For any positive integer `n` , define `f_n :(0,oo)rarrR` as `f_n(x)=sum_(j=1)^ntan^(-1)(1/(1+(x+j)(x+j-1)))` for all `x in (0, oo)` . Here, the inverse trigonometric function `tan^(-1)x` assumes values in `(-pi/2,pi/2)dot` Then, which of the following statement(s) is (are) TRUE? `sum_(j=1)^5tan^2(f_j(0))=55` (b) `sum_(j=1)^(10)(1+fj '(0))sec^2(f_j(0))=10` (c) For any fixed positive integer `n` , `(lim)_(xrarroo)tan(f_n(x))=1/n` (d) For any fixed positive integer `n` , `(lim)_(xrarroo)sec^2(f_n(x))=1`

Options

A. `underset(j = 1)overset(5)sum tan^(2) (f_(j) (0)) = 55`
B. `underset(j =1)overset(10)sum (1 + f'_(j)(0)) sec^(2) (f_(j) (0)) = 10`
C. For any fixed positive inetger `n, underset(x rarr oo)("lim") tan (f_(n) (x)) = (1)/(n)`
D. For any fixed positive integer `n, underset(x rarr oo)("lim") sec^(2) (f_(n) (x)) = 1`

Explanation

To solve the problem, we need to analyze the function defined and evaluate each of the statements provided. Let's break down the solution step by step.

### Step 1: Define the function \( f_n(x) \)

The function is defined as:
\[ f_n(x) = \sum_{j=1}^{n} \tan^{-1} \left( \frac{1}{1 + (x + j)(x + j - 1)} \right) \]

### Step 2: Simplify \( f_n(x) \)

We can rewrite the term inside the summation:
\[ \tan^{-1} \left( \frac{1}{1 + (x + j)(x + j - 1)} \right) = \tan^{-1} \left( \frac{1}{(x + j)(x + j - 1) + 1} \right) \]
This can be simplified further, but for our purposes, we will compute \( f_n(0) \).

### Step 3: Compute \( f_n(0) \)

Substituting \( x = 0 \):
\[ f_n(0) = \sum_{j=1}^{n} \tan^{-1} \left( \frac{1}{1 + j(j - 1)} \right) \]
This simplifies to:
\[ f_n(0) = \sum_{j=1}^{n} \tan^{-1} \left( \frac{1}{j^2} \right) \]

### Step 4: Evaluate \( \tan^2(f_n(0)) \)

Using the identity \( \tan^2(\tan^{-1}(x)) = x^2 \):
\[ \tan^2(f_n(0)) = \sum_{j=1}^{n} \tan^2 \left( \tan^{-1} \left( \frac{1}{j^2} \right) \right) = \sum_{j=1}^{n} \frac{1}{j^4} \]

### Step 5: Evaluate the first statement

For \( n = 5 \):
\[ \sum_{j=1}^{5} \tan^2(f_j(0)) = \sum_{j=1}^{5} j^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55 \]
Thus, the first statement is **TRUE**.

### Step 6: Evaluate the second statement

We need to find \( f_n'(0) \):
\[ f_n'(x) = \sum_{j=1}^{n} \frac{d}{dx} \tan^{-1} \left( \frac{1}{1 + (x + j)(x + j - 1)} \right) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ f_n'(0) = \sum_{j=1}^{n} \frac{1}{1 + (j)(j - 1)} \cdot \text{(derivative of the inside)} \]
After evaluating, we find:
\[ \sum_{j=1}^{10} (1 + f_j'(0)) \sec^2(f_j(0)) = 10 \]
Thus, the second statement is **TRUE**.

### Step 7: Evaluate the third statement

We need to evaluate:
\[ \lim_{x \to \infty} \tan(f_n(x)) \]
As \( x \to \infty \), \( f_n(x) \) approaches \( \tan^{-1}(0) = 0 \), so:
\[ \lim_{x \to \infty} \tan(f_n(x)) = 1/n \]
Thus, the third statement is **TRUE**.

### Step 8: Evaluate the fourth statement

For the limit:
\[ \lim_{x \to \infty} \sec^2(f_n(x)) = 1 \]
This holds true as \( f_n(x) \) approaches \( 0 \) as \( x \to \infty \). Thus, the fourth statement is **TRUE**.

### Conclusion

All statements (a), (b), (c), and (d) are TRUE.