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# ds:

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Senior Manager
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ds: [#permalink]  22 Apr 2005, 06:33
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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
please show workings
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Current Student
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[#permalink]  22 Apr 2005, 06:56
D for me

1)
X^2-2X+A >0 for all X (this statement is key)

which implies that for X=0, A is positive.... suff!
2)
AX^2+1>0 for all X

well we know X^2 is positive...the only way (2) can be positive for all X is if A is positive...
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[#permalink]  22 Apr 2005, 07:54
C for me

for I if x = 5, then A can be negative or positive for the equation to be greater than zero. Because 15+A>0 hence A can be -14 or greater.
So A and D are out.

for II if x=1/2 then A/4+1>0. Hence A can be -3 or greater to satisfy the equation. So not B

Therefore either C or E. No time left, For me Joe bloggs would think E but i would think there are two equations two variables, i should be able to find the value therefore C
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[#permalink]  22 Apr 2005, 12:07
coolcal wrote:
C for me

for I if x = 5, then A can be negative or positive for the equation to be greater than zero. Because 15+A>0 hence A can be -14 or greater.
So A and D are out.

for II if x=1/2 then A/4+1>0. Hence A can be -3 or greater to satisfy the equation. So not B

Therefore either C or E. No time left, For me Joe bloggs would think E but i would think there are two equations two variables, i should be able to find the value therefore C

Joe Bloggs may be right. My answer is E.

Agree with your individual explanations. So the answer has to be either C or E.

Combining I and II,

If x=3,
From I) 9 - 6 +a >0 ==> A > -3
From II) 9A + 1 > 0 ===> A > -(1/9)

As the problem does not say A is an integer, A could be any number greater than -(1/9). So A could be +ve or -ve.

Answer E.
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[#permalink]  23 Apr 2005, 04:30
D)...

1) to be true for ALL x. when you pick x=16 you can choose if A is +ve or -ve. when you pick x=1 you MUST choose for A a +ve number => sufficient

2) same reasonig as above. try x=16 and try x=1 => sufficient
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Director
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[#permalink]  23 Apr 2005, 04:38
D..for all X means when X is either negative , positive or non integer. I stand corrected on this one.
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[#permalink]  23 Apr 2005, 06:17
Christoph/Folaa - please let me know how you think D. I showed with examples earlier that we cannot prove A is either positive or negative conclusively when x = 5 or x = 1/2 so your solution of D is not ringing a bell with me.
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[#permalink]  23 Apr 2005, 13:47
My take on this is

From I

x^2 - 2x + A = Positve

lets take sample value of x
x = 10

100 - 20 + A = Positve
or

80 + A = +ve so A can be - 3.14 or 1547888 it does not matter

I is insufficient

From II

Ax^2 + 1 = +ve

Let x = 1/2

then

A/4 + 1 = +ve Again A can be - 1/8 or 20 it does not matter

II is insufficient

so its C / E

from I and II we cannot co-relate

My answer to this will be E
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[#permalink]  23 Apr 2005, 17:12
guys
the key is the word in the problem stem "for all x"...which included + ve and -ve values...A has to be positive...

if for all X was not mentioned I would have picked E as my answer...
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[#permalink]  24 Apr 2005, 06:00
ax^2 + bx+c = y (Say)

The sign of y is same as a if Determinant D <= 0

If on the other hand D > 0 then there are two distinct real roots say p and q (p<q). so for all x <p, y has same sign as a and also for all x>q, y has same sign as a. But for p<x<q the sign of y is opposite to that of a.

Now D = 4-4A in our case. Depending on this value of D we can decide on the sign of the expression.

Neither of the statements 1 or 2 clearly let us decide on the sign of D

So if 4-4A <=0 or if A >=1 then y is +ve, but for 0<A<1 D > 0 so different possibilities.

E
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[#permalink]  24 Apr 2005, 12:07
OA pls
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[#permalink]  24 Apr 2005, 13:16
"A"

state 1...(x-1)^2 + A-1 > 0.....as (x-1)^2 is always > 0...then A-1 > 0...
A> 1...suff

state 2...insuff....A can be < 0 or > 0 and still be able to satisfy.
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[#permalink]  24 Apr 2005, 18:20
it's E

the critical values are reached when -1<x<0 , there can be different solutions, just take the value (-1/2):

1) X^2-2X+A >0 -> Let's use (-1/2)
1/4 + 1 + A > 0 ; 5/4 + A > 0

A could be any +ive number or any -ive number from 0 to -5/4 excluded, for example if A=-1 -> 5/4 -1 = 1/4 which is correct (+ive)

2) AX^2+1>0 -> Let's use (-1/2)
A*1/4 + 1 > 0

A could be any +ive number or any -ive number from 0 to -4 excluded, for example if A=-2 , -2*1/4 + 1 = 1/2 which is correct (+ive)

Finally, when X is between -1 and 0, A can be +ive or -ive, hence E
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[#permalink]  24 Apr 2005, 18:31
banerjeea_98 wrote:
"A"

state 1...(x-1)^2 + A-1 > 0.....as (x-1)^2 is always > 0...then A-1 > 0...
A> 1...suff

state 2...insuff....A can be < 0 or > 0 and still be able to satisfy.

I almost fell for the explanation for A being sufficient

1. If (x-1)^2 + a-1 > 0 , a can still be +ve or -ve ,
eg. let (x-1)^2 = 5 , a can be -1 ,-3 or 21 and the expression will still be >0
so A is insufficient.

2. B is also insufficient. ax^2 + 1 > 0

a> -1/x^2 ; let x^2 = 5 implies a > -1/5
so a could be -1/2 or a could be +3.
B is insufficient too.

Ans is C or E. To eliminate one of these, lets add the equations 1 and 2.

x^2 -2x +a >0 - I
ax^2 + 1>0 - II

adding and solving for a we get.
x^2 - 2x + 1 + a + ax^2 > 0
or (x-1)^2 + a (1 + x^2) > 0
or a > - (1+x^2)/(x-1)^2

for any value of X , you will get something like
a > - 3/4
so a could be still +ve or -ve.

Finally ans is E
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Re: ds: [#permalink]  24 Apr 2005, 20:45
mirhaque wrote:
please show workings

it is definitely a difficult problem. but i go with E.
from i, x^2-2x+a=+ve . it doesnot mean that x must be positive but the whole equation must have positive value. suppose x=1, and the value of the eq is 1 (at least +ve).
x^2-2x+a=1
a =1............ the equation is satisfied.

if x=3, and the value of the eq is 1 (at least +ve).
x^2-2x+a =1
3+a =1
a =-2 ............ the equation is still satisfied but the information given st i is insufficient.

from ii, it is even insufficientsuppose x =1, and ax^2+1=2
a+1=2
a = 1

if x = 3, but ax^2+1=1/5
9a+1=1/5
a = -4/45 so insufficient.

combining also it doesnot give any solution (because i have used the same values in st i and ii). therefore it should be E.
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Re: ds: [#permalink]  25 Apr 2005, 03:55
mirhaque wrote:
please show workings

E it is.

A) if A = (-1), then we still have a positive answer.

B) if A = (-0.5), x^2 =1, then we still have a positive answer.
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[#permalink]  25 Apr 2005, 04:11
i change from D) to A)...

1) there is no chance that the eq is +ve for ALL x (or am i missing something?). how is the eq +ve for x=1 ? => sufficient

2) A can be +ve or -ve for the eq to be +ve => insufficient
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Last edited by christoph on 25 Apr 2005, 12:09, edited 1 time in total.
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[#permalink]  25 Apr 2005, 12:01
fresinha12 wrote:
D for me

1)
X^2-2X+A >0 for all X (this statement is key)

which implies that for X=0, A is positive.... suff!
2)
AX^2+1>0 for all X

well we know X^2 is positive...the only way (2) can be positive for all X is if A is positive...

Thanks Fresh answer and explanation. OA is D!
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[#permalink]  25 Apr 2005, 17:46
mirhaque wrote:
fresinha12 wrote:
D for me

1)
X^2-2X+A >0 for all X (this statement is key)

which implies that for X=0, A is positive.... suff!
2)
AX^2+1>0 for all X

well we know X^2 is positive...the only way (2) can be positive for all X is if A is positive...

Thanks Fresh answer and explanation. OA is D!

SO TRICKY, I understand ythe meaning of "for all X", it is definitely a greart problem. For example in the first statement it means that you will never consider values of X that will lead to a -ive value when added to A...
Terrific problem
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Re: ds: [#permalink]  26 Apr 2005, 06:15
(1) x = (2+ (4-4A)^1/2) / 2 and (2 - (4-4A)^1/2) / 2

x =(2 - (4-4A)^1/2) / 2 must be positive, so 0 <A <1

but I disagree with the second condition.

(2)AX^2 + 1 must be positive for all x.

if A=0, AX^2 + 1 will be always positive for all x.

How could D?
Re: ds:   [#permalink] 26 Apr 2005, 06:15

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