KCET · Maths · Functions
The domain of the function \(f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}\) is
- A \([-2,0) \cap(0,1)\)
- B \([-2,1)\)
- C \([-2,0)\)
- D \([-2,0) \cup(0,1)\)
Answer & Solution
Correct Answer
(D) \([-2,0) \cup(0,1)\)
Step-by-step Solution
Detailed explanation
Given, function
\[
f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}
\]
\(f(x)\) will be defined if,
\[
1-x>0, \neq 1 \text { and } x+2 \geq 0
\]
or \(\quad 1>x, 1-x \neq 1\) and \(x \geq-2\)
or \(x < 1, x \neq 0\) and \(x \geq-2\)
\[
\Rightarrow \quad x \in[-2,0) \cup(0,1)
\]
\[
f(x)=\frac{1}{\log _{10}(1-x)}+\sqrt{x+2}
\]
\(f(x)\) will be defined if,
\[
1-x>0, \neq 1 \text { and } x+2 \geq 0
\]
or \(\quad 1>x, 1-x \neq 1\) and \(x \geq-2\)
or \(x < 1, x \neq 0\) and \(x \geq-2\)
\[
\Rightarrow \quad x \in[-2,0) \cup(0,1)
\]
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