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S1 is sufficient. x^2 - 2*x + A = (x - 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0. The correct answer is A.

***** I believe the solution is incorrect on this question:

S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious.

S2 does not work, and neither does S1+S2, so the answer should be E.

I think you've misunderstood S1. It says that x^2 - 2*x + A > 0 *for all x*

while x=10 gives you the constraint A>-80, it doesn't mean A can equal -79, because then the above statement would not be true for x=9 since 9^2 - 2*9 - 79 = -16.

Also, if you plot the graph, y = -x^2+2x, the maximum value is y=1 at x=1. So A has to be greater than 1 for S1 to hold true for all x.

Re: m13 q23 for Mmond4 [#permalink]
04 Oct 2008, 14:08

Hello mmond4-

I see you already answered a problem I had a question on and still have the same question. Any of the explanations don't make sense to me. If a statement is true for all x...then I assume what that means is that we can put Any value into x and the resulting equation will also end up positive. Also, what ever value you give for X will automatically change the value of A. I can understand if the question states that A's value doesn't change, but why would we just assume that?

Is the sentense implying that, as in most variable equations, A has one set value and what is that set value of A for all values of x? So, we could actually come up with a value for A, going beyond just confirming it's positive? Thanks-

Re: m13 q23 for Mmond4 [#permalink]
09 Oct 2008, 04:19

Hi, The S1 is given to identify the minimum value of A which makes the \(x^2 - 2*x + A \gt 0\) for all \(x\). If it's given that \(x^2 - 2*x + A \gt 0\) for all \(x\), then we must be sure that \(A\) is such that the inequality holds true. mmond4 provided a good explanation how the range for \(A\) can be found. \(A\) must be greater than 1 for the inequality \(x^2 - 2*x + A \gt 0\) to hold true. If we know that \(A \gt 1\), we can answer the question (\(A\) is positive).

mmond4 +1.

Hope this helps.

dczuchta wrote:

Hello mmond4-

I see you already answered a problem I had a question on and still have the same question. Any of the explanations don't make sense to me. If a statement is true for all x...then I assume what that means is that we can put Any value into x and the resulting equation will also end up positive. Also, what ever value you give for X will automatically change the value of A. I can understand if the question states that A's value doesn't change, but why would we just assume that?

Is the sentense implying that, as in most variable equations, A has one set value and what is that set value of A for all values of x? So, we could actually come up with a value for A, going beyond just confirming it's positive? Thanks-

for S1 - x^2 - 2x + A >= 0 for all X. Eg. if X = -1 your solution now becomes: (sub x = -1 into S1) 1 + 2 + A >=0 therefore A can be negative OR positive hence it should be INSUFFICIENT.

S2 - Ax^2 + 1 >=0 A can be 0 (non-negative or positive) or positive - hence insufficient.

You're misunderstanding S1. It works the other way. The whole expression is positive for ALL \(x\). If you substitute \(x=0\), you'll end up with \(A > 0\) (not \(A \ge 0\) as 0 is neither positive nor negative). This is only possible if \(A\) is positive.

viperm5 wrote:

i believe the solution set to this is wrong.

for S1 - x^2 - 2x + A >= 0 for all X. Eg. if X = -1 your solution now becomes: (sub x = -1 into S1) 1 + 2 + A >=0 therefore A can be negative OR positive hence it should be INSUFFICIENT.

S2 - Ax^2 + 1 >=0 A can be 0 (non-negative or positive) or positive - hence insufficient.

Re: m13 q23 for Mmond4 [#permalink]
02 Aug 2009, 12:45

dzyubam wrote:

Hi, The S1 is given to identify the minimum value of A which makes the \(x^2 - 2*x + A \gt 0\) for all \(x\). If it's given that \(x^2 - 2*x + A \gt 0\) for all \(x\), then we must be sure that \(A\) is such that the inequality holds true. mmond4 provided a good explanation how the range for \(A\) can be found. \(A\) must be greater than 1 for the inequality \(x^2 - 2*x + A \gt 0\) to hold true. If we know that \(A \gt 1\), we can answer the question (\(A\) is positive).

mmond4 +1.

Hope this helps.

dczuchta wrote:

Hello mmond4-

I see you already answered a problem I had a question on and still have the same question. Any of the explanations don't make sense to me. If a statement is true for all x...then I assume what that means is that we can put Any value into x and the resulting equation will also end up positive. Also, what ever value you give for X will automatically change the value of A. I can understand if the question states that A's value doesn't change, but why would we just assume that?

Is the sentense implying that, as in most variable equations, A has one set value and what is that set value of A for all values of x? So, we could actually come up with a value for A, going beyond just confirming it's positive? Thanks-

[color=#408040][/color]

what is x = 3. the equation will be 3^2-2*3+A > 0 so 9-6+A>0 => 3+A>0 A>-3. let us consider the value of A be -1, then the value of the expression will be 2 i.e. positive. So, A can be negative.. we cannot conclude that A is positive by statement 1 alone..

Re: m13 q23 for Mmond4 [#permalink]
03 Aug 2009, 09:11

atomy wrote:

dzyubam wrote:

Hi, The S1 is given to identify the minimum value of A which makes the \(x^2 - 2*x + A \gt 0\) for all \(x\). If it's given that \(x^2 - 2*x + A \gt 0\) for all \(x\), then we must be sure that \(A\) is such that the inequality holds true. mmond4 provided a good explanation how the range for \(A\) can be found. \(A\) must be greater than 1 for the inequality \(x^2 - 2*x + A \gt 0\) to hold true. If we know that \(A \gt 1\), we can answer the question (\(A\) is positive).

mmond4 +1.

Hope this helps.

dczuchta wrote:

Hello mmond4-

I see you already answered a problem I had a question on and still have the same question. Any of the explanations don't make sense to me. If a statement is true for all x...then I assume what that means is that we can put Any value into x and the resulting equation will also end up positive. Also, what ever value you give for X will automatically change the value of A. I can understand if the question states that A's value doesn't change, but why would we just assume that?

Is the sentense implying that, as in most variable equations, A has one set value and what is that set value of A for all values of x? So, we could actually come up with a value for A, going beyond just confirming it's positive? Thanks-

[color=#408040][/color]

what is x = 3. the equation will be 3^2-2*3+A > 0 so 9-6+A>0 => 3+A>0 A>-3. let us consider the value of A be -1, then the value of the expression will be 2 i.e. positive. So, A can be negative.. we cannot conclude that A is positive by statement 1 alone..

thats exactly it Atomy. A is not restricted to ALWAYS be > 0 because x can be ANY number. So for instance you sub in 3, A can be negative OR positive and the equation will still hold true. HENCE S1 is insufficient.

Got it wrong as well. Difficult and tricky question. And as the OE goes, the key is to get the equation in the form (if I go and meditate in the Tibet mountains for few years, then maybe I'll get it in test conditions ): \(x^2 - 2*x + A = (x - 1)^2 + (A - 1)\)

First part of the expression is always positive, For the full expression to be positive, A has to be greater than 1.

In normal terms, I understood this as \(x^2 - 2x + A\) is positive where A is constant. So I need to focus on the case when \(x^2 - 2x\) is NEGATIVE, because then A will HAVE to be a positive and because A is a constant and x varies, A remains positive for other cases too... hence positive-A which is same as OA A

Re: m13 q23 for Mmond4 [#permalink]
29 Mar 2010, 18:58

viperm5 wrote:

viperm5 wrote:

viperm5 wrote:

Hello mmond4-

I see you already answered a problem I had a question on and still have the same question. Any of the explanations don't make sense to me. If a statement is true for all x...then I assume what that means is that we can put Any value into x and the resulting equation will also end up positive. Also, what ever value you give for X will automatically change the value of A. I can understand if the question states that A's value doesn't change, but why would we just assume that?

Is the sentense implying that, as in most variable equations, A has one set value and what is that set value of A for all values of x? So, we could actually come up with a value for A, going beyond just confirming it's positive? Thanks-

[color=#408040][/color]

what is x = 3. the equation will be 3^2-2*3+A > 0 so 9-6+A>0 => 3+A>0 A>-3. let us consider the value of A be -1, then the value of the expression will be 2 i.e. positive. So, A can be negative.. we cannot conclude that A is positive by statement 1 alone..

thats exactly it Atomy. A is not restricted to ALWAYS be > 0 because x can be ANY number. So for instance you sub in 3, A can be negative OR positive and the equation will still hold true. HENCE S1 is insufficient.

Yeah this is tricky and it got me.

However since the stimulus states that the expression is positive for all x, it means that it MUST hold for all values together. This is an AND condition rather than an OR.

If it didn't include the phrase "for all x" then you can say that for a certain x value, A can have a certain value, thus A can be negative here:

If x = 3 then A > -3 OR if x = 10 then A > -80 OR if x = 6 then A > -24, etc... This is what a standard equation says...

But since it holds for ALL x, then we have the following condition:

If x = 3 then A > -3 AND if x = 10 then A > -80 AND if x = 6 then A > -24, etc... so as the x values change the boundary condition for A keeps changing, i.e.

A > -80 AND A > -24 AND A > -3 etc... and ends up holding true for all AND any value of x when A > 0.

Hope that clears things up.. _________________

If you like my post, a kudos is always appreciated

1: suppose x = -1, to have \(x^2 - 2x + A\) positive, A will be +ve but when x = 3, then equation becomes 9-6+A so A can be -1, -2 and still the equation is +ve hence statement insufficient. 2: for all x other than 0 x^2 will be +ve so \(Ax^2 + 1\) to be +ve A has to +ve for x = 0 we are not sure on value of A, as anything multiplied by 0 is 0.

Combining, its sufficient as when x = 0 then from (1) we can say A is +ve and for x > 2 from (2) we can say A is +ve hence C. OA should be wrong.

S1 is sufficient. x^2 - 2*x + A = (x - 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0. The correct answer is A.

***** I believe the solution is incorrect on this question:

S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious.

S2 does not work, and neither does S1+S2, so the answer should be E.

Question: is \(A>0\)?

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

\(f(x)=x^2-2x+A\) is a function of of upward parabola (as coefficient of \(x^2\) is positive). We are told that it's positive for all \(x\) --> \(f(x)=x^2-2x+A>0\), which means that this function 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.

x^2 will be positive irrespective the value of x.So Ax^2 will be + or - based on the value of A.So statement 2 is insuff.

Lets take statement 1 now x^2 -2x + A is +ve for +ve x.

Resolving this Q.E we get x = 1+/- sqrt(1-A) .This will be positive only when A <1.Putting the value A =0,we will get x=0 but given x is positive.So A has be to -ve.So this statement is suff.

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

S1. Option 1: assume x = 10 and A = 1, then S1 = 100-20+1 = 81 which is +ve Option 2: assume x = 10 and A = -1, then S1 = 100-20-1 = 79 which also holds true i.e. S1 is +ve Insufficient

S2. Option 1: assume x = 1 and A = 1, then S2 = 1+1 = 2 which is +ve Option 2: assume x = 1 and A = -1/2, then S2 = -1/2+1 = 1/2 which also holds true i.e. S2 is +ve Insufficient

Here is my preposition. Pardon me if it is over ambitious or too pedantic. Why r we evaluating the sign of A for the min value of the expression x^2-2*x? Should not we be looking at the max value of x^2-2*x to determine whether A is positive or negative? Therefore,I think S1 is insuff.

Any comments or corrections will be much appreciated.

Here is my preposition. Pardon me if it is over ambitious or too pedantic. Why r we evaluating the sign of A for the min value of the expression x^2-2*x? Should not we be looking at the max value of x^2-2*x to determine whether A is positive or negative? Therefore,I think S1 is insuff.

Any comments or corrections will be much appreciated.