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JEE MATHEMATICS MCQ: INCREASING & DECREASING

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Analyzing +4 | -1
If ∫0x2xf2(t)dt=(∫0x2f(xβˆ’t)dt)2\int_0^x 2 x f^2(t) d t=\left(\int_0^x 2 f(x-t) d t\right)^2 for f(1)=1f(1)=1 and f(x)f(x) is continuous function for x>0x>0 and {an}\left\{a_n\right\} is a sequence such that an+1=an+1+an2a_{n+1}=a_n+\sqrt{1+a_n^2} for a=0a=0; if f(x)f(x) is an increasing function, then lim⁑nβ†’βˆžak2nβˆ’1=\lim _{n \rightarrow \infty} \frac{a_k}{2^{n-1}}= (where k=f(n2βˆ’1))\left.k=f\left(n^{\sqrt{2}-1}\right)\right) is
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Correct option: B
4/Ο€4 / \pi
Option B (4/Ο€4 / \pi) is correct. After solving the integral equation for f(x)f(x) and using the recurrence relation for ana_n, we find the limit to be 4/Ο€4/\pi.

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