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 `p =b-4a, q=2a +b,` then pq is:

Options

A. negative
B. positive
C. 0
D. nothing can be said

Answer

To solve the problem step by step, we will analyze the given conditions and derive the required values for \( p \) and \( q \).

### Step 1: Understand the symmetry condition
We are given that \( f(-1 - x) = f(-1 + x) \) for all \( x \in \mathbb{R} \). This implies that the function \( f(x) \) is symmetric about \( x = -1 \).

### Step 2: Rewrite the function
The function can be expressed as:
\[ f(x) = a(x + 1)^2 + k \]
where \( k \) is some constant. This form ensures that the vertex of the parabola is at \( x = -1 \).

### Step 3: Determine the coefficients
Expanding \( f(x) \):
\[ f(x) = a(x^2 + 2x + 1) + k = ax^2 + 2ax + (a + k) \]
From this, we can identify:
- \( b = 2a \)
- \( c = a + k \)

### Step 4: Analyze the condition for no real roots
The condition that \( f(x) = 0 \) has no real roots implies that the discriminant must be less than zero:
\[ b^2 - 4ac < 0 \]
Substituting \( b \) and \( c \):
\[ (2a)^2 - 4a(a + k) < 0 \]
This simplifies to:
\[ 4a^2 - 4a^2 - 4ak < 0 \implies -4ak < 0 \]
Since \( a > 0 \) (as derived later), this implies:
\[ k > 0 \]

### Step 5: Use the condition \( 4a + b > 0 \)
Substituting \( b = 2a \):
\[ 4a + 2a > 0 \implies 6a > 0 \]
This confirms that \( a > 0 \).

### Step 6: Calculate \( p \) and \( q \)
Now, we calculate:
\[ p = b - 4a = 2a - 4a = -2a \]
\[ q = 2a + b = 2a + 2a = 4a \]

### Step 7: Calculate \( pq \)
Now, we find \( pq \):
\[ pq = (-2a)(4a) = -8a^2 \]
Since \( a > 0 \), we have \( -8a^2 < 0 \).

### Conclusion
Thus, \( pq \) is negative.

### Final Answer
\[ pq < 0 \]