For any positive integer n, let Sₙ:(0,∞)arrow{R} be defined by sₙ(x)=_{k=1}ⁿ^{-1}≤ft({1+k(k+1)x²}{x}) where for any x {R},\ ^{-1}x (0,π) and ^{-1}(x) ≤ft(-π/2,π/2). Then which of the following statements is (are) TRUE? [JEE Adv 2021 P1]

$S_{10}(x)=\frac{\pi}{2}-\tan^{-1}\left(\frac{1+11x^{2}}{10x}\right),\;\text{for all}\;x>0$
$\lim\limits_{n\to\infty}\cot(S_{n}(x))=x$ , for all $x>0$
The equation $S_{3}(x)=\frac{\pi}{4}$ has a root in $(0,\infty)$
$\tan(S_{n}(x))\le\frac{1}{2}$ , for all $n\ge1$ and $x>0$

Solution:

Given that

S_{n}(x)=\sum\limits_{k=1}^{n}\cot^{-1}\left(\frac{1+k(k+1)x^{2}}{x}\right)

=\sum\limits_{k=1}^{n}\tan^{-1}\left(\frac{x}{1+kx(kx+x)}\right)

=\sum\limits_{k=1}^{n}\tan^{-1}\left(\frac{(kx+x)-(kx)}{1+(kx+x)(kx)}\right)

\Rightarrow S_{n}(x)=\tan^{-1}(nx+x)-\tan^{-1}x

=\tan^{-1}\left(\frac{nx}{1+(n+1)x^{2}}\right)

(a)

S_{10}(x)=\tan^{-1}\left(\frac{10x}{1+11x^{2}}\right)=\frac{\pi}{2}-\tan^{-1}\left(\frac{1+11x^{2}}{10x}\right)(x>0)

(Option (a) is correct)

(b)

\lim\limits_{n\to\infty}\cot(S_{n}(x))=\lim\limits_{n\to\infty}\cot^{-1}\left(\frac{1+(n+1)x^{2}}{nx}\right)

=\lim\limits_{n\to\infty}\frac{\frac{1}{n}+\left(1+\frac{1}{n}\right)x^{2}}{x}=x\quad(x>0)

(Option (b) is correct)

S_{3}(x)=\tan^{-1}\left(\frac{3x}{1+4x^{2}}\right)=\frac{\pi}{4}

\Rightarrow 4x^{2}-3x+1=0\Rightarrow x\notin\mathbb{R}[\because

D is negative

]

(Option (c) is incorrect)

(d) For

x=1

\tan(S_{n}(x))=\frac{n}{n+2}\ge\frac{1}{2}

\tan(S_{n}(x))=\frac{n}{n+2}\ge\frac{1}{2}

for

n\ge3

. (Option (d) is incorrect)

Go to Question:

Prev Question

Next Question