Is A positive?
1) x^2-2x+A is positive for all x
2) Ax^+1 is positive for all x
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- prachich1987
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Statement 1: (x² - 2x + A) > 0 for all x.prachich1987 wrote:Is A positive?
1) x^2-2x+A is positive for all x
2) Ax^+1 is positive for all x ... I assume it is Ax^2 + 1
=> (x² - 2x + 1 + A - 1) > 0
=> (x - 1)² + (A - 1) > 0
=> (A - 1) > 0 ...................... As (x - 1)² ≥ 0 for all x
=> A > 1
=> A is positive
Sufficient
Statement 2: (Ax² + 1) > 0 for all x
=> Ax² > -1
=> x² > -(1/A)
Now if A is negative -(1/A) is positive. But for x = 0, x² = 0, which cannot be greater than a positive quantity. Hence A has to be positive quantity such that -(1/A) is always negative and x² is always greater than a negative quantity.
Now what if A = 0?
Then the expression (Ax² + 1) is equal to 1 for all x. Thus (Ax² + 1) > 0 for all x.
Hence if (Ax² + 1) > 0 for all x, then A ≥ 0
Not Sufficient.
The correct answer is A.
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- prachich1987
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Thanks Anurag.I understand that B alone is not sufficientAnurag@Gurome wrote:
Statement 1: (x² - 2x + A) > 0 for all x.
=> (x² - 2x + 1 + A - 1) > 0
=> (x - 1)² + (A - 1) > 0
=> (A - 1) > 0 ...................... As (x - 1)² ≥ 0 for all x
=> A > 1
=> A is positive
Sufficient
But according to me also A alone is not sufficient.
Refer to your explanation for statement 1 above
(x - 1)² + (A - 1) > 0
Assume that x=8
So the eqn would be 49+A>0
But it's not necessary that A >O
Assume that A=-7.The inequality is true.
Please advise if I am going wrong anywhere
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Statement 2 says (x² + 2x + A) > 0 for all x.prachich1987 wrote:Thanks Anurag.I understand that B alone is not sufficient
But according to me also A alone is not sufficient.
Refer to your explanation for statement 1 above
(x - 1)² + (A - 1) > 0
Assume that x=8
So the eqn would be 49+A>0
But it's not necessary that A >O
Assume that A=-7.The inequality is true.
Please advise if I am going wrong anywhere
You've assumed A = -7, thus the expression becomes (x² + 2x + A) = (x² + 2x - 7). Now what for x = 0? (x² + 2x - 7) = -7, which is less than zero.
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- prachich1987
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- prachich1987
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What if x=0.5 & A= -0.5Anurag@Gurome wrote:Statement 2 says (x² + 2x + A) > 0 for all x.prachich1987 wrote:Thanks Anurag.I understand that B alone is not sufficient
But according to me also A alone is not sufficient.
Refer to your explanation for statement 1 above
(x - 1)² + (A - 1) > 0
Assume that x=8
So the eqn would be 49+A>0
But it's not necessary that A >O
Assume that A=-7.The inequality is true.
Please advise if I am going wrong anywhere
You've assumed A = -7, thus the expression becomes (x² + 2x + A) = (x² + 2x - 7). Now what for x = 0? (x² + 2x - 7) = -7, which is less than zero.
x² + 2x + A= 0.25+1-0.5=0.75
- anshumishra
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I think there has been a typo here, the equation is (x^2-2x+A) and not (x^2+2x+A).prachich1987 wrote:What if x=0.5 & A= -0.5Anurag@Gurome wrote:Statement 2 says (x² + 2x + A) > 0 for all x.prachich1987 wrote:Thanks Anurag.I understand that B alone is not sufficient
But according to me also A alone is not sufficient.
Refer to your explanation for statement 1 above
(x - 1)² + (A - 1) > 0
Assume that x=8
So the eqn would be 49+A>0
But it's not necessary that A >O
Assume that A=-7.The inequality is true.
Please advise if I am going wrong anywhere
You've assumed A = -7, thus the expression becomes (x² + 2x + A) = (x² + 2x - 7). Now what for x = 0? (x² + 2x - 7) = -7, which is less than zero.
x² + 2x + A= 0.25+1-0.5=0.75
In any case, you want to make sure that for all values of x :
(x - 1)² + (A - 1) > 0
Now the least value of (x - 1)² can be 0 (when x=1)
So, to ensure the sum to be +ve (A-1) > 0 , So if A>1...doesn't matter what value of x you choose, the equation will have a +ve value.
Thanks
Anshu
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Please try to understand the fact that in the expression (x² + 2x + A), x is a variable. You cannot determine the value of A for which (x² + 2x + A) > 0 for all x by fixing a value of x. That destroys the varying nature of x.prachich1987 wrote:What if x=0.5 & A= -0.5
x² + 2x + A= 0.25+1-0.5=0.75
Again in this example of yours, if A = -0.5, for x = 0, (x² + 2x + A) = -0.5, which is less than 0.
In fact for any negative value of A, the expression (x² + 2x + A) will be negative too for x = 0.
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- goyalsau
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Anurag@Gurome wrote:Statement 1: (x² - 2x + A) > 0 for all x.prachich1987 wrote:Is A positive?
1) x^2-2x+A is positive for all x
2) Ax^+1 is positive for all x ... I assume it is Ax^2 + 1
=> (x² - 2x + 1 + A - 1) > 0
=> (x - 1)² + (A - 1) > 0
=> (A - 1) > 0 ...................... As (x - 1)² ≥ 0 for all x
=> A > 1
=> A is positive
Sufficient
(x² - 2x + A) > 0 for all x
Lets Put X = -2 , A = -1
4 + 4 - 1 = 7
What's wrong in this???????:?: : : :
Expression is still positive.......
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As I have mentioned in the previous post (Just previous to you), for any negative value of A, the expression (x² + 2x + A) will be negative too for x = 0.goyalsau wrote:(x² - 2x + A) > 0 for all x
Lets Put X = -2 , A = -1
4 + 4 - 1 = 7
What's wrong in this???????:?: : : :
Expression is still positive.......
Put x = 0 with A = -1, the expression = -1 < 0
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- prachich1987
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Thanks Anurag,anshumishraAnurag@Gurome wrote:As I have mentioned in the previous post (Just previous to you), for any negative value of A, the expression (x² + 2x + A) will be negative too for x = 0.goyalsau wrote:(x² - 2x + A) > 0 for all x
Lets Put X = -2 , A = -1
4 + 4 - 1 = 7
What's wrong in this???????:?: : : :
Expression is still positive.......
Put x = 0 with A = -1, the expression = -1 < 0
It was so stupid of me
Thanks for the nice explanation