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# Is A positive?

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Director
Joined: 30 Nov 2006
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11 Jun 2007, 11:09
Yahooooooooo !!!!

So, what do you think Fig ??

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11 Jun 2007, 12:30
Mishari wrote:
Yahooooooooo !!!!

So, what do you think Fig ??

A it is .... Yes

It was to your question I said (E)

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Director
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11 Jun 2007, 13:01
I got it, thanks Mishari for your patience.

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Senior Manager
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12 Jun 2007, 11:18
How can it be A?
x^2-2*x + A>0
A>2*x-x^2
assume x=1
A>2-1
A>1 --> A is positiv
assume x=3
A>6-9
A>-3
A might be -2, -1, 0 1 and so on...
So (1) is insuff

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Director
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12 Jun 2007, 11:39
You know what guys .. I was trying to remember and think how I got A as an answer .. I failed !!!!!!!

I read my own explaination, it makes no sense to me !!!!!!!!

OH My God ! help

Depending on the value of X, A can be either positive or negative. Nothing is said about whether x is an integer or not.

When 0<x<1> A is positive
when -1<x<0> A could be negative or positive

For either case, the result of the equation is positive

I disagree with the OA

------------------------------------

Fig .. hmmm

I ... you know .. I really .. it's just that ..

aah well .. can you tell me how you got A for an answer ?

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12 Jun 2007, 11:48
ok..Now I will try to get A
x^2-2*x +A is a parabola
x^2-2*x +A >0 as it is given in (1)
it means that for every given x, y>0
here we have a min point
min point of a parabola has coordinates [-b/(2a); c-(b^2/(4a))] ---> (1;A-1)
y-coord of min point of this parabola = A-1
y-coord must be > 0
A-1>0
A>1
A - positive

I don't know
I am lost

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12 Jun 2007, 11:52
Mishari wrote:
You know what guys .. I was trying to remember and think how I got A as an answer .. I failed !!!!!!!

I read my own explaination, it makes no sense to me !!!!!!!!

OH My God ! help

Depending on the value of X, A can be either positive or negative. Nothing is said about whether x is an integer or not.

When 0<x<1> A is positive
when -1<x<0> A could be negative or positive

For either case, the result of the equation is positive

I disagree with the OA

------------------------------------

Fig .. hmmm

I ... you know .. I really .. it's just that ..

aah well .. can you tell me how you got A for an answer ?

It's by using the curves
y = a*(x - x1)*(x-x2)

o If x < x1 or x > x2, then y will be of the sign of a
o If x1 < x < x2, then y will be of the sign of -a

but,
In case of no root exists : b^2 - 4*a*c < 0, y will all time be of the sign of a from y = a*x^2 + b*x + c.

So, here, we search the value A for which 4- 4*1*A < 0. That gives us A > 1 > 0.

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Director
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12 Jun 2007, 12:14
My approach:

Stmt1: x^2 - 2x + A > 0
x^2 - 2x can be +ve or -ve.
Here the -ve value of if x^2 - 2x is the deciding factor value of A
if x^2 - 2x is -ve, the least value x^2 - 2x can have is -1.
So A should be +ve.

Stmt2: A*x^2 + 1 > 0
for the above expression to be +ve for all x. A should be greater than or equal to zero. So INSUFF.

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12 Jun 2007, 19:38
Let's look at stmt 1 : x^2 - 2*x + A
Put X = 0 , A>0 .
Put X = -1, 1+2+A>0 therefore A>-3, does not prove conclusively.
Put X = 1 , 1-2+A>0 A-1>0, A>1, therefore A>0

Statement 1 is not sufficent.

Let's look at stmt 2 : A*x^2 + 1 is positive for all x
Put X = 0 , A*0 + 1 >0, does not prove anything. A can be any value
Put X =-1 , A*-1+ 1 >0, -A+1<0>-1

Statement 2 is not sufficent.

Hence E should be the answer.

Last edited by vc019 on 12 Jun 2007, 20:04, edited 1 time in total.

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12 Jun 2007, 19:58
St1:
A must be at least 1. negative values of x poses no problems, but for values of x like 1 or 2, A must be positive in order for the function to stay positive. Sufficient.

St2:
Sufficient. If A is negative, then A*x^2 -1 would not be positive for all x.
E.g. if x = 9, and A = -1, then Ax^2+1 = -8.

Ans: D

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12 Jun 2007, 23:07
ywilfred wrote:
St1:
A must be at least 1. negative values of x poses no problems, but for values of x like 1 or 2, A must be positive in order for the function to stay positive. Sufficient.

St2:
Sufficient. If A is negative, then A*x^2 -1 would not be positive for all x.
E.g. if x = 9, and A = -1, then Ax^2+1 = -8.

Ans: D

In stat 2, A could be equal to 0 : 0*x^2 + 1 = 1 > 0 ... I have fallen in the trap too

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12 Jun 2007, 23:35
Little help here plz. I too think that A cannot be the answer. I surely dont know about parabolas or curves yet, so i have decided to adopt the strategy of picking numbers. i have decided to pick 3 simple number 0,1,-1 and when plug them into statement 1. for 0 and +1, A>0 but for -1, A >-1/2, which shows that it is not sufficeint. If u could please explain this paradox and explain how it sufficient in simple words.

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12 Jun 2007, 23:46
Fig wrote:
ywilfred wrote:
St1:
A must be at least 1. negative values of x poses no problems, but for values of x like 1 or 2, A must be positive in order for the function to stay positive. Sufficient.

St2:
Sufficient. If A is negative, then A*x^2 -1 would not be positive for all x.
E.g. if x = 9, and A = -1, then Ax^2+1 = -8.

Ans: D

In stat 2, A could be equal to 0 : 0*x^2 + 1 = 1 > 0 ... I have fallen in the trap too

ah yes... then it shoud be A

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Director
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Re: DS: Is A +ve? [#permalink]

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13 Jun 2007, 00:26
Himalayan wrote:
Is A positive?

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

I still think it should be E. It was discussed earlier and posted OE as under:

OE is: St1. can be rewritten as (x - 1)^2 + A – 1 > 0, for this to hold true for all possible values of x, A > 1. So sufficient.

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Re: DS: Is A +ve? [#permalink]

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13 Jun 2007, 00:38
Himalayan wrote:
Himalayan wrote:
Is A positive?

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

I still think it should be E. It was discussed earlier and posted OE as under:

OE is: St1. can be rewritten as (x - 1)^2 + A – 1 > 0, for this to hold true for all possible values of x, A > 1. So sufficient.

so if it's sufficient, how can it still be E? E

E - neither statement is sufficient to answer the question.

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Re: DS: Is A +ve? [#permalink]

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13 Jun 2007, 00:38
Himalayan wrote:
Himalayan wrote:
Is A positive?

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

I still think it should be E. It was discussed earlier and posted OE as under:

OE is: St1. can be rewritten as (x - 1)^2 + A – 1 > 0, for this to hold true for all possible values of x, A > 1. So sufficient.

The OE is right : this approach is excellent too

We know that (x-1)^2 >= 0. Thus, to be sure that (x - 1)^2 + A – 1 > 0, we need to have A - 1 > 0 <=> A > 1.

Indeed, we take the minimum of (x-1)^2 said 0 here and we plug it in the inequation:
0 + (A - 1) > 0
<=> A > 1

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Re: DS: Is A +ve? [#permalink]

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13 Jun 2007, 00:40
ywilfred wrote:
Himalayan wrote:
Himalayan wrote:
Is A positive?

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

I still think it should be E. It was discussed earlier and posted OE as under:

OE is: St1. can be rewritten as (x - 1)^2 + A – 1 > 0, for this to hold true for all possible values of x, A > 1. So sufficient.

so if it's sufficient, how can it still be E? E

E - neither statement is sufficient to answer the question.

I think Himalayan disagree with the OA

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13 Jun 2007, 07:49
Can some explain me
why do you say this?

From 1
x^2 - 2*x + A > 0
Meaning that :
b^2 - 4*a*c < 0 (this is the descriminat of a*x^2 + b*x + c)

How do you know that it has no real solution?

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13 Jun 2007, 07:54
ajisha wrote:
Can some explain me
why do you say this?

From 1
x^2 - 2*x + A > 0
Meaning that :
b^2 - 4*a*c < 0 (this is the descriminat of a*x^2 + b*x + c)

How do you know that it has no real solution?

We want to keep for all x : x^2 - 2*x + A > 0.

If b^2 - 4*a*c < 0, we have Sign(a*x^2+b*x+c) = Sign(a). So, the sign will not flip between roots cause there is no root (the curve stays on y>0 if a>0 or the curve stays on y <0 if a<0)

Hope that helps

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Re: DS: Is A +ve? [#permalink]

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01 Nov 2007, 04:24
Himalayan wrote:
Is A positive?

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

1. y=(x-1)^2 +(A-1) >0

If A is not positive, y could be as small as A - 1, which would be negative.
Thus A must be positive
SUFF

2 A must be positive or 0. NOT SUFF

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Re: DS: Is A +ve?   [#permalink] 01 Nov 2007, 04:24

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