Mathematics - Inverse Trigonometric Function Question with Solution | TestHub

$S_n\left(x\right)=\sum _{k=1}^n{\cot }^{-1}\left(\frac{1+k(k+1)x^2}{x}\right)$

$=\sum _{k=1}^n{\cot }^{-1}\left(\frac{1+kx(k+1)x}{(k+1)x-kx}\right)$

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

$=\sum _{k=1}^n\left[{\tan }^{-1}(k+1)x-{\tan }^{-1}kx\right]$

$=\left[{\tan }^{-1}(n+1)x-{\tan }^{-1}nx\right]+\dots +\left[{\tan }^{-1}3x-{\tan }^{-1}2x\right]+\left[{\tan }^{-1}2x-{\tan }^{-1}x\right]$

$={\tan }^{-1}(n+1)x-{\tan }^{-1}x$

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

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

1. $S_{10}\left(x\right)={\tan }^{-1}\left(\frac{10x}{1+11x^2}\right)$

$=\frac{\pi }{2}-{\cot }^{-1}\left(\frac{10x}{1+11x^2}\right)$

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

2. $\underset{n\to \infty }{\lim }\cot \left(S_n\left(x\right)\right)=\underset{n\to \infty }{\lim }\cot \left({\tan }^{-1}\left(\frac{nx}{1+(n+1)x^2}\right)\right)$

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

$=\underset{n\to \infty }{\lim }\left(\frac{1+(n+1)x^2}{nx}\right)$

$=\frac{x^2}{x}=x$

3. $S_3\left(x\right)={\tan }^{-1}\left(\frac{3x}{1+4x^2}\right)=\frac{\pi }{4}$

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

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

$\Rightarrow 1+4x^2=3x$

$\Rightarrow 4x^2-3x+1=0$

$D<0$

Hence, no real roots.

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

$\left(\frac{nx}{1+(n+1)x^2}\right)\le \frac{1}{2}\Rightarrow 2nx\le 1+(n+1)x^2$

$\Rightarrow 2nx\le (n+1)x^2+1$

$\Rightarrow (n+1)x^2-2nx+1\ge 0 \forall n\ge 1; x>0$

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 $n=1$.