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 find the maximum negative integral value of \( b \) for which the point \( (2b + 3, b^2) \) lies above the line given by the equation \( 3x - 4y - a(a - 2) = 0 \) for all \( a \in \mathbb{R} \), we can follow these steps:

### Step 1: Rewrite the line equation
The line equation can be rewritten in slope-intercept form:
\[ y = \frac{3}{4}x - \frac{a(a - 2)}{4} \]
This shows that the slope of the line is \( \frac{3}{4} \) and the y-intercept depends on \( a \).

### Step 2: Identify the point
The point we are considering is \( P(2b + 3, b^2) \). We need to ensure that the y-coordinate of this point is greater than the y-coordinate of the line for any value of \( a \).

### Step 3: Set up the inequality
For the point \( P \) to be above the line, we need:
\[ b^2 > \frac{3}{4}(2b + 3) - \frac{a(a - 2)}{4} \]
Since \( \frac{a(a - 2)}{4} \) can take any real value, we need to focus on the condition:
\[ b^2 > \frac{3}{4}(2b + 3) \]

### Step 4: Simplify the inequality
Let's simplify the inequality:
\[ b^2 > \frac{3}{4}(2b + 3) \]
Multiplying through by 4 to eliminate the fraction:
\[ 4b^2 > 3(2b + 3) \]
Expanding the right side:
\[ 4b^2 > 6b + 9 \]
Rearranging gives:
\[ 4b^2 - 6b - 9 > 0 \]

### Step 5: Solve the quadratic inequality
To solve the quadratic inequality \( 4b^2 - 6b - 9 > 0 \), we first find the roots using the quadratic formula:
\[ b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot (-9)}}{2 \cdot 4} \]
Calculating the discriminant:
\[ 36 + 144 = 180 \]
Thus, the roots are:
\[ b = \frac{6 \pm \sqrt{180}}{8} = \frac{6 \pm 6\sqrt{5}}{8} = \frac{3 \pm 3\sqrt{5}}{4} \]

### Step 6: Determine intervals
The roots are:
\[ b_1 = \frac{3 - 3\sqrt{5}}{4}, \quad b_2 = \frac{3 + 3\sqrt{5}}{4} \]
The quadratic opens upwards (since the coefficient of \( b^2 \) is positive), so the inequality \( 4b^2 - 6b - 9 > 0 \) holds outside the interval \( (b_1, b_2) \).

### Step 7: Find the maximum negative integral value of \( b \)
Now we need to find the maximum negative integral value of \( b \). The approximate value of \( \sqrt{5} \) is about 2.236, so:
\[ b_1 \approx \frac{3 - 3 \cdot 2.236}{4} \approx \frac{3 - 6.708}{4} \approx \frac{-3.708}{4} \approx -0.927 \]
\[ b_2 \approx \frac{3 + 3 \cdot 2.236}{4} \approx \frac{3 + 6.708}{4} \approx \frac{9.708}{4} \approx 2.427 \]
Thus, the interval where \( 4b^2 - 6b - 9 > 0 \) is \( (-\infty, b_1) \cup (b_2, \infty) \).

The maximum negative integral value of \( b \) that is less than \( -0.927 \) is \( -1 \), but since it is not included, the next maximum negative integral value is \( -2 \).

### Final Answer
Thus, the maximum negative integral value of \( b \) is:
\[ \boxed{-2} \]