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Trigonometry — Hard JEE math MCQ

For any positive integer $n$, define $f_{n}:(0, \infty) \rightarrow \mathbb{R}$ as \[ f_{n}(x)=\sum_{j=1}^{n} \tan ^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right) \text { for all } x \in(0, \infty) \] (Here, the inverse trigonometric function $\tan ^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. ) Then, which of the following statement(s) is (are) TRUE?
  1. A. $\sum_{j=1}^{5} \tan ^{2}\left(f_{j}(0)\right)=55$
  2. B. $\sum_{j=1}^{10}\left(1+f_{j}^{\prime}(0)\right) \sec ^{2}\left(f_{j}(0)\right)=10$
  3. C. For any fixed positive integer $n, \lim _{x \rightarrow \infty} \tan \left(f_{n}(x)\right)=\frac{1}{n}$
  4. D. For any fixed positive integer $n$, $\lim _{x \rightarrow \infty} \sec ^{2}\left(f_{n}(x)\right)=1$

Solution

The correct option is **D**.

MATH

hard • PYQ Reworded
Question
Read carefully, then pick the best option.
For any positive integer nn, define fn:(0,∞)→Rf_{n}:(0, \infty) \rightarrow \mathbb{R} as fn(x)=∑j=1ntan⁡−1(11+(x+j)(x+j−1)) for all x∈(0,∞) f_{n}(x)=\sum_{j=1}^{n} \tan ^{-1}\left(\frac{1}{1+(x+j)(x+j-1)}\right) \text { for all } x \in(0, \infty) (Here, the inverse trigonometric function tan⁡−1x\tan ^{-1} x assumes values in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right). ) Then, which of the following statement(s) is (are) TRUE?
AI hints
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