Let `4x^(2)-4(alpha-2)x + alpha-2=0(alpha in R)` be a quadratic equation. Find the values of `alpha` for which
Exactly one root lies in `(0, (1)/(2))`

Answer

To solve the problem, we need to find the values of \( \alpha \) for which exactly one root of the quadratic equation \( 4x^2 - 4(\alpha-2)x + \alpha-2 = 0 \) lies in the interval \( (0, \frac{1}{2}) \).

### Step 1: Identify the quadratic equation
The given quadratic equation is:
\[ 4x^2 - 4(\alpha - 2)x + (\alpha - 2) = 0 \]

### Step 2: Determine the conditions for the roots
For a quadratic equation \( ax^2 + bx + c = 0 \) to have real and distinct roots, the discriminant \( D \) must be greater than zero:
\[ D = b^2 - 4ac > 0 \]
Here, \( a = 4 \), \( b = -4(\alpha - 2) \), and \( c = \alpha - 2 \).

### Step 3: Calculate the discriminant
Calculating the discriminant:
\[ D = [-4(\alpha - 2)]^2 - 4 \cdot 4 \cdot (\alpha - 2) \]
\[ = 16(\alpha - 2)^2 - 16(\alpha - 2) \]
\[ = 16[(\alpha - 2)^2 - (\alpha - 2)] \]
\[ = 16(\alpha - 2)(\alpha - 3) \]
For the roots to be real and distinct, we need:
\[ 16(\alpha - 2)(\alpha - 3) > 0 \]
This simplifies to:
\[ (\alpha - 2)(\alpha - 3) > 0 \]

### Step 4: Solve the inequality
The critical points are \( \alpha = 2 \) and \( \alpha = 3 \). We analyze the sign of the product in the intervals:
1. \( \alpha < 2 \) : Both factors are negative, so the product is positive.
2. \( 2 < \alpha < 3 \) : One factor is positive and the other is negative, so the product is negative.
3. \( \alpha > 3 \) : Both factors are positive, so the product is positive.

Thus, the solution to the inequality is:
\[ \alpha < 2 \quad \text{or} \quad \alpha > 3 \]

### Step 5: Check the condition for root in the interval (0, 1/2)
We need to ensure that exactly one root lies in the interval \( (0, \frac{1}{2}) \).

1. **Evaluate \( f(0) \)**:
\[ f(0) = \alpha - 2 \]
For \( f(0) > 0 \) (the quadratic opens upwards), we need:
\[ \alpha - 2 > 0 \implies \alpha > 2 \]

2. **Evaluate \( f\left(\frac{1}{2}\right) \)**:
\[ f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^2 - 4(\alpha - 2)\left(\frac{1}{2}\right) + (\alpha - 2) \]
\[ = 1 - 2(\alpha - 2) + (\alpha - 2) \]
\[ = 1 - 2\alpha + 4 + \alpha - 2 \]
\[ = 3 - \alpha \]
For \( f\left(\frac{1}{2}\right) < 0 \):
\[ 3 - \alpha < 0 \implies \alpha > 3 \]

### Step 6: Combine the conditions
From the conditions derived:
1. \( \alpha < 2 \) or \( \alpha > 3 \) (for distinct roots)
2. \( \alpha > 3 \) (for one root in \( (0, \frac{1}{2}) \))

The only valid solution is:
\[ \alpha > 3 \]

### Final Answer
The values of \( \alpha \) for which exactly one root lies in \( (0, \frac{1}{2}) \) are:
\[ \alpha > 3 \]