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

It is currently 16 Dec 2018, 22:02

Close

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.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
  • 10 Keys to nail DS and CR questions

     December 17, 2018

     December 17, 2018

     06:00 PM PST

     07:00 PM PST

    Join our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong.
  • R1 Admission Decisions: Estimated Decision Timelines and Chat Links for Major BSchools

     December 17, 2018

     December 17, 2018

     10:00 PM PST

     11:00 PM PST

    From Dec 5th onward, American programs will start releasing R1 decisions. Chat Rooms: We have also assigned chat rooms for every school so that applicants can stay in touch and exchange information/update during decision period.

For which of the following functions is f(−1/2) > f(2)?

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

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 51230
For which of the following functions is f(−1/2) > f(2)?  [#permalink]

Show Tags

New post 22 Aug 2018, 01:04
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

89% (00:59) correct 11% (01:41) wrong based on 37 sessions

HideShow timer Statistics

Director
Director
User avatar
P
Status: Learning stage
Joined: 01 Oct 2017
Posts: 931
WE: Supply Chain Management (Energy and Utilities)
Premium Member
Re: For which of the following functions is f(−1/2) > f(2)?  [#permalink]

Show Tags

New post 22 Aug 2018, 01:12
Bunuel wrote:
For which of the following functions is f(−1/2) > f(2)?


A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2


Since x^2 is in the denominator in option E, it may yield a greater value of f(-1/2).

f(-1/2)=\(\frac{3}{(-1/2)^2}\)=\(3*2^2=12\)
f(2)=\(\frac{3}{2^2}\)=3/4

Ans. (E)
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

PS Forum Moderator
avatar
D
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82
GMAT ToolKit User Premium Member Reviews Badge
For which of the following functions is f(−1/2) > f(2)?  [#permalink]

Show Tags

New post 22 Aug 2018, 01:13
Bunuel wrote:
For which of the following functions is f(−1/2) > f(2)?


A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2


Move through the options -

A. \(f(x) = 3x^2\) -------------this will lead to positive value but here \(2^2>(-\frac{1}{2})^2\). Reject

B. \(f(x)= 3x\) ------------ this will make \(f(-\frac{1}{2})\) negative and \(f(2)\) positive. Reject

C. \(f(x)= 3 + x^2\) ----------- Same reasoning as Option A. Reject

D. \(f(x)= 3 + \frac{1}{x}\) ------------ Same reasoning as Option B. \(\frac{1}{x}<x\), hence the value for \(f(-\frac{1}{2})<f(2)\). Reject

E. \(f(x)= \frac{3}{x^2}\) ----------- This should be our Answer, we can check it by solving

\(f(x)= \frac{3}{x^2} => f(-\frac{1}{2})=\frac{3}{(-1/2)^2}=>3*4=12\)

and \(f(2)=\frac{3}{2^2}=\frac{3}{4}=0.75\). Clearly \(f(-\frac{1}{2})>f(2)\)

Option E
Senior Manager
Senior Manager
User avatar
D
Joined: 22 Feb 2018
Posts: 413
CAT Tests
For which of the following functions is f(−1/2) > f(2)?  [#permalink]

Show Tags

New post Updated on: 22 Aug 2018, 01:30
Bunuel wrote:
For which of the following functions is f(−1/2) > f(2)?


A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2


OA:E
\(\begin{matrix}
& f(-\frac{{1}}{2}) & & f(2) \\
f(x)= 3x^2 & \frac{3}{4} & {<} & 12 \\
f(x)= 3x & -\frac{3}{2} & {<} & 6 \\
f(x)= 3+x^2 & \frac{13}{4} & {<} & 7 \\
f(x)= 3 + 1/x & 1 & {<} & \frac{7}{2} \\
f(x)= 3/x^2 & 12 & {>} & \frac{3}{4}
\end{matrix}\)
_________________

Good, good Let the kudos flow through you


Originally posted by Princ on 22 Aug 2018, 01:25.
Last edited by Princ on 22 Aug 2018, 01:30, edited 1 time in total.
Senior PS Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 3327
Location: India
GPA: 3.12
Premium Member CAT Tests
For which of the following functions is f(−1/2) > f(2)?  [#permalink]

Show Tags

New post 22 Aug 2018, 01:29
1
Bunuel wrote:
For which of the following functions is f(−1/2) > f(2)?

A. f(x) = 3x^2

B. f(x)= 3x

C. f(x)= 3 + x^2

D. f(x)= 3 + 1/x

E. f(x)= 3/x^2


Of the 5 answer options, you can straight away eliminate Options A, B, and C because all of these
options have an x or x^2 in the numerator. Substituting -\(\frac{1}{2}\) can never be greater than substituting 2.

Now that we are down to two answer options,
we can substitute the values to check in which case, value of f(-\(\frac{1}{2}\)) > value of f(\(2\))

In Option D, f(-\(\frac{1}{2}\)) = 3 + (-2) = 1 (because \(\frac{1}{\frac{-1}{2}}\) = -2) but f(\(2\)) = \(3 + \frac{1}{2} = \frac{7}{2}\). Here, f(-\(\frac{1}{2}\)) < f(\(2\))

Therefore, the only answer option we are left with is f(x) = \(\frac{3}{x^2}\) (Option E) which is our answer!
_________________

You've got what it takes, but it will take everything you've got

GMAT Club Bot
For which of the following functions is f(−1/2) > f(2)? &nbs [#permalink] 22 Aug 2018, 01:29
Display posts from previous: Sort by

For which of the following functions is f(−1/2) > f(2)?

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.