A function f is such that `f(x)=x^3+ax^2+bx+c`, where a, b, c are chosen by throwing a die three times. The probability that f is an increasing function is

Options

A. `1/9`
B. `2/9`
C. `1/3`
D. `4/9`

Answer

To solve the problem, we need to find the probability that the function \( f(x) = x^3 + ax^2 + bx + c \) is an increasing function, given that \( a, b, c \) are chosen by throwing a die three times.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function is given as \( f(x) = x^3 + ax^2 + bx + c \). For this function to be increasing, its derivative \( f'(x) \) must be non-negative for all \( x \).

2. **Finding the Derivative**:
We differentiate \( f(x) \):
\[ f'(x) = 3x^2 + 2ax + b \]

3. **Condition for Increasing Function**:
The function \( f(x) \) is increasing if \( f'(x) \geq 0 \) for all \( x \). This means that the quadratic \( 3x^2 + 2ax + b \) must not have any real roots, or it must touch the x-axis at most once. This is ensured if the discriminant of the quadratic is less than or equal to zero.

4. **Calculating the Discriminant**:
The discriminant \( D \) of the quadratic \( 3x^2 + 2ax + b \) is given by:
\[ D = (2a)^2 - 4 \cdot 3 \cdot b = 4a^2 - 12b \]
For the function to be increasing, we need:
\[ 4a^2 - 12b \leq 0 \]
This simplifies to:
\[ a^2 \leq 3b \]

5. **Choosing Values for a, b, c**:
Since \( a, b, c \) are chosen by throwing a die three times, each can take values from 1 to 6. We need to count the number of valid pairs \( (a, b) \) that satisfy \( a^2 \leq 3b \).

6. **Counting Valid Pairs**:
We will check each possible value of \( a \) (1 through 6) and find the corresponding valid values of \( b \):

- For \( a = 1 \): \( 1^2 \leq 3b \) → \( b \geq \frac{1}{3} \) → valid \( b = 1, 2, 3, 4, 5, 6 \) (6 options)
- For \( a = 2 \): \( 2^2 \leq 3b \) → \( b \geq \frac{4}{3} \) → valid \( b = 2, 3, 4, 5, 6 \) (5 options)
- For \( a = 3 \): \( 3^2 \leq 3b \) → \( b \geq 3 \) → valid \( b = 3, 4, 5, 6 \) (4 options)
- For \( a = 4 \): \( 4^2 \leq 3b \) → \( b \geq \frac{16}{3} \) → valid \( b = 5, 6 \) (2 options)
- For \( a = 5 \): \( 5^2 \leq 3b \) → \( b \geq \frac{25}{3} \) → valid \( b = 6 \) (1 option)
- For \( a = 6 \): \( 6^2 \leq 3b \) → \( b \geq 12 \) → no valid \( b \) (0 options)

Now we sum the valid options:
\[ 6 + 5 + 4 + 2 + 1 + 0 = 18 \]

7. **Total Sample Space**:
The total number of outcomes when throwing a die three times is:
\[ 6 \times 6 \times 6 = 216 \]

8. **Calculating the Probability**:
The probability \( P \) that \( f(x) \) is an increasing function is given by:
\[ P = \frac{\text{Number of valid pairs}}{\text{Total outcomes}} = \frac{18}{216} = \frac{1}{12} \]

Since the problem states that the probability can be expressed as \( \frac{k}{9} \), we can equate:
\[ \frac{1}{12} = \frac{k}{9} \]

Cross-multiplying gives:
\[ 9 = 12k \implies k = \frac{9}{12} = \frac{3}{4} \]

### Final Answer:
Thus, the value of \( k \) is \( \frac{3}{4} \).