JEE Main 2024 — Mathematics Calculus

$\displaystyle \begin{aligned} & \sum_{\mathrm{r}=1}^{\infty} \frac{\mathrm{n}}{\sqrt{\mathrm{n}^4+\mathrm{r}^4}}-\frac{2 \mathrm{nr}^2}{\left(\mathrm{n}^2+\mathrm{r}^2\right) \sqrt{\mathrm{n}^4+\mathrm{r}^4}} \\ & \sum_{\mathrm{r}=1}^{\infty} \frac{\frac{1}{\mathrm{n}}}{\sqrt{1+\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^4}}-\frac{2\left(\frac{1}{\mathrm{n}}\right)\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^2}{\left(1+\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^2\right) \sqrt{1+\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^4}} \\ & \Rightarrow \int_0^1 \frac{\mathrm{dx}}{\sqrt{1+\mathrm{x}^4}}-\frac{2 \mathrm{x}^2 \mathrm{dx}}{\left(1+\mathrm{x}^2\right) \sqrt{1+\mathrm{x}^4}} \\ & \Rightarrow \int_0^1 \frac{1-\mathrm{x}^2}{\left(1+\mathrm{x}^2\right) \sqrt{1+\mathrm{x}^4}} d x \\ & \Rightarrow \int_0^1 \frac{\frac{1}{\mathrm{x}^2}-1}{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right) \sqrt{\mathrm{x}^2+\frac{1}{\mathrm{x}^2}}} \mathrm{dx}\end{aligned}$​r=1∑∞​n4+r4​n​−(n2+r2)n4+r4​2nr2​r=1∑∞​1+(nr​)4​n1​​−(1+(nr​)2)1+(nr​)4​2(n1​)(nr​)2​⇒∫01​1+x4​dx​−(1+x2)1+x4​2x2dx​⇒∫01​(1+x2)1+x4​1−x2​dx⇒∫01​(x+x1​)x2+x21​​x21​−1​dx​ $\displaystyle \begin{aligned} & \Rightarrow-\int_0^1 \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-2}} d x \\ & x+\frac{1}{x}=t \Rightarrow 1-\frac{1}{x^2} d x=d t\end{aligned}$​⇒−∫01​(x+x1​)(x+x1​)2−2​1−x21​​dxx+x1​=t⇒1−x21​dx=dt​ $\displaystyle \begin{aligned} & \Rightarrow-\int_{\infty}^2 \frac{\mathrm{dt}}{\mathrm{t} \sqrt{\mathrm{t}^2-2}} \\ & \Rightarrow-\int_{\infty}^2 \frac{\mathrm{tdt}}{\mathrm{t}^2 \sqrt{\mathrm{t}^2-2}} \\ & \text { take } \mathrm{t}^2-2=\alpha^2 \\ & \mathrm{tdt}=\alpha \mathrm{d} \alpha \\ & \Rightarrow-\int_{\infty}^{\sqrt{2}} \frac{\alpha \mathrm{d} \alpha}{\left(\alpha^2+2\right) \alpha} \\ & \Rightarrow-\int_{\infty}^{\sqrt{2}} \frac{\mathrm{d} \alpha}{\alpha^2+2}\end{aligned}$​⇒−∫∞2​tt2−2​dt​⇒−∫∞2​t2t2−2​tdt​ take t2−2=α2tdt=αdα⇒−∫∞2​​(α2+2)ααdα​⇒−∫∞2​​α2+2dα​​ $\begin{aligned}
& \left.\Rightarrow \frac{-1}{\sqrt{2}} \tan ^{-1} \frac{\alpha}{\sqrt{2}}\right]_{\infty}^{\sqrt{2}} \
& \Rightarrow \frac{-1}{\sqrt{2}}\left{\tan ^{-1} 1\right}+\frac{1}{\sqrt{2}} \tan ^{-1} \infty \
& \Rightarrow \frac{1}{\sqrt{2}}\left{\frac{\pi}{2}-\frac{\pi}{4}\right} \
& \Rightarrow \frac{\pi}{4 \sqrt{2}}=\frac{\pi}{\mathrm{K}}
\end{aligned}$

So $\mathrm{K}=4 \sqrt{2}$K=42​ $\mathrm{K}^2=32$K2=32