Question

The domain of the function \(f(x) = \frac{1}{\log(1-x)} + \sqrt{x+2}\) is

• A. \([-2,0) \cap (0,1)\)
• B. \([-2,0)\)
• C. \([-2,1)\)
• D. \([-2,0) \cup (0,1)\)

Answer 1

The Correct Option is D

Given function: \( f(x) = \frac{1}{\log(1 - x)} + \sqrt{x + 2} \)

Step 1: Domain of \(\sqrt{x + 2}\)
For a square root function to be defined, the expression inside the root must be non-negative:
\[ x + 2 \geq 0 \Rightarrow x \geq -2 \] Step 2: Domain of \(\frac{1}{\log(1 - x)}\)
For the logarithmic term to be defined and non-zero in the denominator:
 

• \(1 - x > 0 \Rightarrow x < 1\)
• \(\log(1 - x) \neq 0 \Rightarrow 1 - x \neq 1 \Rightarrow x \neq 0\)

Step 3: Combine all conditions
\[ x \geq -2,\ x < 1,\ x \neq 0 \Rightarrow x \in [-2, 1) \text{ and } x \neq 0 \] So, domain is: \[ [-2, 0) \cup (0, 1) \]

 

Correct Answer: \([-2, 0) \cup (0, 1)\)

Answer 2

To find the domain of the function \(f(x) = \frac{1}{\log(1-x)} + \sqrt{x+2}\), we need to determine the values of \(x\) for which both terms of the function are defined.

Condition 1: The term \(\sqrt{x+2}\) must be defined.

The expression under a square root must be non-negative.

\[ x+2 \ge 0 \] \[ x \ge -2 \]

In interval notation, this condition is \([-2, \infty)\).

(1-x)} + \sqrt{x+2}\), we need to determine the values of \(x\) for which both terms in the function are defined.

1. Domain of \(\sqrt{x+2}\):

For the square root term to be defined, the expression under the square root must be non-negative:

\[ x + 2 \ge 0 \] \[ \mathbf{x \ge -2} \]

The domain for this part is \([-2, \infty)\).

2. Domain of \(\frac{1}{\log(1-x)}\):

For this term to be defined, two conditions must be met:

a) The argument of the logarithm must be strictly positive:

\[ 1 - x > 0 \] \[ 1 > x \] \[ \mathbf{x < 1} \]

b) The denominator, \(\log(1-x)\), must not be equal to zero. The logarithm of a number is zero if and only if the number is 1 (regardless of the base, assuming the base is valid, e.g., base 10 or base e).

\[ \log(1-x) \neq 0 \] \[ 1 - x \neq 1 \] \[ -x \neq 0 \] \[ \mathbf{x \neq 0} \]

Combining the conditions for the second term, we need \(x < 1\) and \(x \neq 0\). The domain for this part is \((-\infty, 0) \cup (0, 1)\).

3. Overall Domain of \(f(x)\):

The domain of the function \(f(x)\) is the intersection of the domains of its individual parts. We need to satisfy all the conditions simultaneously:

• \(x \ge -2\)
• \(x < 1\)
• \(x \neq 0\) 

Combining \(x \ge -2\) and \(x < 1\) gives the interval \([-2, 1)\).

We must also exclude \(x = 0\) from this interval.

Therefore, the domain of \(f(x)\) is \([-2, 1) \setminus \{0\}\).

In interval notation, this is written as \([-2, 0) \cup (0, 1)\).

Comparing this with the given options, the correct option is:

\([-2,0) \cup (0,1)\)