Suppose two numbers a and b are chosen randomly from the set {1, 2, 3, 4, 5, 6}. The probability that function `f(x) = x^3+ ax^2 + bx` is strictly increasing function on R is:

Options

A. `1/15`
B. `2/15`
C. `13/15`
D. `14/15`

Answer

To solve the problem, we need to find the probability that the function \( f(x) = x^3 + ax^2 + bx \) is strictly increasing for randomly chosen values of \( a \) and \( b \) from the set \( \{1, 2, 3, 4, 5, 6\} \).

### Step 1: Determine the condition for strict increase
A function is strictly increasing if its derivative is always positive. We first find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 + ax^2 + bx) = 3x^2 + 2ax + b \]
We need \( f'(x) > 0 \) for all \( x \).

### Step 2: Analyze the quadratic form
The expression \( 3x^2 + 2ax + b \) is a quadratic function in \( x \). For this quadratic to be always positive, it must not have any real roots. This occurs when the discriminant is less than zero.

### Step 3: Calculate 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 quadratic to be strictly positive, we need:
\[ 4a^2 - 12b < 0 \]
This simplifies to:
\[ a^2 < 3b \]

### Step 4: Count the favorable outcomes
Now, we will count the pairs \( (a, b) \) such that \( a^2 < 3b \) where \( a \) and \( b \) are chosen from \( \{1, 2, 3, 4, 5, 6\} \).

1. **For \( b = 1 \)**: \( 3b = 3 \) → \( a^2 < 3 \) → Possible values of \( a \): None (since \( a \) must be at least 1).
2. **For \( b = 2 \)**: \( 3b = 6 \) → \( a^2 < 6 \) → Possible values of \( a \): 1, 2 (2 values).
3. **For \( b = 3 \)**: \( 3b = 9 \) → \( a^2 < 9 \) → Possible values of \( a \): 1, 2, 3 (3 values).
4. **For \( b = 4 \)**: \( 3b = 12 \) → \( a^2 < 12 \) → Possible values of \( a \): 1, 2, 3 (3 values).
5. **For \( b = 5 \)**: \( 3b = 15 \) → \( a^2 < 15 \) → Possible values of \( a \): 1, 2, 3, 4 (4 values).
6. **For \( b = 6 \)**: \( 3b = 18 \) → \( a^2 < 18 \) → Possible values of \( a \): 1, 2, 3, 4 (4 values).

### Step 5: Total favorable cases
Now, we sum the number of favorable pairs:
- For \( b = 1 \): 0
- For \( b = 2 \): 2
- For \( b = 3 \): 3
- For \( b = 4 \): 3
- For \( b = 5 \): 4
- For \( b = 6 \): 4

Total favorable cases = \( 0 + 2 + 3 + 3 + 4 + 4 = 16 \).

### Step 6: Total possible cases
The total number of ways to choose \( a \) and \( b \) from the set \( \{1, 2, 3, 4, 5, 6\} \) is \( 6 \times 6 = 36 \) (since \( a \) and \( b \) can be the same).

### Step 7: Calculate the probability
The probability \( P \) that the function is strictly increasing is given by:
\[ P = \frac{\text{Total favorable cases}}{\text{Total possible cases}} = \frac{16}{36} = \frac{4}{9} \]

### Final Answer
Thus, the probability that the function \( f(x) \) is strictly increasing is \( \frac{4}{9} \).

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