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Quadratic Equations

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by garuhape » Thu Mar 03, 2011 3:52 am
P3101 wrote:find the minimum value of the expression (p+1/p); p>0.

(a) 1
(b) 0
(c) 2
(d) depends upon the value of p
(e) None of these
It should be p.

Let's try plugging in some values for p.

First let's take p=0.1 -> 0.1+1/0.1 = 0.1+10/1 = 10.1
Now let's take p=1 -> 1+1/1 = 2 (smaller ;) )
Now let's take p=2 -> 2+2/1 = 4 (bigger again)

Therefore we know that the values will decrease if you plug in a number between ]0;1[, that the value will be minimum at 1 and later increase again.
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by garuhape » Thu Mar 03, 2011 3:54 am
garuhape wrote:
It should be p.
Sorry, I meant it should be C
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by rohu27 » Thu Mar 03, 2011 5:59 am
guess it shud be D then :D
garuhape wrote:
garuhape wrote:
It should be p.
Sorry, I meant it should be C
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by garuhape » Thu Mar 03, 2011 6:10 am
rohu27 wrote:guess it shud be D then :D
garuhape wrote:
garuhape wrote:
It should be p.
Sorry, I meant it should be C
Of course it depends on the value of p. So this choice is a bit confusing, but you get the minimum if you plug in 2 for p.

Hope this helps.
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by rohu27 » Thu Mar 03, 2011 6:14 am
the question asks wht is the min value of P, not wht value would give the min. value of p.
so shudnt it be D?
am i missing sumthng
garuhape wrote:
rohu27 wrote:guess it shud be D then :D
garuhape wrote:
garuhape wrote:
It should be p.
Sorry, I meant it should be C
Of course it depends on the value of p. So this choice is a bit confusing, but you get the minimum if you plug in 2 for p.

Hope this helps.
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by garuhape » Thu Mar 03, 2011 6:17 am
rohu27 wrote:the question asks wht is the min value of P, not wht value would give the min. value of p.
so shudnt it be D?
am i missing sumthng
The questions asks for the minimum value of the expression under the condition that p is positive and not for the minimum value of p. Otherwise you didn't even have to look at the expression.
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by Night reader » Thu Mar 03, 2011 6:24 am
I hastened myself Discriminant in (c) can be solved for p
if p is positive (p>0) then the minimum value of (p+1/p) is still positive but close to 0;
Just plug in the values to reveal the answer, because brutal method could be infinite solutions' area :)
a) (p+1/p)=1, p^2+1=p OR p^2-p+1=0 this cannot be defined as D (discriminant is -ve)
b) (p+1/p)=0, p^2+1=0, p^2=-1 :( cannot be defined
c) (p+1/p)=2, p^2+1=2p, actually this can be defined as D=0
d) this is one seems correct, but we have defined above that the range of values for p could be infinite BUT we need one value
e) ---

IOM c opps
P3101 wrote:find the minimum value of the expression (p+1/p); p>0.

(a) 1
(b) 0
(c) 2
(d) depends upon the value of p
(e) None of these








O.A. (c)[/spoiler][/list][/list]
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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by rohu27 » Thu Mar 03, 2011 6:29 am
sorry thts wht i meant. min value of the exp. not the value of p. as it varies wth the value of p, asnwer shud be D.
rohu27 wrote:the question asks wht is the min value of P, not wht value would give the min. value of p.
so shudnt it be D?
am i missing sumthng
garuhape wrote:
rohu27 wrote:guess it shud be D then :D
garuhape wrote:
garuhape wrote:
It should be p.
Sorry, I meant it should be C
Of course it depends on the value of p. So this choice is a bit confusing, but you get the minimum if you plug in 2 for p.

Hope this helps.
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by Night reader » Thu Mar 03, 2011 6:35 am
guys if this GMAT question, then this is one value answer problem. In D you have many answers :( the answer is C
p^-2p+1=0, D(disrmt)=0 and p(1,2)=[-(-2) +- 0]/2=2, we have answer p=1 for min.(p+1/p)
rohu27 wrote:sorry thts wht i meant. min value of the exp. not the value of p. as it varies wth the value of p, asnwer shud be D.
rohu27 wrote:the question asks wht is the min value of P, not wht value would give the min. value of p.
so shudnt it be D?
am i missing sumthng
garuhape wrote:
rohu27 wrote:guess it shud be D then :D
garuhape wrote:
garuhape wrote:
It should be p.
Sorry, I meant it should be C
Of course it depends on the value of p. So this choice is a bit confusing, but you get the minimum if you plug in 2 for p.

Hope this helps.
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com
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by anshumishra » Thu Mar 03, 2011 7:58 am
Exactly like Night reader (except instead of checking the options lets find the minimum value ):

p+ 1/p = k > 0 (as p> 0 , k must be greater than 0)
=> p^2 - kp + 1 = 0
Since, p is a real number the discriminant must be >= 0
So, D = k^2 - 4 >= 0 => k>=2
=> so minimum value of k = p+1/p = 2
Thanks
Anshu

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