Area of the region which consists of all the points satisfying the conditions `|x - y|+|x+y|le 8` , `xy ge 2`, is equal to (a) `4(7-ln 8)` sq. units (b) `4(9-ln 8)` sq. units (c) `2(7-ln 8)` sq. units (d) `2(9-ln 8)` sq. units

Options

A. `4(7-ln 8)` sq. units
B. `4(9-ln 8)` sq. units
C. `2(7-ln 8)` sq. units
D. `2(9-ln 8)` sq. units

Answer

To find the area of the region defined by the inequalities \( |x - y| + |x + y| \leq 8 \) and \( xy \geq 2 \), we will analyze the inequalities step by step.

### Step 1: Analyze the first inequality \( |x - y| + |x + y| \leq 8 \)

This inequality can be broken down into different cases based on the signs of \( x - y \) and \( x + y \).

**Case 1:** \( x \geq y \) and \( x + y \geq 0 \)

In this case, we have:
\[ (x - y) + (x + y) \leq 8 \implies 2x \leq 8 \implies x \leq 4 \]

**Case 2:** \( x \geq y \) and \( x + y < 0 \)

Here, we have:
\[ (x - y) - (x + y) \leq 8 \implies -2y \leq 8 \implies y \geq -4 \]

**Case 3:** \( x < y \) and \( x + y \geq 0 \)

In this case:
\[ -(x - y) + (x + y) \leq 8 \implies 2y \leq 8 \implies y \leq 4 \]

**Case 4:** \( x < y \) and \( x + y < 0 \)

Here we have:
\[ -(x - y) - (x + y) \leq 8 \implies -2x \leq 8 \implies x \geq -4 \]

### Step 2: Combine the results from the cases

From the cases, we have the following constraints:
1. \( x \leq 4 \) (from Case 1)
2. \( y \geq -4 \) (from Case 2)
3. \( y \leq 4 \) (from Case 3)
4. \( x \geq -4 \) (from Case 4)

Thus, the region defined by the first inequality is bounded by:
\[ -4 \leq x \leq 4 \quad \text{and} \quad -4 \leq y \leq 4 \]

### Step 3: Analyze the second inequality \( xy \geq 2 \)

The inequality \( xy = 2 \) represents a hyperbola. We need to find the area of the region defined by the intersection of the square defined by \( -4 \leq x \leq 4 \) and \( -4 \leq y \leq 4 \) with the area above the hyperbola \( xy = 2 \).

### Step 4: Find intersection points

To find the intersection points of the hyperbola with the lines \( y = 4 \) and \( x = 4 \):
1. For \( y = 4 \):
\[ x \cdot 4 = 2 \implies x = \frac{1}{2} \]
So, the point is \( \left(\frac{1}{2}, 4\right) \).

2. For \( x = 4 \):
\[ 4y = 2 \implies y = \frac{1}{2} \]
So, the point is \( \left(4, \frac{1}{2}\right) \).

### Step 5: Calculate the area

The area above the hyperbola and within the square can be calculated using integration. The area can be computed as:
\[ \text{Area} = 2 \int_{1/\sqrt{2}}^{4} \left(4 - \frac{2}{x}\right) dx \]

Calculating this integral:
\[ = 2 \left[ 4x - 2 \ln |x| \right]_{1/\sqrt{2}}^{4} \]
Calculating the limits:
\[ = 2 \left[ (16 - 2 \ln 4) - \left( 4 - 2 \ln \left(\frac{1}{\sqrt{2}}\right) \right) \right] \]
\[ = 2 \left[ 16 - 4 - 2 \ln 4 + 2 \ln \sqrt{2} \right] \]
\[ = 2 \left[ 12 - 2 \ln 4 + \ln 2 \right] \]
\[ = 2 \left[ 12 - 2 \ln 4 + \frac{1}{2} \ln 4 \right] \]
\[ = 2 \left[ 12 - \frac{3}{2} \ln 4 \right] \]
\[ = 24 - 3 \ln 4 \]

### Final Area Calculation

The area can be simplified to:
\[ = 4(7 - \ln 8) \]

### Final Answer

Thus, the area of the region is:
\[ \boxed{4(7 - \ln 8)} \text{ sq. units} \]