The domain of f(x) = `sqrt(log_(1//4)((5x-x^(2))/4))+""^(10)C_(x)` is

Options

A. `(0,1]cup[4,5)`
B. (0,5)
C. {1,4}
D. none of these

Answer

To find the domain of the function \( f(x) = \sqrt{\log_{1/4}\left(\frac{5x - x^2}{4}\right)} + \binom{10}{x} \), we need to ensure that the expression inside the square root is non-negative and that the binomial coefficient is defined.

### Step 1: Analyze the logarithmic part
The expression inside the square root is \( \log_{1/4}\left(\frac{5x - x^2}{4}\right) \). For the square root to be defined, we need:
\[ \log_{1/4}\left(\frac{5x - x^2}{4}\right) \geq 0 \]

### Step 2: Convert the logarithmic inequality
Since the base \( \frac{1}{4} \) is less than 1, the logarithmic function is decreasing. Therefore, we can rewrite the inequality:
\[ \frac{5x - x^2}{4} \leq 1 \]

### Step 3: Simplify the inequality
Multiplying both sides by 4 (which is positive), we get:
\[ 5x - x^2 \leq 4 \]
Rearranging gives:
\[ -x^2 + 5x - 4 \leq 0 \]
or
\[ x^2 - 5x + 4 \geq 0 \]

### Step 4: Factor the quadratic
Factoring the quadratic, we find:
\[ (x - 1)(x - 4) \geq 0 \]

### Step 5: Solve the inequality
To solve \( (x - 1)(x - 4) \geq 0 \), we find the critical points \( x = 1 \) and \( x = 4 \). Testing intervals:
- For \( x < 1 \): both factors are negative, so the product is positive.
- For \( 1 < x < 4 \): one factor is positive and the other is negative, so the product is negative.
- For \( x > 4 \): both factors are positive, so the product is positive.

Thus, the solution is:
\[ x \leq 1 \quad \text{or} \quad x \geq 4 \]

### Step 6: Consider the binomial coefficient
The term \( \binom{10}{x} \) is defined for non-negative integers \( x \) such that \( 0 \leq x \leq 10 \). Therefore, we also have:
\[ 0 \leq x \leq 10 \]

### Step 7: Combine the conditions
Now we combine the conditions:
1. From the logarithmic part: \( x \leq 1 \) or \( x \geq 4 \)
2. From the binomial coefficient: \( 0 \leq x \leq 10 \)

### Step 8: Determine the valid intervals
- From \( x \leq 1 \): valid values are \( x = 0, 1 \).
- From \( x \geq 4 \): valid values are \( x = 4, 5, 6, 7, 8, 9, 10 \).

### Final Domain
Thus, the domain of \( f(x) \) is:
\[ \{0, 1\} \cup \{4, 5, 6, 7, 8, 9, 10\} \]