Let a function `f (x)=` be such that `f(x)=||x^(2)-3|-2|`.
Equation `f (x)=lambda` has 4 solutions if :

Options

A. `lambda=2,0`
B. `lambda ge 2`
C. `1 le lambda le 2`
D. `1 le lambda lt 2`

Answer

To solve the problem, we need to analyze the function \( f(x) = ||x^2 - 3| - 2| \) and determine the conditions under which the equation \( f(x) = \lambda \) has four solutions.

### Step 1: Understand the function structure
The function \( f(x) \) is composed of nested absolute values. We can break it down step by step.

1. **Inner function**: Start with \( g(x) = x^2 - 3 \).
2. **First absolute value**: \( h(x) = |g(x)| = |x^2 - 3| \).
3. **Second step**: \( k(x) = h(x) - 2 = |x^2 - 3| - 2 \).
4. **Final absolute value**: \( f(x) = |k(x)| = ||x^2 - 3| - 2| \).

### Step 2: Analyze the critical points
To find where \( f(x) = \lambda \) has four solutions, we need to analyze the behavior of \( f(x) \).

1. **Find where \( g(x) = 0 \)**:
\[ x^2 - 3 = 0 \implies x = \pm \sqrt{3} \]

2. **Find where \( h(x) = 0 \)**:
\[ |x^2 - 3| = 0 \implies x^2 - 3 = 0 \implies x = \pm \sqrt{3} \]

3. **Find where \( k(x) = 0 \)**:
\[ |x^2 - 3| - 2 = 0 \implies |x^2 - 3| = 2 \]
This gives us two cases:
- Case 1: \( x^2 - 3 = 2 \implies x^2 = 5 \implies x = \pm \sqrt{5} \)
- Case 2: \( x^2 - 3 = -2 \implies x^2 = 1 \implies x = \pm 1 \)

### Step 3: Determine the behavior of \( f(x) \)
Now we have critical points at \( x = -\sqrt{5}, -\sqrt{3}, -1, 1, \sqrt{3}, \sqrt{5} \).

### Step 4: Sketch the graph
1. **Behavior of \( f(x) \)**:
- For \( x < -\sqrt{5} \) and \( x > \sqrt{5} \), \( f(x) \) will be increasing.
- Between \( -\sqrt{5} \) and \( -\sqrt{3} \), \( f(x) \) will decrease to a minimum.
- Between \( -\sqrt{3} \) and \( -1 \), \( f(x) \) will increase.
- Between \( -1 \) and \( 1 \), \( f(x) \) will decrease to a minimum.
- Between \( 1 \) and \( \sqrt{3} \), \( f(x) \) will increase.
- Between \( \sqrt{3} \) and \( \sqrt{5} \), \( f(x) \) will decrease to a minimum.

### Step 5: Find values of \( \lambda \)
To have four solutions, the horizontal line \( y = \lambda \) must intersect the graph of \( f(x) \) at four points.

1. **Minimum values of \( f(x) \)**:
- The minimum value occurs at \( x = -1 \) and \( x = 1 \) where \( f(-1) = f(1) = 1 \).
- The minimum value occurs at \( x = \sqrt{3} \) and \( x = -\sqrt{3} \) where \( f(\sqrt{3}) = f(-\sqrt{3}) = 0 \).

### Conclusion
Thus, for \( f(x) = \lambda \) to have four solutions, \( \lambda \) must be in the range where the graph intersects four times, specifically:
\[ 0 < \lambda < 2 \]
This means that the correct values of \( \lambda \) for which the equation \( f(x) = \lambda \) has four solutions are \( \lambda = 0 \) and \( \lambda = 2 \).

### Final Answer
The equation \( f(x) = \lambda \) has four solutions if \( \lambda \) is in the range \( (0, 2) \).