happypuppy wrote:
Hi
chetan2u ,
SaquibHGMATWhizI tried but couldn't get a simple solution by squaring (A) both sides for ex.
Can you please share a methodical approach to this question?
Hi happypuppy,
I would prefer to go the way Bunuel has solved this one because we can draw the number line directly from the absolute inequality.
However, we can also do it after doing the square.
\(|x - y|^2 < |y|^2\\
\)
\(x^2 - 2xy + y^2 < y^2 \)
\(x^2 - 2xy < 0\)
\(x(x - 2y) < 0\)
Now if you try to form the number line this can be true in two ways,
Scenario 1 x < 0 and
x - 2y > 0 or x > 2y
So we can easily draw x < 0 here on a number line but we need to understand how we write x > 2y.
If y is less than x (a negative number) here, then y (is negative here) can take any values as 2y is going to be smaller. So in such a case, y can go till - infinity
_y___________x______ 0
If y > x (a negative number) here, then y needs to be negative and y should be greater than x and less than x/2.
This part is a bit tricky and also important to understand The idea here is that y should have such a value so that y falls towards right or x but 2y falls towards left and for that y must be between x and x/2.
So, the second scenario from here looks like
________x______y__(x/2)________0
The point x/2 is given to make sure that you understand exactly where 'y' can be.
You can approach a similar way for
scenario 2. x > 0 and x - 2y < 0 or x < 2y.
You should get the possibilities as,
________0____________________x______________y (y can be anywhere towards right of x)
________0______(x/2)___y______x_____________ (y in between x/2 and x)
Once we are done with this, I always suggest considering the zero cases separately. By zero case, I mean that x = y could also be a possibility.
Since I am already getting contradicting answers so I am not really bothered about the x = y scenario as it will not change my answers.
Hope this helps.
_________________
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