JEE Advanced 2021 · Paper 1, Mathematics question 15

Telescoping. For $x>0$, each argument is positive, so
$$\cot^{-1}\frac{1+k(k+1)x^2}{x}=\tan^{-1}\frac{(k+1)x-kx}{1+(k+1)x\cdot kx}=\tan^{-1}((k+1)x)-\tan^{-1}(kx).$$
Summing gives $S_n(x)=\tan^{-1}((n+1)x)-\tan^{-1}x$.

Check each option.

• (A) $S_{10}=\tan^{-1}\dfrac{10x}{1+11x^2}=\dfrac\pi2-\tan^{-1}\dfrac{1+11x^2}{10x}$. True.
• (B) $S_n\to\frac\pi2-\tan^{-1}x$, so $\cot S_n\to\tan(\tan^{-1}x)=x$. True.
• (C) $S_3=\frac\pi4$ means $\frac{3x}{1+4x^2}=1$, i.e. $4x^2-3x+1=0$, which has no real root. False.
• (D) $\tan S_n=\dfrac{nx}{1+(n+1)x^2}$ has maximum $\dfrac n{2\sqrt{n+1}}$, which is $\frac34$ for $n=3$. False.

(A), (B)