MCQ
The set $A = \{ x:x \in R,\,{x^2} = 16$ and $2x = 6\} $ equals
- ✓$\phi $
- B$\{14, 3, 4\}$
- C$\{3\}$
- D$\{4\}$
$2x = 6$ ==> $x = 3$
There is no value of $x$ which satisfies both the above equations. Thus, $A = \phi $.
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