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 will first analyze the given conditions and then derive the relationships between α, β, and γ.

### Step 1: Understanding the function and symmetry condition
Given the function:
\[ f(x) = ax^2 + bx + c \]
where \( a \neq 0 \). The condition \( f(-1-x) = f(-1+x) \) implies that the function is symmetric about \( x = -1 \).

### Step 2: Finding the relationship between coefficients
To satisfy the symmetry condition, we can substitute \( x = -1 - x \) and \( x = -1 + x \) into the function:
\[ f(-1-x) = a(-1-x)^2 + b(-1-x) + c \]
\[ = a(1 + 2x + x^2) - b(1 + x) + c \]
\[ = ax^2 + (2a - b)x + (a - b + c) \]

And for \( f(-1+x) \):
\[ f(-1+x) = a(-1+x)^2 + b(-1+x) + c \]
\[ = a(1 - 2x + x^2) - b(1 - x) + c \]
\[ = ax^2 + (-2a + b)x + (a + b + c) \]

Setting these two expressions equal gives us:
\[ (2a - b) = (-2a + b) \quad \text{(coefficient of } x\text{)} \]
\[ (a - b + c) = (a + b + c) \quad \text{(constant term)} \]

From the first equation:
\[ 2a - b = -2a + b \implies 4a = 2b \implies b = 2a \]

From the second equation:
\[ a - b + c = a + b + c \implies -b = b \implies 2b = 0 \implies b = 0 \]
This contradicts our previous finding, so we need to check the conditions again.

### Step 3: Analyzing the conditions
We know that:
1. \( f(x) = 0 \) has no real roots, which implies the discriminant \( D = b^2 - 4ac < 0 \).
2. Given \( 4a + b > 0 \).

Substituting \( b = 2a \) into \( 4a + b > 0 \):
\[ 4a + 2a > 0 \implies 6a > 0 \implies a > 0 \]

### Step 4: Finding α, β, and γ
Now we can express α, β, and γ:
- \( \alpha = 4a - 2b + c = 4a - 2(2a) + c = 4a - 4a + c = c \)
- \( \beta = 9a + 3b + c = 9a + 3(2a) + c = 9a + 6a + c = 15a + c \)
- \( \gamma = 9a - 3b + c = 9a - 3(2a) + c = 9a - 6a + c = 3a + c \)

### Step 5: Comparing α, β, and γ
Now we compare:
- \( \alpha = c \)
- \( \beta = 15a + c \)
- \( \gamma = 3a + c \)

Since \( a > 0 \):
- \( \beta > \alpha \) (because \( 15a + c > c \))
- \( \gamma > \alpha \) (because \( 3a + c > c \))
- To compare β and γ:
\[ \beta - \gamma = (15a + c) - (3a + c) = 12a > 0 \implies \beta > \gamma \]

### Conclusion
Thus, the relationships are:
\[ \beta > \gamma > \alpha \]