left(0\right)=0$, 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 $f\left(\frac{1}{3}\right)>\frac{1}{3}$
• 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 $f\left(\frac{1}{3}\right)<\frac{1}{3}$

${\int }_0^x\sqrt{1-\left(f^{\prime }\right(t){)}^2}dt={\int }_0^{x}f\left(t\right)dt$
Differentiating both sides using Leibnitz rule,
$f\left(x\right)=\sqrt{1-{\left(f^{\prime }\right(x\left)\right)}^2}$
$\Rightarrow f\left(x\right)=\sin x$
And we know , $x>\sin x\forall x>0$.
$\Rightarrow f\left(x\right)<x$
Hence option 'C' is correct choice.