The maximum negative integral value of b for which the point `(2b+3, b^(2))` lies above the line
`3x-4y-a(a-2)=0, AA ain R` is

Options

A. `-1`
B. `-3`
C. `-2`
D. `-4`

Answer

To solve the problem, we need to determine the maximum negative integral value of \( b \) such that the point \( (2b + 3, b^2) \) lies above the line given by the equation \( 3x - 4y - a(a - 2) = 0 \) for all \( a \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Rearranging the Line Equation**:
The line equation can be rewritten as:
\[ 3x - 4y - a(a - 2) = 0 \implies 4y = 3x - a(a - 2) \implies y = \frac{3}{4}x - \frac{1}{4}a(a - 2) \]

2. **Identifying the Point**:
The point we are considering is \( (2b + 3, b^2) \).

3. **Substituting the Point into the Line Equation**:
For the point to lie above the line, we need:
\[ 4(b^2) > 3(2b + 3) - a(a - 2) \]
Rearranging gives:
\[ 4b^2 > 6b + 9 - a(a - 2) \]
This simplifies to:
\[ 4b^2 - 6b - 9 + a(a - 2) > 0 \]

4. **Analyzing the Quadratic in \( a \)**:
The expression \( a(a - 2) \) is a quadratic in \( a \) and opens upwards. The minimum value occurs at \( a = 1 \):
\[ a(a - 2) = 1(1 - 2) = -1 \]
Thus, the maximum value of \( 4b^2 - 6b - 9 + a(a - 2) \) occurs when \( a(a - 2) \) is minimized, which is \( -1 \).

5. **Setting Up the Inequality**:
We need:
\[ 4b^2 - 6b - 9 - 1 > 0 \implies 4b^2 - 6b - 10 > 0 \]

6. **Finding the Roots of the Quadratic**:
Using the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):
\[ A = 4, B = -6, C = -10 \]
\[ b = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot (-10)}}{2 \cdot 4} = \frac{6 \pm \sqrt{36 + 160}}{8} = \frac{6 \pm \sqrt{196}}{8} = \frac{6 \pm 14}{8} \]
This gives us:
\[ b = \frac{20}{8} = 2.5 \quad \text{and} \quad b = \frac{-8}{8} = -1 \]

7. **Determining the Intervals**:
The quadratic \( 4b^2 - 6b - 10 \) is positive outside the roots:
- \( b < -1 \) or \( b > 2.5 \)

8. **Finding the Maximum Negative Integral Value**:
The maximum negative integral value of \( b \) that satisfies \( b < -1 \) is \( b = -2 \).

### Conclusion:
Thus, the maximum negative integral value of \( b \) for which the point \( (2b + 3, b^2) \) lies above the line for all \( a \in \mathbb{R} \) is:
\[ \boxed{-2} \]