MHT CET · Maths · Functions
The domain of the function \(f(x)=\frac{1}{\sqrt{x+|x|}}\) is
- A \((-\infty, 0)\)
- B \((2,5)\)
- C \((0, \infty)\)
- D \((-\infty, \infty)\)
Answer & Solution
Correct Answer
(C) \((0, \infty)\)
Step-by-step Solution
Detailed explanation
\(
f(x)=\frac{1}{\sqrt{x+|x|}}
\)
Here \(x+|x| \geq 0\) and \(\sqrt{x+|x|} \neq 0\)
\(
\therefore \mathrm{x}+|\mathrm{x}|>0
\)
Now when \(x>0, x+|x|=x+x \Rightarrow 2 x>0\).
When \(\mathrm{x} < 0, \mathrm{x}+|\mathrm{x}|=\mathrm{x}-\mathrm{x}=0\)
\(\therefore \mathrm{x}>0\) is the required domain.
f(x)=\frac{1}{\sqrt{x+|x|}}
\)
Here \(x+|x| \geq 0\) and \(\sqrt{x+|x|} \neq 0\)
\(
\therefore \mathrm{x}+|\mathrm{x}|>0
\)
Now when \(x>0, x+|x|=x+x \Rightarrow 2 x>0\).
When \(\mathrm{x} < 0, \mathrm{x}+|\mathrm{x}|=\mathrm{x}-\mathrm{x}=0\)
\(\therefore \mathrm{x}>0\) is the required domain.
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