Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
One can make out that for expression (X - A)^2 + (X - B)^2 to be smallest we need to select value for X such that the squares are having least value
For X=A we get (A-A)^2 + (A-A-4)^2 = 16 For X = A+2 we get (A+2 -A)^2 + (A+2 -A-4)^2 = 8 For X= A+3 we get (A+3 -A)^2 + (A+3 -A-4)^2 = 10 for expression A-1 and B+1 we will get value more than 8
Re: Very Tough Algebra II Question --- Please Help [#permalink]
19 Oct 2009, 20:04
I go along with the explanation above, which is the easiest way to SOLVE the problem but not to UNDERSTAND it.
My understanding is like this:
Since B=X+4, the expression (X-A)^2+(X-B)^2 =(X-A)^2+[(X-A)-4)]^2 =(X-A)^2+(X-A)^2-8(X-A)+16 =2[(X-A)^2-4(X-A)+8] =2[(X-A)^2-4(X-A)+4+4] =2[(X-A-2)^2+4]
As (X-A-2)^2 can not be less than 0, the expression is smallest when (X-A-2)^2 =0, Therefore, option C: X=A+2 is selected
The reasoning process is only for understanding the problem, I still recommend the method by asterixmatrix for problem-solving purpose in GMAT testing
Given that: (X - A)^2 + (X - B)^2 = (X - A)^2 + (X - a - 4)^2
If x = a-1, (X - A)^2 part becomes 1 but (X - a - 4)^2 part becomes 25. So total = 26. If x = a, (X - A)^2 part becomes 0 but (X - a - 4)^2 part becomes 16. So total = 16. If x = a+2, (X - A)^2 part becomes 4 but (X - a - 4)^2 part becomes 4. So total = 8. If x = a+3, (X - A)^2 part becomes 9 but (X - a - 4)^2 part becomes 1. So total = 10. If x = b+1=a+4+1=a+5, (X - A)^2 part becomes 25 and (X - a - 4)^2 = 1 So total = 26
18 seconds method, take the derivatives of both polynomials of order two to get 2 ( 2x - A - B) = 0, now substitute for B as given in the question and get the value of X which is X = A + 2
Re: Very Tough Algebra II Question --- Please Help [#permalink]
20 Oct 2009, 20:52
LUGO wrote:
18 seconds method, take the derivatives of both polynomials of order two to get 2 ( 2x - A - B) = 0, now substitute for B as given in the question and get the value of X which is X = A + 2
Lugo, thanks for reminding me about derivatives!
Such a basic concept, completely overlooked on gmat prepping.
To find the minimum/maximum value of an equation, find the point where its derivative is 0.
Given that this is a convex (upwards function), the value obtained will be the minimum value.
Re: Very Tough Algebra II Question --- Please Help [#permalink]
06 Oct 2010, 12:45
2
This post received KUDOS
I solved it using derivative. But do not forget to check the second derivative. If second derivate is >0 then only the first derivate can be simplified to get the min value.
Another solution:
\((x-a)^2 + (x-b)^2 =\) square of the distance between (x,x) and (a,b)
Since (x,x) lies on the line x=y the min distance between the line x=y and (a,b) is our required result.
This distance will be minimum when the perpendicular from the point a,b intersects the x=y at point x,x
since slope of x=y is 1, the slope of perpendicular from a,b to line x=y is -1
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?
1. A-1 2. A 3. A+2 4. A+3 5. B+1
Nice question and it can be answered in 10 sec - Answer is 3. A+2 for the expression to be smallest, Integer x must reside between B and A, hence (B+A)/2 = A+2. However, plug and play can also be applied to resolve this question.
Hope it helps. Aj. _________________
----------------------------------------------------------------------------------------- What you do TODAY is important because you're exchanging a day of your life for it! -----------------------------------------------------------------------------------------
how you know this "for the expression to be smallest, Integer x must reside between B and A", could u plz explain.. is it bcuz this especial expression? thnx
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?
1. A-1 2. A 3. A+2 4. A+3 5. B+1
since (x-A)^2 + (x-B)^2 and B=A+4 we have (x-A)^2 + (x-(A+4))^2 or (x-A)^2 + (x-A-4)^2 plug in each answ.choice and u get the answ i.e. x=a-1 then ((A-1)-A)^2 + ((A-1)-A-4)^2 =1+25=26 and so on ...each answ choise... _________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth
how you know this "for the expression to be smallest, Integer x must reside between B and A", could u plz explain.. is it bcuz this especial expression? thnx
Yes it is only applicable to this special expression. If Integer x doesn't reside between B and A than one of the value (either (x-A)^2 or (x-B)^2) in expression (x-A)^2 + (x-B)^2 will explode.
Hope it helps. Cheers! _________________
----------------------------------------------------------------------------------------- What you do TODAY is important because you're exchanging a day of your life for it! -----------------------------------------------------------------------------------------
(I've posted another approach to this problem below this post) Hi My friends, (sorry for my bad English) This problem will be much easier and only take 10 seconds to be solved if you know this inequality : a^2+b^2>= 2 a.b (proof: we have (a-b)^2 >=0 so please expand it and then we have a^2 +b^2 >= 2 a.b). ( this is AM-GM inequality-very popular and basic inequality) So the minimum of a^2 +b^2 will be 2 a.b only when a=b. .....and applied to this problem, we will have the minimum value when x-A=x-B so we have x= (A+B)/2= A+2 The answer is C _________________
Life is not easy I knew that and now I don't even expect life to be easy
Last edited by retailingvnsupernova on 27 Apr 2013, 11:21, edited 1 time in total.
Hi my friends, let me try another approach to this problem: let's \(C=(x-A)^2 + (x-B)^2\) so \(C= 2x^2 +A^2+B^2 - 2 x. (A+B)\) and then \(C = (\sqrt{2}x)^2 + \frac{(A+B)^2}{2} + \frac{(A-B)^2}{2} - 2.\sqrt{2}.x. \frac{(A+B)}{\sqrt{2}}\) Why I do this? cause I want to simplified this to \(x^2+P >= P\) now we have \(C= (\sqrt{2}x-\frac{(A+B)}{\sqrt{2}})^2 + \frac{(A-B)^2}{2}\) clearly \(C>= \frac{(A-B)^2}{2}=8\) \(C=8\) (minimum) only when \(\sqrt{2}.x=\frac{(A+B)}{\sqrt{2}}\) and then we have \(x=\frac{(A+B)}{2} = A+2\) The answer: C _________________
Life is not easy I knew that and now I don't even expect life to be easy