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Source: — Data Sufficiency |

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by Anurag@Gurome » Sat Jan 01, 2011 5:45 am
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
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

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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by prachich1987 » Sat Jan 01, 2011 6:06 am
Anurag@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
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
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by Anurag@Gurome » Sat Jan 01, 2011 6:17 am
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
Statement 2 says (x² + 2x + A) > 0 for all x.
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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by prachich1987 » Sat Jan 01, 2011 6:39 am
Thanks!
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by prachich1987 » Sat Jan 01, 2011 6:43 am
Anurag@Gurome wrote:
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
Statement 2 says (x² + 2x + A) > 0 for all x.
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.
What if x=0.5 & A= -0.5
x² + 2x + A= 0.25+1-0.5=0.75
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by anshumishra » Sat Jan 01, 2011 6:47 am
prachich1987 wrote:
Anurag@Gurome wrote:
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
Statement 2 says (x² + 2x + A) > 0 for all x.
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.
What if x=0.5 & A= -0.5
x² + 2x + A= 0.25+1-0.5=0.75
I think there has been a typo here, the equation is (x^2-2x+A) and not (x^2+2x+A).
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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by Anurag@Gurome » Sat Jan 01, 2011 6:48 am
prachich1987 wrote:What if x=0.5 & A= -0.5
x² + 2x + A= 0.25+1-0.5=0.75
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.

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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by goyalsau » Sat Jan 01, 2011 10:23 am
Anurag@Gurome wrote:
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
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


(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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by Anurag@Gurome » Sat Jan 01, 2011 11:27 am
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.......
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.

Put x = 0 with A = -1, the expression = -1 < 0
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by prachich1987 » Sat Jan 01, 2011 10:51 pm
Anurag@Gurome wrote:
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.......
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.

Put x = 0 with A = -1, the expression = -1 < 0
Thanks Anurag,anshumishra
It was so stupid of me
Thanks for the nice explanation
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