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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
(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
(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
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
(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.
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).
(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....
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
(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.
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.
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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