Responding to a pm:

|x| > 3 implies that x is a point whose distance from 0 is more than 3. So x could be greater than 3 or less than -3. Before you move further, think about the values x can take: 3.00001, 3.5, 4.2, 5.7, 67, 1000, -3.45, -4, -8, -100 etc. The only values it cannot take are -3 <= x <= 3

Which of the following must be true?

I. x > 3

For every value that x can take, must x be greater than 3? No. e.g. if x takes -3.45, -4 etc, it will not be greater than 3 so this is not true.

II. X^2 > 9
This is the same as |x| > 3 so it must be true

III. |x-1|>2
This implies that the distance of x from 1 must be greater than 2. So x is either greater than 3 or less than -1. Now, recall all the values that x can take. For each value, can be say that x is either greater than 3 or less than -1? Yes.
3.00001 - x is greater than 3
3.5 : x is greater than 3
4.2 : x is greater than 3
5.7 : x is greater than 3
67 : x is greater than 3
1000 : x is greater than 3
-3.45 : x is less than -1
-4 : x is less than -1
-8 : x is less than -1
-100 : x is less than -1

For every value that x can take, x will be either greater than 3 or less than -1. Note that we are not saying that every value less than -1 must be valid for x. We are saying that every value that is valid for x (found by using |x| > 3) will be either greater than 3 or less than -1. Hence |x-1|>2 must be true for every value that x can take.
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Now if I take x=-2 will it satisfy the condition?
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From|x|>3, x will always have to be >3 or <-3, so "our" x will never be between -3 and 3, making the III statement always true.

lol. you're doing the same error in reasoning I was doing. You have to prove statement III as true an not the other way around (prove |x|>3 true). If you know that your x will always be more than 3 OR always less than -3 (from |x|>3) then you are restricted to these values when looking to prove statement III as true. Hope you understand.

Edit: Upppss..You got it. Cheers.

Good explanation thanks . generally I do not this silly but crucial error but this time I could have sworn tha D was false. Indeed, III was tricky especially if I think under time pressure. The latter changes all things, all prespective.

@Corvinis
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I just came across this one and it is clear that this is a tough cookie.

Given that |x| > 3, so that means x > 3 or x < -3. We can also test values here, only 3,4,5,6 or -3, -4, -5 etc work.

Now, let's tackle the statements:
I. x > 3

We know that x > 3 or x < -3, but it is not ALWAYS the case that x > 3. We do have values of x that are less than 3, i.e., when x is -4 or -5. So, I is false.

II. x^2 > 9.

This means x>3 or x < -3. That is what we have above so this is golden. True.

III. |x - 1| > 2

When x > 0
x -1 > 2
x > 3

x < 0
-x + 1 > 2
-x > 1
x < -1

So , III says IF x > 3 or x < -3, THEN x > 3 OR x < -1. True. Test values to prove this x > 3 or x < -3 means x can be -4 and that hits x < -1

i am clearly unclear how option 3 is true, since if i take the value of x as -2. It holds true for |x-1|>2. However the value of -2 does not hold true for |x| > 3. Please clarify.

if x<-3 then sure x<-1 ....

If some numbers confuse you, don't fixate on them. Go ahead and take some other easier examples.
Let's keep the wording of the question same but make it simple.

If n < 6, which of the following must be true?

I.

II.

III. n < 8

Can we say that III must be true? Yes!
If n is less than 6 then obviously it is less than 8 too.
If n is less than 6, it will take values such as -20, 2, 5 etc. All of these values will be less than 8 too.

Values 6 and 7 are immaterial because n cannot take these values. You are given that n is less than 6 so you only need to worry about values that n CAN take. Those should satisfy n < 8.

Similarly, your question says that x > 3 or x < -3

Then we can say that x > 3 or x < -1. All values that will be less than -3 will be less than -1 too.

Check out my post on a similar tricky question : http://www.veritasprep.com/blog/2012/07 ... -and-sets/
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|x| > 3 means x>3 or x<-3
so I is false because x<-3 is true.


II. x^2 > 9 satisfies both equations (use x=-4 or x=5 )

III. is also true for x=-4 and x=5.

so D
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