JEE Advanced 2018 — Paper 2 — Question 28

Use the identity $\tan^{-1}(a)-\tan^{-1}(b)=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$. For each term, set $a=x+j$ and $b=x+j-1$, so $\tan^{-1}(x+j)-\tan^{-1}(x+j-1)=\tan^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right)$. The sum telescopes: $f_n(x)=\sum_{j=1}^n [\tan^{-1}(x+j)-\tan^{-1}(x+j-1)] = \tan^{-1}(x+n)-\tan^{-1}(x)$. Further, $\tan(f_n(x)) = \frac{(x+n)-x}{1+x(x+n)} = \frac{n}{1+x(x+n)}$. Check options: For A and B, $x=0$ is not in the domain $(0,\infty)$, so $f_j(0)$ is not defined. Hence A and B are false. For C, $\lim_{x\to\infty} \tan(f_n(x)) = \lim_{x\to\infty} \frac{n}{1+x(x+n)} = 0$, not $\frac{1}{n}$. So C is false. For D, $\sec^2(f_n(x)) = 1 + \tan^2(f_n(x)) = 1 + \left(\frac{n}{1+x(x+n)}\right)^2$. As $x\to\infty$, $\tan^2(f_n(x)) \to 0$, so $\sec^2(f_n(x)) \to 1$. Hence D is true.