Question

Let $f$ be a non-negative function in $[0,1]$ and twice differentiable in $(0,1).$ If $\underset{0}{\overset{x}{\int }}\sqrt{1-{\left(f^{\prime }(t)\right)}^2}dt=\underset{0}{\overset{x}{\int }}f(t)dt,0\le x\le 1$ and $f(0)=0$, then $\underset{x\to 0}{\lim }\frac{1}{x^2}\underset{0}{\overset{x}{\int }}f(t)dt:$

• A. equals 0
• B. equals 1
• C. does not exist
• D. equals $\frac{1}{2}$

Answer 1

Answer: (D)

${\int }_0^x\sqrt{1-{\left(f^{\prime }(t)\right)}^2}dt=\underset{0}{\overset{x}{\int }}f(t)dt$
$0\le x\le 1$
differentiating both the sides
$\sqrt{1-{\left(f^{\prime }(x)\right)}^2}=f(x)$
$\Rightarrow 1-{\left(f^{\prime }(x)\right)}^2=f^2(x)$
$\frac{f^{\prime }(x)}{\sqrt{1-f^2(x)}}=1$
${\sin }^{-1}f(x)=x+C$
$\because f(0)=0$
$\Rightarrow C=0$
$\Rightarrow f(x)=\sin x$
Now $\underset{x\to 0}{\lim }\frac{\underset{0}{\overset{x}{\int }}\sin tdt}{x^2}\left(\frac{0}{0}\right)=\frac{1}{2}$