If the area bounded by `y=2-|2-x| and y=3/|x|` is `(k-3ln3)/(2)`, then `k` is equal to _____.

Answer

To solve the problem of finding 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 start by defining the functions based on the given equations.

1. **For \( y = 2 - |2 - x| \)**:
- When \( x < 2 \): \( |2 - x| = 2 - x \), so \( y = 2 - (2 - x) = x \).
- When \( x \geq 2 \): \( |2 - x| = x - 2 \), so \( y = 2 - (x - 2) = 4 - x \).

Therefore, we can express the function as:
\[ f(x) = \begin{cases} x & \text{if } x < 2 \\ 4 - x & \text{if } x \geq 2 \end{cases} \]

2. **For \( y = \frac{3}{|x|} \)**:
- When \( x > 0 \): \( y = \frac{3}{x} \).
- When \( x < 0 \): \( y = -\frac{3}{x} \).

Thus, we can express the function as:
\[ g(x) = \begin{cases} \frac{3}{x} & \text{if } x > 0 \\ -\frac{3}{x} & \text{if } x < 0 \end{cases} \]

### Step 2: Find Points of Intersection
Next, we need to find the points of intersection of the two functions.

1. **Intersection in the first quadrant**:
Set \( f(x) = g(x) \):
\[ x = \frac{3}{x} \implies x^2 = 3 \implies x = \sqrt{3} \]
Thus, one point of intersection is \( A(\sqrt{3}, \sqrt{3}) \).

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

### Step 3: Calculate the Area
The area between the curves can be calculated by integrating the functions over the appropriate intervals.

1. **Area from \( A \) to \( B \) (from \( \sqrt{3} \) to \( 2 \))**:
\[ \text{Area}_1 = \int_{\sqrt{3}}^2 f(x) \, dx = \int_{\sqrt{3}}^2 x \, dx \]
\[ = \left[ \frac{x^2}{2} \right]_{\sqrt{3}}^2 = \frac{2^2}{2} - \frac{(\sqrt{3})^2}{2} = 2 - \frac{3}{2} = \frac{1}{2} \]

2. **Area from \( B \) to \( C \) (from \( 2 \) to \( 3 \))**:
\[ \text{Area}_2 = \int_{2}^{3} f(x) \, dx = \int_{2}^{3} (4 - x) \, dx \]
\[ = \left[ 4x - \frac{x^2}{2} \right]_{2}^{3} = \left( 12 - \frac{9}{2} \right) - \left( 8 - 2 \right) = 12 - 4.5 - 6 = 1.5 \]

3. **Subtract the area under \( g(x) \) from \( A \) to \( C \)**:
\[ \text{Area}_3 = \int_{\sqrt{3}}^{3} g(x) \, dx = \int_{\sqrt{3}}^{3} \frac{3}{x} \, dx \]
\[ = 3 \left[ \ln |x| \right]_{\sqrt{3}}^{3} = 3 \left( \ln 3 - \frac{1}{2} \ln 3 \right) = 3 \cdot \frac{1}{2} \ln 3 = \frac{3}{2} \ln 3 \]

### Step 4: Total Area Calculation
The total area \( A \) is given by:
\[ A = \text{Area}_1 + \text{Area}_2 - \text{Area}_3 = \frac{1}{2} + 1.5 - \frac{3}{2} \ln 3 \]
\[ = 2 - \frac{3}{2} \ln 3 \]

### Step 5: Relate to Given Area Expression
According to the problem, the area is given as:
\[ A = \frac{k - 3 \ln 3}{2} \]
Setting the two expressions for area equal gives:
\[ 2 - \frac{3}{2} \ln 3 = \frac{k - 3 \ln 3}{2} \]
Multiplying through by 2:
\[ 4 - 3 \ln 3 = k - 3 \ln 3 \]
Thus, we find:
\[ k = 4 \]

### Final Answer
The value of \( k \) is \( \boxed{4} \).