Let `f(x) = sqrt(log (2x-x^(2)).` Then dom (f )=?

Options

A. `(0,2)`
B. `[1,2]`
C. `(-oo,1]`
D. `{1}`

Explanation

To find the domain of the function \( f(x) = \sqrt{\log(2x - x^2)} \), we need to ensure that the expression inside the square root is non-negative. This means we need to satisfy two conditions:

1. The argument of the logarithm must be positive: \( 2x - x^2 > 0 \).
2. The logarithm itself must be non-negative: \( \log(2x - x^2) \geq 0 \), which implies \( 2x - x^2 \geq 1 \).

### Step 1: Solve \( 2x - x^2 > 0 \)

We can rearrange this inequality:

\[ -x^2 + 2x > 0 \]

Factoring gives:

\[ -x(x - 2) > 0 \]

This inequality holds when \( x \) is between the roots of the equation \( -x(x - 2) = 0 \). The roots are \( x = 0 \) and \( x = 2 \). To determine the intervals where this inequality is satisfied, we can test the intervals:

- For \( x < 0 \): Choose \( x = -1 \) → \( -(-1)(-1 - 2) = -(-1)(-3) = -3 < 0 \) (not satisfied)
- For \( 0 < x < 2 \): Choose \( x = 1 \) → \( -(1)(1 - 2) = -1 < 0 \) (satisfied)
- For \( x > 2 \): Choose \( x = 3 \) → \( -(3)(3 - 2) = -3 < 0 \) (not satisfied)

Thus, the solution to \( 2x - x^2 > 0 \) is:

\[ 0 < x < 2 \]

### Step 2: Solve \( 2x - x^2 \geq 1 \)

Rearranging gives:

\[ -x^2 + 2x - 1 \geq 0 \]

Factoring gives:

\[ -(x^2 - 2x + 1) \geq 0 \]

This can be rewritten as:

\[ -(x - 1)^2 \geq 0 \]

The only time this inequality holds is when \( (x - 1)^2 = 0 \), which gives:

\[ x = 1 \]

### Step 3: Combine the results

From Step 1, we found that \( 0 < x < 2 \). From Step 2, we found that \( x = 1 \) is the only point where \( 2x - x^2 \geq 1 \).

Thus, the only value that satisfies both conditions is:

\[ x = 1 \]

### Conclusion

The domain of the function \( f(x) = \sqrt{\log(2x - x^2)} \) is:

\[ \text{dom}(f) = \{1\} \]