Q42 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
If $f$ is a real valued function from $A$ onto $B$ defined by $f(x)=\frac{1}{\sqrt{|x-|x||}}$ then $A \cap B=$
- Correct answerA$\phi$
- B$(-\infty, 0)$
- C$(0, \infty)$
- D$(-\infty, \infty)$
Solution
$\quad f(x)=\frac{1}{\sqrt{|x-|x||}}$
Domain of $f(x)=x \in(-\infty, 0) \Rightarrow A \in(-\infty, 0)$ Range of $f(x)=y \in(0, \infty) \Rightarrow B \in(0, \infty)$ $A \cap B=\phi .$
Domain of $f(x)=x \in(-\infty, 0) \Rightarrow A \in(-\infty, 0)$ Range of $f(x)=y \in(0, \infty) \Rightarrow B \in(0, \infty)$ $A \cap B=\phi .$
Hence, the correct answer is option: 'A'.