Domain of `(sqrt(x^(2)-4x+3)+1) log_(5)""((x)/(5))+(1)/(x)(sqrt(8x-2x^(2)-6)+1) le 1` is

Options

A. `(-oo,1] cup [3,oo)`
B. `[1,3]`
C. {1,3}
D. {1}

Answer

To find the domain of the expression

\[ \sqrt{x^2 - 4x + 3} + 1 + \log_5\left(\frac{x}{5}\right) + \frac{1}{x}\sqrt{8x - 2x^2 - 6} + 1 \leq 1, \]

we need to analyze each component of the expression to ensure they are defined and satisfy the inequality.

### Step 1: Analyze the square root \(\sqrt{x^2 - 4x + 3}\)

The expression inside the square root must be non-negative:

\[ x^2 - 4x + 3 \geq 0. \]

Factoring the quadratic:

\[ (x - 1)(x - 3) \geq 0. \]

To find the intervals where this inequality holds, we can use a number line:

- The roots are \(x = 1\) and \(x = 3\).
- Test intervals:
- For \(x < 1\), choose \(x = 0\): \((0 - 1)(0 - 3) = 3 > 0\) (true).
- For \(1 < x < 3\), choose \(x = 2\): \((2 - 1)(2 - 3) = -1 < 0\) (false).
- For \(x > 3\), choose \(x = 4\): \((4 - 1)(4 - 3) = 3 > 0\) (true).

Thus, the solution to this inequality is:

\[ (-\infty, 1] \cup [3, \infty). \]

### Step 2: Analyze the logarithm \(\log_5\left(\frac{x}{5}\right)\)

The argument of the logarithm must be positive:

\[ \frac{x}{5} > 0 \implies x > 0. \]

### Step 3: Analyze the term \(\frac{1}{x}\)

This term is defined for \(x \neq 0\). Since we already have \(x > 0\) from the logarithm, this condition is satisfied.

### Step 4: Analyze the square root \(\sqrt{8x - 2x^2 - 6}\)

The expression inside this square root must also be non-negative:

\[ 8x - 2x^2 - 6 \geq 0. \]

Rearranging gives:

\[ -2x^2 + 8x - 6 \geq 0 \implies 2x^2 - 8x + 6 \leq 0. \]

Factoring gives:

\[ 2(x^2 - 4x + 3) \leq 0 \implies (x - 1)(x - 3) \leq 0. \]

Using a similar analysis as before, we find:

- The roots are \(x = 1\) and \(x = 3\).
- The solution to this inequality is:

\[ [1, 3]. \]

### Step 5: Combine all conditions

Now we combine the intervals obtained from each step:

1. From \(\sqrt{x^2 - 4x + 3}\): \((- \infty, 1] \cup [3, \infty)\).
2. From \(\log_5\left(\frac{x}{5}\right)\): \((0, \infty)\).
3. From \(\sqrt{8x - 2x^2 - 6}\): \([1, 3]\).

The intersection of these intervals is:

- From \((- \infty, 1] \cup [3, \infty)\) and \((0, \infty)\), we get \((0, 1] \cup [3, \infty)\).
- Now intersecting this with \([1, 3]\), we find that the only valid point is \(x = 1\) and \(x = 3\).

### Final Domain

Thus, the domain of the expression is:

\[ \{1, 3\}. \]