For any positive integer n , let S n : ( 0 , ∞ ) → R be defined by S n x = ∑ k = 1 n cot -…

For any positive integer n, let Sn:(0,∞)→R be defined by
Snx=∑k=1ncot-11+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 statements is (are) TRUE ?
  1. S10x=π2-tan-11+11x210x, for all x>0
  2. limn→∞cotSn(x)=x, for all x>0
  3. The equation S3(x)=π4 has a root in (0,∞)
  4. tanSn(x)≤12, for all n≥1 and x>0

Solution

Snx=∑k=1ncot-11+k(k+1)x2x

=∑k=1ncot-11+kx(k+1)x(k+1)x-kx

=∑k=1ntan-1(k+1)x-kx1+kx(k+1)x

=∑k=1ntan-1(k+1)x-tan-1kx

=tan-1(n+1)x-tan-1nx+…+tan-13x-tan-12x+tan-12x-tan-1x

=tan-1(n+1)x-tan-1x

=tan-1(n+1)x-x1+(n+1)x2

=tan-1nx1+(n+1)x2

$1. S_{10}(x) = \tan^{-1}\left(\frac{10x}{1+11x^2}\right)$ $= \frac{\pi}{2} - \cot^{-1}\left(\frac{10x}{1+11x^2}\right)$ $= \frac{\pi}{2} - \tan^{-1}\left(\frac{1+11x^2}{10x}\right)$ $2. \lim_{n \to \infty} \cot\left(S_n(x)\right) = \lim_{n \to \infty} \cot\left(\tan^{-1}\left(\frac{nx}{1+(n+1)x^2}\right)\right)$ $= \lim_{n \to \infty} \cot\left(\cot^{-1}\left(\frac{1+(n+1)x^2}{nx}\right)\right)$ $= \lim_{n \to \infty} \left(\frac{1+(n+1)x^2}{nx}\right)$ $= \frac{x^2}{x} = x$ $3. S_{3}(x) = \tan^{-1}\left(\frac{3x}{1+4x^2}\right) = \frac{\pi}{4}$ $\Rightarrow \left(\frac{3x}{1+4x^2}\right) = \tan\left(\frac{\pi}{4}\right)$ $\Rightarrow \frac{3x}{1+4x^2} = 1$ $\Rightarrow 1+4x^2 = 3x$ $\Rightarrow 4x^2 - 3x + 1 = 0$ $D < 0$ Hence, no real roots. $4. \tan\left(S_n(x)\right) = \tan\left(\tan^{-1}\left(\frac{nx}{1+(n+1)x^2}\right)\right) = \left(\frac{nx}{1+(n+1)x^2}\right)$ $\left(\frac{nx}{1+(n+1)x^2}\right) \leq \frac{1}{2} \Rightarrow 2nx \leq 1+(n+1)x^2$ $\Rightarrow 2nx \leq (n+1)x^2 + 1$ $\Rightarrow (n+1)x^2 - 2nx + 1 \geq 0 \quad \forall \quad n \geq 1, x > 0$ Let, $y = (n+1)x^2 - 2nx + 1$ $D = 4n^2 - 4(n+1)$ and $n \in N$ $D < 0$ for $n = 1$ Hence, no solution if $n = 1$.

Asked in: JEE Advanced 2021 (Paper 1)

Practice more Trigonometric Functions questions on Aicharya