Question

Let $f$ be a non-negative function in $[0,1]$ and twice differentiable in $(0,1) .$ If $\int\limits_{0}^{x} \sqrt{1-\left(f'(t)\right)^{2}} dt =\int\limits_{0}^{x} f ( t ) dt, 0 \leq x \leq 1$ and $f(0)=0$, then $\displaystyle\lim _{x \rightarrow 0} \frac{1}{x^{2}} \int\limits_{0}^{x} f(t) d t:$

• 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}} d t=\int\limits_{0}^{x} f(t) d t $
$0 \leq x \leq 1$
differentiating both the sides
$\sqrt{1-\left(f'(x)\right)^{2}}=f(x)$
$\Rightarrow 1-\left(f'(x)\right)^{2}=f^{2}(x)$
$\frac{f'(x)}{\sqrt{1-f^{2}(x)}}=1$
$\sin ^{-1} f(x)=x+C$
$\because f(0)=0 $
$\Rightarrow C=0 $
$\Rightarrow f(x)=\sin x$
Now $\displaystyle\lim _{x \rightarrow 0} \frac{\int\limits_{0}^{x} \sin t d t}{x^{2}}\left(\frac{0}{0}\right)=\frac{1}{2}$