Solution set of `3^x (3^x –1) + 2 = |3^x –1| + |3^x – 2|` contains

Options

A. singleton set
B. two elements
C. at least four elements
D. infinite elements

Explanation

To solve the equation \(3^x (3^x - 1) + 2 = |3^x - 1| + |3^x - 2|\), we will follow these steps:

### Step 1: Substitute \(3^x\) with \(t\)
Let \(t = 3^x\). The equation then becomes:
\[ t(t - 1) + 2 = |t - 1| + |t - 2| \]

### Step 2: Analyze the absolute values
We need to consider different cases based on the value of \(t\) to handle the absolute values.

#### Case 1: \(t < 1\)
In this case, both \(t - 1\) and \(t - 2\) are negative:
\[ |t - 1| = -(t - 1) = -t + 1 \]
\[ |t - 2| = -(t - 2) = -t + 2 \]
Substituting these into the equation gives:
\[ t(t - 1) + 2 = -t + 1 - t + 2 \]
This simplifies to:
\[ t^2 - t + 2 = -2t + 3 \]
Rearranging gives:
\[ t^2 + t - 1 = 0 \]

### Step 3: Solve the quadratic equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{5}}{2} \]
Calculating the roots:
\[ t_1 = \frac{-1 + \sqrt{5}}{2}, \quad t_2 = \frac{-1 - \sqrt{5}}{2} \]
Since \(t\) must be positive (as \(t = 3^x\)), we only consider \(t_1 = \frac{-1 + \sqrt{5}}{2}\).

### Step 4: Check if \(t_1 < 1\)
We need to check if \(\frac{-1 + \sqrt{5}}{2} < 1\):
\[ -1 + \sqrt{5} < 2 \implies \sqrt{5} < 3 \text{ (True)} \]
Thus, \(t_1\) is valid for this case.

### Step 5: Case 2: \(1 \leq t < 2\)
Here, \(t - 1\) is non-negative and \(t - 2\) is negative:
\[ |t - 1| = t - 1, \quad |t - 2| = -t + 2 \]
Substituting gives:
\[ t(t - 1) + 2 = (t - 1) + (-t + 2) \]
This simplifies to:
\[ t^2 - t + 2 = 1 \]
Rearranging gives:
\[ t^2 - t + 1 = 0 \]
The discriminant is:
\[ (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \text{ (No real roots)} \]

### Step 6: Case 3: \(t \geq 2\)
In this case, both \(t - 1\) and \(t - 2\) are non-negative:
\[ |t - 1| = t - 1, \quad |t - 2| = t - 2 \]
Substituting gives:
\[ t(t - 1) + 2 = (t - 1) + (t - 2) \]
This simplifies to:
\[ t^2 - t + 2 = 2t - 3 \]
Rearranging gives:
\[ t^2 - 3t + 5 = 0 \]
The discriminant is:
\[ (-3)^2 - 4 \cdot 1 \cdot 5 = 9 - 20 = -11 \text{ (No real roots)} \]

### Conclusion
The only solution comes from Case 1, which gives us a single valid solution:
\[ t = \frac{-1 + \sqrt{5}}{2} \]
Thus, the solution set of the original equation contains only one element.

### Final Answer
The solution set of the equation is a singleton set.