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M60-01

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M60-01  [#permalink]

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New post 11 Jun 2018, 02:00
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The function \(y=px^2-4x+q\) in the x-y plane attains a minimum value. What is the value of \(x\)?

1) \(p = 2\)
2) \(q = 5\)

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New post 11 Jun 2018, 02:00
Official Solution:

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

\(y=px^2-4x+q\) has a minimum value when \(x = -(-4)/2p = 2/p.\) Thus, the question asks for the value of p.

Since only condition 1) gives us information about p, only condition 1) is sufficient.

Therefore, A is the answer.



Answer: A
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Re: M60-01  [#permalink]

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New post 31 Jul 2018, 11:50
Can someone please explain the logic behind reducing y=px(x2-4x)+Q to x=-(-4)/2P ?
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Re: M60-01  [#permalink]

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New post 03 Aug 2018, 23:18
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Paugustin90 wrote:
Can someone please explain the logic behind reducing y=px(x2-4x)+Q to x=-(-4)/2P ?


For maxima and minima,

Take derivative of the equation and equate to 0 (rate of change is o at this point).

so dy/dx = 2px -4 = 0

x = 4/2p

Hope this helps!
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Re: M60-01  [#permalink]

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New post 10 Aug 2018, 13:05
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Hi everyone,
Just wanted to say that the question here states " The function in the x-y plane attains a minimum value. What is THE value of x ? "
I believe the question should be rephrased to " What is that value of x" so that it becomes clear that we are talking about that x which will make this a minimum valued function.
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New post 17 Aug 2018, 22:22
If the question had asked what is the value of x if the Function attained a Maximum value in the XY Plane? Please, advise.
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New post 18 Aug 2018, 01:30
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Sharvil93 wrote:
If the question had asked what is the value of x if the Function attained a Maximum value in the XY Plane? Please, advise.



Sharvil93

zondice

The function y=px2−4x+q in the x-y plane attains a minimum value. What is the value of x?

1) p=2
2) q=5

Please note that y=px2−4x+q represents Parabola.

If p is positive then function will attain minima at some value of X (i.e. X= -b/2a) (Parabola open towards +ve Y- axis)

If p is negative then funcation will attain maxima at some value of X (i.e. X = -b/2a) ((Parabola open towards -ve Y- axis)

Hope this helps!!
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Re M60-01  [#permalink]

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New post 19 Oct 2018, 11:43
Hi Everyone ,
Can someone pls elaborate the solution of this problem . I am unable to understand

Thanks
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New post 19 Oct 2018, 18:14
NCRanjan wrote:
Hi Everyone ,
Can someone pls elaborate the solution of this problem . I am unable to understand

Thanks


NCRanjan

https://gmatclub.com/forum/algebra-tips ... 75003.html

https://www.askiitians.com/iit-jee-alge ... ssion.aspx

https://gmatclub.com/forum/what-is-the- ... 26302.html

Please go through above resources and my earlier post.

I think that will help.

Posted from my mobile device
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New post 19 Oct 2018, 18:28
NCRanjan wrote:
Hi Everyone ,
Can someone pls elaborate the solution of this problem . I am unable to understand

Thanks


This question relies on understanding that derivatives provide a maximum value

d/dx y = px^2 - 4x + q


0 = 2px - 4

p = 4/2x

If we know p we can figure out x.

For parabola equation

ax^2 + bx + c = y

We can find he value of x to be = -b/2a

Where b and a are the coeffecients in the formula

In this particular question it is like this

y = px^2 - 4x + q

b = -4
a = p

So x = -(-4)/2p = 4/2p

Hope it clears it out.

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New post 13 Nov 2018, 01:20
I think this the explanation isn't clear enough, please elaborate.
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New post 19 Nov 2018, 06:26
Why everyone is solving through derivative (calculus) if it is not tested on GMAT? Not all are comfortable with it.
Can someone explain it in a simple way?

chetan2u VeritasKarishma pls help :please
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New post 21 Dec 2018, 15:58
topper97
It's simply fancy language. Finding the maximum/minimum of a parabola that has the equation ax^2+bx+c follows the equation x = -b / 2a. I don't know calculus. I don't know how that equation is derived. It's simply one of the equations taught when learning quadratics in algebra class.

In saying that, it would probably help me to know how it is derived...

As a side note, I concur with zondice that the wording in this question is poor.
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New post 15 Jan 2019, 11:20
I think this is a high-quality question and the explanation isn't clear enough, please elaborate.
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New post 18 Jan 2019, 02:00
I think this is a high-quality question. Though the explanation is correct but requires more clarity on concepts
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New post 22 Apr 2019, 20:07
Few things to know to undertsand the solution:
1. The derivative of a function equated to 0, can be solved to find the data point on the function at which it has slope=0.
2. The given quadratic equation is a parabola, hence it will have a slope =0 at only one point due to the parabolic curve shape.
3. Also the point of the function where slope =0 will be minimal if the parabola points upwards. And maximum if it points downwards.

Solving :
Apply dy/dx = 0 with the following derivative formulae:
d(x^2)/dx = 2x;
d(x)/dx = 1 ;
d(constant)/dx = 0

We get d(px^2-4x+q)/dx = 2px-4 = 0
=> x=4/2p = 2/p

Statement 1
p = 2
=> A parabola pointed upwards. As we have a positive value for the coefficient of x^2.
=> We can now find a value of x, which will be the minimal value of the curve in the x,y plane.

Hence A is the answer.
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M60-01   [#permalink] 22 Apr 2019, 20:07
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