If `a ,b ,c in {1,2,3,4,5,6,}` find the number of ways `a, b, c` can be selected if `f(x)=x^(3)+a x^2+b x+c` is an increasing function.

Options

A. 90
B. 95
C. 96
D. 98

Answer

To solve the problem of finding the number of ways to select \( a, b, c \) from the set \( \{1, 2, 3, 4, 5, 6\} \) such that the function \( f(x) = x^3 + ax^2 + bx + c \) is an increasing function, we will follow these steps:

### Step 1: Understand the condition for \( f(x) \) to be increasing
A function is increasing if its derivative \( f'(x) \) is non-negative for all \( x \).

### Step 2: Find the derivative \( f'(x) \)
The derivative of the function is given by:
\[ f'(x) = 3x^2 + 2ax + b \]
This is a quadratic function in \( x \).

### Step 3: Determine the conditions for \( f'(x) \) to be non-negative
For the quadratic \( f'(x) = 3x^2 + 2ax + b \) to be non-negative for all \( x \), its discriminant must be less than or equal to zero. The discriminant \( D \) of a quadratic \( Ax^2 + Bx + C \) is given by \( D = B^2 - 4AC \).

### Step 4: Calculate the discriminant
Here, \( A = 3 \), \( B = 2a \), and \( C = b \). Thus, the discriminant is:
\[ D = (2a)^2 - 4 \cdot 3 \cdot b = 4a^2 - 12b \]
For \( f'(x) \) to be non-negative, we need:
\[ 4a^2 - 12b \leq 0 \]
This simplifies to:
\[ a^2 \leq 3b \]

### Step 5: Analyze the inequality for different values of \( a \)
Now we will analyze the inequality \( a^2 \leq 3b \) for each possible value of \( a \) from the set \( \{1, 2, 3, 4, 5, 6\} \):

- **If \( a = 1 \)**:
\[ 1^2 \leq 3b \implies 1 \leq 3b \implies b \geq \frac{1}{3} \quad \text{(valid for all } b = 1, 2, 3, 4, 5, 6\text{)} \]
Possible values for \( b \): \( 1, 2, 3, 4, 5, 6 \) (6 options)

- **If \( a = 2 \)**:
\[ 2^2 \leq 3b \implies 4 \leq 3b \implies b \geq \frac{4}{3} \quad \text{(valid for } b = 2, 3, 4, 5, 6\text{)} \]
Possible values for \( b \): \( 2, 3, 4, 5, 6 \) (5 options)

- **If \( a = 3 \)**:
\[ 3^2 \leq 3b \implies 9 \leq 3b \implies b \geq 3 \quad \text{(valid for } b = 3, 4, 5, 6\text{)} \]
Possible values for \( b \): \( 3, 4, 5, 6 \) (4 options)

- **If \( a = 4 \)**:
\[ 4^2 \leq 3b \implies 16 \leq 3b \implies b \geq \frac{16}{3} \quad \text{(valid for } b = 6\text{)} \]
Possible values for \( b \): \( 6 \) (1 option)

- **If \( a = 5 \)**:
\[ 5^2 \leq 3b \implies 25 \leq 3b \implies b \geq \frac{25}{3} \quad \text{(not valid for any } b\text{)} \]
Possible values for \( b \): None (0 options)

- **If \( a = 6 \)**:
\[ 6^2 \leq 3b \implies 36 \leq 3b \implies b \geq 12 \quad \text{(not valid for any } b\text{)} \]
Possible values for \( b \): None (0 options)

### Step 6: Count the total combinations
Now we can summarize the valid combinations:
- For \( a = 1 \): 6 options for \( b \)
- For \( a = 2 \): 5 options for \( b \)
- For \( a = 3 \): 4 options for \( b \)
- For \( a = 4 \): 1 option for \( b \)
- For \( a = 5 \): 0 options for \( b \)
- For \( a = 6 \): 0 options for \( b \)

Thus, the total number of valid pairs \( (a, b) \) is:
\[ 6 + 5 + 4 + 1 + 0 + 0 = 16 \]

### Step 7: Consider the selection of \( c \)
Since \( c \) can be any value from the set \( \{1, 2, 3, 4, 5, 6\} \), there are 6 options for \( c \).

### Step 8: Calculate the total number of ways to select \( a, b, c \)
The total number of ways to select \( a, b, c \) such that \( f(x) \) is an increasing function is:
\[ 16 \times 6 = 96 \]

### Final Answer
The total number of ways to select \( a, b, c \) such that \( f(x) \) is an increasing function is **96**.