The number of integral values of k for which the equation `|x^(2)-5|x|+6|=k` has four solution is

Options

A. 5
B. 2
C. 1
D. none of these

Answer

To find the number of integral values of \( k \) for which the equation \( |x^2 - 5|x| + 6| = k \) has four solutions, we will analyze the function step by step.

### Step 1: Understand the Function
The equation we are dealing with is:
\[ |x^2 - 5|x| + 6| = k \]
We need to analyze the function \( f(x) = |x^2 - 5|x| + 6| \).

### Step 2: Break Down the Absolute Value
The expression \( |x^2 - 5|x| \) can be broken down based on the value of \( x \):
1. For \( x \geq 0 \), \( |x| = x \), so \( f(x) = |x^2 - 5x + 6| \).
2. For \( x < 0 \), \( |x| = -x \), so \( f(x) = |x^2 + 5x + 6| \).

### Step 3: Find the Roots of the Quadratics
#### Case 1: \( x \geq 0 \)
The quadratic \( x^2 - 5x + 6 = 0 \) can be factored:
\[ (x - 2)(x - 3) = 0 \]
Thus, the roots are \( x = 2 \) and \( x = 3 \).

#### Case 2: \( x < 0 \)
The quadratic \( x^2 + 5x + 6 = 0 \) can be factored:
\[ (x + 2)(x + 3) = 0 \]
Thus, the roots are \( x = -2 \) and \( x = -3 \).

### Step 4: Analyze the Function Behavior
- For \( x \geq 0 \), the function \( f(x) \) changes at \( x = 2 \) and \( x = 3 \).
- For \( x < 0 \), the function \( f(x) \) changes at \( x = -2 \) and \( x = -3 \).

### Step 5: Determine the Minimum Value
To find the minimum value of \( f(x) \):
- For \( x \geq 0 \), evaluate \( f(0) = |0^2 - 5(0) + 6| = 6 \).
- For \( x = 2 \) and \( x = 3 \):
- \( f(2) = |2^2 - 5(2) + 6| = |4 - 10 + 6| = 0 \)
- \( f(3) = |3^2 - 5(3) + 6| = |9 - 15 + 6| = 0 \)

- For \( x < 0 \), evaluate \( f(-2) = |-2^2 + 5(-2) + 6| = |4 - 10 + 6| = 0 \).
- For \( x = -3 \):
- \( f(-3) = |-3^2 + 5(-3) + 6| = |9 - 15 + 6| = 0 \).

Thus, the minimum value of \( f(x) \) is \( 0 \).

### Step 6: Determine the Range for \( k \)
The function \( f(x) \) achieves a minimum value of \( 0 \) and goes to infinity as \( |x| \) increases. We need to find values of \( k \) such that the equation has exactly four solutions.

1. For \( k < 0 \): No solutions.
2. For \( 0 < k < 6 \): The function intersects the line \( y = k \) at four points.
3. For \( k = 6 \): The function intersects at two points.
4. For \( k > 6 \): The function intersects at two points.

### Step 7: Count Integral Values of \( k \)
The integral values of \( k \) that satisfy \( 0 < k < 6 \) are \( 1, 2, 3, 4, 5 \).

Thus, the number of integral values of \( k \) for which the equation has four solutions is:
\[ \boxed{5} \]