The minimum possible distnace between the points `A(a, a-1)` and `B(b, b^(2)+b+1)AA a, b in R` is D units, then the value of `D^(2)` is

Answer

To find the minimum possible distance between the points \( A(a, a-1) \) and \( B(b, b^2 + b + 1) \), we will follow these steps:

### Step 1: Write the distance formula
The distance \( D \) between the points \( A(a, a-1) \) and \( B(b, b^2 + b + 1) \) can be expressed using the distance formula:
\[ D = \sqrt{(b - a)^2 + \left((b^2 + b + 1) - (a - 1)\right)^2} \]

### Step 2: Simplify the expression
We can simplify the expression for \( D \):
\[ D = \sqrt{(b - a)^2 + (b^2 + b + 1 - a + 1)^2} \]
\[ D = \sqrt{(b - a)^2 + (b^2 + b - a + 2)^2} \]

### Step 3: Find the critical points
To minimize \( D \), we can minimize \( D^2 \) instead:
\[ D^2 = (b - a)^2 + (b^2 + b - a + 2)^2 \]
Let \( f(a, b) = (b - a)^2 + (b^2 + b - a + 2)^2 \).

### Step 4: Differentiate with respect to \( a \) and \( b \)
To find the minimum, we need to take partial derivatives and set them to zero:
1. Differentiate \( f \) with respect to \( a \):
\[ \frac{\partial f}{\partial a} = -2(b - a) - 2(b^2 + b - a + 2) = 0 \]
2. Differentiate \( f \) with respect to \( b \):
\[ \frac{\partial f}{\partial b} = 2(b - a) + 2(b^2 + b - a + 2)(2b + 1) = 0 \]

### Step 5: Solve the equations
From the first equation, we can express \( a \) in terms of \( b \):
\[ -2(b - a) - 2(b^2 + b - a + 2) = 0 \implies b - a + b^2 + b - a + 2 = 0 \]
This simplifies to:
\[ b^2 + 2b + 2 - 2a = 0 \implies a = \frac{b^2 + 2b + 2}{2} \]

Substituting \( a \) into the second equation and solving for \( b \) will yield the critical points.

### Step 6: Calculate the minimum distance
Once we find the values of \( a \) and \( b \) that minimize \( D^2 \), we can substitute them back into the distance formula to find \( D \).

### Step 7: Compute \( D^2 \)
After finding the optimal values of \( a \) and \( b \), we compute \( D^2 \).

### Final Result
After performing the calculations, we find that the minimum distance \( D \) is \( \sqrt{2} \), therefore:
\[ D^2 = 2 \]