JEE Advanced 2018 · Paper 2, Mathematics question 1

Telescope. Since $(x+j)-(x+j-1)=1$,
$$\tan^{-1}\frac1{1+(x+j)(x+j-1)}=\tan^{-1}(x+j)-\tan^{-1}(x+j-1),$$
so
$$f_n(x)=\tan^{-1}(x+n)-\tan^{-1}x,\qquad\tan f_n(x)=\frac n{1+x(x+n)}.$$

• (C) As $x\to\infty$, $\tan f_n(x)\to0$, not $\dfrac1n$. False.

• (D) So $\sec^2f_n(x)=1+\tan^2f_n(x)\to1$. True.

• (A), (B) These are statements about $f_j(0)$ and $f_j'(0)$, but $f_n$ is defined only on $(0,\infty)$. The official key treats them as not true and accepts only (D).

  (Extending $f_j$ continuously to 0 would give $\tan f_j(0)=j$, so $\sum_1^5j^2=55$. It would also give $(1+f_j'(0))\sec^2f_j(0)=\frac1{1+j^2}(1+j^2)=1$, summing to 10.)

(D)