The solution set of the inequation
`3^x+3^(1-x) -4 lt 0`, is

Options

A. (1, 3)
B. (0,1)
C. (1, 2)
D. (0, 2)

Answer

To solve the inequation \( 3^x + 3^{1-x} - 4 < 0 \), we will follow these steps:

### Step 1: Rewrite the Inequation
We start with the given inequation:
\[ 3^x + 3^{1-x} - 4 < 0 \]
We can rewrite \( 3^{1-x} \) as \( \frac{3}{3^x} \):
\[ 3^x + \frac{3}{3^x} - 4 < 0 \]

### Step 2: Substitute \( t = 3^x \)
Let \( t = 3^x \). Then the inequation becomes:
\[ t + \frac{3}{t} - 4 < 0 \]
Multiplying through by \( t \) (since \( t > 0 \)):
\[ t^2 - 4t + 3 < 0 \]

### Step 3: Factor the Quadratic
We can factor the quadratic expression:
\[ t^2 - 4t + 3 = (t - 1)(t - 3) < 0 \]

### Step 4: Determine the Intervals
To find the intervals where the product is negative, we analyze the sign of the factors:
- The roots of the equation are \( t = 1 \) and \( t = 3 \).
- The intervals to test are \( (-\infty, 1) \), \( (1, 3) \), and \( (3, \infty) \).

Testing these intervals:
1. For \( t < 1 \) (e.g., \( t = 0 \)): \( (0 - 1)(0 - 3) = 3 > 0 \)
2. For \( 1 < t < 3 \) (e.g., \( t = 2 \)): \( (2 - 1)(2 - 3) = -1 < 0 \)
3. For \( t > 3 \) (e.g., \( t = 4 \)): \( (4 - 1)(4 - 3) = 3 > 0 \)

Thus, the solution set for \( (t - 1)(t - 3) < 0 \) is:
\[ 1 < t < 3 \]

### Step 5: Substitute Back for \( x \)
Recall that \( t = 3^x \). Therefore, we have:
\[ 1 < 3^x < 3 \]
Taking logarithms (base 3):
\[ \log_3(1) < x < \log_3(3) \]
This simplifies to:
\[ 0 < x < 1 \]

### Final Answer
Thus, the solution set of the inequation \( 3^x + 3^{1-x} - 4 < 0 \) is:
\[ (0, 1) \]