The area bounded by `y = 2- abs(2 -x),y =3/absx` is

Options

A. `((5 - 4 ln 2)/3)` sq unit
B. `((2-ln 3)/2)` sq unit
C. `((4-3 ln3)/2)` sq unit
D. none of these

Explanation

To find the area bounded by the curves \( y = 2 - |2 - x| \) and \( y = \frac{3}{|x|} \), we will follow these steps:

### Step 1: Redefine the equations based on the absolute values
The equation \( y = 2 - |2 - x| \) can be split into two cases:
- For \( x < 2 \): \( |2 - x| = 2 - x \) so \( y = 2 - (2 - x) = x \)
- For \( x \geq 2 \): \( |2 - x| = x - 2 \) so \( y = 2 - (x - 2) = 4 - x \)

The equation \( y = \frac{3}{|x|} \) can also be split:
- For \( x > 0 \): \( |x| = x \) so \( y = \frac{3}{x} \)
- For \( x < 0 \): \( |x| = -x \) so \( y = -\frac{3}{x} \)

### Step 2: Identify the intersection points
To find the area, we need to determine the intersection points of the curves. We will consider the cases for \( x \geq 0 \) and \( x < 0 \).

1. **For \( x \geq 0 \)**:
- Set \( x = \frac{3}{x} \) (since \( y = x \) and \( y = \frac{3}{x} \))
- This leads to \( x^2 = 3 \) or \( x = \sqrt{3} \) (we discard the negative root since we are considering \( x \geq 0 \)).

2. **For \( x \geq 2 \)**:
- Set \( 4 - x = \frac{3}{x} \)
- Rearranging gives \( 4x - x^2 = 3 \) or \( x^2 - 4x + 3 = 0 \)
- Factoring gives \( (x - 1)(x - 3) = 0 \), so \( x = 1 \) and \( x = 3 \).

### Step 3: Determine the area between the curves
The area can be calculated by integrating the upper curve minus the lower curve between the intersection points.

1. **From \( \sqrt{3} \) to \( 2 \)**:
- Upper curve: \( y = x \)
- Lower curve: \( y = \frac{3}{x} \)
- Area \( A_1 = \int_{\sqrt{3}}^{2} \left( x - \frac{3}{x} \right) dx \)

2. **From \( 2 \) to \( 3 \)**:
- Upper curve: \( y = 4 - x \)
- Lower curve: \( y = \frac{3}{x} \)
- Area \( A_2 = \int_{2}^{3} \left( (4 - x) - \frac{3}{x} \right) dx \)

### Step 4: Calculate the integrals
1. **Calculate \( A_1 \)**:
\[ A_1 = \int_{\sqrt{3}}^{2} \left( x - \frac{3}{x} \right) dx = \left[ \frac{x^2}{2} - 3 \ln |x| \right]_{\sqrt{3}}^{2} \]

Evaluating this gives:
\[ A_1 = \left( \frac{2^2}{2} - 3 \ln 2 \right) - \left( \frac{(\sqrt{3})^2}{2} - 3 \ln \sqrt{3} \right) = \left( 2 - 3 \ln 2 \right) - \left( \frac{3}{2} - \frac{3}{2} \ln 3 \right) \]

2. **Calculate \( A_2 \)**:
\[ A_2 = \int_{2}^{3} \left( (4 - x) - \frac{3}{x} \right) dx = \left[ 4x - \frac{x^2}{2} - 3 \ln |x| \right]_{2}^{3} \]

Evaluating this gives:
\[ A_2 = \left( 12 - \frac{9}{2} - 3 \ln 3 \right) - \left( 8 - 2 - 3 \ln 2 \right) = \left( \frac{15}{2} - 3 \ln 3 \right) - \left( 6 - 3 \ln 2 \right) \]

### Step 5: Combine the areas
The total area \( A \) is:
\[ A = A_1 + A_2 \]

### Final Area Calculation
After performing the calculations and simplifications, we find:
\[ A = 4 - \frac{3}{2} \ln 3 \]

Thus, the area bounded by the curves is:
\[ \boxed{4 - \frac{3}{2} \ln 3} \]