If `y=tan^(-1).(1)/(1+x+x^(2))+tan^(-1).(1)/(x^(2)+3x+3)` upto `+tan^(-1).(1)/(x^(2)+5x+7)+….+2n` terms `(AA x ge0),` then y(0) is

Options

A. `tan^(-1)(n)`
B. `tan^(-1)(2n)`
C. `2tan^(-1)(n)`
D. 0

Answer

To solve the problem, we need to evaluate the expression for \( y \) given by:

\[ y = \tan^{-1}\left(\frac{1}{1+x+x^2}\right) + \tan^{-1}\left(\frac{1}{x^2 + 3x + 3}\right) + \tan^{-1}\left(\frac{1}{x^2 + 5x + 7}\right) + \ldots + \tan^{-1}\left(\frac{1}{x^2 + (2n-1)x + (2n-1)}\right) \]

for \( x \geq 0 \) and find \( y(0) \).

### Step 1: Evaluate \( y(0) \)
Substituting \( x = 0 \) into the expression for \( y \):

\[ y(0) = \tan^{-1}\left(\frac{1}{1+0+0^2}\right) + \tan^{-1}\left(\frac{1}{0^2 + 3 \cdot 0 + 3}\right) + \tan^{-1}\left(\frac{1}{0^2 + 5 \cdot 0 + 7}\right) + \ldots + \tan^{-1}\left(\frac{1}{0^2 + (2n-1) \cdot 0 + (2n-1)}\right) \]

This simplifies to:

\[ y(0) = \tan^{-1}(1) + \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \ldots + \tan^{-1}\left(\frac{1}{2n-1}\right) \]

### Step 2: Recognize the pattern
The general term in the series is:

\[ \tan^{-1}\left(\frac{1}{2k-1}\right) \quad \text{for } k = 1, 2, \ldots, n \]

Thus, we can express \( y(0) \) as:

\[ y(0) = \sum_{k=1}^{n} \tan^{-1}\left(\frac{1}{2k-1}\right) \]

### Step 3: Use the addition formula for \( \tan^{-1} \)
Using the property of the tangent inverse function, we know:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \quad \text{if } ab < 1 \]

We can apply this iteratively, but it is simpler to recognize that:

\[ \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 4: Calculate the sum
The sum \( y(0) \) can be computed as follows:

\[ y(0) = \tan^{-1}(1) + \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \ldots + \tan^{-1}\left(\frac{1}{2n-1}\right) \]

This series can be evaluated or approximated based on known results for sums of arctangents.

### Final Result
The final result for \( y(0) \) is:

\[ y(0) = \tan^{-1}(2n) \]

Thus, the answer is:

\[ \boxed{\tan^{-1}(2n)} \]