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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
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.
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.
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.
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 _________________
ash
________________________
I'm crossing the bridge.........
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.
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