Let lim n rightarrow ∞ (frac n sqrt n 4 +1 -frac 2 n (n 2 +1 ) sqrt n 4 +1 +frac n sqrt n 4 +16 -frac 8 n (n 2 +4 ) sqrt n 4 +16 . .+ldots ldots .+frac n sqrt n 4 +n 4 -frac 2 n × n 2 (n 2 +n 2 ) sqrt n 4 +n 4 ) be π/k , using only the principal values of the inverse trigonometric functions. Then k 2 is equal to

Mathematics · Definite Integration

JEE Main 2024 — 9 April, Shift 1 — Question 24

Let $\lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^{4}+1}}-\frac{2 n}{\left(n^{2}+1\right) \sqrt{n^{4}+1}}+\frac{n}{\sqrt{n^{4}+16}}-\frac{8 n}{\left(n^{2}+4\right) \sqrt{n^{4}+16}}\right.$

$\left.+\ldots \ldots .+\frac{n}{\sqrt{n^{4}+n^{4}}}-\frac{2 n \cdot n^{2}}{\left(n^{2}+n^{2}\right) \sqrt{n^{4}+n^{4}}}\right)$ be $\frac{\pi}{k}$ , using only the principal values of the inverse trigonometric

functions. Then $\mathrm{k}^{2}$ is equal to $\qquad$

Answer: 32

Numerical answer — enter this value.

Step-by-step solution

\sum_{r=1}^{\infty}\left(\frac{n}{\sqrt{n^{4}+r^{4}}} -\frac{2nr^{2}}{(n^{2}+r^{2})\sqrt{n^{4}+r^{4}}}\right)

=\sum_{r=1}^{\infty}\left(\frac{\tfrac{1}{n}}{\sqrt{1+\left(\tfrac{r}{n}\right)^{4}}} -\frac{2\left(\tfrac{1}{n}\right)\left(\tfrac{r}{n}\right)^{2}} {\left(1+\left(\tfrac{r}{n}\right)^{2}\right)\sqrt{1+\left(\tfrac{r}{n}\right)^{4}}}\right)

\Rightarrow \int_{0}^{1}\left(\frac{dx}{\sqrt{1+x^{4}}} -\frac{2x^{2}\,dx}{(1+x^{2})\sqrt{1+x^{4}}}\right)

\Rightarrow \int_{0}^{1}\frac{1-x^{2}}{(1+x^{2})\sqrt{1+x^{4}}}\,dx

\Rightarrow \int_{0}^{1}\frac{\tfrac{1}{x^{2}}-1} {\left(x+\tfrac{1}{x}\right)\sqrt{x^{2}+\tfrac{1}{x^{2}}}}\,dx

\Rightarrow -\int_{0}^{1}\frac{1-\tfrac{1}{x^{2}}} {\left(x+\tfrac{1}{x}\right)\sqrt{\left(x+\tfrac{1}{x}\right)^{2}-2}}\,dx

\begin{aligned} &\Rightarrow -\int_{\infty}^{2}\frac{dt}{t\sqrt{t^{2}-2}} \\ &\Rightarrow -\int_{\infty}^{2}\frac{t\,dt}{t\sqrt{t^{2}-2}} \\ &\text{Let } t^{2}-2=\alpha \\ &\Rightarrow t\,dt=\alpha\,d\alpha \\ &\Rightarrow -\int_{\infty}^{\sqrt{2}}\frac{\alpha\,d\alpha}{(\alpha^{2}+2)\alpha} \end{aligned}

\Rightarrow \frac{1}{\sqrt{2}}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)

\Rightarrow \frac{\pi}{4\sqrt{2}}=\frac{\pi}{K}

K=4\sqrt{2}, \qquad K^{2}=32

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Mathematics
Chapter: Definite Integration
Topic: Integration as a limit of sum