Let `4x^(2)-4(alpha-2)x + alpha-2=0(alpha in R)` be a quadratic equation. Find the values of `alpha` for which
One root is smaller than `(1)/(2)` and other root is greater than `(1)/(2)`

Answer

To solve the given quadratic equation \(4x^2 - 4(\alpha - 2)x + \alpha - 2 = 0\) and find the values of \(\alpha\) for which one root is smaller than \(\frac{1}{2}\) and the other root is greater than \(\frac{1}{2}\), we can follow these steps:

### Step 1: Identify the conditions for the roots
Since one root is smaller than \(\frac{1}{2}\) and the other is greater, it implies that \(\frac{1}{2}\) lies between the two roots. This means that the quadratic must be positive at \(x = \frac{1}{2}\).

### Step 2: Calculate the discriminant
For the roots to be real, the discriminant \(D\) of the quadratic must be greater than zero. The discriminant is given by:
\[ D = b^2 - 4ac \]
For our equation, \(a = 4\), \(b = -4(\alpha - 2)\), and \(c = \alpha - 2\). Thus, we have:
\[ D = [-4(\alpha - 2)]^2 - 4 \cdot 4 \cdot (\alpha - 2) \]
Calculating this gives:
\[ D = 16(\alpha - 2)^2 - 16(\alpha - 2) \]
Factoring out \(16(\alpha - 2)\):
\[ D = 16(\alpha - 2)[(\alpha - 2) - 1] = 16(\alpha - 2)(\alpha - 3) \]
For the roots to be real, we need:
\[ 16(\alpha - 2)(\alpha - 3) > 0 \]

### Step 3: Analyze the inequality
The inequality \(16(\alpha - 2)(\alpha - 3) > 0\) implies that the product \((\alpha - 2)(\alpha - 3)\) must be positive. This occurs in two intervals:
1. \(\alpha < 2\)
2. \(\alpha > 3\)

### Step 4: Evaluate the quadratic at \(x = \frac{1}{2}\)
Next, we need to ensure that the quadratic is negative at \(x = \frac{1}{2}\):
\[ f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^2 - 4(\alpha - 2)\left(\frac{1}{2}\right) + (\alpha - 2) \]
Calculating this gives:
\[ f\left(\frac{1}{2}\right) = 4 \cdot \frac{1}{4} - 2(\alpha - 2) + (\alpha - 2) = 1 - 2(\alpha - 2) + (\alpha - 2) \]
Simplifying:
\[ f\left(\frac{1}{2}\right) = 1 - 2\alpha + 4 + \alpha - 2 = 3 - \alpha \]
For \(f\left(\frac{1}{2}\right) < 0\):
\[ 3 - \alpha < 0 \implies \alpha > 3 \]

### Step 5: Combine the conditions
From the analysis:
1. The condition from the discriminant gives \(\alpha < 2\) or \(\alpha > 3\).
2. The condition from evaluating \(f\left(\frac{1}{2}\right)\) gives \(\alpha > 3\).

Thus, the only valid solution is:
\[ \alpha > 3 \]

### Final Answer
The values of \(\alpha\) for which one root is smaller than \(\frac{1}{2}\) and the other root is greater than \(\frac{1}{2}\) are:
\[ \alpha > 3 \]