Let `4x^(2)-4(alpha-2)x + alpha-2=0(alpha in R)` be a quadratic equation. Find the values of `alpha` for which
Both the roots are smaller than `(1)/(2)`

Answer

To solve the quadratic equation \(4x^2 - 4(\alpha - 2)x + \alpha - 2 = 0\) and find the values of \(\alpha\) for which both roots are smaller than \(\frac{1}{2}\), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation can be rewritten in standard form \(ax^2 + bx + c = 0\), where:
- \(a = 4\)
- \(b = -4(\alpha - 2) = -4\alpha + 8\)
- \(c = \alpha - 2\)

### Step 2: Condition for real roots
For the roots to be real, the discriminant \(D\) must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Calculating the discriminant:
\[ D = (-4\alpha + 8)^2 - 4 \cdot 4 \cdot (\alpha - 2) \]
Expanding this:
\[ D = (16\alpha^2 - 64\alpha + 64) - 16(\alpha - 2) \]
\[ D = 16\alpha^2 - 64\alpha + 64 - 16\alpha + 32 \]
\[ D = 16\alpha^2 - 80\alpha + 96 \]
Setting the discriminant greater than or equal to zero:
\[ 16\alpha^2 - 80\alpha + 96 \geq 0 \]
Dividing through by 16:
\[ \alpha^2 - 5\alpha + 6 \geq 0 \]
Factoring the quadratic:
\[ (\alpha - 2)(\alpha - 3) \geq 0 \]
This inequality holds when:
\[ \alpha \leq 2 \quad \text{or} \quad \alpha \geq 3 \]

### Step 3: Condition for roots to be less than \(\frac{1}{2}\)
For both roots to be less than \(\frac{1}{2}\), we need the following conditions:
1. The vertex of the parabola, given by \(x = -\frac{b}{2a}\), must be less than \(\frac{1}{2}\).
2. The value of the quadratic at \(x = \frac{1}{2}\) must be positive.

Calculating the vertex:
\[ x = -\frac{-4\alpha + 8}{2 \cdot 4} = \frac{4\alpha - 8}{8} = \frac{\alpha - 2}{2} \]
Setting this less than \(\frac{1}{2}\):
\[ \frac{\alpha - 2}{2} < \frac{1}{2} \]
Multiplying through by 2:
\[ \alpha - 2 < 1 \implies \alpha < 3 \]

### Step 4: Value of the quadratic at \(x = \frac{1}{2}\)
Calculating \(f\left(\frac{1}{2}\right)\):
\[ f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^2 - 4(\alpha - 2)\left(\frac{1}{2}\right) + (\alpha - 2) \]
\[ = 4 \cdot \frac{1}{4} - 2(\alpha - 2) + (\alpha - 2) \]
\[ = 1 - 2\alpha + 4 + \alpha - 2 \]
\[ = 3 - \alpha \]
Setting this greater than 0:
\[ 3 - \alpha > 0 \implies \alpha < 3 \]

### Step 5: Combine conditions
From Step 2, we have \(\alpha \leq 2\) or \(\alpha \geq 3\). From Steps 3 and 4, we have \(\alpha < 3\). The only overlapping condition that satisfies all inequalities is:
\[ \alpha \leq 2 \]

### Final Answer
Thus, the values of \(\alpha\) for which both roots of the quadratic equation are smaller than \(\frac{1}{2}\) are:
\[ \alpha \in (-\infty, 2] \]