If `f(x)=ax^(2)+bx+c, f(-1) gt (1)/(2), f(1) lt -1` and `f(-3)lt -(1)/(2)`, then

Options

A. `a=0`
B. `alt0`
C. `agt 0`
D. Sign of a can not be determined

Answer

To solve the problem step by step, we start with the function \( f(x) = ax^2 + bx + c \) and the given inequalities:

1. \( f(-1) > \frac{1}{2} \)
2. \( f(1) < -1 \)
3. \( f(-3) < -\frac{1}{2} \)

### Step 1: Set up the inequalities

We will evaluate the function at the specified points:

- For \( f(-1) \):
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c > \frac{1}{2} \]
This gives us our first inequality:
\[ a - b + c > \frac{1}{2} \quad \text{(Inequality 1)} \]

- For \( f(1) \):
\[ f(1) = a(1)^2 + b(1) + c = a + b + c < -1 \]
This gives us our second inequality:
\[ a + b + c < -1 \quad \text{(Inequality 2)} \]

- For \( f(-3) \):
\[ f(-3) = a(-3)^2 + b(-3) + c = 9a - 3b + c < -\frac{1}{2} \]
This gives us our third inequality:
\[ 9a - 3b + c < -\frac{1}{2} \quad \text{(Inequality 3)} \]

### Step 2: Rearranging the inequalities

Now we have three inequalities:
1. \( a - b + c > \frac{1}{2} \)
2. \( a + b + c < -1 \)
3. \( 9a - 3b + c < -\frac{1}{2} \)

### Step 3: Manipulating the inequalities

To eliminate \( c \), we can manipulate these inequalities.

**Multiply Inequality 1 by 3:**
\[ 3(a - b + c) > 3 \cdot \frac{1}{2} \implies 3a - 3b + 3c > \frac{3}{2} \quad \text{(Inequality 4)} \]

**Now, add Inequality 4 to Inequality 3:**
\[ (9a - 3b + c) + (3a - 3b + 3c) < -\frac{1}{2} + \frac{3}{2} \]
This simplifies to:
\[ 12a - 6b + 4c < 1 \quad \text{(Inequality 5)} \]

### Step 4: Further manipulation

Now, let's manipulate Inequality 2.

**Multiply Inequality 2 by 3:**
\[ 3(a + b + c) < 3 \cdot (-1) \implies 3a + 3b + 3c < -3 \quad \text{(Inequality 6)} \]

**Now, add Inequality 6 to Inequality 3:**
\[ (9a - 3b + c) + (3a + 3b + 3c) < -\frac{1}{2} - 3 \]
This simplifies to:
\[ 12a + 4c < -\frac{7}{2} \quad \text{(Inequality 7)} \]

### Step 5: Combine Inequalities

Now we have two new inequalities:
1. \( 12a - 6b + 4c < 1 \) (from Inequality 5)
2. \( 12a + 4c < -\frac{7}{2} \) (from Inequality 7)

### Step 6: Solve for \( a \)

Now, we can isolate \( a \) from these inequalities.

From Inequality 7:
\[ 4c < -\frac{7}{2} - 12a \implies c < -\frac{7}{8} - 3a \]

Substituting \( c \) in Inequality 5:
\[ 12a - 6b + 4\left(-\frac{7}{8} - 3a\right) < 1 \]
This simplifies to:
\[ 12a - 6b - \frac{7}{2} - 12a < 1 \implies -6b < \frac{9}{2} \implies b > -\frac{3}{4} \]

### Conclusion

From the inequalities, we can conclude that:
\[ a < -\frac{15}{48} \]
Thus, the value of \( a \) must be less than 0.

### Final Answer

The value of \( a \) is less than 0.