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x^2-2x+A is positive for all x Ax^2+1 is positive for all x

OA is A

Is \(A>0\)?

(1) \(x^2-2x+A\) is positive for all \(x\):

Quadratic expression \(x^2-2x+A\) is a function of of upward parabola (it's upward as coefficient of \(x^2\) is positive). We are told that this expression is positive for all \(x\) --> \(x^2-2x+A>0\), which means that this parabola is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation \(x^2-2x+A=0\) has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> \(D=2^2-4A=4-4A<0\) --> \(1-A<0\) --> \(A>1\).

Sufficient.

(2) \(Ax^2+1\) is positive for all \(x\):

\(Ax^2+1>0\) --> when \(A\geq0\) this expression is positive for all \(x\). So \(A\) can be zero too.

I dont get it sorry... I mean I understand your equations Bunuel, but I tried first with picking numbers:

If I pick -0.5 for x --> x^2-2x+A>0 will hold for A > -1.25

...

Where is my mistake??

The point here is that \(x^2-2x+A>0\) for all \(x-es\).

Let's do this in another way:

We have \((x^2-2x)+A>0\) for all \(x-es\). The sum of 2 quantities (\(x^2-2x\) and \(A\)) is positive for all \(x-es\). So for the least value of \(x^2-2x\), \(A\) must make the whole expression positive.

So what is the least value of \(x^2-2x\)? The least value of quadratic expression \(ax^2+bx+c\) is when \(x=-\frac{b}{2a}\), so in our case the least value of \(x^2-2x\) is when \(x=-\frac{-2}{2}=1\) --> \(x^2-2x=-1\) --> \(-1+A>0\) --> \(A>1\).

OR:

\(x^2-2x+A>0\) --> \(x^2-2x+1+A-1>0\) --> \((x-1)^2+A-1>0\) --> least value of \((x-1)^2\) is zero thus \(A-1\) must be positive (\(0+A-1>0\))--> \(A-1>0\) --> \(A>1\).

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

\(x^2 - 2x + A = 9 +6 +A = 15 + A >0\)

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers

Not every question can be solved by number picking.

For all \(x-es\) means that no matter what \(x\) you pick \(x^2 - 2x + A\) must be positive. So it must be positive even for the lowest value of \(x^2 - 2x\) which is -1 --> so \(-1+A\) must be positive hence A must be more than 1.

Now again: if A>1 then for any \(x\) expression \(x^2 - 2x + A\) is positive.

But if A=-15 (or any other number less than 1) we can find some \(x-es\) for which expression \(x^2 - 2x + A\) is not positive, so theese values of A (values of \(A\leq{1}\)) are not valid.

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

\(x^2 - 2x + A = 9 +6 +A = 15 + A >0\)

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers

Not every question can be solved by number picking.

For all \(x-es\) means that no matter what \(x\) you pick \(x^2 - 2x + A\) must be positive. So it must be positive even for the lowest value of \(x^2 - 2x\) which is -1 --> so \(-1+A\) must be positive hence A must be more than 1.

Now again: if A>1 then for any \(x\) expression \(x^2 - 2x + A\) is positive.

But if A=-15 (or any other number less than 1) we can find some \(x-es\) for which expression \(x^2 - 2x + A\) is not positive, so theese values of A (values of \(A\leq{1}\)) are not valid.

Hope it's clear.

Thanks for the reply Bunuel,

I understood the approach but the fact which is baffling me is that say the equation after subsituting value of x=-3 i.e 15+ A > 0 now we can have a value of A=-3 or may be -4 etc and still have the value of the equation in statement 1 as +ve

Am I thinking too much or just lacking some thing basic concept.

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

\(x^2 - 2x + A = 9 +6 +A = 15 + A >0\)

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers

Not every question can be solved by number picking.

For all \(x-es\) means that no matter what \(x\) you pick \(x^2 - 2x + A\) must be positive. So it must be positive even for the lowest value of \(x^2 - 2x\) which is -1 --> so \(-1+A\) must be positive hence A must be more than 1.

Now again: if A>1 then for any \(x\) expression \(x^2 - 2x + A\) is positive.

But if A=-15 (or any other number less than 1) we can find some \(x-es\) for which expression \(x^2 - 2x + A\) is not positive, so theese values of A (values of \(A\leq{1}\)) are not valid.

Hope it's clear.

Thanks for the reply Bunuel,

I understood the approach but the fact which is baffling me is that say the equation after subsituting value of x=-3 i.e 15+ A > 0 now we can have a value of A=-3 or may be -4 etc and still have the value of the equation in statement 1 as +ve

Am I thinking too much or just lacking some thing basic concept.

I appreciate your patience.

I think you just don't understand one thing in statement (1): \(x^2-2x+A>0\) FOR ALL \(x-es\).

You say that if \(x=-3\) then \(A\) can be for example -10 (or any number more than -15) and \(x^2-2x+A\) will be positive, \(but\) if \(x=1\) does \(A=-10\) makes \(x^2-2x+A\) positive? NO!

So you should find such value of \(A\) (such range) for which \(x^2-2x+A\) is positive no matter what value of \(x\) you'll plug. And the way how to find this range is shown in my previous posts.
_________________

1) X^2-2X+A is positive for all X I think A could not be the answer, for example, if A = 0, and X = 4, then also the expression is positive, but A = 0 is neither positive nor negative Again, if A = 1, and and X = 4, then also the expression is positive 2) AX^2 + 1 is positive for all X Same logic as above, if A is 0, then the expression is positive, and the expression is also postive for any value of X where A > 0

In a nutshell, I too think the answer is E. _________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

1) X^2-2X+A is positive for all X 2) AX^2 + 1 is positive for all X

given answer as A...but i thought it should be E.. source: hard problems from gmatclub tests number properties I

1) X^2-2X+A is positive for all X

For all values of X,\(X^2-2X+A > 0\) This means, for X = 0, \(X^2-2X+A > 0\); for X = 1, \(X^2-2X+A > 0\); for X = -2, \(X^2-2X+A > 0\) etc etc etc

Let's put X = 0. \(0^2-2*0+A > 0\) should hold. Therefore, A > 0 should hold. Sufficient.

2) AX^2 + 1 is positive for all X

For all X, \(AX^2 + 1 > 0\) Here, A could be positive or A could be 0 (since, when A = 0, we get 1 > 0 which holds no matter what the value of X.) Since A can be 0, we cannot say whether A is positive. Not Sufficient.

x^2-2x+A is positive for all x Ax^2+1 is positive for all x

OA is A

(2) \(Ax^2+1\) is positive for all \(x\):

\(Ax^2+1>0\) --> when \(A\geq0\) this expression is positive for all \(x\). So \(A\) can be zero too.

Not sufficient.

Answer: A.

Why didn't you use the discriminant formula to assess statement 2?

I tried the discriminant rule and got a>0. I had 0-4a<0 which turns to a>0.

What am I missing here?

Thanks, Diana

You are right: if we use the same approach for (2) then we'll get A>0 BUT if A=0 then Ax^2+1 won't be a quadratic function anymore. So this approach will work only if A doesn't equal to zero, but we can not eliminate this case and if A=0 then Ax^2+1=1 is also always positive. Hence Ax^2+1 is positive for A>0 (if we use quadratic function approach) as well as for A=0, so for \(A\geq0\).

Is A positive? 1. x^2 -2x +A is positive for all x. 2. A*x^2 +1 is positive for all x.

I got E and my way of solving is as below: St1. x^2 - 2x +A > 0 Let x=0, so A>0. Let x=-1, so A>-3. In this case A can be negative or positive. Insufficient.

St2. A*x^2 +1 > 0 Let x=-1, so A>-1. Again A can be positive or negative. Insufficient.

St1+St2: Let x=-1, so A > -1. Again A can be positive or negative. Insufficient.

So it's E. However the OA is not E. Please advise.

Is A positive? 1. x^2 -2x +A is positive for all x. 2. A*x^2 +1 is positive for all x.

I got E and my way of solving is as below: St1. x^2 - 2x +A > 0 Let x=0, so A>0. Let x=-1, so A>-3. In this case A can be negative or positive. Insufficient.

St2. A*x^2 +1 > 0 Let x=-1, so A>-1. Again A can be positive or negative. Insufficient.

St1+St2: Let x=-1, so A > -1. Again A can be positive or negative. Insufficient.

So it's E. However the OA is not E. Please advise.

Merging similar topics. Please refer to the solutions above.
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Re: Is A positive? x^2-2x+A is positive for all x Ax^2+1 is [#permalink]

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21 Aug 2012, 15:51

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best way to deal this problem is to bet on A more than X.. it wud b yes if A>0 Or No ,if A<0 .... then first assume A>0 , then check whether statement 1 & 2 is true or not for all value of X.... then assume A<0 ,then check whether statement 1 & 2 is true or not for all value of X....
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