For any positive integer n, define $f_{n} : (0, \infty) \to ℝ$ as $f_{n} (x) = \sum_{j = 1}^{n} \tan^{- 1} (\frac{1}{1 + (x + j) (x + j - 1)})$ for all $x \in (0, \infty)$ (Here, the inverse trigonometric function $\tan^{- 1} x$ assume values in $(- \frac{π}{2}, \frac{π}{2})$ Then, which of the following statement(s) is (are) TRUE?

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$\lim_{x \to \infty} \tan (f_{n} (x)) = \frac{1}{n}$ for a fixed number n

$\sum_{j = 1}^{5} \tan^{2} (f_{1} (0)) = 55$

$\sum_{j = 1}^{10} (1 + f'_{1} (0)) \sec^{2} (f_{j} (0)) = 10$

$\lim_{x \to \infty} \sec^{2} (f_{n} (x)) = 1$ for a fixed number n

answer is D.

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Detailed Solution

$\begin{array}{l} f_{n} (x) = \sum_{j = 1}^{n} \tan^{- 1} (\frac{(x + j) - (x + j - 1)}{1 + (x + j) (x + j - 1)}) \\ f_{n} (x) = \sum_{j = 1}^{n} [\tan^{- 1} (x + j) - \tan^{- 1} (x + j - 1)] \\ f_{n} (x) = \tan^{- 1} (x + n) - \tan^{- 1} x \\ ∴ \tan (f_{n} (x)) = \tan [\tan^{- 1} (x + n) - \tan^{- 1} x] \end{array}$

$\begin{array}{l} \tan (f_{n} (x)) = \frac{(x + n) - x}{1 + x (x + n)} \tan (f_{n} (x)) = \frac{n}{1 + x^{2} + n x} \\ ∴ \sec^{2} (f_{n} (x)) = 1 + \tan^{2} (f_{n} (x)) \\ \lim_{x \to \infty} \sec^{2} (f_{n} (x)) = \lim_{x \to \infty} 1 + {(\frac{n}{1 + x^{2} + n x})}^{2} = 1 \end{array}$