The area bounded by `y = 2-|2-x| and y=3/|x|` is:

Options

A. (a) `(4+3ln3)/(2)`
B. (b) `19/8-3 ln 2`
C. (c) `3/2+ln3`
D. (c) `1/2+ln3`

Answer

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

### Step 1: Define the functions
We have two functions:
1. \( f(x) = 2 - |2 - x| \)
2. \( g(x) = \frac{3}{|x|} \)

### Step 2: Analyze \( f(x) \)
The function \( f(x) = 2 - |2 - x| \) can be expressed in piecewise form:
- For \( x < 2 \):
\[ f(x) = 2 - (2 - x) = x \]
- For \( x \geq 2 \):
\[ f(x) = 2 - (x - 2) = 4 - x \]

### Step 3: Analyze \( g(x) \)
The function \( g(x) = \frac{3}{|x|} \) can also be expressed in piecewise form:
- For \( x > 0 \):
\[ g(x) = \frac{3}{x} \]
- For \( x < 0 \):
\[ g(x) = -\frac{3}{x} \]
- Note: \( g(x) \) is undefined at \( x = 0 \).

### Step 4: Find points of intersection
To find the area between the curves, we need to find the points where \( f(x) = g(x) \).

1. **For \( x > 0 \)**:
Set \( f(x) = g(x) \):
\[ x = \frac{3}{x} \implies x^2 = 3 \implies x = \sqrt{3} \]
Thus, point \( A \) is \( \left(\sqrt{3}, \sqrt{3}\right) \).

2. **For \( x \geq 2 \)**:
Set \( f(x) = g(x) \):
\[ 4 - x = \frac{3}{x} \implies 4x - x^2 = 3 \implies x^2 - 4x + 3 = 0 \]
Factoring gives:
\[ (x - 3)(x - 1) = 0 \implies x = 3 \text{ or } x = 1 \]
Since we are considering \( x \geq 2 \), point \( C \) is \( (3, 1) \).

### Step 5: Identify the vertex
The vertex \( B \) of \( f(x) \) is at \( (2, 2) \).

### Step 6: Set up the area calculation
The area \( A \) between the curves from \( A \) to \( C \) can be calculated as:
\[ \text{Area} = \int_{\sqrt{3}}^{2} f(x) \, dx + \int_{2}^{3} f(x) \, dx - \int_{\sqrt{3}}^{3} g(x) \, dx \]

### Step 7: Calculate the integrals
1. **First integral**:
\[ \int_{\sqrt{3}}^{2} f(x) \, dx = \int_{\sqrt{3}}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{\sqrt{3}}^{2} = \frac{2^2}{2} - \frac{(\sqrt{3})^2}{2} = 2 - \frac{3}{2} = \frac{1}{2} \]

2. **Second integral**:
\[ \int_{2}^{3} f(x) \, dx = \int_{2}^{3} (4 - x) \, dx = \left[ 4x - \frac{x^2}{2} \right]_{2}^{3} = \left( 12 - \frac{9}{2} \right) - \left( 8 - 2 \right) = 12 - 4.5 - 6 = 1.5 \]

3. **Third integral**:
\[ \int_{\sqrt{3}}^{3} g(x) \, dx = \int_{\sqrt{3}}^{3} \frac{3}{x} \, dx = 3 \left[ \ln |x| \right]_{\sqrt{3}}^{3} = 3 \left( \ln 3 - \frac{1}{2} \ln 3 \right) = 3 \cdot \frac{1}{2} \ln 3 = \frac{3}{2} \ln 3 \]

### Step 8: Combine the areas
Putting it all together:
\[ \text{Area} = \frac{1}{2} + 1.5 - \frac{3}{2} \ln 3 = 2 - \frac{3}{2} \ln 3 \]

### Final Answer
The area bounded by the curves is:
\[ \text{Area} = 4 + 3 \cdot \frac{\ln 3}{2} \]