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Re M1323 [#permalink]
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15 Sep 2014, 23:49
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Official Solution: Question: is \(a \gt 0\)? (1) \(x^22x+a\) is positive for all \(x\). \(f(x)=x^22x+a\) 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 \(f(x)=x^22x+a \gt 0\), which means that this function is "above" Xaxis OR in other words parabola has no intersections with X axis OR equation \(x^22x+a=0\) has no real roots. In order for a quadratic equation to have no real roots its discriminant must be negative: \(D=2^24a=44a \lt 0\), which simplifies to \(1a \lt 0\) 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. Answer: A
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Re: M1323 [#permalink]
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27 Sep 2014, 11:00
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I used another approach for S1 and got inequality a>4. Is it correct?
x^22x+44+a>0 (x2)^24+a>0 for all x. Min of (x2)^2 if x=2. Therefore a>4
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Re: M1323 [#permalink]
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25 Oct 2014, 15:33
Boycot wrote: I used another approach for S1 and got inequality a>4. Is it correct?
x^22x+44+a>0 (x2)^24+a>0 for all x. Min of (x2)^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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Re: M1323 [#permalink]
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12 Nov 2014, 08:26
in (1) it is said that x^2 2x+a>0 for all x if we substitute x=10 then 10020+a= 80+a>0 therefore a>80 hence a insufficient
can someone please tell where i have gone wrong



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Re: M1323 [#permalink]
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jacobneroth wrote: in (1) it is said that x^2 2x+a>0 for all x if we substitute x=10 then 10020+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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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^22x. sufficient (2) statement can hold true with any value of a. consider 0 or 0.5 or 1. ANSWER A
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Re M1323 [#permalink]
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29 Aug 2016, 07:33
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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Re: M1323 [#permalink]
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06 Sep 2016, 19:18
Hi Bunuel , My approach is as below. Please let me know if this is a valid way of solving the same. 1) x^22x + 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. Answer A. Please advice. Thanks, Arun



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Re: M1323 [#permalink]
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08 Oct 2016, 03:55
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, 06:28



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Re: M1323 [#permalink]
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05 Dec 2016, 21:32
Bunuel wrote: Official Solution:
Question: is \(a \gt 0\)? (1) \(x^22x+a\) is positive for all \(x\). \(f(x)=x^22x+a\) 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 \(f(x)=x^22x+a \gt 0\), which means that this function is "above" Xaxis OR in other words parabola has no intersections with X axis OR equation \(x^22x+a=0\) has no real roots. In order for a quadratic equation to have no real roots its discriminant must be negative: \(D=2^24a=44a \lt 0\), which simplifies to \(1a \lt 0\) 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.
Answer: A I have trouble solving this question to get the correct statement. If x = 3, then 92(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. Please help. Thanks !



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Re: M1323 [#permalink]
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05 Dec 2016, 22:47
shubham1985 wrote: Bunuel wrote: Official Solution:
Question: is \(a \gt 0\)? (1) \(x^22x+a\) is positive for all \(x\). \(f(x)=x^22x+a\) 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 \(f(x)=x^22x+a \gt 0\), which means that this function is "above" Xaxis OR in other words parabola has no intersections with X axis OR equation \(x^22x+a=0\) has no real roots. In order for a quadratic equation to have no real roots its discriminant must be negative: \(D=2^24a=44a \lt 0\), which simplifies to \(1a \lt 0\) 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.
Answer: A I have trouble solving this question to get the correct statement. If x = 3, then 92(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. Please help. Thanks ! I think discussion HERE should clear your doubts.
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Re: M1323 [#permalink]
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11 Dec 2016, 06:29
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" Xaxis 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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Re: M1323 [#permalink]
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08 Jan 2017, 04:23
please clarify x^2−2x+a is positive for all x a>2xx^2 if x =1 ..........a positive but if x=4 a negative



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Re: M1323 [#permalink]
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11 May 2017, 03:48
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Bunuel Please clarify this doubt 1) \(x^22x+a > 0\) If x = 1 then a>1......YES If x = 2 then a>8......NO How is this sufficient? Thanks in advance
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