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# ds:

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Senior Manager
Joined: 02 Feb 2004
Posts: 345
Followers: 1

Kudos [?]: 23 [0], given: 0

ashkg wrote:
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

Sorry...I goofed. Correct answer is E.
VP
Joined: 30 Sep 2004
Posts: 1490
Location: Germany
Followers: 4

Kudos [?]: 91 [0], given: 0

ok. if in 1) x=1, its 1-2+A=+ve => -1+A=+ve

in the case above, how can A be negative or positive for the solution to be +ve number ? IMO A must be positve. where am i wrong ?
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