The values of `'a'` for which the quadraic expression `ax^(2)+(a-2)x-2` is negative for exactly two integral values of `x`, belongs to

Options

A. `[-1,1]`
B. `[1,2)`
C. `[-1,1]`
D. `[-2,-1])

Explanation

To solve the problem, we need to find the values of \( a \) for which the quadratic expression \( ax^2 + (a-2)x - 2 \) is negative for exactly two integral values of \( x \).

### Step 1: Identify the quadratic expression
The given quadratic expression is:
\[ f(x) = ax^2 + (a-2)x - 2 \]

### Step 2: Determine the conditions for negativity
For the quadratic expression to be negative for exactly two integral values of \( x \), the parabola must open upwards. This means that the coefficient of \( x^2 \) (which is \( a \)) must be greater than 0:
\[ a > 0 \]

### Step 3: Find the roots of the quadratic equation
To find the roots of the quadratic equation, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = a \), \( b = a - 2 \), and \( c = -2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (a-2)^2 - 4a(-2) = (a-2)^2 + 8a \]
Expanding \( (a-2)^2 \):
\[ (a-2)^2 = a^2 - 4a + 4 \]
Thus,
\[ b^2 - 4ac = a^2 - 4a + 4 + 8a = a^2 + 4a + 4 = (a + 2)^2 \]

### Step 4: Calculate the roots
Now substituting back into the quadratic formula:
\[ x = \frac{-(a-2) \pm \sqrt{(a+2)^2}}{2a} \]
This simplifies to:
\[ x = \frac{-a + 2 \pm (a + 2)}{2a} \]
Calculating the two roots:
1. For the positive sign:
\[ x_1 = \frac{-a + 2 + a + 2}{2a} = \frac{4}{2a} = \frac{2}{a} \]
2. For the negative sign:
\[ x_2 = \frac{-a + 2 - (a + 2)}{2a} = \frac{-2a}{2a} = -1 \]

### Step 5: Determine the interval for \( a \)
The roots are \( x_1 = \frac{2}{a} \) and \( x_2 = -1 \). For the quadratic to be negative between these two roots, we need:
\[ -1 < x < \frac{2}{a} \]
The quadratic will be negative for exactly two integral values of \( x \) when:
\[ \frac{2}{a} - (-1) = \frac{2}{a} + 1 \text{ is an integer} \]
This means \( \frac{2}{a} + 1 \) must be an integer.

Let \( n = \frac{2}{a} + 1 \), then:
\[ \frac{2}{a} = n - 1 \implies a = \frac{2}{n - 1} \]

### Step 6: Find valid integer values for \( n \)
Since \( a > 0 \), \( n - 1 > 0 \) implies \( n > 1 \). The integral values of \( n \) can be \( 2, 3, 4, \ldots \).

Calculating \( a \) for these values:
- For \( n = 2 \): \( a = 2 \)
- For \( n = 3 \): \( a = 1 \)
- For \( n = 4 \): \( a = \frac{2}{3} \) (not valid since \( a \) must be greater than 1)

### Conclusion
Thus, the valid range for \( a \) that satisfies the condition is:
\[ 1 < a < 2 \]
Therefore, the values of \( a \) for which the quadratic expression is negative for exactly two integral values of \( x \) belong to the interval:
\[ \boxed{(1, 2)} \]