Thank goodness for this thread. Must be true questions are really confusing for me. Here's an example.
If \(|x| > 3\), which of the following must be true?
I.\(x > 3\)
II. \(x^2 > 9\)
III. \(|x – 1| > 2\)
(A) I only
(B) II only
(C) I and II only
(D) II and III only
(E) I, II, and III
NOTE : This is a question from the link
VeritasKarishma shared earlier on the thread.
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2017/0 ... -question/The final answer is D.
As I solved this, I had no problem with Statement I and II. But when statement III came along, I was stuck. The range of \(x\) for \(|x – 1| > 2\) is \(x<-1\) and \(x>3\).
So \(x\) could be anywhere less than \(-1\) and anywhere beyond \(3\). I want to state specially that It could have values such as \(-1.0001\), \(-1.5\), \(-1.75\), \(-2\), \(-2.5\), \(-2.75\) and etc. (This range has values between \(-1\) and \(-3\)) I'll tell you the reason why in a minute.
Looking at the original inequality, \(|x| > 3\). After solving, you'll realize that x could be anywhere less than \(-3\) and anywhere greater than \(3\). Look at the values less than \(-3\). numbers between \(-1\) and \(-3\) are not a part of this range.
I didn't understand the explanation on the link for this specific part of the question on the link.
\(x<-3 ∪ x<3\) is a subset of \(x<-1 ∪ x>3\) isn't it? So will that make it sufficient to say that \(|x – 1| > 2\) must be true?
PS - A clarification for this will help a lot of other people like myself. Kindly do help! Slight margins between choosing the correct option and a wrong one!