If |x| > 3

OA is D
Thanks

|x| > 3
From the definition of mod, if |x| > 3, x>3 or x<-3

1) x > 3 - Is not a 'must-be-true' condition as x can be less than -3
2) x^2 > 9 - Is a 'must-be-true condition as if x>3, then x^2>9 and if x<-3, x^2>9
3) |x - 1| > 2 => x-1>2 or x-1<-2 => x>3 and x<-1 - Is not a 'must-be-true' condition as x can be a value between -1 and -3, say -2.

IMO [spoiler]B)II only[/spoiler]

Only II and III are correct.

|x|>3. That is, x>3 or x<-3
but it is still must be true question.
I. x>3 but what if x=-4? |-4|=4<3 but x is not bigger than 3
II. it is true. if x=4 x^2=16 - true. if x=4, x^2=16, true.
III. two possibilities x<-1 and x>3 (Drow 2 lines to see that if x<-3 x will always be < -1)

If x =-2, |-2 - 1| = 3 > 2. So x=-2 satisfies the equation III but |x| = 2<3 and doesn't satisfy the inequation.

neelgandham wrote:If x =-2, |-2 - 1| = 3 > 2. So x=-2 satisfies the equation III but |x| = 2<3 and doesn't satisfy the inequation.

Look carefully at the stimulus. x is less than 3 or more than 3. It can`t be -2.

katrin - I

Option 3 reads |x - 1| > 2,So x = -2 is one of the solutions for this in-equation. Agreed?
So is x=-2 a solution of the stimulus |x|>3 ?

massi2884 wrote:If |x| > 3, which of the following must be true?

1) x > 3
2) x^2 > 9
3) |x - 1| > 2

I only
II only
I and II only
II and III only
I, II, and III

Plug in a value that satisfies the condition that |x| > 3.
Let x=-4.
Eliminate any statement that is not valid for x=-4.

Statement I: x > 3
-4 > 3
Not true.
Eliminate any answer choice that includes I.
Eliminate A, C and E.

Compare the remaining answer choices.
B and D each include II.
Thus, II must be true, since it is included in both of the remaining answer choices.
Thus, we need to evaluate only statement III.

Statement III: |x-1| > 2
|x-1| > 2 implies that the distance between x and 1 is more than 2 units.
|x| > 3 implies that x is more that 3 units from 0.
Since x is more than 3 units from 0 -- in other words, x<-3 or x>3 -- the distance between x and 1 must be more than 2 units.
Thus, III must be true.
Eliminate B, which doesn't include III.

The correct answer is D.

Please note the following:
If |x| > 3, then statement III must be true: |x-1| > 2.
This does NOT imply the reverse.
If |x-1| > 2, it does NOT have to be true that |x| > 3.
For example, x=-2 satisfies |x-1| > 2 but not |x| > 3.

Thanks for the explanation Mitch! I was solving the other way.

IMO (B) only x^2>9 is correct.

I went by substitution method and i get II and III as answer