JEE Advanced 2021 — Paper 1 — Question 27

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

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

$t_{2}=\cot ^{-1}(2 x)-\cot ^{-1}(3 x)$ $t_{3}=\cot ^{-1}(3 x)-\cot ^{-1}(4 x)$ $\vdots$ $t_{n}=\cot ^{-1}(n x)-\cot ^{-1}((n+1) x)$

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

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

(B) $\lim _{n \rightarrow \infty} \cot \left(S_{n}(x)\right)=\lim _{n \rightarrow \infty} \cot \left(\cot ^{-1}\left(\frac{(n+1) x^{2}+1}{n x}\right)\right)=\lim _{n \rightarrow \infty} \frac{n x^{2}+x^{2}+1}{n x}=x$

(C) $\mathrm{S}_{3}(\mathrm{x})=\cot ^{-1}\left(\frac{4 \mathrm{x}^{2}+1}{3 \mathrm{x}}\right)=\frac{\pi}{4} \Rightarrow \frac{1+4 \mathrm{x}^{2}}{3 \mathrm{x}}=1$ $\Rightarrow 4 \mathrm{x}^{2}-3 \mathrm{x}+1=0$ have imaginary roots

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