Let A = \begin{bmatrix} 2b & 1 \ b & b^2+1 \end{bmatrix} where b > 0. Then the minimum value of det(A)b is:

The correct option is D
-√3/2
A = \begin{bmatrix} 2b & 1 \ b & b^2+1 \end{bmatrix} (b>0)
|A| = 2b(b^2+1) - b(1)
|A| = 2b^3 + 2b - b
|A| = 2b^3 + b
Let f(b) = |A|/b = 2b^2 + 1
For minimum value, we take the derivative and equate it to zero.
f'(b) = 4b = 0
b = 0
However, b > 0, so we check the boundaries.
If b approaches 0, then f(b) approaches 1.
If b approaches infinity, then f(b) approaches infinity.
Let's find the second derivative to check for concavity.
f''(b) = 4 > 0
Thus, the function is concave up, implying a minimum at b = 0.
However, since b > 0, we must consider the limit as b approaches 0 from the positive side.
lim (b→0+) f(b) = lim (b→0+) (2b^2 + 1) = 1
However, this is not one of the options. Let's reconsider the determinant calculation.
|A| = 2b(b^2 + 1) - b = 2b^3 + b
|A|/b = 2b^2 + 1
The minimum value occurs when b is close to 0, so the minimum value is approximately 1.
Let's check the options again. There must be a mistake in the question or options.
Let's re-examine the determinant calculation. It appears there was an error in the original calculation.
|A| = 2b(b^2 + 1) - 1(b) = 2b^3 + 2b - b = 2b^3 + b
Then det(A)/b = 2b^2 + 1. This is minimized when b is close to 0, giving a value close to 1.
There must be a mistake in the provided options or the question statement.