Let a quadratic equation `P(x)=x^2+ax+1` If `P(x)` is increasing in [1,2] then minimum value of a is A and if `P(x)` is decreasing in [1,2] then maximum value of a is B then `abs(A-B)` is

Answer

To solve the problem, we need to analyze the quadratic function \( P(x) = x^2 + ax + 1 \) and determine the conditions under which it is increasing or decreasing over the interval \([1, 2]\).

### Step-by-Step Solution

1. **Differentiate the Function**:
We start by finding the derivative of \( P(x) \):
\[ P'(x) = \frac{d}{dx}(x^2 + ax + 1) = 2x + a \]

2. **Condition for Increasing Function**:
For \( P(x) \) to be increasing on the interval \([1, 2]\), the derivative must be non-negative:
\[ P'(x) \geq 0 \quad \text{for } x \in [1, 2] \]
This gives us:
\[ 2x + a \geq 0 \]

3. **Finding Minimum Value of \( a \)**:
We need to find the minimum value of \( a \) such that \( 2x + a \geq 0 \) for all \( x \) in \([1, 2]\).
- At \( x = 1 \):
\[ 2(1) + a \geq 0 \implies 2 + a \geq 0 \implies a \geq -2 \]
- At \( x = 2 \):
\[ 2(2) + a \geq 0 \implies 4 + a \geq 0 \implies a \geq -4 \]
The more restrictive condition is \( a \geq -2 \). Thus, the minimum value of \( a \) is:
\[ A = -2 \]

4. **Condition for Decreasing Function**:
For \( P(x) \) to be decreasing on the interval \([1, 2]\), the derivative must be non-positive:
\[ P'(x) \leq 0 \quad \text{for } x \in [1, 2] \]
This gives us:
\[ 2x + a \leq 0 \]

5. **Finding Maximum Value of \( a \)**:
We need to find the maximum value of \( a \) such that \( 2x + a \leq 0 \) for all \( x \) in \([1, 2]\).
- At \( x = 1 \):
\[ 2(1) + a \leq 0 \implies 2 + a \leq 0 \implies a \leq -2 \]
- At \( x = 2 \):
\[ 2(2) + a \leq 0 \implies 4 + a \leq 0 \implies a \leq -4 \]
The more restrictive condition is \( a \leq -4 \). Thus, the maximum value of \( a \) is:
\[ B = -4 \]

6. **Calculating \( |A - B| \)**:
Now we need to find \( |A - B| \):
\[ |A - B| = |-2 - (-4)| = |-2 + 4| = |2| = 2 \]

### Final Answer
Thus, the value of \( |A - B| \) is:
\[ \boxed{2} \]