Question

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

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

Answer 1

The correct option is D Equals $\frac{1}{2}$

$\underset{0}{\overset{x}{\int }}\sqrt{1-\left(f^{\prime }\right(t){)}^2}dt=\underset{0}{\overset{x}{\int }}f\left(t\right)dt,0\le x\le 1$
On differentiating both sides, we get
$\sqrt{1-\left(f^{\prime }\right(x){)}^2}=f\left(x\right)\Rightarrow 1-\left(f^{\prime }\right(x){)}^2=(f\left(x\right){)}^2\Rightarrow 1-\left(f\right(x){)}^2=(f^{\prime }\left(x\right){)}^2\Rightarrow \frac{dy}{dx}=\sqrt{1-y^2}\Rightarrow \int \frac{dy}{\sqrt{1-y^2}}=\int dx\Rightarrow {\sin }^{-1}y=x+c\because f\left(0\right)=0\Rightarrow c=0\therefore y=f\left(x\right)=\sin x$
Now,
$\underset{x\to 0}{lim}\frac{1}{x^2}\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
Using L'Hospital's Rule
$=\underset{x\to 0}{lim}\frac{f\left(x\right)}{2x}=\underset{x\to 0}{lim}\frac{\sin x}{2x}=\frac{1}{2}$