The solution set contained in `R`of the following inequation`3^x+3^(1-x)-4<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 \(t\)
Let \(t = 3^x\). Then the inequation becomes:
\[ t + \frac{3}{t} - 4 < 0 \]

### Step 3: Multiply through by \(t\)
To eliminate the fraction, multiply the entire inequation by \(t\) (note that \(t > 0\) since \(3^x\) is always positive):
\[ t^2 + 3 - 4t < 0 \]

### Step 4: Rearrange the inequality
Rearranging gives us:
\[ t^2 - 4t + 3 < 0 \]

### Step 5: Factor the quadratic
Next, we factor the quadratic expression:
\[ (t - 1)(t - 3) < 0 \]

### Step 6: Determine the intervals
To find the solution set for the inequality \((t - 1)(t - 3) < 0\), we identify the critical points where the expression equals zero:
- \(t = 1\)
- \(t = 3\)

Now we test the intervals:
1. For \(t < 1\): Choose \(t = 0\) → \((0 - 1)(0 - 3) = 3 > 0\)
2. For \(1 < t < 3\): Choose \(t = 2\) → \((2 - 1)(2 - 3) = -1 < 0\)
3. For \(t > 3\): Choose \(t = 4\) → \((4 - 1)(4 - 3) = 3 > 0\)

Thus, the solution to the inequality is:
\[ 1 < t < 3 \]

### Step 7: Substitute back for \(t\)
Recalling that \(t = 3^x\), we substitute back:
\[ 1 < 3^x < 3 \]

### Step 8: Solve for \(x\)
Taking logarithms (base 3) gives:
\[ 0 < x < 1 \]

### Final Solution
The solution set contained in \(\mathbb{R}\) is:
\[ (0, 1) \]