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Is A positive?

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Is A positive? [#permalink] New post 08 Jun 2007, 22:45
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A
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Question Stats:

26% (02:35) correct 74% (01:52) wrong based on 65 sessions
Is A positive?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-a-positive-x-2-2x-a-is-positive-for-all-x-ax-2-1-is-97302.html
[Reveal] Spoiler: OA
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Director
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 [#permalink] New post 09 Jun 2007, 08:46
Should be 'E'.

Stmt1: x^2 - 2x +A > 0
(x-1)^2 + (A-1) > 0
(x-1)^2 > 1-A
A can be +ve or -ve. So INSUFF

Stmt2: Ax^2 + 1 > 0
A can be +ve or -ve. So INSUFF

Taking them together:
I can't establish whether A is +ve or -ve.

So INSUFF.
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 [#permalink] New post 09 Jun 2007, 09:12
Go with A

St1

x^2 - 2x + A > 0 for ALL x,

x = 0 should satisfy the above eqn => A > 0


St2

Ax^2 + 1 > 0 for ALL x,

A can be -ve or +ve

Insuff
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 [#permalink] New post 09 Jun 2007, 09:22
(1) when x is negative, A must be positive; when x is positive, A can either be positive or negative. Since the value of A has to result in a positive outcome for all values of x, A is positive

Statement 1 is sufficient

(2) A can be either positive or zero.

statement 2 is insufficient


Answer: A
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 [#permalink] New post 09 Jun 2007, 09:35
Mishari wrote:
(1) when x is negative, A must be positive; when x is positive, A can either be positive or negative. Since the value of A has to result in a positive outcome for all values of x, A is positive

Statement 1 is sufficient

(2) A can be either positive or zero.

statement 2 is insufficient


Answer: A


Take x =4 and A = 2 the equation is satisfied.
Take x =4 and A = -1 the equation is satisfied.
So A can be +ve or -ve. So stmt1 is INSUFF
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 [#permalink] New post 09 Jun 2007, 09:49
But A = -1 does not satisfy the condition when X is negative.

You only used a positive value for X. The value of A should yield into a positive outcome for ALL values of X.

I still think it's A
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 [#permalink] New post 09 Jun 2007, 09:49
vshaunak@gmail.com wrote:
Mishari wrote:
(1) when x is negative, A must be positive; when x is positive, A can either be positive or negative. Since the value of A has to result in a positive outcome for all values of x, A is positive

Statement 1 is sufficient

(2) A can be either positive or zero.

statement 2 is insufficient


Answer: A


Take x =4 and A = 2 the equation is satisfied.
Take x =4 and A = -1 the equation is satisfied.
So A can be +ve or -ve. So stmt1 is INSUFF



ST1 must be true for ALL x,

so when x =0 , A must be +ve , hence suff
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 [#permalink] New post 09 Jun 2007, 13:34
Mishari wrote:
But A = -1 does not satisfy the condition when X is negative.

You only used a positive value for X. The value of A should yield into a positive outcome for ALL values of X.

I still think it's A


Let's take x negative:
x= -5 A = -10
x = -5 A = +10

x^2 - 2x + A is +ve in both of the cases. How can you say A will always be +ve.
Am I missing something here?
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 [#permalink] New post 09 Jun 2007, 13:44
(D) for me :) ..... Excellent question that goes deep in the concept of curves :)

A > 0 ?

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)
<=> 4 - 4*A*1 < 0
<=> 4*A > 4
<=> A > 1 > 0

SUFF.

From 2
A*x^2 + 1 > 0

Meaning that :
b^2 - 4*a*c < 0 (this is the descriminat of a*x^2 + b*x + c) and A must be positive to create a valley shape for the curve A*x^2 + 1.

SUFF.
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 [#permalink] New post 09 Jun 2007, 13:55
vshaunak
------------
The value of A must span all values of X .. ALL values .. small negative fractions, large positive integers .. etc

try x = -0.5 --> A must be positive in order for the condition to apply.
The value of A that results in a positive value of the equation for ALL values of X .. [ all and any value of X ] is a positive A


Fig
----
Holly %#@$ ..
curves ? you INSEAD guys .. grrrrr :x

But I still think it's A
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 [#permalink] New post 09 Jun 2007, 13:59
Fig wrote:
(D) for me :) ..... Excellent question that goes deep in the concept of curves :)

A > 0 ?

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)
<=> 4 - 4*A*1 < 0
<=> 4*A > 4
<=> A > 1 > 0

SUFF.

From 2
A*x^2 + 1 > 0

Meaning that :
b^2 - 4*a*c < 0 (this is the descriminat of a*x^2 + b*x + c) and A must be positive to create a valley shape for the curve A*x^2 + 1.

SUFF.


Fig, there is a mistake in your solution
b^2 - 4*a*c > 0 for the real roots.
4-4A > 0
A < 1
Hence this holds true for +ve as well as for -ve values of A.
Stmt1 is NOT suff.
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 [#permalink] New post 09 Jun 2007, 14:03
The case A=0 for the second case is possible... My mistake... the equation of a*x^2+b*x+c was an assertion without having checked the special case A=0 that works perfectly (0*x^2 + 1 = 1 > 0). :)

So, answer A :)
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 [#permalink] New post 09 Jun 2007, 14:04
Himalayan. . HELP .. what's the OA ? save the paradox
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 [#permalink] New post 09 Jun 2007, 14:05
vshaunak@gmail.com wrote:
Fig wrote:
(D) for me :) ..... Excellent question that goes deep in the concept of curves :)

A > 0 ?

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)
<=> 4 - 4*A*1 < 0
<=> 4*A > 4
<=> A > 1 > 0

SUFF.

From 2
A*x^2 + 1 > 0

Meaning that :
b^2 - 4*a*c < 0 (this is the descriminat of a*x^2 + b*x + c) and A must be positive to create a valley shape for the curve A*x^2 + 1.

SUFF.


Fig, there is a mistake in your solution
b^2 - 4*a*c > 0 for the real roots.
4-4A > 0
A < 1
Hence this holds true for +ve as well as for -ve values of A.
Stmt1 is NOT suff.


Yes... agree :).... I should not come to reply so late after an exhausting day :p.... Completly wrong here....

It's so (E)... Agree :).... Excellent catch ! :D :)
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 [#permalink] New post 09 Jun 2007, 14:11
Mishari wrote:
vshaunak
------------
The value of A must span all values of X .. ALL values .. small negative fractions, large positive integers .. etc

try x = -0.5 --> A must be positive in order for the condition to apply.
The value of A that results in a positive value of the equation for ALL values of X .. [ all and any value of X ] is a positive A


Fig
----
Holly %#@$ ..
curves ? you INSEAD guys .. grrrrr :x

But I still think it's A


Are u insulting me, a futur HEC MBA, by calling me INSEAD guy or.... are u insulting the INSEAD guys by considering me like one of them ? ;)
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 [#permalink] New post 09 Jun 2007, 14:21
loool .. sorry Fig .. My bad.

I guess I was insulting HEC guys :lol: :lol:
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 [#permalink] New post 09 Jun 2007, 14:28
Fig wrote:
vshaunak@gmail.com wrote:
Fig wrote:
(D) for me :) ..... Excellent question that goes deep in the concept of curves :)

A > 0 ?

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)
<=> 4 - 4*A*1 < 0
<=> 4*A > 4
<=> A > 1 > 0

SUFF.

From 2
A*x^2 + 1 > 0

Meaning that :
b^2 - 4*a*c < 0 (this is the descriminat of a*x^2 + b*x + c) and A must be positive to create a valley shape for the curve A*x^2 + 1.

SUFF.


Fig, there is a mistake in your solution
b^2 - 4*a*c > 0 for the real roots.
4-4A > 0
A < 1
Hence this holds true for +ve as well as for -ve values of A.
Stmt1 is NOT suff.


Yes... agree :).... I should not come to reply so late after an exhausting day :p.... Completly wrong here....

It's so (E)... Agree :).... Excellent catch ! :D :)


No... After 1 shower to wake me up, I have read again your sign at start.... I was thinking I had flipped one... but no :)

Actually, I really wanted to say b^2 - 4 * a * c < 0.... to force the curve to be always of the sign of a... Here a=1 and so always positive. :)

Sorry guys.... the answer must be (A) :)

Last edited by Fig on 09 Jun 2007, 14:31, edited 1 time in total.
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 [#permalink] New post 09 Jun 2007, 14:29
Mishari wrote:
loool .. sorry Fig .. My bad.

I guess I was insulting HEC guys :lol: :lol:


;)... So it's not kind anyway ! :)
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 [#permalink] New post 09 Jun 2007, 14:37
A side question Fig: really why do you still participate in GMAT topics and posts in the forum ? you're already waay beyond GMAT .. You're mastered it.
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 [#permalink] New post 10 Jun 2007, 01:21
Mishari, I think the answer to this question is.... (E) non of the statments are sufficient ;)

More seriously, I feel it as a win win situation. By my side, I have time and the will to keep me uptodate and studying. By the community side, I try to clarify the points that I can and if it means some guys learn something from it to kill the GMAT.... that's an excellent plus ! :)

Also, it's a question of personnality & appreciation. I like Math for a while and I like to give back what I recieved :)... The GMATClub was an amazing source of information and learnings during the whole application process, a process very demanding in energy and time. It was so not so much possible for me to compensate in a certain way what I recieved.... It's a kind of pay back now with high interest (rate;)) and entertainements :)..... I must admit also that I'm a kind of handy guy, but do not repeat it ;)

That said, I will probably be much less accessible after December 2007. It will be the time for me to move on into the School. :)
  [#permalink] 10 Jun 2007, 01:21
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