For any positive integer n, let Sn:(0,∞)→ℝ be defined by Sn(x)=∑k=1n cot−1⁡1+k(k+1)x2x where for any x∈ℝ,cot−1⁡(x)∈(0,π) and tan−1⁡(x)∈−π2,π2 Then which of the following statement is (are) TRUE?

Sn:(0,∞)→RSn(x)=∑k=1n tan−1⁡x1+k(k+1)x2Sn(x)=∑k=1n tan−1⁡(k+1)x−kx1+k(kH)x2=∑k=1n tan−1⁡(KH)xSn=tan−1⁡(n+1)x−tan−1⁡xS10(x)=tan−1⁡11x−tan−1⁡x=tan−1⁡10x1+11x2=π2−cot−1⁡10x1+11x2(A)=π2−tan−1⁡1+11x210x (B) limn→∞ cot⁡tan−1⁡(n+1)x−tan−1⁡x=limn→∞ cot⁡tan−1⁡nx1+(n+1)x2=cot⁡tan−1⁡xx2=cot⁡cot⁡x=x (C) S3(x)=π4⇒tan−1⁡4xtan−1⁡x=π4tan−1⁡3x1+4x2=π4⇒3x1+4x2=1⇒4x2−3x+1=09−4×4<0 (D) nx1+(n+1)x2=11nx+x+xn=n1+n×1=1

For any positive integer n, let Sn:(0,∞)→ℝ be defined by Sn(x)=∑k=1n cot−1⁡1+k(k+1)x2x where for any x∈ℝ,cot−1⁡(x)∈(0,π) and tan−1⁡(x)∈−π2,π2 Then which of the following statement is (are) TRUE?

Sn:(0,∞)→RSn(x)=∑k=1n tan−1⁡x1+k(k+1)x2Sn(x)=∑k=1n tan−1⁡(k+1)x−kx1+k(kH)x2=∑k=1n tan−1⁡(KH)xSn=tan−1⁡(n+1)x−tan−1⁡xS10(x)=tan−1⁡11x−tan−1⁡x=tan−1⁡10x1+11x2=π2−cot−1⁡10x1+11x2(A)=π2−tan−1⁡1+11x210x (B) limn→∞ cot⁡tan−1⁡(n+1)x−tan−1⁡x=limn→∞ cot⁡tan−1⁡nx1+(n+1)x2=cot⁡tan−1⁡xx2=cot⁡cot⁡x=x (C) S3(x)=π4⇒tan−1⁡4xtan−1⁡x=π4tan−1⁡3x1+4x2=π4⇒3x1+4x2=1⇒4x2−3x+1=09−4×4<0 (D) nx1+(n+1)x2=11nx+x+xn=n1+n×1=1

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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 statement is (are) TRUE?

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$S_{10} (x) = \frac{π}{2} - \tan^{- 1} (\frac{1 + 11 x^{2}}{10 x}), for all x > 0$

$lim_{n \to \infty} \cot (S_{n} (x)) = x, for all x > 0$

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

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

answer is A, B.

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Detailed Solution

$\begin{array}{l} S_{n} : (0, \infty) \to R \\ S_{n} (x) = \sum_{k = 1}^{n} \tan^{- 1} (\frac{x}{1 + k (k + 1) x^{2}}) \\ S_{n} (x) = \sum_{k = 1}^{n} \tan^{- 1} (\frac{(k + 1) x - kx}{1 + k (kH) x^{2}}) = \sum_{k = 1}^{n} \tan^{- 1} (KH) x \\ S_{n} = \tan^{- 1} (n + 1) x - \tan^{- 1} x \\ S_{10} (x) = \tan^{- 1} 11 x - \tan^{- 1} x \\ = \tan^{- 1} (\frac{10 x}{1 + 11 x^{2}}) = \frac{π}{2} - \cot^{- 1} (\frac{10 x}{1 + 11 x^{2}}) \\ (A) = \frac{π}{2} - \tan^{- 1} (\frac{1 + 11 x^{2}}{10 x}) \end{array}$

$(B) lim_{n \to \infty} \cot (\tan^{- 1} (n + 1) x - \tan^{- 1} x)$

$\begin{matrix} = lim_{n \to \infty} \cot (\tan^{- 1} (\frac{nx}{1 + (n + 1) x^{2}})) = \cot \tan^{- 1} (\frac{x}{x^{2}}) \\ = \cot \cot x = x \end{matrix}$

$\begin{array}{l} (C) S_{3} (x) = \frac{π}{4} \Rightarrow \tan^{- 1} 4 {xtan}^{- 1} x = \frac{π}{4} \\ \tan^{- 1} (\frac{3 x}{1 + 4 x^{2}}) = \frac{π}{4} \Rightarrow \frac{3 x}{1 + 4 x^{2}} = 1 \Rightarrow 4 x^{2} - 3 x + 1 = 0 \\ 9 - 4 \times 4 < 0 \\ (D) \frac{nx}{1 + (n + 1) x^{2}} = \frac{1}{\frac{1}{nx} + x + \frac{x}{n}} \\ = \frac{n}{1 + n \times 1} = 1 \end{array}$