Number of ordered pair (a,b) the set `A = {1,2,3,4,5}` so that the functon `f(x)=(x^(3))/(3)+a/2 x ^(2)+ bx+10` is an injective mapping `AA x in R:`

Options

A. `13
B. 14
C. 15
D. 16

Answer

To solve the problem of finding the number of ordered pairs \((a, b)\) from the set \(A = \{1, 2, 3, 4, 5\}\) such that the function

\[ f(x) = \frac{x^3}{3} + \frac{a}{2} x^2 + bx + 10 \]

is an injective mapping, we will follow these steps:

### Step 1: Differentiate the Function
First, we need to find the derivative of the function \(f(x)\):

\[ f'(x) = x^2 + ax + b \]

### Step 2: Analyze the Nature of the Derivative
For \(f(x)\) to be injective, \(f'(x)\) must not change sign. This means that \(f'(x)\) should either be always positive or always negative. A polynomial of degree 2 (like \(f'(x)\)) will not change sign if its discriminant is less than or equal to zero.

### Step 3: Calculate the Discriminant
The discriminant \(D\) of the quadratic \(f'(x) = x^2 + ax + b\) is given by:

\[ D = a^2 - 4b \]

For \(f'(x)\) to not change sign, we need:

\[ D \leq 0 \implies a^2 - 4b \leq 0 \implies a^2 \leq 4b \]

### Step 4: Find Valid Pairs \((a, b)\)
Now we will find the ordered pairs \((a, b)\) from the set \(A\) such that \(a^2 \leq 4b\).

1. **For \(a = 1\)**:
- \(1^2 \leq 4b \implies 1 \leq 4b \implies b \geq \frac{1}{4}\)
- Possible values of \(b\): \(1, 2, 3, 4, 5\) (5 pairs)

2. **For \(a = 2\)**:
- \(2^2 \leq 4b \implies 4 \leq 4b \implies b \geq 1\)
- Possible values of \(b\): \(1, 2, 3, 4, 5\) (5 pairs)

3. **For \(a = 3\)**:
- \(3^2 \leq 4b \implies 9 \leq 4b \implies b \geq \frac{9}{4} = 2.25\)
- Possible values of \(b\): \(3, 4, 5\) (3 pairs)

4. **For \(a = 4\)**:
- \(4^2 \leq 4b \implies 16 \leq 4b \implies b \geq 4\)
- Possible values of \(b\): \(4, 5\) (2 pairs)

5. **For \(a = 5\)**:
- \(5^2 \leq 4b \implies 25 \leq 4b \implies b \geq \frac{25}{4} = 6.25\)
- No valid values of \(b\) since \(b\) must be in \(A\).

### Step 5: Count the Total Ordered Pairs
Now we will sum the valid pairs:

- For \(a = 1\): 5 pairs
- For \(a = 2\): 5 pairs
- For \(a = 3\): 3 pairs
- For \(a = 4\): 2 pairs
- For \(a = 5\): 0 pairs

Total pairs = \(5 + 5 + 3 + 2 + 0 = 15\).

### Final Answer
The total number of ordered pairs \((a, b)\) such that the function \(f(x)\) is an injective mapping is **15**.

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