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Director
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26% (02:17) correct
73% (01:55) wrong based on 5 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
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Director
Joined: 14 Jan 2007
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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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Director
Joined: 13 Mar 2007
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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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Director
Joined: 30 Nov 2006
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(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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Director
Joined: 14 Jan 2007
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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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Director
Joined: 30 Nov 2006
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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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Director
Joined: 13 Mar 2007
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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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Director
Joined: 14 Jan 2007
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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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SVP
Joined: 01 May 2006
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(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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Director
Joined: 30 Nov 2006
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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 
But I still think it's A
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Director
Joined: 14 Jan 2007
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Fig wrote: (D) for me ..... Excellent question that goes deep in the concept of curves A > 0 ? From 1x^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 2A*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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SVP
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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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Director
Joined: 30 Nov 2006
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Himalayan. . HELP .. what's the OA ? save the paradox
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SVP
Joined: 01 May 2006
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vshaunak@gmail.com wrote: Fig wrote: (D) for me ..... Excellent question that goes deep in the concept of curves A > 0 ? From 1x^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 2A*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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SVP
Joined: 01 May 2006
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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 But I still think it's AAre 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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Director
Joined: 30 Nov 2006
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loool .. sorry Fig .. My bad.
I guess I was insulting HEC guys 
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SVP
Joined: 01 May 2006
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Fig wrote: vshaunak@gmail.com wrote: Fig wrote: (D) for me ..... Excellent question that goes deep in the concept of curves A > 0 ? From 1x^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 2A*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, 15:31, edited 1 time in total.
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SVP
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Mishari wrote: loool .. sorry Fig .. My bad. I guess I was insulting HEC guys ... So it's not kind anyway !
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Director
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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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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.
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close
