For any positive integer n , let S _ n : ( 0, ∞ ) → R be defined by S _ n (x) = ∑ _ k = 1 ^ n ^ - 1 ( 1 + k ( k + 1 ) x ^ 2 x ) where for any x ∈ R, ^ - 1 ( x ) ∈ ( 0, π ) and ^ - 1 ( x ) ∈ (- (π)/(2), (π)/(2)) . Then

Correct answer: A. S _ 10 (x) = (π)/(2) - ^ - 1 ( 1 + 11 x ^ 2 10x ) , for all x > 0 S _ n (x) = ∑ _ k = 1 ^ n ^ - 1 ( 1 + k ( k + 1 ) x ^ 2 x ) = ∑ _ k = 1 ^ n ^ - 1 ((1 + kx ( k + 1 ) x)/(( k + 1 ) x - kx)) = ∑ _ k = 1 ^ n ^ - 1 ((( k + 1 ) x - kx)/(1 + kx ( k + 1 ) x)) = ∑ _ k = 1 ^ n [ ^ - 1 ( k + 1 ) x - ^ - 1 kx] = [ ^ - 1 ( n + 1 ) x - ^ - 1 nx] + + [ ^ - 1 3x - ^ - 1 2x] + [ ^ - 1 2x - ^ - 1 x] = ^ - 1 ( n + 1 ) x - ^ - 1 x = ^ - 1 ( ( n + 1 ) x - x 1 + ( n + 1 ) x ^ 2 ) = ^ - 1 ( nx 1 + ( n + 1 ) x ^ 2 ) 1. S _ 10 (x) = ^ - 1 ( 10x 1 + 11 x ^ 2 ) = (π)/(2) - ^ - 1 ( 10x 1 + 11 x ^ 2 ) = (π)/(2) - ^ - 1 ( 1 + 11 x ^ 2 10x ) 2. _ n → ∞ ( S _ n ( x )) = _ n → ∞ ( ^ - 1 ( nx 1 + ( n + 1 ) x ^ 2 )) = _ n → ∞ ( ^ - 1 ( 1 + ( n + 1 ) x ^ 2 nx )) = _ n → ∞ ( 1 + ( n + 1 ) x ^ 2 nx ) = x ^ 2 x = x 3. S _ 3 (x) = ^ - 1 ( 3x 1 + 4 x ^ 2 ) = (π)/(4) ⇒ ( 3x 1 + 4 x ^ 2 ) = tan(π)/(4) ⇒ 3x 1 + 4 x ^ 2 = 1 ⇒ 1 + 4 x ^ 2 = 3x ⇒ 4 x ^ 2 - 3x + 1 = 0 D Hence, no real roots. 4. ( S _ n ( x )) = ( ^ - 1 ( nx 1 + ( n + 1 ) x ^ 2 )) = ( nx 1 + ( n + 1 ) x ^ 2 ) ( nx 1 + ( n + 1 ) x ^ 2 ) ≤ (1)/(2) ⇒ 2nx ≤ 1 + ( n + 1 ) x ^ 2 ⇒ 2nx ≤ ( n + 1 ) x ^ 2 + 1 ⇒ ( n + 1 ) x ^ 2 - 2nx + 1 ≥ 0 n ≥ 1; x > 0 Let, y = ( n + 1 ) x ^ 2 - 2nx + 1 D = 4 n ^ 2 - 4 ( n + 1 ) and n ∈ N D Hence, no solution if n = 1 .

JEE Advanced 2021Paper 1

Exam: JEE Advanced
Year: 2021
Paper: Paper 1
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Advanced 2021 · Paper 1 · Mathematics · Inverse Trigonometric Functions

For any positive integer n , let S _ n : ( 0, ∞ ) → R be defined by...

For any positive integer

n

, let

{S}_{n}: ( 0, \infty ) \to R

be defined by

{S}_{n}\left(x\right) = \mathop{∑}_{k = 1}^{n}\cot^{- 1}\left(\dfrac{1 + k ( k + 1 ) {x}^{2}}{x}\right)

where for any

x \in R, \cot^{- 1} ( x ) \in ( 0, \pi )

and

\tan^{- 1} ( x ) \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)

. Then which of the following statements is (are) TRUE ?

Options

A ${S}_{10}\left(x\right) = \dfrac{\pi}{2} - \tan^{- 1}\left(\dfrac{1 + 11{x}^{2}}{10x}\right)$ , for all $x > 0$
B $\lim_{n \to \infty}\cot \left({S}_{n} ( x )\right) = x$ , for all $x > 0$
CThe equation ${S}_{3} ( x ) = \dfrac{\pi}{4}$ has a root in $( 0, \infty )$
D $\tan \left({S}_{n} ( x )\right) \le \dfrac{1}{2}$ , for all $n \ge 1$ and $x > 0$

Correct answer

A. ${S}_{10}\left(x\right) = \dfrac{\pi}{2} - \tan^{- 1}\left(\dfrac{1 + 11{x}^{2}}{10x}\right)$ , for all $x > 0$

Step-by-step solution

{S}_{n}\left(x\right) = \mathop{∑}_{k = 1}^{n}\cot^{- 1}\left(\dfrac{1 + k ( k + 1 ) {x}^{2}}{x}\right) $$ = \mathop{∑}_{k = 1}^{n}\cot^{- 1}\left(\dfrac{1 + kx ( k + 1 ) x}{( k + 1 ) x - kx}\right) $$ = \mathop{∑}_{k = 1}^{n}\tan^{- 1}\left(\dfrac{( k + 1 ) x - kx}{1 + kx ( k + 1 ) x}\right) $$ = \mathop{∑}_{k = 1}^{n}\left[\tan^{- 1} ( k + 1 ) x - \tan^{- 1}kx\right] $$ = \left[\tan^{- 1} ( n + 1 ) x - \tan^{- 1}nx\right] + \ldots + \left[\tan^{- 1}3x - \tan^{- 1}2x\right] + \left[\tan^{- 1}2x - \tan^{- 1}x\right] $$ = \tan^{- 1} ( n + 1 ) x - \tan^{- 1}x $$ = \tan^{- 1}\left(\dfrac{( n + 1 ) x - x}{1 + ( n + 1 ) {x}^{2}}\right) $$ = \tan^{- 1}\left(\dfrac{nx}{1 + ( n + 1 ) {x}^{2}}\right) $1.$ {S}_{10}\left(x\right) = \tan^{- 1}\left(\dfrac{10x}{1 + 11{x}^{2}}\right) $$ = \dfrac{\pi}{2} - \cot^{- 1}\left(\dfrac{10x}{1 + 11{x}^{2}}\right) $$ = \dfrac{\pi}{2} - \tan^{- 1}\left(\dfrac{1 + 11{x}^{2}}{10x}\right) $2.$ \lim_{n \to \infty}\cot \left({S}_{n} ( x )\right) = \lim_{n \to \infty}\cot \left(\tan^{- 1}\left(\dfrac{nx}{1 + ( n + 1 ) {x}^{2}}\right)\right) $$ = \lim_{n \to \infty}\cot \left(\cot^{- 1}\left(\dfrac{1 + ( n + 1 ) {x}^{2}}{nx}\right)\right) $$ = \lim_{n \to \infty}\left(\dfrac{1 + ( n + 1 ) {x}^{2}}{nx}\right) $$ = \dfrac{{x}^{2}}{x} = x $3.$ {S}_{3}\left(x\right) = \tan^{- 1}\left(\dfrac{3x}{1 + 4{x}^{2}}\right) = \dfrac{\pi}{4} $$ ⇒ \left(\dfrac{3x}{1 + 4{x}^{2}}\right) = tan\dfrac{\pi}{4} $$ ⇒ \dfrac{3x}{1 + 4{x}^{2}} = 1 $$ ⇒ 1 + 4{x}^{2} = 3x $$ ⇒ 4{x}^{2} - 3x + 1 = 0 $$ D < 0 $Hence, no real roots.4.$ \tan \left({S}_{n} ( x )\right) = \tan \left(\tan^{- 1}\left(\dfrac{nx}{1 + ( n + 1 ) {x}^{2}}\right)\right) = \left(\dfrac{nx}{1 + ( n + 1 ) {x}^{2}}\right) $\left(\dfrac{nx}{1 + ( n + 1 ) {x}^{2}}\right) \le \dfrac{1}{2} ⇒ 2nx \le 1 + ( n + 1 ) {x}^{2}$

$⇒ 2nx \le ( n + 1 ) {x}^{2} + 1$

$⇒ ( n + 1 ) {x}^{2} - 2nx + 1 \ge 0 \forall n \ge 1; x > 0$

Let, $y = ( n + 1 ) {x}^{2} - 2nx + 1$

$D = 4{n}^{2} - 4 ( n + 1 )$ and $n \in N$

$D < 0$ for $n = 1$

Hence, no solution if $n = 1$ .

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