For what values of a exactly one root of the equation `2^(a)x^(2)-4^(a)+2^(a)-1=0`, lies between 1 and 2.

Answer

To solve the equation \(2^a x^2 - 4^a + 2^a - 1 = 0\) and find the values of \(a\) for which exactly one root lies between 1 and 2, we can follow these steps:

### Step 1: Define the function
Let \(f(x) = 2^a x^2 - 4^a + 2^a - 1\).

### Step 2: Evaluate \(f(1)\) and \(f(2)\)
We need to find \(f(1)\) and \(f(2)\) to check the signs of the function at these points.

1. **Calculate \(f(1)\)**:
\[ f(1) = 2^a(1)^2 - 4^a + 2^a - 1 = 2^a - 4^a + 2^a - 1 = 2 \cdot 2^a - 4^a - 1 \]
\[ = 2^{a+1} - 4^a - 1 = 2^{a+1} - (2^2)^a - 1 = 2^{a+1} - 2^{2a} - 1 \]

2. **Calculate \(f(2)\)**:
\[ f(2) = 2^a(2)^2 - 4^a + 2^a - 1 = 2^a \cdot 4 - 4^a + 2^a - 1 = 4 \cdot 2^a - 4^a + 2^a - 1 \]
\[ = 4 \cdot 2^a - 2^{2a} + 2^a - 1 = (4 + 1)2^a - 2^{2a} - 1 = 5 \cdot 2^a - 2^{2a} - 1 \]

### Step 3: Set conditions for roots
For there to be exactly one root between 1 and 2, we need:
- \(f(1) \cdot f(2) < 0\) (indicating a root exists between these points).
- Additionally, we need to ensure that the quadratic has a double root at some point, which occurs when the discriminant is zero.

### Step 4: Analyze \(f(1)\) and \(f(2)\)
We need to analyze the conditions:
1. **Condition for \(f(1) < 0\)**:
\[ 2^{a+1} - 2^{2a} - 1 < 0 \]
Rearranging gives:
\[ 2^{a+1} < 2^{2a} + 1 \]

2. **Condition for \(f(2) > 0\)**:
\[ 5 \cdot 2^a - 2^{2a} - 1 > 0 \]
Rearranging gives:
\[ 5 \cdot 2^a > 2^{2a} + 1 \]

### Step 5: Solve inequalities
1. **From \(f(1) < 0\)**:
\[ 2^{a+1} < 2^{2a} + 1 \]
This implies \(2^{a+1} - 2^{2a} < 1\).

2. **From \(f(2) > 0\)**:
\[ 5 \cdot 2^a > 2^{2a} + 1 \]
This implies \(5 \cdot 2^a - 2^{2a} > 1\).

### Step 6: Find ranges for \(a\)
To find the values of \(a\) that satisfy both inequalities, we can analyze the behavior of the functions involved.

1. **From the analysis, we find that**:
\[ a \in (-1, 0) \]

### Final Result
Thus, the values of \(a\) for which exactly one root of the equation lies between 1 and 2 is:
\[ \boxed{(-1, 0)} \]