Area formed between lines y` = nx +(1)/(n), y = -(1)/(n)x-n(n epsilon I)` and y axis is mixmum for

Options

A. `n=1`
B. two values of n
C. `n=-1`
D. only one value of n

Explanation

To find the maximum area formed between the lines \( y = nx + \frac{1}{n} \), \( y = -\frac{1}{n}x - n \), and the y-axis, we will follow these steps:

### Step 1: Find the Points of Intersection
We need to find the intersection point of the two lines given by their equations.

1. Set the equations equal to each other:
\[ nx + \frac{1}{n} = -\frac{1}{n}x - n \]

2. Rearranging gives:
\[ nx + \frac{1}{n} + \frac{1}{n}x + n = 0 \]
\[ (n + \frac{1}{n})x + \left(\frac{1}{n} + n\right) = 0 \]

3. Solving for \( x \):
\[ (n + \frac{1}{n})x = -\left(\frac{1}{n} + n\right) \]
\[ x = -\frac{\frac{1}{n} + n}{n + \frac{1}{n}} = -1 \]

4. Substitute \( x = -1 \) back into either line to find \( y \):
\[ y = n(-1) + \frac{1}{n} = -n + \frac{1}{n} = \frac{1}{n} - n \]

Thus, the intersection point \( D \) is \( (-1, \frac{1}{n} - n) \).

### Step 2: Identify the Vertices of the Triangle
The vertices of the triangle formed by the lines and the y-axis are:
- Point \( A(0, \frac{1}{n}) \) from the first line (y-intercept).
- Point \( B(0, -n) \) from the second line (y-intercept).
- Point \( D(-1, \frac{1}{n} - n) \) from the intersection of the two lines.

### Step 3: Calculate the Area of the Triangle
The area \( A \) of the triangle formed by points \( A \), \( B \), and \( D \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

1. The base \( AB \) is the distance between points \( A \) and \( B \):
\[ AB = \left| \frac{1}{n} - (-n) \right| = \left| \frac{1}{n} + n \right| = \frac{1 + n^2}{n} \]

2. The height \( OD \) is the x-coordinate of point \( D \):
\[ OD = |-1| = 1 \]

3. Therefore, the area \( A \) becomes:
\[ A = \frac{1}{2} \times \left(\frac{1 + n^2}{n}\right) \times 1 = \frac{1 + n^2}{2n} \]

### Step 4: Maximize the Area
To maximize the area, we need to differentiate \( A \) with respect to \( n \) and set the derivative equal to zero.

1. Differentiate \( A \):
\[ A(n) = \frac{1 + n^2}{2n} \]
Using the quotient rule:
\[ A' = \frac{(2n)(2n) - (1+n^2)(2)}{(2n)^2} = \frac{2n^2 - 2 - 2n^2}{4n^2} = \frac{-2}{4n^2} = -\frac{1}{2n^2} \]

2. Set the derivative equal to zero:
\[ -\frac{1}{2n^2} = 0 \]
This does not yield any critical points, but we can analyze the function directly.

3. The area function \( A(n) \) is maximized when \( n^2 = 1 \), giving \( n = \pm 1 \).

### Conclusion
The maximum area occurs at \( n = 1 \) and \( n = -1 \).