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

(A) $S_{10}(x)=\frac{\pi}{2}-\tan^{-1}\left(\frac{1+11x^{2}}{10x}\right)$, for all $x>0$; (B) $\lim_{n \rightarrow \infty} \cot(S_{n}(x))=x$, for all $x>0$

(C) The equation $S_{3}(x)=\frac{\pi}{4}$ has a root in $(0, \infty)$

(D) $\tan(S_{n}(x)) \leq \frac{1}{2}$, for all $n \geq 1$ and $x>0$

Inverse Trigonometric Functions — JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 Question with Solution

For any positive integer n, let

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

be defined by

S_{n} (x) = \sum_{k = 1}^{n} cot^{- 1} (\frac{1 + k ( k + 1 ) x ^{2}}{x})

,
where for any

x \in R, cot^{- 1} x \in (0, π)

and

tan^{- 1} (x) \in (- \frac{π}{2}, \frac{π}{2})

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

(A) $S_{10} (x) = \frac{π}{2} - tan^{- 1} (\frac{1 + 11 x ^{2}}{10 x})$ , for all $x > 0$ ✓ Correct
(B) $lim_{n \to \infty} cot (S_{n} (x)) = x$ , for all $x > 0$ ✓ Correct
(C)The equation $S_{3} (x) = \frac{π}{4}$ has a root in $(0, \infty)$
(D) $tan (S_{n} (x)) \leq \frac{1}{2}$ , for all $n \geq 1$ and $x > 0$

Answer: (A), (B)

Solution

S_{n} (x) = \sum_{k = 1}^{n} tan^{- 1} (\frac{x}{1 + k x ( k x + x )}) = \sum_{k = 1}^{n} tan^{- 1} (\frac{( k + 1 ) x - k x}{1 + ( k + 1 ) x \cdot k x}) = tan^{- 1} ((n + 1) x) - tan^{- 1} (x) = tan^{- 1} (\frac{n x}{1 + ( n + 1 ) x ^{2}})

. (A)

S_{10} (x) = tan^{- 1} \frac{10 x}{1 + 11 x ^{2}} = \frac{π}{2} - tan^{- 1} (\frac{1 + 11 x ^{2}}{10 x})

for

x > 0

. (B)

lim_{n \to \infty} cot (S_{n} (x)) = lim_{n \to \infty} \frac{1 + ( 1 + \frac{1}{n} ) x ^{2}}{x} = x

for

x > 0

. (C)

S_{3} (x) = \frac{π}{4}

gives

\frac{3 x}{1 + 4 x ^{2}} = 1

, so

4 x^{2} - 3 x + 1 = 0

, which has no real roots. (D) We need

\frac{n x}{1 + ( n + 1 ) x ^{2}} \leq \frac{1}{2}

for all

n \geq 1, x > 0

, i.e.,

(n + 1) x^{2} - 2 n x + 1 \geq 0

. For

n = 1

this holds, but for

n \geq 2

the discriminant is positive and both roots are positive, so the inequality fails for some

x > 0

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · One or more correct options

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Inverse Trigonometric Functions

Solution

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