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# m13 q23

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Re: m13 q23 [#permalink]  17 Oct 2010, 10:40
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dzyubam wrote:
Sure I can change it, but I don't see a difference. Why is "any X" wording better? Sorry if I'm missing something obvious .

Hi
I am non-native as you probably, so I may be wrong as well.
These are examples form Oxford dictionary:
all - consisting or appearing to consist of one thing only,Ex: The magazine was all advertisements.
all- the whole number of, Ex: All horses are animals, but not all animals are horses.

any - used with singular countable nouns to refer to one of a number of things or people, when it does not matter which one, Ex: Take any book you like.

тоесть, я думаю есть различие между словами любой и все(любой, какой бы из Х не был бы выбран выражение будет "+", когда как "для всех Х" - некоторые могут понять как - "а вот я нашел такой х который дает А отрицательным и ответ положительным" - я именно так и подумал.....). - хотя я уже сам запутался. и может не прав...

Well, I would like native speakers to express their point of view.
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Re: m13 q23 [#permalink]  25 Apr 2011, 13:44
Sorry, hopefully it's a badly worded question - but if we're asking if statement one definitively says whether A is positive or not, I'm not sure about the answer!

statement 1) says x^2 - 2x + A must be positive...
inserting x=3....
(3^2) - (3*2) + A must be positive...
9 - 6 + A must be positive
3 + A must be positive

A can be -2, -1, 0 and any value greater than 0... in usual DS-type questions, this suggests 'insufficient' to me...

as for statement 2) A(x^2) + 1 must be positive.... we know x^2 will be positive for any value of x, but since A can be 0 and still satisfy this statement, this is insufficient also.

If someone can explain why the range of values A can take where x=3 in statement 1) definitively answers the 'is A positive?' question, it'd be really appreciated
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Re: m13 q23 [#permalink]  25 Apr 2011, 14:27
bleemgame wrote:
Sorry, hopefully it's a badly worded question - but if we're asking if statement one definitively says whether A is positive or not, I'm not sure about the answer!

statement 1) says x^2 - 2x + A must be positive...
inserting x=3....
(3^2) - (3*2) + A must be positive...
9 - 6 + A must be positive
3 + A must be positive

A can be -2, -1, 0 and any value greater than 0... in usual DS-type questions, this suggests 'insufficient' to me...

as for statement 2) A(x^2) + 1 must be positive.... we know x^2 will be positive for any value of x, but since A can be 0 and still satisfy this statement, this is insufficient also.

If someone can explain why the range of values A can take where x=3 in statement 1) definitively answers the 'is A positive?' question, it'd be really appreciated

Question is correct IMO.
x is the variable and A a constant.
St1 says that A is some magical value such that the expression will always be +ve for any x. No matter what value you assign to x, the expression will always be +ve.

Say, A=-2
For x=3; the expression is +ve. Great!!!
for x=1; the expression is -ve(falsifies the statement)
Means; A must not be -2.
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Re: m13 q23 [#permalink]  25 Apr 2011, 16:55
fluke wrote:
x is the variable and A a constant.

St1 says that A is some magical value such that the expression will always be +ve for any x. No matter what value you assign to x, the expression will always be +ve.

thanks - error in interpretation (treating 'a' as variable that will alter to satisfy the equation accordingly and not a constant).
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Re: m13 q23 [#permalink]  26 Apr 2011, 12:29
I will go with option A .

As I understood...minimum value of A which makes the expression xsq2-2*x+A for all positive. If it's given that x2-x*x+a> 0 for all x , then we must be sure that is such that the inequality holds true.

If we know that a>1 , we can answer the question ( is positive).
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Re: m13 q23 [#permalink]  06 Jul 2011, 03:58
After reading this, I definitely see why stmt 1 is sufficient and stmt 2 is not.
However, I am wondering why can't we evaluate statement 2 by finding the range of the discriminant, as we did with statement 1. I tried doing so and came up with A>0, but we know that A is greater than or equal to 0.

Bunuel wrote:
CrushTheGMAT wrote:
Is $$A$$ positive?

1. $$x^2 - 2x + A$$ is positive for all $$x$$
2. $$Ax^2 + 1$$ is positive for all $$x$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

S1 is sufficient. x^2 - 2*x + A = (x - 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0.

*****
I believe the solution is incorrect on this question:

S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious.

S2 does not work, and neither does S1+S2, so the answer should be E.

Question: is $$A>0$$?

(1) $$x^2-2x+A$$ is positive for all $$x$$:

$$f(x)=x^2-2x+A$$ is a function of of upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$ --> $$f(x)=x^2-2x+A>0$$, which means that this function is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$.

Sufficient.

(2) $$Ax^2+1$$ is positive for all $$x$$:

$$Ax^2+1>0$$ --> when $$A\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too.

Not sufficient.

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Re: m13 q23 [#permalink]  23 Jan 2012, 17:45
In statement 1 why can't we consider x = 10000 (some insane large number), if we do then we can the square of x will be much greater than -2x and in that case I think we can have A = -1 and still have a positive value for the expression

I still believe statement 1 is insuff. I don't get it why suff?
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Re: m13 q23 [#permalink]  23 Jan 2012, 18:11
Expert's post
teal wrote:
In statement 1 why can't we consider x = 10000 (some insane large number), if we do then we can the square of x will be much greater than -2x and in that case I think we can have A = -1 and still have a positive value for the expression

I still believe statement 1 is insuff. I don't get it why suff?

Sure you can find SOME x for which A will be negative, BUT (1) says that x^2-2x+A is positive FOR ALL x-es. This expression to be positive for ALL x-es A must be more than 1. Refer to my post here to see why: m13-q23-70269.html#p716027

Hope it helps.
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Re: m13 q23 [#permalink]  23 Jan 2012, 18:24
Expert's post
Yalephd wrote:
Bunuel wrote:
Question: is $$A>0$$?

(1) $$x^2-2x+A$$ is positive for all $$x$$:

$$f(x)=x^2-2x+A$$ is a function of of upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$ --> $$f(x)=x^2-2x+A>0$$, which means that this function is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$.

Sufficient.

(2) $$Ax^2+1$$ is positive for all $$x$$:

$$Ax^2+1>0$$ --> when $$A\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too.

Not sufficient.

After reading this, I definitely see why stmt 1 is sufficient and stmt 2 is not.
However, I am wondering why can't we evaluate statement 2 by finding the range of the discriminant, as we did with statement 1. I tried doing so and came up with A>0, but we know that A is greater than or equal to 0.

You are right: if we use the same approach for (2) then we'll get A>0 BUT if A=0 then Ax^2+1 won't be a quadratic function anymore. So this approach will work only if A doesn't equal to zero, but we can not eliminate this case and if A=0 then Ax^2+1=1 is also always positive. Hence Ax^2+1 is positive for A>0 (if we use quadratic function approach) as well as for A=0, so for $$A\geq0$$.

Hope it's clear.
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Re: m13 q23 [#permalink]  02 Aug 2012, 00:22
why can't we apply the negative discriminant logic in statement 2; the parabola in stat 2 will also not cut x axis ( in all possible values of a, which only defines the slope of parabola). But discriminant would be

0-4*a*1<0

But that does not give us 0 for A and lead us to suff. Why does this not work in stat 2?
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Re: m13 q23 [#permalink]  02 Aug 2012, 00:26
Expert's post
teal wrote:
why can't we apply the negative discriminant logic in statement 2; the parabola in stat 2 will also not cut x axis ( in all possible values of a, which only defines the slope of parabola). But discriminant would be

0-4*a*1<0

But that does not give us 0 for A and lead us to suff. Why does this not work in stat 2?

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Re: m13 q23 [#permalink]  30 Apr 2013, 05:03
It is A for the simple reason that we have to find a value for A that works for every value of x. So A has to be > 1.

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Re: m13 q23 [#permalink]  30 Apr 2013, 09:44
Bunuel wrote:
CrushTheGMAT wrote:
Is $$A$$ positive?

1. $$x^2 - 2x + A$$ is positive for all $$x$$
2. $$Ax^2 + 1$$ is positive for all $$x$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

S1 is sufficient. x^2 - 2*x + A = (x - 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0.

*****
I believe the solution is incorrect on this question:

S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious.

S2 does not work, and neither does S1+S2, so the answer should be E.

Question: is $$A>0$$?

(1) $$x^2-2x+A$$ is positive for all $$x$$:

$$f(x)=x^2-2x+A$$ is a function of of upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$ --> $$f(x)=x^2-2x+A>0$$, which means that this function is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$.

Sufficient.

(2) $$Ax^2+1$$ is positive for all $$x$$:

$$Ax^2+1>0$$ --> when $$A\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too.

Not sufficient.

Your explanations are always so elegant.
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Re: m13 q23 [#permalink]  30 Apr 2013, 14:13
Is $$A$$ positive?

1. $$x^2 - 2x + A$$ is positive for all $$x$$
2. $$Ax^2 + 1$$ is positive for all $$x$$

Basically the question is, when $$x^2 - 2x + A$$ is positive for all $$x$$ --> can we say under that prerequisitions that $$A$$ is positive?
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Re: m13 q23 [#permalink]  30 Apr 2013, 20:15
Ultimor wrote:
Is $$A$$ positive?

1. $$x^2 - 2x + A$$ is positive for all $$x$$
2. $$Ax^2 + 1$$ is positive for all $$x$$

Basically the question is, when $$x^2 - 2x + A$$ is positive for all $$x$$ --> can we say under that prerequisitions that $$A$$ is positive?

Look for the Bunuel's Post $$\Rightarrow$$ m13-q23-70269.html#p716027 . Your all doubts will be cleared completely.
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Re: m13 q23 [#permalink]  30 Apr 2013, 23:54
teal wrote:
why can't we apply the negative discriminant logic in statement 2; the parabola in stat 2 will also not cut x axis ( in all possible values of a, which only defines the slope of parabola). But discriminant would be

0-4*a*1<0

But that does not give us 0 for A and lead us to suff. Why does this not work in stat 2?

hi,

a general quadratic equation is like this:

y=ax^2+bx+c

now your doubt is "why can't we apply the negative discriminant logic in statement 2".

you can apply the logic of negative discriminant only when you are sure that coefficient of x^2 is positive...i.e "a" is positive.

now in our question equation is AX^2+1

here we dont know whether A is positive or negative now if we use the discriminant rule in this then we are already assuming that "A" is positive...

and we are neglecting the possibility of A < 0...

so TAKEAWAYS:

use the discriminant concept only when coefficient of X^2 is positive.

hope it helps.

SKM
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Re: m13 q23   [#permalink] 30 Apr 2013, 23:54

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# m13 q23

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