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Is A positive?

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Re: [#permalink] New post 21 Apr 2011, 08:39
grad_mba wrote:
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



For equation 1) x^2-2x+A >0
when x = 0 A>0
but however when x=3 the expression would be 3^2-2*3 +A >0 resulting in A> -3 hence A can still have -2, -1 as values to keep the expression positive. So 1) alone is insufficient.
Similarly with 2) A could be greater than -1 to keep expression positive

Both together also does not yield new info. So E
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Re: Re: [#permalink] New post 21 Apr 2011, 08:52
PallavTandon wrote:
grad_mba wrote:
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



For equation 1) x^2-2x+A >0
when x = 0 A>0
but however when x=3 the expression would be 3^2-2*3 +A >0 resulting in A> -3 hence A can still have -2, -1 as values to keep the expression positive. So 1) alone is insufficient.
Similarly with 2) A could be greater than -1 to keep expression positive

Both together also does not yield new info. So E


I disagree with my own approach.
Explanation -->
A is a constant in the equation.
For example the condition should be satisfying for all values of x and one and only one value of A.
If A > 1, the equation is positive for all values of X.

However if A < 1 the expression becomes negative for any value of A so we cannot even consider those scenarios.
So 1) is sufficient

Answer will be A
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Re: DS: Is A +ve? [#permalink] New post 06 Sep 2011, 04:19
**
Quote:
Is A positive?

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


took a long while for me to get my head around this...very tricky question! :shock:

From statement 1
x^2 - 2*x + A > 0, for all x-values
x^2-2x+1+A-1>0
[highlight](x-1)^2[/highlight] + A - 1>0
For all values of x, highlighted portion will always be positive.
In order for whole equation to hold true for all values of x, A has to be at least 1 i.e. A>0(as this will render A-1 to be zero)
If A were to be zero or negative, it will not suffice the equation(as it will yield positive values only for some values of x).
--->sufficient


Quote:
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.


From statement 2
Ax^2 + 1 > 0, for all x values
for this statement to stand true, A must be more than or equals to zero (i.e. not necessarily positive).
If A were to be negative(i.e. <0), the statement may not stand true for some values of x.
--->Insufficient

Answer: A
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Re: DS: Is A +ve? [#permalink] New post 06 Sep 2011, 06:35
Where in the question does it mention integers?

Consider x = 0.1, -0.1
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Re: DS: Is A +ve? [#permalink] New post 06 Sep 2011, 07:47
Here is what I got; please tell me where my reasoning is flawed...if it is

1)x^2-2x+A>0
Just factor the equation: because we know that the two unknown factors multiplied together = +A, but when added together = -2x, we have (x-_ ) (x- _ ), therefore A must be positive. Sufficient

2) Ax^2+1>0
A could be 0 or positive; by definition 0 is neither positive nor negative, so we have two contradicting outcomes; Insufficient

I'm going with A
Re: DS: Is A +ve?   [#permalink] 06 Sep 2011, 07:47
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