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How to solve this the non-differentiation way?

by rahulvsd » Fri Apr 20, 2012 7:13 am
If A and B are integers and B = A + 4 , which of the following represents integer x for which the expression (x - A)^2 + (x - B)^2 is the smallest?

A. A - 1
B. A
C. A + 2
D. A + 3
E. B + 1

[spoiler]OA: C. I tried this problem by differentiating the given equation wrt x.

(x-A)^2 + (x -B)^ 2

Differentiating wrt x, we get
2(x-A) + 2(x-B) = 0

x - A + x - A - 4 = 0
2x = A + 4
x = A + 2. Is the method correct, how do I approach this problem the non-differentiation way?[/spoiler]
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by GMATGuruNY » Fri Apr 20, 2012 10:01 am
rahulvsd wrote:If A and B are integers and B = A + 4 , which of the following represents integer x for which the expression (x - A)^2 + (x - B)^2 is the smallest?

A. A - 1
B. A
C. A + 2
D. A + 3
E. B + 1
Let A=0 and B=4.
Substituting these values into (x - A)² + (x - B)², we get:
(x-0)² + (x-4)²
x² + (x-4)².

From here we can plug in the answers, which represent the value of x.
The correct answer will yield the smallest result when it is plugged into the expression above.
Plugging A=0 and B=4 into the answers, our options for x are as follows:
-1, 0, 2, 3, 5.

A: If x=-1, then x² + (x-4)² = (-1)² + (-1-4)² = 26.

B: If x=0, then x² + (x-4)² = (0)² + (0-4)² = 16.
Eliminate A, since B is smaller.

C: If x=2, then x² + (x-4)² = (2)² + (2-4)² = 8.
Eliminate B, since C is smaller.

Since answer choice D will include 3²=9, and answer choice E will include 5²=25, C yields the smallest value.

The correct answer is C.
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by aneesh.kg » Fri Apr 20, 2012 12:11 pm
Without Differentiation, the best method the solve such questions conclusively is to reduce the given function in the form of f(x) = A(x - P)^2 + Q

In the function given above,
(x - P)^2 >= 0 for all values of x.

So, the minimum value of f(x) is Q when the term A(x - P)^2 vanishes, i.e. x = P.

Using the above concept to solve this:

y = (x - A)^2 + (x - B)^2
y = 2x^2 + (A^2 + B^2) - 2(A + B)
y = 2[x^2 - 2*((A -B)/2) + (A^2 + B^2 - 2AB)/4] + (A^2 + B^2 + AB)/2
y = 2[x - (A + B)/2]^2 + (A^2 + B^2 + AB)/2
So, the minimum value of y is when
x = (A + B)/2
Substituting B = A + 4 in the above,
x = A + 2

(C) is the answer.
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by Shalabh's Quants » Sun Apr 22, 2012 6:45 am
rahulvsd wrote:If A and B are integers and B = A + 4 , which of the following represents integer x for which the expression (x - A)^2 + (x - B)^2 is the smallest?

A. A - 1
B. A
C. A + 2
D. A + 3
E. B + 1

[spoiler]OA: C. I tried this problem by differentiating the given equation wrt x.

(x-A)^2 + (x -B)^ 2

Differentiating wrt x, we get
2(x-A) + 2(x-B) = 0

x - A + x - A - 4 = 0
2x = A + 4
x = A + 2. Is the method correct, how do I approach this problem the non-differentiation way?[/spoiler]
Let A=0 and B=4.
Substituting these values into (x - A)² + (x - B)², we get:
(x-0)² + (x-4)²
x² + (x-4)².

=> f(x) = 2x^2-8x+16 = 2(x^2-4x+8) = 2[(x-2)^2+4]

=> for 2[(x-2)^2+4] to be minimum, (x-2)^2 should be 0 as it will attain +ive value only. So min. value is 0 or (x-2)^2 = 0 => x = 2. Ans. [spoiler]C. x=A+2.[/spoiler]
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by simone88 » Sun Apr 22, 2012 7:25 am
Stay careful rahulvsd. Since you are asked to minimize the function among integers, differentiation ways could lead you, a priori, to wrong answers! For example if the function were f(x)=x^2-x+1 by differentiating you would be left with x=1/2 with the minimum being 3/4, but the minimum among the integers would be 1.
So I think that GMATGuruNY's one is the best way to approach this problem.
Good luck!
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by simone88 » Sun Apr 22, 2012 7:42 am
Ah... only to be sure you understood. The reason why GMATGuruNY put A=0 (and so B=4) is that in the function (x-A)^2+(x-B)^2, since that B is tied to A in this particular way (B=A+4), A doesn't really matter: in fact if you call y=x-A you are left with y^2+(y-4)^2 so A doesn't affect the maximum (In other words by shifting the variables the distances remain the same).
Alternatively you can think that since the right answer must be true for every A it must be true also for A=0.
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by aneesh.kg » Sun Apr 22, 2012 9:00 am
Simone: Differentiation will never give you a wrong answer. Even in the example stated by you, it does not. The laws of mathematics will get defied if it gives a wrong answer.

Differentiation is the best method, without doubt. But, since it is out of scope of GMAT, the method discussed by me and Shalabh (Quadratic Equation) are the second- best methods because it gives you a conclusive answer. With all due respect to all the experts here, I don't like 'plugging in values' and/or eliminating the answer. That should be the last resort, when nothing else works out.

We have a better method here, and I will like everyone to understand that method.

To everyone reading this and preparing for GMAT: Please learn Math properly and avoid short-cuts. When you learn and practice Math conceptually, you acquire so many skills in your repertoire that every question takes very less time to solve.
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by Brent@GMATPrepNow » Sun Apr 22, 2012 9:40 am
aneesh.kg wrote:With all due respect to all the experts here, I don't like 'plugging in values' and/or eliminating the answer. That should be the last resort, when nothing else works out.

To everyone reading this and preparing for GMAT: Please learn Math properly and avoid short-cuts. When you learn and practice Math conceptually, you acquire so many skills in your repertoire that every question takes very less time to solve.
I respectfully disagree.

I think the biggest mistake people make is treating "GMAT math" the same way that they treated "school math" for most of their lives.
In school math, your teacher typically wanted to see all of your work, and in school math, your teacher typically expected a high degree of precision (e.g., answers given to 3 decimal points).

If people carry their "school math" habits over to "GMAT math," their scores will likely suffer. Remember, your goal on the GMAT is not to please your former math teachers; your goal is to maximize your score. The best way to do this is to locate the correct answer as quickly as possible. This will give you more time to work on other questions.

If you believe that "shortcuts" should be used only as a last resort, then you have the potential of disregarding the fastest approach.

With each question, you should assess the different approaches you have at your disposal (plugging in, estimating, eliminating, algebra, listing outcomes, drawing a diagram, working backwards, etc.) and select the approach that will lead you to the correct answer in the shortest amount of time.

Now, I'm not saying that students shouldn't strive to develop a deep understanding of all the required concepts and skills. I'm saying that, in addition to this, students must possess a wide variety of GMAT-specific skills and strategies to help them maximize their score on test day.

Cheers,
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by minhchau1986 » Sun Apr 22, 2012 10:49 am
Subtitute is the best method, I did not quite think to solve the problem above.

(x-A)^2+(x-a-4)^2=2(x-a)^2-8(x-4)+16=2(x-A-2)^2+8

Because a is integer, the expression is minimum when x=a+2. This is doable but can easily make mistake on calculation. Thanks GmatguruNY I agree with Brent that using conceptual method alone will suck out so much time for other problem.
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by aneesh.kg » Sun Apr 22, 2012 11:16 am
Right. I agree with you guys. I went a little astray.
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by simone88 » Sun Apr 22, 2012 1:09 pm
aneesh.kg wrote:Simone: Differentiation will never give you a wrong answer. Even in the example stated by you, it does not.
I don't understand why It doesn't. Maybe there was something wrong with my computations? If the function is f(x)=x^2-x+1, f'(x)=2x-1=0 if x=1/2. Plugging x=1/2 into f we get f(1/2)=3/4. But 1/2 is not an integer. So let's search the minimum among the integers:if x=0,1 f(x)=1; but f(x)=(x-1/2)^2+3/4 so if x is not 0,1 then |x-1/2|>1/2 therefore f(x)>1. So the minimum is 1, a different result from the minimum among all real numbers.
I wanted to explain that the minimum among integers does not always coincide with the minimum among real numbers (generally it is greater or equal than that) even though on the Gmat often it does coincide: but stay carueful! gmat contain tricky questions so you cannot go too far with the "it is often like that"!
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by aneesh.kg » Sun Apr 22, 2012 8:21 pm
Simone: Agreed. The computations are perfect. I didn't read the 'minimum among integers' part. My bad.

Well, if that happens we've to be a little careful and know the nature of the curve.
f(x) = x^2 - x + 1 is parabola pointing downward. If the minima among Real Numbers is x = 1/2, and 1/2 is equidistant from the integers closest to it, we will have two integral minima - 0 and 1.
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by [email protected] » Tue Apr 24, 2012 10:11 pm
The biggest mistake that I have committed in so many mocks in GMAT trying to solve the overall question...

That is a fool's strategy to ace the gmat because you do not know what question is going to be asked in the real test.

Try solving this question the other way i.e put in the value of each and every option in the place of x and see which option turns out to be the least...

Option 1: 26

Option 2: 16

Option 3: 8

Option 4: 10

Option 5: 26


I have not solved the substitution steps but try solving and you get the answer in around 2 mins...

The concept the Aneesh gave initially, is correct but it is complex and might not click in the exam. There are 'n' number of ways of solving the same question especially in the quant section, but always take the approach that is the simplest and least time consuming.

You can also try the above question using the Derivatives. But that is beyond syllabus in the gmat.

Substitution is the best method according to me.

I hope this post must have helped you guyzzzzzz...
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by Stuart@KaplanGMAT » Tue Apr 24, 2012 11:22 pm
rahulvsd wrote:If A and B are integers and B = A + 4 , which of the following represents integer x for which the expression (x - A)^2 + (x - B)^2 is the smallest?

A. A - 1
B. A
C. A + 2
D. A + 3
E. B + 1
Although we don't normally do so when there are variables in the choices, you can also solve by backsolving.

Let's start by subbing in for B to get:

(x - A)^2 + (x - (A + 4))^2
= (x - A)^2 + (x - A - 4)^2

If we Sub in x = A, we get:

(A - A)^2 + (A - A - 4)^2
= 0 + 16

Now let's use some logic.

If x = A - 1, that will make both brackets a bigger negative number, increasing the overall value: eliminate choice (A).

If x = A + 2, that will make the first bracket positive and the second a smaller negative; let's try it out:

(A + 2 - A)^2 + (A + 2 - A - 4)^2
= 4 + 4 = 8... smaller than 16, so eliminate choice (B).

If x = A + 3, the first bracket will be 9, so that's already bigger than 8; eliminate choice (D).

If x = B + 1, then x = A + 4 + 1 = A + 5, giving us a first bracket of 25, which is already bigger than 8; eliminate (E).

Choose (C)!
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by ronnie1985 » Wed Apr 25, 2012 10:05 am
Convert it into a sum of perfect square and a constant
the constant is the maxima or minima as the case may be and the solution of the perfect square is the value of x at which the equation will acquire its maxima or minima.
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