JEE Advanced 2021 Paper 1 Q15 Mathematics Trigonometry Inverse Trigonometric Functions Hard

JEE Advanced 2021 Paper 1 · Q15 · Inverse Trigonometric Functions

For any positive integer $n$, let $S_n : (0, \infty) \to \mathbb{R}$ be defined by $S_n(x) = \sum_{k=1}^{n} \cot^{-1}\left(\dfrac{1 + k(k+1)x^2}{x}\right)$, where for any $x \in \mathbb{R}$, $\cot^{-1}(x) \in (0, \pi)$ and $\tan^{-1}(x) \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$. Then which of the following statements is (are) TRUE ?

  1. A. $S_{10}(x) = \dfrac{\pi}{2} - \tan^{-1}\left(\dfrac{1 + 11x^2}{10x}\right)$, for all $x > 0$
  2. B. $\lim_{n \to \infty} \cot(S_n(x)) = x$, for all $x > 0$
  3. C. The equation $S_3(x) = \dfrac{\pi}{4}$ has a root in $(0, \infty)$
  4. D. $\tan(S_n(x)) \leq \dfrac{1}{2}$, for all $n \geq 1$ and $x > 0$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B

Solution

Using $\cot^{-1}\dfrac{1+k(k+1)x^2}{x}=\cot^{-1}(kx)-\cot^{-1}((k+1)x)$, the sum telescopes: $S_n(x)=\cot^{-1}(x)-\cot^{-1}((n+1)x)=\cot^{-1}\dfrac{(n+1)x^2+1}{nx}$. (A) $S_{10}(x)=\cot^{-1}\dfrac{11x^2+1}{10x}=\dfrac{\pi}{2}-\tan^{-1}\dfrac{1+11x^2}{10x}$. TRUE. (B) $\cot(S_n(x))=\dfrac{(n+1)x^2+1}{nx}\to x$ as $n\to\infty$. TRUE. (C) $S_3(x)=\pi/4\Rightarrow \cot^{-1}\dfrac{4x^2+1}{3x}=\pi/4\Rightarrow 4x^2+1=3x$, i.e., $4x^2-3x+1=0$ has discriminant $9-16<0$, no real root. FALSE. (D) $\tan(S_n(x))=\dfrac{nx}{(n+1)x^2+1}$; by AM-GM $(n+1)x^2+1\ge 2\sqrt{n+1}\cdot x$, so $\tan(S_n)\le \dfrac{n}{2\sqrt{n+1}}$ which exceeds $1/2$ for $n\ge 2$. FALSE.

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