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If f(x)= 4x  1 + x3 + x + 1 (Question of the DayII) [#permalink]
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29 Oct 2010, 11:18
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If f(x) = 4x  1 + x3 + x + 1, what is the minimum value of f(x)? (A) 3 (B) 4 (C) 5 (D) 21/4 (E) 7 (Still high on mods! Next week, will make questions on some other topic.)
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Last edited by Bunuel on 06 Jul 2013, 02:33, edited 1 time in total.
Added the OA.



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Re: Question of the Day  II [#permalink]
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VeritasPrepKarishma wrote: Q. If f(x) = 4x  1 + x3 + x + 1, what is the minimum value of f(x)?
(A) 3 (B) 4 (C) 5 (D) 21/4 (E) 7
(Still high on mods! Next week, will make questions on some other topic.) put x=0 we get f(x) = 5..rule out D and E put x=1/4 we get f(x) = 0 + 11/4 + 5/4 = 16/4 = 4.. rule out C D and E Now the answer is either 3 or 4. Reason for above checking of values: for every value of x> 3 the f(x) is quite big because of 4x1 for every value of x < 1 the f(x) if bigger than 3 and 4. for x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x1 + 3x + x+1 = 4x+3 > 3 => only 4 is the probable answer. Thus we only need to check x>=1 and x<= 1/4 f(x) in this domain is = 14x + 3x + x+1 = 54x => for f(x) to be minimum the x should be +ve => for x = 1/4 , f(x) = 4. This can be solved using graph as well by plotting the f(x) in different domains.
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Re: Question of the Day  II [#permalink]
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30 Oct 2010, 01:59
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Let g(x) = 4x  1 & h(x)=x3 + x + 1 ... So f(x)=g(x)+h(x) Now h(x) is equal to 4 between 1 and 3 and is higher everywhere else, so it minimizes between 1 & 3 g(x) is minimum at x=(1/4) where it is equal to 0. Since 1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point f(1/4)=g(1/4)+h(1/4)=0+4 Hence answer is 4
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Re: Question of the Day  II [#permalink]
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30 Oct 2010, 08:53
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That's right! I can count on you guys to always give the correct answer and some innovative ways of solving... Let me give you the method I would use to solve it too... Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend... So x3 is the distance from 3, x + 1 is the distance from 1 and 4x  1 is 4 times the distance from 1/4 So 4x  1 + x3 + x + 1 is the sum of distances from 3, 1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible? Attachment:
Ques.jpg [ 4.56 KiB  Viewed 14954 times ]
Can I say that minimum total distance will be covered from point 1/4? Attachment:
Ques1.jpg [ 3.43 KiB  Viewed 14937 times ]
The logic being that the distance from 3 to 1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all?
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Re: Question of the Day  II [#permalink]
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In other words, as my mentor says, assume there is one guy on point 3, one on point 1 and 4 guys on point 1/4. If they have to meet up, but cover minimum distance, they should meet at point 1/4. What happens when there are 4 such points? Lets add another term (2x  3) to f(x)...
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Re: Question of the Day  II [#permalink]
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30 Oct 2010, 09:29
would the answer still remain 1/4? based on your 'distance' method?
The 3 guys standing at the new 3/2 post + the existing persons at 1 and 3 can again meet at 1/4 as that would be the shortest distance



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Re: Question of the Day  II [#permalink]
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30 Oct 2010, 16:17
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gmat1011 wrote: would the answer still remain 1/4? based on your 'distance' method?
The 3 guys standing at the new 3/2 post + the existing persons at 1 and 3 can again meet at 1/4 as that would be the shortest distance Yes, that is right! The answer still remains 1/4. Since 2x  3 = 2x  3/2 it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at 1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4. Attachment:
Ques.jpg [ 5.2 KiB  Viewed 14965 times ]
Try some other combinations. e.g. f(x) = x  1 + x3 + x + 1 + x + 6 f(x) = 2x  3 + 4x + 7 etc
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Re: Question of the Day  II [#permalink]
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09 Nov 2010, 23:23
Great Method Karishma !!
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Re: Question of the Day  II [#permalink]
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14 Jun 2011, 02:14
for x = 0, f(x) = 5. POE options C,D and E. for x = 1/4, f(x) = 4. B it is.
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Re: Question of the Day  II [#permalink]
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14 Jun 2011, 03:47
Very good question thanks for posting



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Re: Question of the Day  II [#permalink]
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15 Mar 2012, 11:07
VeritasPrepKarishma wrote: gmat1011 wrote: would the answer still remain 1/4? based on your 'distance' method?
The 3 guys standing at the new 3/2 post + the existing persons at 1 and 3 can again meet at 1/4 as that would be the shortest distance Yes, that is right! The answer still remains 1/4. Since 2x  3 = 2x  3/2 it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at 1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4. Attachment: Ques.jpg Try some other combinations. e.g. f(x) = x  1 + x3 + x + 1 + x + 6 f(x) = 2x  3 + 4x + 7 etc Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ? In example : x  1 + x3 + x + 1 + x + 6 .. the posts on the number line are : 6, 1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ? Please correct me if I am wrong.



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ficklehead wrote: Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ? In example : x  1 + x3 + x + 1 + x + 6 .. the posts on the number line are : 6, 1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?
Please correct me if I am wrong. Forget these numbers. Think logically. My house is 10 miles away from your house. If we have to meet up, how much distance do we need to cover together? In any case, we need to cover 10 miles together at least, right? Either you come down to my place (you cover 10 miles) or I come down to yours (I cover 10 miles) or we meet mid way (10 miles covered together) or we meet up at a nice coffee place 2 miles further down from my house in the opposite direction in which case we will need to cover more than 10 miles (i.e. we cover 2 + 12 = 14 miles) Now say, another friend is at my place. In which case will people cover minimum distance together? If two of us come down to your place, we cover 10+10 = 20 miles together but if you come down to our place, you cover only 10 miles. If instead, we meet midway, we cover 5+5 and you cover 5 miles so in all 15 miles. So less number of people should travel the entire distance. If there are 4 people at point A and 2 at point B, minimum distance will be covered if people at point B travel to point A. So people at 3/2 should come down to 1/4.
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Re: Question of the Day  II [#permalink]
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15 Mar 2012, 11:33
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ficklehead wrote: In example : x  1 + x3 + x + 1 + x + 6 .. the posts on the number line are : 6, 1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?
Please correct me if I am wrong. x is that point on the number line whose sum of distances from 6, 1, 1 and 3 is minimum. So basically there is a person each at points 6, 1, 1 and 3. You need to make them all meet by covering minimum distance. Distance between 6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: 6, 1, 0, 1 or 3 etc they will cover a distance of 9 together. If 1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at 1 travels down to 1 and 6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet. If they instead meet at 1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between 1 and 1, the total distance covered will be 11. To check, put x = 1. you get x  1 + x3 + x + 1 + x + 6 = 11 put x = 1, you get x  1 + x3 + x + 1 + x + 6 = 11 put x = 0, you get x  1 + x3 + x + 1 + x + 6 = 11
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Re: Question of the Day  II [#permalink]
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15 Mar 2012, 14:48
VeritasPrepKarishma wrote: ficklehead wrote: In example : x  1 + x3 + x + 1 + x + 6 .. the posts on the number line are : 6, 1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?
Please correct me if I am wrong. x is that point on the number line whose sum of distances from 6, 1, 1 and 3 is minimum. So basically there is a person each at points 6, 1, 1 and 3. You need to make them all meet by covering minimum distance. Distance between 6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: 6, 1, 0, 1 or 3 etc they will cover a distance of 9 together. If 1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at 1 travels down to 1 and 6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet. If they instead meet at 1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between 1 and 1, the total distance covered will be 11. To check, put x = 1. you get x  1 + x3 + x + 1 + x + 6 = 11 put x = 1, you get x  1 + x3 + x + 1 + x + 6 = 11 put x = 0, you get x  1 + x3 + x + 1 + x + 6 = 11 Thanks Karishma for this detailed explanation. I got it now. I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11. To seal the concept, for 2x3+4x+7, minimum distance be : 3/2+7/4=13/4 ?



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Re: Question of the Day  II [#permalink]
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15 Mar 2012, 22:03
ficklehead wrote: I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.
To seal the concept, for 2x3+4x+7, minimum distance be : 3/2+7/4=13/4 ? Look at the diagram below. Make a number line in such questions. Attachment:
Ques3.jpg [ 5.1 KiB  Viewed 13919 times ]
For x and y to meet, they have to cover a distance of 9 together. For p and q to meet, they have to cover a distance of 2 together. They can meet anywhere between 1 and 1 and they will cover a total distance of 11 only. So x can take any value 1 < x < 1 and the value of the expression will be 11. 2x3+4x+7 = 2x3/2 + 4x+7/4 Attachment:
Ques4.jpg [ 5.35 KiB  Viewed 13934 times ]
There are 4 people at 7/4 and 2 people at 3/2. Distance between the two points is 7/4 + 3/2 = 13/4 For these people to meet covering the minimum distance, the 2 people X and Y should travel to point 7/4. (Make minimum people travel). So minimum distance that needs to be covered = 2*13/4 = 13/2 (because 2 people travel 13/4 each) which is the minimum value of the expression. The value of x when the expression takes minimum value is 7/4. Check by putting x = 7/4. You get 2x3+4x+7 = 13/2 Also see that when you put x = 0 or 3/2 etc, the value of the expression is higher.
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Re: Question of the Day  II [#permalink]
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16 Mar 2012, 09:01
Thanks a lot, Karishma.



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Re: Question of the Day  II [#permalink]
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16 Mar 2012, 19:48
I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : x+6x1 ?



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Re: Question of the Day  II [#permalink]
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20 Mar 2012, 14:33
I got C = 5
f(x) = 4x1 + x3 + x+1
=(4x1) + (x3) + (x+1) or =  (4x1)  (x3)  (x+1) so, x= 1/2
then i used the value of x=1/2
f(x) = 4x1 + x3 + x+1
f(1/2) = 4(1/2)1 + (1/2)3 + (1/2)+1 = 1 + 5/2 + 3/2 = 1 + 5/2 + 3/2 = 5
Is that correct?



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Re: Question of the Day  II [#permalink]
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20 Mar 2012, 21:07
kuttingchai wrote: I got C = 5
f(x) = 4x1 + x3 + x+1
=(4x1) + (x3) + (x+1) or =  (4x1)  (x3)  (x+1) so, x= 1/2
then i used the value of x=1/2
f(x) = 4x1 + x3 + x+1
f(1/2) = 4(1/2)1 + (1/2)3 + (1/2)+1 = 1 + 5/2 + 3/2 = 1 + 5/2 + 3/2 = 5
Is that correct? Put x = 1/4 and the minimum value you will get is 4. How did you get x = 1/2? I would suggest you to check out one of the approaches mentioned above.
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ficklehead wrote: I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : x+6x1 ? You can do it with a negative sign too. You want to find the minimum value of (distance from 6)  (distance from 1) Make a number line with 6 and 1 on it. (6)..........................(1) Think of a point in the center of 6 and 1. Its distance from 6 is equal to distance from 1 and hence (distance from 6)  (distance from 1) = 0 . What if instead, the point x is at 6? Distance from 6 is 0 and distance from 1 is 7 so (distance from 6)  (distance from 1) = 0  7 = 7 If you keep moving to the left, (distance from 6)  (distance from 1) will remain 7 so the minimum value is 7.
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