Let `f (x) =ax ^(2) +bx + c,a ne 0,` such the `f (-1-x)=f (-1+ x) AA x in R.` Also given that `f (x) =0` has no real roots and `4a + b gt 0.`
Let `alpha =4a -2b+c, beta =9a+3b+c, gamma =9a -3b+c,` then which of the following is correct ?

Options

A. `beta lt alpha lt gamma`
B. `gamma lt alpha lt beta`
C. `alpha lt gamma lt beta`
D. `alpha lt beta lt gamma`

Answer

To solve the problem, we start with the function given:

\[ f(x) = ax^2 + bx + c \]

where \( a \neq 0 \). We are given that:

\[ f(-1 - x) = f(-1 + x) \quad \forall x \in \mathbb{R} \]

### Step 1: Analyzing the Symmetry Condition

The condition \( f(-1 - x) = f(-1 + x) \) implies that the function is symmetric about \( x = -1 \). This means that the vertex of the parabola represented by \( f(x) \) lies on the line \( x = -1 \).

### Step 2: Finding the Vertex

The vertex \( x_v \) of a quadratic function \( f(x) = ax^2 + bx + c \) is given by:

\[ x_v = -\frac{b}{2a} \]

Setting this equal to -1 for symmetry, we have:

\[ -\frac{b}{2a} = -1 \implies b = 2a \]

### Step 3: Substituting \( b \) into the Function

Now substituting \( b = 2a \) into the function:

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

### Step 4: Condition for No Real Roots

The condition that \( f(x) = 0 \) has no real roots implies that the discriminant must be negative:

\[ D = b^2 - 4ac < 0 \]

Substituting \( b = 2a \):

\[ (2a)^2 - 4ac < 0 \implies 4a^2 - 4ac < 0 \implies a^2 - ac < 0 \implies a(a - c) < 0 \]

This means either \( a > 0 \) and \( c < a \) or \( a < 0 \) and \( c > a \). Given \( 4a + b > 0 \):

\[ 4a + 2a > 0 \implies 6a > 0 \implies a > 0 \]

Thus, we conclude \( a > 0 \) and \( c < a \).

### Step 5: Calculating \( \alpha, \beta, \gamma \)

Now we calculate \( \alpha, \beta, \gamma \):

1. **For \( \alpha \)**:
\[ \alpha = 4a - 2b + c = 4a - 2(2a) + c = 4a - 4a + c = c \]

2. **For \( \beta \)**:
\[ \beta = 9a + 3b + c = 9a + 3(2a) + c = 9a + 6a + c = 15a + c \]

3. **For \( \gamma \)**:
\[ \gamma = 9a - 3b + c = 9a - 3(2a) + c = 9a - 6a + c = 3a + c \]

### Step 6: Comparing \( \alpha, \beta, \gamma \)

Now we compare \( \alpha, \beta, \gamma \):

- We know \( \alpha = c \)
- \( \gamma = 3a + c \)
- \( \beta = 15a + c \)

Since \( a > 0 \), we have:

\[ \alpha < \gamma < \beta \quad \text{(because \( c < a \))} \]

### Conclusion

Thus, the correct order is:

\[ \alpha < \gamma < \beta \]

The correct option is **(C) \( \alpha < \gamma < \beta \)**.