JEE Main 2022 · 30 Jul · Shift 2 · Mathematics Q72

← JEE Main 2022 · 30 Jul · Shift 2

MathematicsSingle correct+4 / -1Question 72 of 90

$\displaystyle\lim_{n\to\infty}\sum_{k=1}^n\frac{n(n^2+k^2)}{(n^4+n^2k^2+k^4)}$ is equal to

\dfrac\pi{\sqrt3}

\dfrac{2\pi}{3\sqrt3}

\dfrac\pi{2\sqrt3}

\dfrac\pi3

Answer and worked solution

Official answer: C

Riemann sum. $\displaystyle\int_0^1\frac{1+x^2}{1+x^2+x^4}dx=\int_0^1\frac{1+1/x^2}{(x-1/x)^2+3}dx=\frac1{\sqrt3}\left[\tan^{-1}\frac{x-1/x}{\sqrt3}\right]_0^1=\frac\pi{2\sqrt3}$ .

(C)

Topics:Limit of a sum

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JEE Main 2022 · 30 Jul · Shift 2, Mathematics question 72