Is x^2 < x - |y|?

IMO answer is B D

is x^2 < x - |y|?

From Stmt 1) y > x

Lets use numbers y = 1 and x = 0 when applied to x^2 < x - |y| results as 0 < - 1 Valid Not Valid
y = 5 x = 4 => when applied to x^2 < x - |y| results 16 < -1 Not Valid

Hence - Sufficient

From Stmt 2) x < 0

Substituting various values for x and any value for y in x^2 < x - |y| results x^2 is NOT < x - |y|

Hence - Sufficient.

What is the OA?

Hi nt2010,

they are both not valid

Hi Zarrolou!

Could you run me through this:

(i) y >x
or -y<-x (we need to have the same operator to sum inequalities). Sum this to each case:
I)x^2+(-y)<(-x)+x-y or x^2<0 Never true
II)x^2+(-y)<(-x)+x+y or x^2<2y, but since in this case we are considering y<0 that equation is never true
x^2<-ve Never true.
Sufficient

Firstly, why do we plug in -y<-x? I get where that comes from, but why do we use it here?

Secondly, where do we get x^2+(-y)<(-x)+x-y from? In other words, how does x^2 < x - |y| translate to x^2+(-y)<(-x)+x-y?

Thanks!

The most important passage is the first one: \(x^2 < x - |y|\) from this I divide the function into 2 cases:
\(y>0\) => \(x^2<x-y\)
\(y<0\) => \(x^2<x+y\)
I want to make this clear, so now:

1)It's an useful technique! When you have 2 inequalities, you know what they say taken on their own, but what do they say together?
You can sum(not subtract) inequalities as long as they have the same operator (> or <)

2)I divided the first equation into two cases : y>0 and y<0
Now I take case \(y>0\):
\(x^2<+x-y\), sum (literally sum) the equation of statement 1
\(-y<-x\), and I obtain this:
\(x^2+(-y)<(-x)+x-y\), now I semplify its form into \(x^2<0\). So in this scenario the equation is never true, 0 cannot be greater than a squared number.
Now I repeat the same passages for the other case \(y<0\):
\(x^2<+x+y\), sum the equation of statement 1
\(-y<-x\), and I obtain this:
\(x^2+(-y)<(-x)+x+y\), and I get \(x^2<2y\). But since in this scenario I am considering \(y<0\) the right term \(2y\) will be negative.
So once more, can a negative number be greater than a squared number? \(x^2<-ve\)? No, so also in this scenario the answer to the main question is NO.
To summarize what we did here: we have divided the main question into two cases, we have studied each case on its own, and since the answer to the main question is the same (NO in both) we can say that statement 1 is sufficient

I like D.

Point 1 is sufficient because X^2 is always positive, and x - |y| if y > x is always negative.
Point 2 is also sufficient for pretty much the same reason.

That took my a while but I think I got it. Kudos to you! Thanks!

\(x^2-x >|y|\)

\(x(1-x)>0\) since RHS is always > 0

\(1>X>0\)

Statement 2 seems sufficient but I am unable to use statement 1 to conclude this logic

One more thing...

I think I solved for #II but I ran into a snag.

I used the method Zarrolou showed my by summing up the inequalities. So, for #2...

x^2<x-|y| So...

I. x^2 < x- y
OR
II. x^2 < x + y

AND

x < 0

Thus:

x^2 < x-y,
x<0

x^2 + x < x-y ==> x^2 < -y which isn't possible because x^2 will always be positive or zero

AND

x^2 < x+y
x<0

(using the same process as above) ==> x^2 < y Which is possible. For example, what if x^2 = 2 and y =4? Then x^2 could in fact be less than y.

Where did I go wrong?

That is not possible.
Remember that you are in the case \(y<0\)

\(x^2 < y\) here y will be negative so \(x^2<-ve\)=> not possible

Hope it's clear

Ok, I think I understand. We have two cases for |y|...y and -y. In the case of x^2 < y... it is derived from the inequality x^2 < x -(-y) right?

The passages above were correct, so I think you got the point.

Yes, from \(x^2<x+y\) you sum \(x<0\) and obtain \(x^2<y\), that's correct

Bunuel sama,

I had to ask this....
Why are u so awesome!!!!
Respect.

Posted from my mobile device

How I solved the problem.

Is x^2 < x - |y|?

(1) y > x
(2) x < 0


Is x^2 < x - |y|

Is |y| < x - x^2

|y| >= 0 , and x - x^2 is only >= 0 when 1<= x <= 0


(1) y>x - Does x - x^2 decrease or increase for some value 1< a < 0?

Test:
x = 1/2 , then x - x^2 = 1/4. Since y > 1/2, answer NO, sufficient

(2) x < 0 - Any value x < 0 produces a negative value for x - x^2. Since |y| > 0, answer NO, sufficient.

Is x^2 < x – |y|?
(1) y > x
(2) x < 0

OE


This is good one.

Merging similar topics. Please refer to the solutions above.

Sol: We need to find whether x^2< x- |y| or x^2+|y|<x
Note that LHS is always positive no matter what is value of x or y

For any negative value of x the inequality will not hold because LHS >/ 0 and hence B statement itself becomes sufficient

St 1. y >x (Considering only values of x>0 )

y=3, x=2 LHS>RHS
y=1/2 x= 1/10 again LHS >RHS
y=3,x= 0 agains LHS>RHS
Hence the inequality will not be true under given set of conditions.

Ans D

I GOT IT WRONG TWICE BEFORE GETTING IT RIGHT

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