Let _ n arrow ∞ ( n n^4+1 - 2 n (n^2+1) n^4+1 + n n^4+16 - 8 n (n^2+4) n^4+16 . .+ + n n^4+n^4 - 2 n · n^2 (n^2+n^2) n^4+n^4 ) be (π)/(k) , using only the principal values of the inverse trigonometric functions. Then

Correct answer: 0 aligned & ∑_ r=1 ^ ∞ n n^4+r^4 - 2 nr^2 (n^2+r^2) n^4+r^4 & ∑_ r=1 ^ ∞ 1 n 1+( r n )^4 - 2( 1 n )( r n )^2 (1+( r n )^2) 1+( r n )^4 & ∫_0^1 dx 1+x^4 - 2 x^2 dx (1+x^2) 1+x^4 & ∫_0^1 1-x^2 (1+x^2) 1+x^4 d x & ∫_0^1 1 x^2 -1 (x+ 1 x ) x^2+ 1 x^2 dx aligned aligned & -∫_0^1 1-(1)/(x^2) (x+(1)/(x)) (x+(1)/(x))^2-2 d x & x+(1)/(x)=t 1-(1)/(x^2) d x=d t aligned aligned & -∫_ ∞ ^2 dt t t^2-2 & -∫_ ∞ ^2 tdt t^2 t^2-2 & take t^2-2=α^2 & tdt=α d α & -∫_ ∞ ^ 2 α d α (α^2+2) α & -∫_ ∞ ^ 2 d α α^2+2 aligned aligned & . -1 2 ^ -1 α 2 ]_ ∞ ^ 2 & -1 2 \ ^ -1 1\ + 1 2 ^ -1 ∞ & 1 2 \ (π)/(2)-(π)/(4)\ & π 4 2 = π K aligned So K=4 2 K^2=32

JEE Main 20249 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 9 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2024 · 9 April · Morning Shift · Mathematics · Limits

Let _ n arrow ∞ ( n n^4+1 - 2 n (n^2+1) n^4+1 + n n^4+16 - 8 n...

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+\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)

\frac{\pi}{k}

, using only the principal values of the inverse trigonometric functions. Then

\mathrm{k}^2

is equal to ________

Correct answer

Step-by-step solution

\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}

\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}

\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}

\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} \mathrm{K}^2=32$

Practice Limits on Quantrex Academy →

More from Limits

Let _ x → 2 (( (x-2))(rx^2 + (p-2)x - 2p))/((x-2)^2) = 5 for some r, p ∈ ℝ . If the set… JEE Main 2026 The value of _ x → 0 ((x^2 ^2 x)/(x^2 - ^2 x)) is: JEE Main 2026 Let f(x) = _ y → 0 ((1 - (xy)) (xy))/(y^3) . Then the number of solutions of the equation… JEE Main 2026 The product of all possible values of α , for which _ x → 0 ((1 - (α x) ((α+1)x)… JEE Main 2026 If _ x → 2 (x^3 - 5x^2 + ax + b) ( x-1 - 1) _e(x-1) = m , then a + b + m is equal to : JEE Main 2026 The value of _ x arrow 0 _ e ( (e x) · (e^ 2 x) · … · (e^ 10 x)) e^ 2 -e^ 2 x is equal to JEE Main 2026 If _ x arrow 0 e^ (a-1) x +2 ~b x+(c-2) e^ -x x x- _ e (1+x) =2 , then a^ 2 +b^ 2 +c^ 2… JEE Main 2026 Let [·] denote the greatest integer function and f(x)= _ n arrow ∞ 1 n^ 3 ∑_ k=1 ^ n [ k^… JEE Main 2026 Let f: R arrow(0, ∞) be a twice differentiable function such that f(3)=18, f^ (3)=0 and… JEE Main 2026 Let α and β be the real numbers such that _ x arrow 0 (1)/(x^3)((α)/(2) ∫_0^x (1)/(1-t^2)… JEE Advanced 2025 Let x_0 be the real number such that e^ x_0 +x_0=0 . For a given real number α , define… JEE Advanced 2025 Given below are two statements : Statement I : _ x arrow 0 ( ^ -1 x+ _e (1+x)/(1-x) -2 x… JEE Main 2025 If the function f(x)=( ( x)- ( x))/( x- x) is continuous at x=0 , then f(0) is equal to… JEE Main 2025 _ x arrow 0^ + (5(x)^ (1)/(3) ) _e(1+3 x^2) ( ^ -1 3 x )^2(e^ 5(x)^ (4)/(3) -1) is equal… JEE Main 2025 If _ x arrow 1^ + ((x-1)(6+λ (x-1))+μ (1-x))/((x-1)^3)=-1 , where λ, μ ∈ ℝ , then λ+μ is… JEE Main 2025 If Lim_ x arrow 0 (( x)/(x))^ (1)/(x^2) =p , then 96 _e p is equal to ______ JEE Main 2025 Full Limits list Limits JEE Maths PYQs JEE Mathematics PYQs All JEE Main PYQs