The equation `2^(2x) + (a - 1)2^(x+1) + a = 0` has roots of opposite
sing, then exhaustive set of values of a is

Options

A. ` a in (- 1, 0 )`
B. `a lt 0`
C. ` a in (-infty, 1//3)`
D. ` a in (0, 1//3)`

Explanation

To solve the equation \(2^{2x} + (a - 1)2^{x+1} + a = 0\) for values of \(a\) such that the roots are of opposite signs, we will follow these steps:

### Step 1: Substitute \(2^x\) with \(t\)
Let \(t = 2^x\). Then, \(2^{2x} = t^2\) and \(2^{x+1} = 2t\). Substituting these into the equation gives us:
\[ t^2 + (a - 1)(2t) + a = 0 \]
This simplifies to:
\[ t^2 + (2a - 2)t + a = 0 \]

### Step 2: Identify the coefficients of the quadratic equation
The quadratic equation can be expressed as:
\[ t^2 + (2a - 2)t + a = 0 \]
Here, the coefficients are:
- \(A = 1\)
- \(B = 2a - 2\)
- \(C = a\)

### Step 3: Condition for roots of opposite signs
For the roots of the quadratic equation to be of opposite signs, the product of the roots must be negative. The product of the roots \( \alpha \) and \( \beta \) is given by:
\[ \alpha \beta = \frac{C}{A} = a \]
For the roots to be of opposite signs, we need:
\[ a < 0 \]

### Step 4: Ensure the quadratic opens upwards
Since the coefficient of \(t^2\) (which is \(A\)) is positive, the parabola opens upwards. This means that the vertex of the parabola will be the minimum point.

### Step 5: Find the vertex condition
The vertex \(t\) value can be found using:
\[ t = -\frac{B}{2A} = -\frac{2a - 2}{2} = 1 - a \]
Now, we evaluate the function at \(t = 0\):
\[ f(0) = 0^2 + (2a - 2)(0) + a = a \]
For the minimum point to be below the x-axis, we need:
\[ f(1 - a) < 0 \]

### Step 6: Evaluate \(f(1 - a)\)
Substituting \(t = 1 - a\) into the quadratic:
\[ f(1 - a) = (1 - a)^2 + (2a - 2)(1 - a) + a \]
Expanding this:
\[ = (1 - 2a + a^2) + (2a - 2 - 2a + 2a^2) + a \]
Simplifying:
\[ = 1 - 2a + a^2 - 2 + 2a^2 + a = 3a^2 - a - 1 \]
Setting this less than 0 gives us:
\[ 3a^2 - a - 1 < 0 \]

### Step 7: Solve the quadratic inequality
To find the roots of \(3a^2 - a - 1 = 0\), we use the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 12}}{6} = \frac{1 \pm \sqrt{13}}{6} \]
Calculating the roots:
\[ a_1 = \frac{1 + \sqrt{13}}{6}, \quad a_2 = \frac{1 - \sqrt{13}}{6} \]
The approximate values are:
\[ a_1 \approx 0.767, \quad a_2 \approx -0.267 \]

### Step 8: Determine the intervals
The quadratic \(3a^2 - a - 1\) opens upwards, so it is negative between its roots:
\[ \frac{1 - \sqrt{13}}{6} < a < \frac{1 + \sqrt{13}}{6} \]
Since we also have the condition \(a < 0\), we take the intersection of the intervals:
\[ a \in \left( \frac{1 - \sqrt{13}}{6}, 0 \right) \]

### Final Answer
Thus, the exhaustive set of values of \(a\) for which the equation has roots of opposite signs is:
\[ a \in \left( \frac{1 - \sqrt{13}}{6}, 0 \right) \]