If `f(x)=ax^(2)+bx+c` and `f(-1) ge -4`, `f(1) le 0` and `f(3) ge 5`, then the least value of `a` is

Options

A. `1//4`
B. `1//8`
C. `1//3`
D. `-1//3`

Explanation

To solve the problem, we need to analyze the given conditions for the quadratic function \( f(x) = ax^2 + bx + c \).

### Step-by-Step Solution:

1. **Evaluate \( f(-1) \)**:
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]
Given that \( f(-1) \geq -4 \):
\[ a - b + c \geq -4 \quad \text{(1)} \]

2. **Evaluate \( f(1) \)**:
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]
Given that \( f(1) \leq 0 \):
\[ a + b + c \leq 0 \quad \text{(2)} \]

3. **Evaluate \( f(3) \)**:
\[ f(3) = a(3)^2 + b(3) + c = 9a + 3b + c \]
Given that \( f(3) \geq 5 \):
\[ 9a + 3b + c \geq 5 \quad \text{(3)} \]

4. **Set up the inequalities**:
We now have three inequalities:
- From (1): \( a - b + c \geq -4 \)
- From (2): \( a + b + c \leq 0 \)
- From (3): \( 9a + 3b + c \geq 5 \)

5. **Express \( c \) in terms of \( a \) and \( b \)**:
From inequality (2):
\[ c \leq -a - b \quad \text{(4)} \]
Substitute (4) into (1):
\[ a - b - a - b \geq -4 \implies -2b \geq -4 \implies b \leq 2 \quad \text{(5)} \]

6. **Substitute \( c \) into (3)**:
Substitute (4) into (3):
\[ 9a + 3b - a - b \geq 5 \implies 8a + 2b \geq 5 \]
Rearranging gives:
\[ 4a + b \geq \frac{5}{2} \quad \text{(6)} \]

7. **Combine inequalities (5) and (6)**:
We have:
- \( b \leq 2 \)
- \( 4a + b \geq \frac{5}{2} \)

Substitute \( b = 2 \) into (6):
\[ 4a + 2 \geq \frac{5}{2} \implies 4a \geq \frac{5}{2} - 2 \implies 4a \geq \frac{1}{2} \implies a \geq \frac{1}{8} \]

8. **Conclusion**:
The least value of \( a \) that satisfies all the inequalities is:
\[ a = \frac{1}{8} \]

### Final Answer:
The least value of \( a \) is \( \frac{1}{8} \).