Question

The domain of the function $ y = \frac{1}{\log_{10}(3 - x)} + \sqrt{x + 7} $ is:

• A. \( [-7, 3] \setminus \{1\} \)
• B. \( (-7, 3) \setminus \{0\} \)
• C. \( [-7, 3] \setminus \{2\} \)
• D. \( (-7, 3) \)

Hint:

When finding the domain of functions involving logarithms and square roots, always consider the restrictions for both. For logarithms, ensure that the argument is positive, and for square roots, the argument must be non-negative.

Answer 1

The Correct Option is C

We are given the function \( y = \frac{1}{\log_{10}(3 - x)} + \sqrt{x + 7} \). We need to find the domain of the function, i.e., the set of all possible values of \( x \) for which the function is defined.
1. For the logarithmic term \( \frac{1}{\log_{10}(3 - x)} \), the argument of the logarithm must be positive, so: \[ 3 - x > 0 \quad \Rightarrow \quad x < 3 \] This implies that \( x \) must be less than 3. 
2. For the square root term \( \sqrt{x + 7} \), the argument of the square root must be non-negative, so: \[ x + 7 \geq 0 \quad \Rightarrow \quad x \geq -7 \] This implies that \( x \) must be greater than or equal to -7. Thus, the domain of the function is restricted to \( -7 \leq x < 3 \). 
3. Additionally, the logarithmic term has a restriction that it cannot be zero because division by zero is undefined. Therefore, we must ensure that: \[ \log_{10}(3 - x) \neq 0 \quad \Rightarrow \quad 3 - x \neq 1 \quad \Rightarrow \quad x \neq 2 \] Hence, \( x = 2 \) must be excluded from the domain. So, the domain of the function is: \[ [-7, 3] \setminus \{2\} \] Thus, the correct answer is (C).