Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2001:30 PM EST
-02:30 PM IST
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Nov 2206:30 AM PST
-08:30 AM PST
Let’s dive deep into advanced CR to ace GMAT Focus! Join this webinar to unlock the secrets to conquering Boldface and Paradox questions with expert insights and strategies. Elevate your skills and boost your GMAT Verbal Score now!
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2407:00 PM PST
-08:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
28 May 2020, 02:04
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
56% (02:36) correct
44%
(02:17)
wrong
based on 385
sessions
History
Date
Time
Result
Not Attempted Yet
Function \(f(x)\) is defined as follows:
If \(x\) is positive or 0, then \(f(x) = 1 - x\);
If \(x\) is negative, then \(f(x) = x - 1\).
Which of the following statements must be true?
I. \(f(|x|) = |f(x)|\)
II. \(f(x) = f(-x)\)
III. \(|f(x + 1)| = |x|\)
A. I only
B. II only
C. III only
D. I and III only
E. None of the above.
M11-24
If \(x\) is positive or 0, then \(f(x) = 1 - x\);
If \(x\) is negative, then \(f(x) = x - 1\).
Which of the following statements must be true?
I. \(f(|x|) = |f(x)|\)
II. \(f(x) = f(-x)\)
III. \(|f(x + 1)| = |x|\)
A. I only
B. II only
C. III only
D. I and III only
E. None of the above.
M11-24
Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
28 May 2020, 02:07
Kudos
Bookmarks
Official Solution:
Function \(f(x)\) is defined as follows:
If \(x\) is positive or 0, then \(f(x) = 1 - x\);
If \(x\) is negative, then \(f(x) = x - 1\).
Which of the following statements must be true?
I. \(f(|x|) = |f(x)|\)
II. \(f(x) = f(-x)\)
III. \(|f(x + 1)| = |x|\)
A. I only
B. II only
C. III only
D. I and III only
E. None of the above.
I. \(f(|x|) = |f(x)|\).
This option is not necessarily true. Consider \(x = -1\). \(f(|x|) = f(1) =1 - 1 =0\) does not equal \(|f(x)| =|f(-1)| = |-1-1|= |-2|= 2\).
II. \(f(x) = f(-x)\).
This option is not necessarily true. Consider \(x = 1\). \(f(x) =f(1) = 1 - 1 = 0\) does not equal \(f(-x) = f(-1) = -1 - 1 = -1 - 1 = -2\).
III. \(|f(x + 1)| = |x|\).
This option is always true:
If \(x + 1\) is non-negative, then \(|f(x + 1)| = |1 - (x + 1)| = |-x| = |x|\);
If \(x + 1\) is negative, then \(|f(x + 1)| = |(x + 1) - 1| = |x|\)
Answer: C
Function \(f(x)\) is defined as follows:
If \(x\) is positive or 0, then \(f(x) = 1 - x\);
If \(x\) is negative, then \(f(x) = x - 1\).
Which of the following statements must be true?
I. \(f(|x|) = |f(x)|\)
II. \(f(x) = f(-x)\)
III. \(|f(x + 1)| = |x|\)
A. I only
B. II only
C. III only
D. I and III only
E. None of the above.
I. \(f(|x|) = |f(x)|\).
This option is not necessarily true. Consider \(x = -1\). \(f(|x|) = f(1) =1 - 1 =0\) does not equal \(|f(x)| =|f(-1)| = |-1-1|= |-2|= 2\).
II. \(f(x) = f(-x)\).
This option is not necessarily true. Consider \(x = 1\). \(f(x) =f(1) = 1 - 1 = 0\) does not equal \(f(-x) = f(-1) = -1 - 1 = -1 - 1 = -2\).
III. \(|f(x + 1)| = |x|\).
This option is always true:
If \(x + 1\) is non-negative, then \(|f(x + 1)| = |1 - (x + 1)| = |-x| = |x|\);
If \(x + 1\) is negative, then \(|f(x + 1)| = |(x + 1) - 1| = |x|\)
Answer: C
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 17 Jul 2025
Posts: 3,086
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
28 May 2020, 02:27
Kudos
Bookmarks
If x is +/0 = F(x) = 1-x
If x is - = F(x) = x-1
1) F(|x|) = |F(x)|
Case 1: x=1
F(|x|) = |F(x)|
F(|1|) = |F(1)|
1-|x| = |1-x|
1-|1| = |1-1|
0 = 0 (sufficient)
Case 2: x=-1
F(|x|) = |F(x)|
F(|-1|) = |F(-1)|
|x|-1 = |x-1|
|-1|-1 = |-1-1|
0 = 2 (Insufficient)
2) F(x) = F(-x)
Case 1: x=1
F(x) = F(-x)
F(1) = F(-1)
1-x = 1-(-1)
1-1 = 1+1
0 = 2 (Insufficient)
Case 2: x=-1
F(x) = F(-x)
F(-1) = F[-(-1)]
x-1 = x-1
-1-1 = +1-1
-2 = 0 (Insufficient)
3) |F(x+1)| = |x|
Case 1: x=1
|F(x+1)| = |x|
|F(1+1)| = |1|
|1-x| = 1
|1-2| = 1
1 = 1 (Sufficient)
Case 2: x=-1
|F(x+1)| = |x|
|F(-1+1)| = |-1|
|0-1| = |-1|
+1 = +1 (Sufficient)
So only option 3 is sufficient.
Answer is C
Posted from my mobile device
If x is - = F(x) = x-1
1) F(|x|) = |F(x)|
Case 1: x=1
F(|x|) = |F(x)|
F(|1|) = |F(1)|
1-|x| = |1-x|
1-|1| = |1-1|
0 = 0 (sufficient)
Case 2: x=-1
F(|x|) = |F(x)|
F(|-1|) = |F(-1)|
|x|-1 = |x-1|
|-1|-1 = |-1-1|
0 = 2 (Insufficient)
2) F(x) = F(-x)
Case 1: x=1
F(x) = F(-x)
F(1) = F(-1)
1-x = 1-(-1)
1-1 = 1+1
0 = 2 (Insufficient)
Case 2: x=-1
F(x) = F(-x)
F(-1) = F[-(-1)]
x-1 = x-1
-1-1 = +1-1
-2 = 0 (Insufficient)
3) |F(x+1)| = |x|
Case 1: x=1
|F(x+1)| = |x|
|F(1+1)| = |1|
|1-x| = 1
|1-2| = 1
1 = 1 (Sufficient)
Case 2: x=-1
|F(x+1)| = |x|
|F(-1+1)| = |-1|
|0-1| = |-1|
+1 = +1 (Sufficient)
So only option 3 is sufficient.
Answer is C
Posted from my mobile device
