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Let $$\int_0^x f^2(t)\;dt=\left( \int_0^x 2f(x-t)\;dt \right)^2, \quad f(1)=1$$ function $f(x)$ is continuous for $x>0$, $\{a_n\}$ is a sequence such that $a_{n+1}=a_n+\sqrt{1+a_n^2}$ for $a_0=0$. If $f(x)$ is an increasing function, then $$\lim_{n\to \infty} \frac{a_k}{2^{n-1}}$$ where $k=f(n^{\sqrt{2}-1})$ equals to $$(\text{A}) \frac{\pi}{4}\qquad (\text{B}) \frac{\pi}{8}\qquad (\text{C}) \frac{4}{\pi}\qquad (\text{D}) \frac{8}{\pi}$$

Please give some hint. How to approach to solve the question? I tried differentiating the function on both sides by applying the Newton-Leibniz formula but I was not able to solve further.

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    Consider posting how did you approach or what all have you tried, so that other users will be able to help you better. – 2017-01-29
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    Note that $k = f(n^{\sqrt{2} - 1})$ is not necessarily an integer… But, at least, you can figure out the unique function $f$ which satisfy the condition on the integrals. (tip: differential equation of the first order, differentiate and look for relations) – 2017-01-29
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    I tried to solve but not able to find any good relations among those functions.Please can you provide any other hint ? – 2017-02-02
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    @Raito plz give me some sort of strong hint. I found $f(x)$=$e^{8(x-1)}$ is that correct.plz Tell me how to find the sequence. – 2017-02-18
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    I think it is not correct. You can write the calculations to check them. – 2017-02-20

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