Question

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

Differentiating using Libnitz formula, we get

$\sqrt{1-\left(f^{{}^{\prime }}\right(x){)}^2}=f\left(x\right)$

$\sqrt{1-\left(f\right(x){)}^2}=f^{{}^{\prime }}\left(x\right)=\frac{d\left(f\right(x\left)\right)}{dx}$

$dx=\frac{d\left(f\right(x\left)\right)}{\sqrt{1-\left(f\right(x){)}^2}}$

Now, on integarting we get

$x+C={\sin }^{-1}\left(f\right(x\left)\right)$

Now for x to be real, we can say that

$0\le f\left(x\right)<1orf\left(x\right)<\frac{1}{2}andf\left(x\right)<\frac{1}{3}$