The value of 'a' for which the quadratic expression `ax^(2)+|2a-3|x-6` is positive for exactly two integral values of x is

Options

A. `(-(3)/(4), -(3)/(5))`
B. `(-(3)/(4), -(3)/(5)]`
C. `[-(3)/(4), -(3)/(5)]`
D. `(-(3)/(4),-(7)/(10))`

Answer

To find the value of 'a' for which the quadratic expression \( ax^2 + |2a - 3|x - 6 \) is positive for exactly two integral values of \( x \), we can follow these steps:

### Step 1: Analyze the quadratic expression
The expression is of the form \( ax^2 + bx + c \), where:
- \( a = a \)
- \( b = |2a - 3| \)
- \( c = -6 \)

For the quadratic expression to be positive for exactly two integral values of \( x \), the quadratic must touch the x-axis at two points (roots) and be negative between these roots.

### Step 2: Determine the conditions for the quadratic
1. The quadratic must have real roots, which means the discriminant \( D \) must be greater than or equal to zero:
\[ D = b^2 - 4ac \geq 0 \]
Substituting \( b \) and \( c \):
\[ D = (|2a - 3|)^2 - 4a(-6) \geq 0 \]
\[ D = (|2a - 3|)^2 + 24a \geq 0 \]

### Step 3: Analyze the discriminant
We need to consider two cases for \( |2a - 3| \):

**Case 1:** \( 2a - 3 \geq 0 \) (i.e., \( a \geq \frac{3}{2} \))
- Here, \( |2a - 3| = 2a - 3 \):
\[ D = (2a - 3)^2 + 24a \geq 0 \]
Expanding this:
\[ D = 4a^2 - 12a + 9 + 24a \geq 0 \]
\[ D = 4a^2 + 12a + 9 \geq 0 \]
This is always true since it is a perfect square:
\[ D = (2a + 3)^2 \geq 0 \]

**Case 2:** \( 2a - 3 < 0 \) (i.e., \( a < \frac{3}{2} \))
- Here, \( |2a - 3| = -(2a - 3) = 3 - 2a \):
\[ D = (3 - 2a)^2 + 24a \geq 0 \]
Expanding this:
\[ D = 9 - 12a + 4a^2 + 24a \geq 0 \]
\[ D = 4a^2 + 12a + 9 \geq 0 \]
This is also always true since it is a perfect square:
\[ D = (2a + 3)^2 \geq 0 \]

### Step 4: Determine the nature of roots
For the quadratic to be positive for exactly two integral values of \( x \), we need the roots to be distinct and the quadratic must be negative between the roots.

The condition for this is that \( a < 0 \) (the parabola opens downwards).

### Step 5: Find the range of 'a'
From the analysis:
- For \( a < 0 \), we need to ensure that the quadratic has roots. Since the discriminant is always non-negative, we need to check the values of \( a \):
1. The quadratic must be negative at \( x = 0 \) and \( x = 1 \) (the two integral values).
2. Set \( f(0) < 0 \):
\[ f(0) = -6 < 0 \quad \text{(always true)} \]
3. Set \( f(1) < 0 \):
\[ f(1) = a + |2a - 3| - 6 < 0 \]
Depending on the value of \( a \):
- If \( a < \frac{3}{2} \):
\[ f(1) = a + (3 - 2a) - 6 < 0 \]
\[ -a - 3 < 0 \Rightarrow a > -3 \]

### Step 6: Combine conditions
From the conditions:
- \( a < 0 \)
- \( a > -3 \)

Thus, the value of \( a \) must satisfy:
\[ -3 < a < 0 \]

### Conclusion
The value of \( a \) for which the quadratic expression \( ax^2 + |2a - 3|x - 6 \) is positive for exactly two integral values of \( x \) is:
\[ \boxed{(-3, 0)} \]