GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 25 May 2019, 00:07 ### GMAT Club Daily Prep

#### 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. ### Request Expert Reply # If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9239
Location: Pune, India
Re: Question of the Day - II  [#permalink]

### Show Tags

1
ankitbansal85 wrote:
Hi Karishma...

Can you please explain the example which has negative between the modulus (could not understand fully the explaination given by you earlier). Also if in a question we have a combination of + and - then how to go about it?

When you add two mods, you try to add up the distances e.g.
|x - 1| + |x - 5| = 10
you try to find the point where distance from 1 and distance from 5 adds up to give you 10.

When you subtract two mods, you subtract out the distances e.g.
|x - 1| - |x - 5| = 3
you try to find the point where distance from 1 and distance from 5 have a difference of 3. You know that at x = 3, distances from 1 and from 5 are equal (distance of 1 from 3 is 2 and distance of 5 from 3 is also 2). At x = 4, the difference between the distances will be 2. At x = 4.5, the difference between the distances will be 3.

The subtraction is a little less intuitive and will take more practice. Questions with both + and - would be too complicated though do-able. Most people will probably not get any mods question with more than 2 terms.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Senior Manager  Joined: 13 Aug 2012
Posts: 418
Concentration: Marketing, Finance
GPA: 3.23
Re: Question of the Day - II  [#permalink]

### Show Tags

I love this question. One must understand that f(x) = |4x-1| + |x-3| + |x+1| means the sum of the distances of x to 1/4, 3 and -1.
The best way to minimize is to zero out the distance in the middle. ==========(-1)==========(0)======(1/4)=========(3)=======
So if x = 1/4
|4x-1| = 0
|x+1| = 1 1/4
|x-3| = 2 3/4

Answer: 4
_________________
Impossible is nothing to God.
Manager  Joined: 24 Mar 2010
Posts: 63
Re: Question of the Day - II  [#permalink]

### Show Tags

Karishma,

So from your method, I infer

That minimum of a function will always be at either its critical points or zero.

Await your valued response.
_________________
- Stay Hungry, stay Foolish -
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9239
Location: Pune, India
Re: Question of the Day - II  [#permalink]

### Show Tags

eaakbari wrote:
Karishma,

So from your method, I infer

That minimum of a function will always be at either its critical points or zero.

Await your valued response.

The minimum could also be in an entire range. Take this question for example.

f(x) = |3x + 1| + |2x-3| + |x - 7|

For what value(s) of x will f(x) have the minimum value?
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Retired Moderator B
Joined: 05 Jul 2006
Posts: 1700
Re: Question of the Day - II  [#permalink]

### Show Tags

What if the question asks for max value of f(x) ??? disregarding the answer choices ??
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9239
Location: Pune, India
Re: Question of the Day - II  [#permalink]

### Show Tags

yezz wrote:
What if the question asks for max value of f(x) ??? disregarding the answer choices ??

Not every function will have a minimum and a maximum value. The greater the value of x, the greater the function will become. It is an infinitely increasing function.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern  Joined: 24 Mar 2013
Posts: 23
Re: Question of the Day - II  [#permalink]

### Show Tags

1
Many Thanks to all of you for sharing such amazing techniques. I was overwhelmed with mod questions when I started, but your explanations and techniques have helped me build confidence. Bunuel, Karishma, Gurpreet, Shrouded1....Awesome!

How about this approach:

F(x) will be minimum when each individual term in the function has the lowest possible value. So, I get x = 3, -1 and 1/4.
Now, substituting each value of x in F(x), I can easily see that x=1/4 gives me the smallest possible value for F(x) = 4

Thanks,
Rohit
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9239
Location: Pune, India
Re: Question of the Day - II  [#permalink]

### Show Tags

eaakbari wrote:
Karishma,

What about a function with evenly spaced numbers?

for instance

f(x) = | 4x + 1| + | 2x + 1| + | 4x + 3| + | x |

What will be the min. value of x for this?

The function will take a minimum value for a range of values of x (between the second and the third values). Think 'Why?'
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9239
Location: Pune, India
Re: Question of the Day - II  [#permalink]

### Show Tags

1
rohitd80 wrote:
Many Thanks to all of you for sharing such amazing techniques. I was overwhelmed with mod questions when I started, but your explanations and techniques have helped me build confidence. Bunuel, Karishma, Gurpreet, Shrouded1....Awesome!

How about this approach:

F(x) will be minimum when each individual term in the function has the lowest possible value. So, I get x = 3, -1 and 1/4.
Now, substituting each value of x in F(x), I can easily see that x=1/4 gives me the smallest possible value for F(x) = 4

Thanks,
Rohit

The technique is fine but the logic is not sound. Why should we say that the function will take minimum value only when it takes one of these three values? For one of these values, sure one mod will be 0 but the other two could be much greater.
The reason why this works is because the minimum value will be at one of the transition points - the middle point (logic explained in the post on previous page) in case there are odd number of terms OR at two points (and for every value in between) in case there are even number of terms.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Senior Manager  Joined: 13 May 2013
Posts: 420
Re: Question of the Day - II  [#permalink]

### Show Tags

Hello all.

I am wondering why we cannot take the positive and negative cases of f(x) = |4x - 1| + |x-3| + |x + 1| and solve for x that way?

In other words, f(x) = |4x - 1| + |x-3| + |x + 1|

I. f(x) = (4x-1) + (x-3) + (x+1)

II. f(x) = -(4x-1) + -(x-3) + -(x-1)

Thanks!
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1051
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: Question of the Day - II  [#permalink]

### Show Tags

1
WholeLottaLove wrote:
Hello all.

I am wondering why we cannot take the positive and negative cases of f(x) = |4x - 1| + |x-3| + |x + 1| and solve for x that way?

In other words, f(x) = |4x - 1| + |x-3| + |x + 1|

I. f(x) = (4x-1) + (x-3) + (x+1)

II. f(x) = -(4x-1) + -(x-3) + -(x-1)

Thanks!

It's not that easy!

if you wanna study the absolute value, more math is required.
You have to study each $$abs>0$$ so
$$4x-1>0$$ and $$x-3>0$$ and $$x+1>0$$
$$x>\frac{1}{4}$$ and $$x>3$$ and $$x>-1$$
so 4x-1 is positive for x>1/4, x-3 is +ve for x>3 and x-1 is +ve for x>-1

Now you have to split the original function into the areas defined above:
$$x<-1$$ all functions are negative
$$f(x) = -(4x-1) + -(x-3) + -(x-1)$$
if $$-1<x<\frac{1}{4}$$ the third term is positive,the others negative
$$f(x) = -(4x-1) + -(x-3) +(x-1)$$
and so on...

You cannot take all positive or all negative, you have to study each function in all possible intervals
Below there is the graph of F(x) that I hope will make thing clear. As you see there are 4 functions, each one defined in the intervals above, so your way of studying the abs value (reducing all to 2 functions) is incomplete

Hope it's clear
Attachments Untitled.png [ 5.15 KiB | Viewed 2127 times ]

_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
Senior Manager  Joined: 13 May 2013
Posts: 420
Re: Question of the Day - II  [#permalink]

### Show Tags

Ok, I get that for f(x) = (4x-1) + (x-3) + (x+1), x must be greater than 1/4, 3 and -1 respectively. But that's where I get lost.

I'm sorry for being so dense on this topic!
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1051
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: Question of the Day - II  [#permalink]

### Show Tags

1
1
WholeLottaLove wrote:
Ok, I get that for f(x) = (4x-1) + (x-3) + (x+1), x must be greater than 1/4, 3 and -1 respectively. But that's where I get lost.

I'm sorry for being so dense on this topic!

Study of the absolute value:

1)take each term into the "| |" and define where it's positive
2)draw a number line with each "edge value" (where each term changes sign)
3)Split the function according to those intervals

Refer to the image

So if x>3 all terms are positive =>f(x) = (4x-1) + (x-3) + (x+1)
if 1/4<x<3 for example you see that 4x-1 is positive, x+1 is positive BUT x-3 is negative => f(x) = (4x-1) + (-)(x-3) + (x+1)

Repeat this operation for each interval and you'll have all possible combinations

Remeber that the each function is valid only in that interval

What I mean is that f(x) = (4x-1) + (-)(x-3) + (x+1) is valid only in the 1/4<x<3 area. Each area has its own function
Attachments Untitled.png [ 1.55 KiB | Viewed 2325 times ]

_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
Senior Manager  Joined: 13 May 2013
Posts: 420
Re: Question of the Day - II  [#permalink]

### Show Tags

Haha - I am lost (not your fault...I am just very slow with mathematical concepts, unfortunately)

I understand that for certain values of x,(say if x =2) (4x-1) and (x+1) would be positive but (x-3) would be negative. But why would we bother finding what is positive and what is negative? I almost feel as if testing a series of integers and fractions, both positive and negative, would be a quicker way to figure out the right answer. Still, I am trying to get the concepts buttoned down.
VP  Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1051
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: Question of the Day - II  [#permalink]

### Show Tags

1
WholeLottaLove wrote:
Haha - I am lost (not your fault...I am just very slow with mathematical concepts, unfortunately)

I understand that for certain values of x,(say if x =2) (4x-1) and (x+1) would be positive but (x-3) would be negative. But why would we bother finding what is positive and what is negative? I almost feel as if testing a series of integers and fractions, both positive and negative, would be a quicker way to figure out the right answer. Still, I am trying to get the concepts buttoned down.

What I explained above is how to study an abs value from a theoretical point of view, because you original methos is wrong.

Of course this is not required to answer the question, you can try with real number and see what you find out.
But am I trying to explain how an abs function works, for instance your original method

I. f(x) = (4x-1) + (x-3) + (x+1)

II. f(x) = -(4x-1) + -(x-3) + -(x-1)

does not work to find the answer. The function cannot be reduced to that form!

"But why would we bother finding what is positive and what is negative? "
This is required to study an abs value. In my original post you see the graph of F(X), and you notice that is defined into sections.
Each section is one of the intervals above, and in each one of those the fucntion has a different equation.

The concept that I apply here is the same as the one that you would apply to solve
$$y=|x|$$
How would you study this?
for x>0 => y=x
for x<0 => y=-x

Define where the funct is positive, treat each part as a separate equation. The concept in the question is the same, only involves more intervals.
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]
Manager  Joined: 04 Mar 2013
Posts: 66
Location: India
Concentration: General Management, Marketing
GPA: 3.49
WE: Web Development (Computer Software)
Re: Question of the Day - II  [#permalink]

### Show Tags

VeritasPrepKarishma wrote:
Q. If f(x) = |4x - 1| + |x-3| + |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.)

keep x = 0 and eliminate d, e.
now put x = 1, 1/2 we end up with some dis satisfing valye now keep x =1/4 u get ur answer

what i did was, i took even 1/8 but pattern reverses and u get to know 1/4 is correct and confirmed

logic and basic = magic at gmat Senior Manager  Joined: 13 May 2013
Posts: 420
Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)  [#permalink]

### Show Tags

If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)?

x=1/4, x=3, x=-1

x<-1, -1<x<1/4, 1/4<x<3, x>3

x<-1
|4x - 1| + |x-3| + |x + 1|
-(4x-1) + -(x-3) + -(x+1)
-4x+1 + -x+3 + -x-1
-6x+3
If x<-1 then minimally, -6x+3 would be greater than 9

-1<x<1/4
|4x - 1| + |x-3| + |x + 1|
-(4x-1) + -(x-3) + (x+1)
-4x+1 + -x+3 + x + 1
-4x+5
If x is between -1 and 1/4...try three values of -1, 0, 1/4
a)-4x+5
-4(-1)+5 = 9
-4(0)+5 = 5
-4(1/4)+5 = 4

1/4<x<3
|4x - 1| + |x-3| + |x + 1|
(4x-1) + -(x-3) + (x+1)
4x-1 + -x+3 + x+1
4x+3

For 4x+3, any value 1/4<x<3 will produce a value greater than 4

x>3
|4x - 1| + |x-3| + |x + 1|
(4x-1) + (x-3) + (x-1)
6x-5

For 6x-5, any value x>3 will produce a value greater than 13

The smallest number possible for this particular function is f(x) = 4

(B)
Intern  Joined: 29 May 2013
Posts: 6
Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)  [#permalink]

### Show Tags

hI Gurpreet,

Why did you change the value of |x-3| to 3-x,

For x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x-1 + 3-x + x+1 = 4x+3 > 3

Kriti
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9239
Location: Pune, India
Re: If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)  [#permalink]

### Show Tags

Kriti2013 wrote:
hI Gurpreet,

Why did you change the value of |x-3| to 3-x,

For x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x-1 + 3-x + x+1 = 4x+3 > 3

Kriti

Because when x is less than 3, (x - 3) is negative.
Therefore, |x-3| = - (x - 3) = 3 - x
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Intern  Status: Finance Analyst
Affiliations: CPA Australia
Joined: 10 Jul 2012
Posts: 16
Location: Australia
Concentration: Finance, Healthcare
Schools: AGSM '16 (A\$)
GMAT 1: 470 Q38 V19 GMAT 2: 600 Q44 V34 GPA: 3.5
WE: Accounting (Health Care)
Re: Question of the Day - II  [#permalink]

### Show Tags

VeritasPrepKarishma wrote:
eaakbari wrote:
Karishma,

So from your method, I infer

That minimum of a function will always be at either its critical points or zero.

Await your valued response.

The minimum could also be in an entire range. Take this question for example.

f(x) = |3x + 1| + |2x-3| + |x - 7|

For what value(s) of x will f(x) have the minimum value?

Thank you for your intuitive explanation and clearing the concept for us. So using your analogy of people at every point, shouldn't -1/3 have the minimum value for x (or distance so to speak)?

To be precise, you said 1/4 point has the minimum distance for x or gives the minimum value because of the denominator 4 (my assumption - as you did not state it specifically). This is why we chose 1/4 over 3/2!
_________________
Our deepest fear is not that we are inadequate. Our deepest fear is that we are powerful beyond measure. It is our light not our darkness that most frightens us.

Your playing small does not serve the world. There's nothing enlightened about shrinking so that other people won't feel insecure around you.

It's not just in some of us; it's in everyone. And as we let our own light shine, we unconsciously give other people permission to do the same.

As we are liberated from our own fear, our presence automatically liberates others.

—Marianne Williamson Re: Question of the Day - II   [#permalink] 06 Aug 2013, 21:46

Go to page   Previous    1   2   3    Next  [ 53 posts ]

Display posts from previous: Sort by

# If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

#### MBA Resources  