Quadratic Equations

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.

garuhape wrote:
It should be p.

Sorry, I meant it should be C

guess it shud be D then

garuhape wrote:

garuhape wrote:
It should be p.

Sorry, I meant it should be C

rohu27 wrote:guess it shud be D then

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.

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

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.

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.

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]

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

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.

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

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.

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