Question

What is the domain of definition of \( f(x)=\frac{\sqrt{(15-x)(\log_{10}x-1)}}{x-24 \log_{4}2} \)?

• A. A union of more than three open intervals
• B. A union of two open intervals
• C. A union of three open intervals
• D. An open interval

Hint:

Always check the domain of logarithmic functions (\(x>0\)) and zeros of the denominator first. For inequalities involving products, use the "wavy curve" or sign scheme method.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
To find the domain of the function \( f(x) \), we need to determine the values of \( x \) for which the function is defined. This involves ensuring the expression under the square root is non-negative and the denominator is non-zero. Also, the argument of the logarithm must be positive. Step 2: Key Conditions:
1. Logarithm Condition: The term \( \log_{10}x \) requires \( x>0 \).
2. Denominator Condition: The denominator cannot be zero. \[ x - 24 \log_{4}2 \neq 0 \] 3. Square Root Condition: The term under the square root must be non-negative. \[ (15-x)(\log_{10}x - 1) \geq 0 \] Step 3: Detailed Explanation:
\textit{Condition 2 (Denominator):}
Simplify \( \log_{4}2 \): \[ \log_{4}2 = \log_{2^2}2 = \frac{1}{2}\log_{2}2 = \frac{1}{2} \] So, \( x - 24(0.5) \neq 0 \Rightarrow x - 12 \neq 0 \Rightarrow x \neq 12 \). \textit{Condition 3 (Square Root):}
Find the critical points where the expression is zero: \[ 15 - x = 0 \Rightarrow x = 15 \] \[ \log_{10}x - 1 = 0 \Rightarrow \log_{10}x = 1 \Rightarrow x = 10 \] Now, test the intervals defined by these critical points (keeping \( x>0 \)):
Interval \((0, 10)\): Let \( x = 1 \). \[ (15-1)(\log_{10}1 - 1) = (14)(0-1) = -14 \quad (\text{Negative, Invalid}) \] Interval \((10, 15)\): Let \( x = 11 \). \[ (15-11)(\log_{10}11 - 1) = (4)(\text{positive}) \quad (\text{Positive, Valid}) \] Interval \((15, \infty)\): Let \( x = 20 \). \[ (15-20)(\log_{10}20 - 1) = (-5)(\text{positive}) \quad (\text{Negative, Invalid}) \] The expression is non-negative for \( x \in [10, 15] \). \textit{Combining all conditions:}
We have the interval \( [10, 15] \), but we must exclude \( x = 12 \) (from the denominator condition). \[ \text{Domain} = [10, 12) \cup (12, 15] \] Step 4: Final Conclusion:
The domain is the union of the intervals \( [10, 12) \) and \( (12, 15] \). While these are technically semi-open intervals, option (B) "a union of two open intervals" is the closest and intended description of the set structure compared to the other options.