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

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

### Step 2: Substitute \(k\)
Let \(k = 3^x\). Then, the inequality becomes:
\[ k + \frac{3}{k} - 4 < 0 \]
To eliminate the fraction, multiply through by \(k\) (noting that \(k > 0\)):
\[ k^2 + 3 - 4k < 0 \]
This simplifies to:
\[ k^2 - 4k + 3 < 0 \]

### Step 3: Factor the Quadratic
Now we need to factor the quadratic:
\[ k^2 - 4k + 3 = (k - 3)(k - 1) < 0 \]

### Step 4: Analyze the Sign of the Factors
Next, we find the critical points by setting the factors to zero:
\[ k - 3 = 0 \quad \Rightarrow \quad k = 3 \]
\[ k - 1 = 0 \quad \Rightarrow \quad k = 1 \]
We will test the intervals determined by these points: \( (-\infty, 1) \), \( (1, 3) \), and \( (3, \infty) \).

- For \(k < 1\) (e.g., \(k = 0\)): \((0 - 3)(0 - 1) > 0\) (positive)
- For \(1 < k < 3\) (e.g., \(k = 2\)): \((2 - 3)(2 - 1) < 0\) (negative)
- For \(k > 3\) (e.g., \(k = 4\)): \((4 - 3)(4 - 1) > 0\) (positive)

### Step 5: Determine the Solution Set for \(k\)
The inequality \((k - 3)(k - 1) < 0\) holds true in the interval:
\[ 1 < k < 3 \]

### Step 6: Substitute Back for \(x\)
Recall that \(k = 3^x\). Therefore, we have:
\[ 1 < 3^x < 3 \]
Taking logarithms (base 3) of all parts:
\[ \log_3(1) < x < \log_3(3) \]
This simplifies to:
\[ 0 < x < 1 \]

### Final Answer
Thus, the solution set contained in \(R\) is:
\[ \boxed{(0, 1)} \]