If `f(x)=x^(3)-12x+p,p in {1,2,3,…,15}` and for each 'p', the number of real roots of equation f(x)=0 is denoted by `theta`, the `1/5 sum theta` is equal to ……….. .

Answer

To solve the problem, we need to analyze the function \( f(x) = x^3 - 12x + p \) and determine the number of real roots it has for each integer value of \( p \) in the set \( \{1, 2, 3, \ldots, 15\} \). We will denote the number of real roots for each \( p \) as \( \theta \) and then calculate \( \frac{1}{5} \sum \theta \).

### Step-by-Step Solution:

1. **Find the derivative of \( f(x) \)**:
\[ f'(x) = 3x^2 - 12 \]
This derivative will help us find the critical points.

2. **Set the derivative to zero to find critical points**:
\[ 3x^2 - 12 = 0 \implies x^2 = 4 \implies x = -2, 2 \]
So, the critical points are \( x = -2 \) and \( x = 2 \).

3. **Determine the behavior of \( f(x) \) at the critical points**:
- Calculate \( f(-2) \):
\[ f(-2) = (-2)^3 - 12(-2) + p = -8 + 24 + p = 16 + p \]
- Calculate \( f(2) \):
\[ f(2) = (2)^3 - 12(2) + p = 8 - 24 + p = -16 + p \]

4. **Analyze the values of \( f(-2) \) and \( f(2) \)**:
- For \( f(-2) \geq 0 \):
\[ 16 + p \geq 0 \implies p \geq -16 \quad \text{(always true since \( p \in \{1, 2, \ldots, 15\} \))} \]
- For \( f(2) \leq 0 \):
\[ -16 + p \leq 0 \implies p \leq 16 \quad \text{(always true since \( p \in \{1, 2, \ldots, 15\} \))} \]

5. **Determine the number of real roots based on the critical points**:
- Since \( f(x) \) is a cubic polynomial, it can have either 1 or 3 real roots.
- The function changes from positive to negative or vice versa at the critical points, which indicates the presence of real roots.

6. **Evaluate the number of roots for \( p = 1, 2, \ldots, 15 \)**:
- For \( p = 1 \): \( f(-2) = 17 \) (positive), \( f(2) = -15 \) (negative) → 3 real roots.
- For \( p = 2 \): \( f(-2) = 18 \) (positive), \( f(2) = -14 \) (negative) → 3 real roots.
- Continuing this way, for \( p = 3 \) to \( p = 15 \), \( f(-2) \) remains positive and \( f(2) \) remains negative → 3 real roots for each \( p \).

7. **Count the total number of roots**:
- Since there are 15 values of \( p \) and each gives 3 real roots, we have:
\[ \sum \theta = 15 \times 3 = 45 \]

8. **Calculate \( \frac{1}{5} \sum \theta \)**:
\[ \frac{1}{5} \sum \theta = \frac{1}{5} \times 45 = 9 \]

### Final Answer:
\[ \frac{1}{5} \sum \theta = 9 \]