Let `f (x) = || x^(2)-10x+21|-p|,` then the exhausive set of values of for which f (x) has exactly 6 points of non-derivability, is:

Options

A. `(4,oo)`
B. `(0,4)`
C. `[0,4]`
D. `(-4,4)`

Answer

To solve the problem, we need to analyze the function \( f(x) = ||x^2 - 10x + 21| - p| \) and determine the values of \( p \) for which the function has exactly 6 points of non-differentiability.

### Step 1: Analyze the inner function
First, we simplify the inner function:
\[ g(x) = x^2 - 10x + 21 \]
This is a quadratic function. We can factor it:
\[ g(x) = (x - 3)(x - 7) \]
The roots of this equation are \( x = 3 \) and \( x = 7 \). The quadratic opens upwards (the coefficient of \( x^2 \) is positive), so it has a minimum point between these roots.

### Step 2: Find the vertex of the quadratic
To find the vertex (minimum point), we use the formula for the x-coordinate of the vertex:
\[ x = -\frac{b}{2a} = \frac{10}{2} = 5 \]
Now we can find \( g(5) \):
\[ g(5) = 5^2 - 10 \cdot 5 + 21 = 25 - 50 + 21 = -4 \]

### Step 3: Analyze the absolute value
Now we consider the absolute value:
\[ |g(x)| = |x^2 - 10x + 21| \]
The function \( g(x) \) is negative between its roots (from \( x = 3 \) to \( x = 7 \)) and positive outside this interval. Thus, the absolute value function will reflect the negative part:
- For \( x < 3 \) and \( x > 7 \): \( |g(x)| = g(x) \)
- For \( 3 \leq x \leq 7 \): \( |g(x)| = -g(x) \)

### Step 4: Determine points of non-differentiability
Next, we analyze the function \( f(x) = ||g(x)| - p| \). The points of non-differentiability occur at:
1. The points where \( g(x) = 0 \) (i.e., \( x = 3 \) and \( x = 7 \)).
2. The points where \( |g(x)| = p \).

### Step 5: Set conditions for non-differentiability
To have exactly 6 points of non-differentiability, we need to find \( p \) such that \( |g(x)| = p \) intersects the graph of \( |g(x)| \) at 4 additional points (2 from \( g(x) = p \) and 2 from \( g(x) = -p \)).

### Step 6: Analyze the values of \( p \)
The minimum value of \( |g(x)| \) is \( 4 \) (at \( x = 5 \)). Thus, for \( p < 4 \), the equation \( |g(x)| = p \) will have 4 intersections:
- 2 intersections from \( g(x) = p \) (above the x-axis).
- 2 intersections from \( g(x) = -p \) (below the x-axis).

For \( p = 4 \), there will be 2 intersections (just touching the graph at \( x = 5 \)). For \( p > 4 \), the number of intersections decreases.

### Conclusion
Thus, the exhaustive set of values of \( p \) for which \( f(x) \) has exactly 6 points of non-differentiability is:
\[ 0 < p < 4 \]