Re: If x/|x|

There was a lot of confusion between options (A) and (B). Therefore, I would like to explain why option (B) is correct using diagrams.

Forget this question for a minute. Say instead you have this question:

Question 1: x > 2 and x < 7. What integral values can x take?
I guess most of you will come up with 3, 4, 5, 6. That’s correct. I can represent this on the number line.

You see that the overlapping area includes 3, 4, 5 and 6.

Now consider this:

Question 2: x > 2 or x > 5. What integral values can x take?
Let’s draw that number line again.

So is the solution again the overlapping numbers i.e. all integers greater than 5? No. This question is different. x is greater than 2 OR greater than 5. This means that if x satisfies at least one of these conditions, it is included in your answer. Think of sets. AND means it should be in both the sets (i.e. overlapping). OR means it should be in at least one of the sets. Hence, which values can x take? All integral values starting from 3 onwards i.e. 3, 4, 5, 6, 7, 8, 9 …

Now go back to this question. The solution is a one liner.

If \frac{x}{|x|}<x which of the following must be true about x?
(A) x>1
(B) x>-1
(C) |x|<1
(D) |x|=1
(E) |x|^2>1

\frac{x}{|x|} is either 1 or -1.
So x > 1 or x > -1
So which values can x take? All values that are included in at least one of the sets. Therefore, x > -1.