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# M13-23

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Math Expert
Joined: 02 Sep 2009
Posts: 59622

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16 Sep 2014, 00:49
00:00

Difficulty:

95% (hard)

Question Stats:

40% (01:58) correct 60% (02:14) wrong based on 99 sessions

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Is $$a$$ positive?

(1) $$x^2 - 2x + a$$ is positive for all $$x$$

(2) $$ax^2 + 1$$ is positive for all $$x$$

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Joined: 26 Dec 2017
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13 Jan 2018, 04:44
4
1
Hi Bunnel ,

I too have same doubt that why imaginary roots discriminant method is not used in S2 if used you will get a>0 but also actually it is not true as a=0 also satisfies the equation which is the reason for insufficiency.

So m main doubt is Can I know whether discriminant method is suffice(S1) or not (S2) for checking whether statements suffice or not
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27 Sep 2014, 12:00
3
I used another approach for S1 and got inequality a>4. Is it correct?

x^2-2x+4-4+a>0
(x-2)^2-4+a>0 for all x. Min of (x-2)^2 if x=2.
Therefore a>4

Sufficient
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16 Sep 2014, 00:49
2
5
Official Solution:

Question: is $$a \gt 0$$?

(1) $$x^2-2x+a$$ is positive for all $$x$$.

$$f(x)=x^2-2x+a$$ is a function of an upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$, so $$f(x)=x^2-2x+a \gt 0$$, 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.

In order for a quadratic equation to have no real roots its discriminant must be negative: $$D=2^2-4a=4-4a \lt 0$$, which simplifies to $$1-a \lt 0$$ and finally to $$a \gt 1$$. Sufficient.

(2) $$ax^2+1$$ is positive for all $$x$$:

$$ax^2+1 \gt 0$$. Now, when $$a \ge 0$$ this expression is positive for all $$x$$. So, $$a$$ can be zero too. Not sufficient.

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25 Oct 2014, 16:33
1
Boycot wrote:
I used another approach for S1 and got inequality a>4. Is it correct?

x^2-2x+4-4+a>0
(x-2)^2-4+a>0 for all x. Min of (x-2)^2 if x=2.
Therefore a>4

Sufficient

I did it with another approach, question says that X^2+2x+a is positive for all values of X so we can take any value of X
if he take X=0 then X^2 and 2x will also be -ve and remaining portion should be positive which is a
1) sufficient
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15 Nov 2014, 01:38
1
jacobneroth wrote:
in (1) it is said that x^2 -2x+a>0 for all x
if we substitute x=10 then 100-20+a= 80+a>0
therefore a>-80
hence a insufficient

can someone please tell where i have gone wrong

Hi jacobneroth,

Statement 1 says the expression must be positive for 'all' X...x^2 -2x+a>0 ..So we want to make sure that the expression stays positive for any value of X..In your substitution, the expression is positive for 10, when a is negative; however, if we substitute 0 for X, the expression will be positive only if a is positive...for the expression to hold positive for all values of X, 'a' must be positive.

Hope this helps..
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12 Dec 2015, 07:55
1
can we approach this with number picking?

(1) if we pick x=1 then the inequality would become (1)^2 - 2*(1) + a ---> 1 - 2 + a. for the statement 1 to hold true and remain positive a has to be positive - so to say a has to compensate for whatever negative result may come out of x^2-2x. sufficient

(2) statement can hold true with any value of a. consider 0 or -0.5 or 1.

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11 May 2017, 04:48
1
Bunuel

1) $$x^2-2x+a > 0$$
If x = 1 then a>1......YES
If x = -2 then a>-8......NO

How is this sufficient?

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12 Nov 2014, 09:26
in (1) it is said that x^2 -2x+a>0 for all x
if we substitute x=10 then 100-20+a= 80+a>0
therefore a>-80
hence a insufficient

can someone please tell where i have gone wrong
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Posts: 59622

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12 Nov 2014, 09:38
jacobneroth wrote:
in (1) it is said that x^2 -2x+a>0 for all x
if we substitute x=10 then 100-20+a= 80+a>0
therefore a>-80
hence a insufficient

can someone please tell where i have gone wrong

I think discussion HERE should clear your doubts.
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29 Aug 2016, 08:33
I don't agree with the explanation. Second statement says that ax^2+1 is positive for all x----- this can be written as ax^2 +1 > 0 => ax^2>-1 => x^2> -1/a => a can't be zero, it has to be some -ve number to make the rhs positive. Pls explain if I am wrong.
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29 Aug 2016, 09:19
I don't agree with the explanation. Second statement says that ax^2+1 is positive for all x----- this can be written as ax^2 +1 > 0 => ax^2>-1 => x^2> -1/a => a can't be zero, it has to be some -ve number to make the rhs positive. Pls explain if I am wrong.

You cannot divide by a the way you did. You artificially exclude a = 0 this way. If a = 0, $$ax^2+1$$ is positive for all $$x$$, so a can be 0.
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06 Sep 2016, 20:18
Hi Bunuel ,

My approach is as below. Please let me know if this is a valid way of solving the same.

1) x^2-2x + a> 0 for all x. So if X=0 then a has to be positive. So a is positive
2) a(x^2)+1 > 0 as you have discussed if a is zero the equation is still valid so cannot be determined.

Thanks,
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08 Oct 2016, 04:55
for 2nd statement : ax^2+1 >0
same can be done as in 1.

discriminant <0 ; here a=a ; b=0 ; c=1

so b^2 - (4*a*c)< 0
0- (4a)<0
-4a<0
a>0

therefore even 2. is sufficient
whats wrong in this analysis??
Quote:
Bunuel
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08 Oct 2016, 07:28
deepak268 wrote:
for 2nd statement : ax^2+1 >0
same can be done as in 1.

discriminant <0 ; here a=a ; b=0 ; c=1

so b^2 - (4*a*c)< 0
0- (4a)<0
-4a<0
a>0

therefore even 2. is sufficient
whats wrong in this analysis??
Quote:
Bunuel

If a = 0, $$ax^2+1$$ is positive for all $$x$$, so a can be 0.
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05 Dec 2016, 22:32
Bunuel wrote:
Official Solution:

Question: is $$a \gt 0$$?

(1) $$x^2-2x+a$$ is positive for all $$x$$.

$$f(x)=x^2-2x+a$$ is a function of an upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$, so $$f(x)=x^2-2x+a \gt 0$$, 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.

In order for a quadratic equation to have no real roots its discriminant must be negative: $$D=2^2-4a=4-4a \lt 0$$, which simplifies to $$1-a \lt 0$$ and finally to $$a \gt 1$$. Sufficient.

(2) $$ax^2+1$$ is positive for all $$x$$:

$$ax^2+1 \gt 0$$. Now, when $$a \ge 0$$ this expression is positive for all $$x$$. So, $$a$$ can be zero too. Not sufficient.

I have trouble solving this question to get the correct statement.
If x = 3,
then 9-2(3)+a > 0 --> 3+a > 0 (per statement 1). Now , here a can be positive or negative and the expression will still be positive.
Just wondering, how can this statement be sufficient then.
Thanks !
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Joined: 02 Sep 2009
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05 Dec 2016, 23:47
shubham1985 wrote:
Bunuel wrote:
Official Solution:

Question: is $$a \gt 0$$?

(1) $$x^2-2x+a$$ is positive for all $$x$$.

$$f(x)=x^2-2x+a$$ is a function of an upward parabola (as coefficient of $$x^2$$ is positive). We are told that it's positive for all $$x$$, so $$f(x)=x^2-2x+a \gt 0$$, 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.

In order for a quadratic equation to have no real roots its discriminant must be negative: $$D=2^2-4a=4-4a \lt 0$$, which simplifies to $$1-a \lt 0$$ and finally to $$a \gt 1$$. Sufficient.

(2) $$ax^2+1$$ is positive for all $$x$$:

$$ax^2+1 \gt 0$$. Now, when $$a \ge 0$$ this expression is positive for all $$x$$. So, $$a$$ can be zero too. Not sufficient.

I have trouble solving this question to get the correct statement.
If x = 3,
then 9-2(3)+a > 0 --> 3+a > 0 (per statement 1). Now , here a can be positive or negative and the expression will still be positive.
Just wondering, how can this statement be sufficient then.
Thanks !

I think discussion HERE should clear your doubts.
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11 Dec 2016, 07:29
f(x)=x2−2x+af(x)=x2−2x+a is a function of an upward parabola (as coefficient of x2x2 is positive). We are told that it's positive for all xx, so f(x)=x2−2x+a>0f(x)=x2−2x+a>0, which means that this function is "above" X-axis OR in other words parabola has no intersections with X -axis OR equation x2−2x+a=0x2−2x+a=0 has no real roots.

How do we know that it is upward parabola?

Is there any theory to go through to solve such questions?
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11 Dec 2016, 10:35
dsheth7 wrote:
f(x)=x2−2x+af(x)=x2−2x+a is a function of an upward parabola (as coefficient of x2x2 is positive). We are told that it's positive for all xx, so f(x)=x2−2x+a>0f(x)=x2−2x+a>0, which means that this function is "above" X-axis OR in other words parabola has no intersections with X -axis OR equation x2−2x+a=0x2−2x+a=0 has no real roots.

How do we know that it is upward parabola?

Is there any theory to go through to solve such questions?

Check here: math-coordinate-geometry-87652.html

GMAT Club's questions are mostly quite difficulty. One should not attempt them if the fundamentals are not strong enough.
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08 Jan 2017, 05:23
x^2−2x+a is positive for all x
a>2x-x^2
if x =1 ..........a positive but if x=4 a negative
Re: M13-23   [#permalink] 08 Jan 2017, 05:23

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# M13-23

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