M60-01

Question

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

MathRevolution · Answer

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

Paugustin90 · Answer

Can someone please explain the logic behind reducing y=px(x2-4x)+Q to x=-(-4)/2P ?

Helium · Answer

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!

zondice · Answer

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.

Sharvil93 · Answer

If the question had asked what is the value of x if the Function attained a Maximum value in the XY Plane? Please, advise.

Helium · Answer

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!!

NCRanjan · Answer

Hi Everyone ,
Can someone pls elaborate the solution of this problem . I am unable to understand

Thanks

Helium · Answer

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

Salsanousi · Answer

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.

Posted from my mobile device

Priya650 · Answer

I think this the explanation isn't clear enough, please elaborate.

EncounterGMAT · Answer

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

philipssonicare · Answer

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.

meridelsa135 · Answer

I think this is a  high-quality  question and the explanation isn't clear enough, please elaborate.

manisha1234 · Answer

I think this is a  high-quality  question. Though the explanation is correct but requires more clarity on concepts

kgmat12 · Answer

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.

Farina · Answer

Hi  ScottTargetTestPrep

Can you please explain this question?

ScottTargetTestPrep · Answer

This question can indeed be solved using calculus and the solution is indeed quite simple IF you happen to know calculus; however, calculus is beyond the scope of this test and I would not recommend anyone to learn calculus just for a few questions such as this one. It would be a huge overkill. If you already know calculus, you can use it to solve the question or verify your answer but if you don't, here's how you can solve it without any calculus:

The minimum value of a parabola y = ax^2 + bx + c occurs at the vertex and the x-coordinate of the vertex is given by x = -b/(2a). For the given function y = px^2 - 4x + q; we have a = p, b = -4 and c = q. Since we already know that b = -4, we only need the value of a = p in order to be able to determine the x-coordinate of the vertex.

Statement One Alone: p = 2

Since we know the value of a = p, we can determine the x-coordinate of the vertex (which is x = -(-4)/(2*2) = 1). Statement one alone is sufficient.

Statement Two Alone: q = 5

Since we have no information about a = p, we cannot determine the x-coordinate of the vertex. Statement two alone is not sufficient.

Answer: A

Farina · Answer

Yes I have studied calculus but I have forgotten it now. Thank you for your reply. Is there any other way we can do such questions?

ScottTargetTestPrep · Answer

Yes, just follow my solution:). I showed how to solve it without using calculus.

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