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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

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

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.

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.

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.

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

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.
Also, Fig due, if u could please tell us all about curves and parabolas stuff or perhaps upload o document or a link where we can find details about this stuff, that will be a great help.

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

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)

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