Let f be a non-negative function in [0,1] and twice
Question:

Let $\mathrm{f}$ be a non-negative function in $[0,1]$ and twice differentiable in $(0,1) .$ If $\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} d t=\int_{0}^{x} f(t) d t$

$0 \leq x \leq 1$ and $f(0)=0$, then $\lim _{x \rightarrow 0} \frac{1}{x^{2}} \int_{0}^{x} f(t) d t:$

  1. equals 0

  2. equals 1

  3. does not exist

  4. equals $\frac{1}{2}$


Correct Option: , 4

Solution:

$\int_{0}^{x} \sqrt{1-\left(f^{\prime}(t)\right)^{2}} \mathrm{dt}=\int_{0}^{x} f(t)$ dt $\quad 0 \leq x \leq 1$

differentiating both the sides

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

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

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

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

$\because f(0)=0 \Rightarrow \mathrm{C}=0 \Rightarrow f(\mathrm{x})=\sin \mathrm{x}$

Now $\lim _{x \rightarrow 0} \frac{\int_{0}^{x} \sin t d t}{x^{2}}\left(\frac{0}{0}\right)=\frac{1}{2}$

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