BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Function

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1325
Joined: Sun Nov 01, 2009 6:24 amFollowed by: 14 members
Thanked: 105 times

by vikram4689 » Sat Jun 04, 2011 4:57 am
General Rule of | |:
|x| is +ve for x>0
|x| is -ve for x<0

f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-0.5)]
=>f(1.5)=1
Premise: If you like my post
Conclusion : Press the Thanks Button ;)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members
Location: India
Thanked: 375 times

by Frankenstein » Sat Jun 04, 2011 5:10 am
vikram4689 wrote:General Rule of | |:
|x| is +ve for x>0
|x| is -ve for x<0

f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-0.5)]
=>f(1.5)=1
Hi,
|x| is never negative.
f(1.5) = |1.5|+|1.5-2| = 1.5+0.5 = 2.
In fact, for any value of x satisfying 0 ≤ x ≤ 2, f(x) = |x| + |x - 2| =2
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 407
Joined: Tue Jan 25, 2011 9:19 amFollowed by: 7 members
Thanked: 25 times

by Ozlemg » Sat Jun 04, 2011 5:16 am
I agree with Frank!

Because 0 ≤ x ≤ 2,

f(x) = |x| + |x - 2| =2
First part of the equation (red) is (+)ve and second part of (blue) it is (-)ve
so x-x+2=2
The more you suffer before the test, the less you will do so in the test! :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1325
Joined: Sun Nov 01, 2009 6:24 amFollowed by: 14 members
Thanked: 105 times

by vikram4689 » Sat Jun 04, 2011 7:33 am
Sorry pals I DO NOT agree on this ( though i made a silly mistake on a -ve sign but concept is correct)

General Rule of | |: Google MODULUS FUNCTION GRAPH, it is symm. about y-axis
|x| is +ve for x>0
|x| is -ve for x<0

f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-(-0.5)] mistook 1.5-2 as 0.5 , correcting it to -0.5
=2
=>f(1.5)=2
Premise: If you like my post
Conclusion : Press the Thanks Button ;)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1309
Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members
Location: India
Thanked: 310 times
GMAT Score:750

by cans » Sat Jun 04, 2011 9:17 am
If f(x) = |x| + |x - 2| then find the value of F(1.5) for 0 ≤ x ≤ 2
f(1.5) = |1.5| + |1.5-2| = 1.5 + |-.5| = 1.5 + .5 =2
If my post helped you- let me know by pushing the thanks button ;)

Contact me about long distance tutoring!
[email protected]

Cans!!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Sat Jun 04, 2011 9:38 am
vikram4689 wrote:General Rule of | |: Google MODULUS FUNCTION GRAPH, it is symm. about y-axis
Yes, the modulus graph is symmetrical about y-axis but that does not mean
vikram4689 wrote:|x| is +ve for x>0
|x| is -ve for x<0
|x| is always non-negative for any value of x.

By definition of |x|,
  • |x| = x for x ≥ 0
    |x| = -x for x < 0
Hence, if x is positive, |x| = x = +ve
and if x is negative, |x| = -x = -(-ve) = +ve
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members
Location: India
Thanked: 375 times

by Frankenstein » Sat Jun 04, 2011 9:39 am
vikram4689 wrote:Sorry pals I DO NOT agree on this ( though i made a silly mistake on a -ve sign but concept is correct)

General Rule of | |: Google MODULUS FUNCTION GRAPH, it is symm. about y-axis
|x| is +ve for x>0
|x| is -ve for x<0
Hi,
You might be having a good picture of modulus but what you have written is still wrong.
Can you give an example for |x| being -ve. By definition modulus is non-negative.
It would be correct to say
|x| is x for x>0
|x| is -x for x<0 --> As x is -ve, -x will be positive.
I think this is what you have intended to write.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 222
Joined: Mon Oct 13, 2008 4:04 pmFollowed by: 2 members
Thanked: 3 times

by venmic » Sun Jun 05, 2011 2:03 pm
why did you convert the modulus to [-(x-2)] negaive ???????
i disagree
vikram4689 wrote:General Rule of | |:
|x| is +ve for x>0
|x| is -ve for x<0

f(1.5)= |x| + |x-2|
= x + [-(x-2)]
= 1.5 + [-0.5)]
=>f(1.5)=1
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1325
Joined: Sun Nov 01, 2009 6:24 amFollowed by: 14 members
Thanked: 105 times

by vikram4689 » Sun Jun 05, 2011 6:29 pm
Actually i confused the whole scenario ;).... what i intended was |x| is -ve for x<0 and when that value of x is substituted in -x than it becomes +ve.

For |x-2|, it is -ve for x<2 i.e.-(x-2) and when i substitute x=1.5 it becomes +ve

Apologies for creating unintended confusion.
Premise: If you like my post
Conclusion : Press the Thanks Button ;)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 13
Joined: Wed May 25, 2011 9:40 am

by TuanNguyen87 » Sun Jun 05, 2011 8:48 pm
I think we just need to replace x by 1.5 and because 1.5 > 0. Thus,

f(1.5)= |1.5| + |1.5 - 2|= 1.5 + (2 - 1.5)= 2 :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 99
Joined: Sat Oct 16, 2010 1:55 pm
Thanked: 1 times

by aftableo2006 » Sun Jun 05, 2011 11:14 pm
f(x)=|x|+|x-2|
for x>2 f(x)=x+x-2=2x-2
for x<0 f(x)=-x-x+2=2-2x
for 0<x>2 f(x)=x-x+2=2
s0 f(1.5)=2
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 934
Joined: Tue Nov 09, 2010 5:16 amFollowed by: 14 members
Location: AAMCHI MUMBAI LOCAL
Thanked: 63 times

by [email protected] » Thu Jun 09, 2011 7:51 pm
OA is 2
IT IS TIME TO BEAT THE GMAT

LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!

Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 401
Joined: Tue May 24, 2011 1:14 amFollowed by: 5 members
Thanked: 37 times

by MBA.Aspirant » Thu Jun 09, 2011 7:57 pm
f(1.5) = |1.5| + |1.5-2| = 1.5 + 0.5 = 2
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1