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

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Is A positive? 1. x^2 - 2x + A is positive for all x 2. [#permalink] New post 20 Dec 2007, 09:24
Is A positive?

1. x^2 - 2x + A is positive for all x
2. A*x^2 + 1 is positive for all x
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 [#permalink] New post 20 Dec 2007, 16:42
I guess A

1. y=x^2 - 2x +A>0. min(y)=1-2+A>0, A>1. SUFF.
2. A*x^2 + 1 >0. A>=0. 0 - is not positive number. (?) INSUFF.

I'm not sure about bold statement.
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 [#permalink] New post 20 Dec 2007, 20:21
Tough, but my guess will be between C or E:

I. Insuff: if x = 1: 1 - 2 + A >0
A > 1
if x = 0: a > 0
if x = 3: 9 - 6 + A > 0, where A could be -ve

II. Insuff:
A * x^2 > -1
A > -1/x^2
where A could be -ve and +ve

At the end I try educated guess =) and vote for E
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Re: positive A [#permalink] New post 21 Dec 2007, 22:30
bmwhype2 wrote:
Is A positive?

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


From (i)
A > 2x-x^2

say,x=5
A>-15

A might be negative.

And for other values of X it might be positive.

(i) is insufficient.

From (ii)

A > -1/x^2

A can be positive as well as negative.

Thus answer should be "E".
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 [#permalink] New post 22 Dec 2007, 01:35
1. x^2 - 2x + A is positive for all x

I think you forget about "all x"
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 [#permalink] New post 22 Dec 2007, 20:51
i felt this was a hard one.


OA is A
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 [#permalink] New post 22 Dec 2007, 20:57
walker wrote:
I guess A

1. y=x^2 - 2x +A>0. min(y)=1-2+A>0, A>1. SUFF.
2. A*x^2 + 1 >0. A>=0. 0 - is not positive number. (?) INSUFF.

I'm not sure about bold statement.


in min(y), how do u know that x is an integer?
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 [#permalink] New post 23 Dec 2007, 00:47
bmwhype2 wrote:
in min(y), how do u know that x is an integer?


I use fast method (out of GMAT) to find extreme (minimum) of parabola:

f(x)=x^2 - 2x +A
f'(x)=2x-2=0 ==> x=1
f(x=1)=1-2+A=A-1, So A-1 must be positive and A>1
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 [#permalink] New post 23 Dec 2007, 02:00
Another way can be to search for Delta and to make it negative (no roots) :)

x^2 - 2x + A

Delta = b^2 - 4*a*c
= (-2)^2 -4*(1)*A
= 4*(1-A)

x^2 - 2x + A remains positive, of the sign of 1*x^2, if and only if Delta < 0

Thus,
4*(1-A) < 0
<=> A > 1

:)
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 [#permalink] New post 23 Dec 2007, 02:28
walker wrote:
Fig wrote:
Another way can be to search for Delta and to make it negative (no roots) :)


Nice way!


Your's as well :)
  [#permalink] 23 Dec 2007, 02:28
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