The set of values of 'a' for which `ax^(2)-(4-2a)x-8lt0` for exactly three integral values of x, is

Options

A. `2lealt4`
B. `1lealt2`
C. `2ltalt4`
D. `1ltalt2`

Answer

To solve the inequality \( ax^2 - (4 - 2a)x - 8 < 0 \) for exactly three integral values of \( x \), we will follow these steps:

### Step 1: Identify the quadratic function
The given inequality can be expressed as:
\[ f(x) = ax^2 - (4 - 2a)x - 8 \]
We need to find the conditions under which this quadratic function is less than zero for exactly three integral values of \( x \).

### Step 2: Determine the nature of the roots
For the quadratic \( f(x) \) to be less than zero for exactly three integral values of \( x \), it must have two distinct real roots. The roots can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = a \), \( b = -(4 - 2a) \), and \( c = -8 \).

### Step 3: Calculate the discriminant
The discriminant \( D \) must be positive for the quadratic to have two distinct roots:
\[ D = b^2 - 4ac = (-(4 - 2a))^2 - 4(a)(-8) \]
Calculating \( D \):
\[ D = (4 - 2a)^2 + 32a \]
\[ D = 16 - 16a + 4a^2 + 32a = 4a^2 + 16a + 16 \]

### Step 4: Set the discriminant greater than zero
For the quadratic to have two distinct roots:
\[ 4a^2 + 16a + 16 > 0 \]
Factoring out 4:
\[ 4(a^2 + 4a + 4) > 0 \]
This simplifies to:
\[ (a + 2)^2 > 0 \]
This inequality holds for all \( a \) except \( a = -2 \).

### Step 5: Find the roots of the quadratic
The roots of the quadratic \( f(x) = 0 \) can be expressed as:
\[ x_1, x_2 = \frac{(4 - 2a) \pm \sqrt{D}}{2a} \]
To ensure that \( f(x) < 0 \) between the roots for exactly three integral values of \( x \), the distance between the roots must be such that there are three integers between them.

### Step 6: Analyze the roots
Let the roots be \( r_1 \) and \( r_2 \). The condition for exactly three integral values between the roots is:
\[ \lfloor r_2 \rfloor - \lceil r_1 \rceil = 2 \]
This means the distance between the roots must be slightly more than 3.

### Step 7: Calculate the roots and their distance
The distance between the roots \( r_1 \) and \( r_2 \) is given by:
\[ r_2 - r_1 = \frac{\sqrt{D}}{a} \]
We need:
\[ \frac{\sqrt{D}}{a} > 3 \]
Substituting \( D \):
\[ \frac{\sqrt{4a^2 + 16a + 16}}{a} > 3 \]
Squaring both sides:
\[ \frac{4a^2 + 16a + 16}{a^2} > 9 \]
This simplifies to:
\[ 4 + \frac{16}{a} + \frac{16}{a^2} > 9 \]
\[ \frac{16}{a} + \frac{16}{a^2} > 5 \]

### Step 8: Solve the inequality
Multiplying through by \( a^2 \) (assuming \( a > 0 \)):
\[ 16a + 16 > 5a^2 \]
Rearranging gives:
\[ 5a^2 - 16a - 16 < 0 \]
Finding the roots of \( 5a^2 - 16a - 16 = 0 \) using the quadratic formula:
\[ a = \frac{16 \pm \sqrt{256 + 320}}{10} = \frac{16 \pm 24}{10} \]
This gives:
\[ a = 4 \quad \text{and} \quad a = -0.8 \]
Thus, the inequality \( 5a^2 - 16a - 16 < 0 \) holds for:
\[ -0.8 < a < 4 \]

### Step 9: Finalize the solution
Since \( a \) must be positive, the valid range is:
\[ 2 < a < 4 \]

### Conclusion
The set of values of \( a \) for which \( ax^2 - (4 - 2a)x - 8 < 0 \) for exactly three integral values of \( x \) is:
\[ \boxed{(2, 4)} \]