Last visit was: 01 Jun 2026, 17:34 It is currently 01 Jun 2026, 17:34
Home
Messages
MBA Fair
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
avatar
CaptainLevi
avatar
Joined: 18 Jul 2019
Last visit: 26 Nov 2019
Posts: 34
Own Kudos:
503
 [28]
Given Kudos: 143
Products:
My Reviews
Posts: 34
Kudos: 503
 [28]
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 07:24
1
Kudos
Add Kudos
27
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

26% (02:07) correct 74% (01:52) wrong based on 262 sessions

History

Date
Time
Result
Not Attempted Yet
Given f(x) = \(a|x| − bx^2\), where a and b are constants. Then at x = 0, f (x) is

a) maximized whenever a > 0, b > 0.
b) minimized whenever a > 0, b > 0.
c) minimized whenever a < 0, b > 0.
d) maximized whenever a < 0, b > 0.
e) minimized whenever a < 0 , b < 0.
ShowHide Answer
Official Answer
D
avatar
ManjariMishra
avatar
Joined: 10 May 2018
Last visit: 11 Jul 2021
Posts: 59
Own Kudos:
41
 [2]
Given Kudos: 99
Posts: 59
Kudos: 41
 [2]
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 09:04
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel chetan2u Can you please explain this
avatar
Cheeku2023
avatar
Joined: 13 Oct 2019
Last visit: 17 Sep 2024
Posts: 9
Own Kudos:
4
Given Kudos: 5
Location: Pakistan
Concentration: Finance, Technology
GMAT 1: 620 Q44 V31
GPA: 3.53
GMAT 1: 620 Q44 V31
Posts: 9
Kudos: 4
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 09:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I am not clear on how to solve. If x=0, then a and b will also become 0. Do we substitute +ve and -ve values to figure out which way the function goes? That would require 6 combinations of a and b as +ve and -ve. Please help
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 01 Jun 2026
Posts: 11,241
Own Kudos:
45,302
 [3]
Given Kudos: 338
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,241
Kudos: 45,302
 [3]
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 09:49
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
CaptainLevi
Given f(x) = \(a|x| − bx^2\), where a and b are constants. Then at x = 0, f (x) is

a) maximized whenever a > 0, b > 0.
b) minimized whenever a > 0, b > 0.
c) minimized whenever a < 0, b > 0.
d) maximized whenever a < 0, b > 0.
e) minimized whenever a < 0 , b < 0.
Show more

This can be done in two ways..

(I) Parabola..
f(x) = \(a|x| − bx^2\)
Since we have a NEGATIVE coefficient of x^2, the parabola will open downwards and the value of y will be at the vertex, that is \(\frac{-a}{2b}\)
This will be maximum when a is negative and b is positive
D

(II) Determinants
Determinant of f(x) is -2bx+a=0.......x=-a/2b
Next determinant .. -2b, As this is negative, the value will be maximized at x=-a/2b
Again D as above

D

Not likely to be tested in GMAT
avatar
ManjariMishra
avatar
Joined: 10 May 2018
Last visit: 11 Jul 2021
Posts: 59
Own Kudos:
41
 [1]
Given Kudos: 99
Posts: 59
Kudos: 41
 [1]
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 25 Nov 2019, 11:52
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u Thanks for the explanation .

But I am unable to understand why the max value is at -a/2b

for ax^2 +bx +c when a<0 the max value will be at x=-b/2a

as per this f(x) = -bx^2 + ax

the max value will be at x= -a/2*-b
x= a/2b


Also for the second method
II) Determinants
Determinant of f(x) is -2bx+a=0.......x=-a/2b ---> Isn't it should be -2bx +a= 0

-2bx=-a
x=a/2b


Could you please guide where I am going wrong
User avatar
gvij2017
User avatar
Joined: 09 Aug 2017
Last visit: 18 Jun 2024
Posts: 663
Own Kudos:
510
 [1]
Given Kudos: 778
Posts: 663
Kudos: 510
 [1]
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 01 Dec 2019, 03:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I have few doubts here.
There are two cases at first.
b>0 parabola opens in downward direction. Means expression will have maximum value.
b<0 parabola opens in upward direction. Means expression will have minimum value.
From above two points I choose a, d, e,.

Now |x| = a/2b
If b<0, a<0 or if b>0, a>0
after this point, I am stuck.
Now need your help!



chetan2u
CaptainLevi
Given f(x) = \(a|x| − bx^2\), where a and b are constants. Then at x = 0, f (x) is

a) maximized whenever a > 0, b > 0.
b) minimized whenever a > 0, b > 0.
c) minimized whenever a < 0, b > 0.
d) maximized whenever a < 0, b > 0.
e) minimized whenever a < 0 , b < 0.

This can be done in two ways..

(I) Parabola..
f(x) = \(a|x| − bx^2\)
Since we have a NEGATIVE coefficient of x^2, the parabola will open downwards and the value of y will be at the vertex, that is \(\frac{-a}{2b}\)
This will be maximum when a is negative and b is positive
D

(II) Determinants
Determinant of f(x) is -2bx+a=0.......x=-a/2b
Next determinant .. -2b, As this is negative, the value will be maximized at x=-a/2b
Again D as above

D

Not likely to be tested in GMAT
Show more
User avatar
sambitspm
User avatar
Joined: 05 Aug 2019
Last visit: 13 Jan 2022
Posts: 317
Own Kudos:
314
 [1]
Given Kudos: 130
Location: India
Concentration: Leadership, Technology
GMAT 1: 600 Q50 V22
GMAT 2: 670 Q50 V28 (Online)
GPA: 4
GMAT 2: 670 Q50 V28 (Online)
Posts: 317
Kudos: 314
 [1]
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 07 Aug 2020, 05:59
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Kindly see the attachment.

I tried the 2 approaches and man they are difficult. This question involves a lot of thinking.
I preferred the Plotting method.
Attachments

File comment: Method 1. Graph and Plotting
Method 1.png
Method 1.png [ 592.84 KiB | Viewed 6357 times ]

File comment: Method 2. Determinants
Method 2.png
Method 2.png [ 624.81 KiB | Viewed 6351 times ]

User avatar
hadimadi
User avatar
Joined: 26 Oct 2021
Last visit: 03 Dec 2022
Posts: 113
Own Kudos:
31
Given Kudos: 94
Posts: 113
Kudos: 31
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 15 Dec 2021, 05:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi,

here is a simple solution to this:

Check the function for x>=0 and x<0. You can see directly that the graph is mirrored on the Y-Axis. So we can continue to work on this problem just by looking at x>=0. Now we know that f(0)=0.

To know if f(0) is a min / max, we have to know if there is any point x in the graph with f(x) smaller than or greater than 0.

If we take the first and second derivatives, we get f_1(x)=a-2bx and f_2=-2b.

Now, plug in the conditions from (A) to (E):

(A) It follows f_1(x)=lal-2lblx. For x=0, f_1(x)=lal>0, so f(x)=0 can't be a maximum.
(B) We have a downwards sloping parabola, so 0 can't be a minimum.
(E) It follows f_1(x)=-lal+2lblx. For x=0 we get a negative slope, so 0 can't be the minimum.
(C) and (D) It follows f_1(x)=-lal+2lblx. Plug in 0, and see again that we get a negative first derivative. So 0 can't be a minimum. This leaves us with only (D) as possible choice.

Please correct me if I'm wrong.

Thanks
User avatar
StacyArko
User avatar
Joined: 24 Jan 2020
Last visit: 01 Jun 2026
Posts: 24
Own Kudos:
7
Given Kudos: 156
Location: Ghana
GMAT 1: 600 Q39 V34
Products:
Forum Quiz
GMAT 1: 600 Q39 V34
Posts: 24
Kudos: 7
Given f(x) = a|x| − bx^2, where a and b are constants. Then at x = 0, 14 Jan 2022, 11:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u
CaptainLevi
Given f(x) = \(a|x| − bx^2\), where a and b are constants. Then at x = 0, f (x) is

a) maximized whenever a > 0, b > 0.
b) minimized whenever a > 0, b > 0.
c) minimized whenever a < 0, b > 0.
d) maximized whenever a < 0, b > 0.
e) minimized whenever a < 0 , b < 0.

This can be done in two ways..

(I) Parabola..
f(x) = \(a|x| − bx^2\)
Since we have a NEGATIVE coefficient of x^2, the parabola will open downwards and the value of y will be at the vertex, that is \(\frac{-a}{2b}\)
This will be maximum when a is negative and b is positive
D

(II) Determinants
Determinant of f(x) is -2bx+a=0.......x=-a/2b
Next determinant .. -2b, As this is negative, the value will be maximized at x=-a/2b
Again D as above

D

Not likely to be tested in GMAT
Show more

chetan2u, thank you very much for your explanation.

A quick question please. Could you kindly highlight why we chose D over A. This is my confusion: given the coefficient of x^2 is negative, I thought the vertex is at a/-2b, which equals - a/2b, which holds when a and b are both positive . Why then do we choose the case where a<0 and b>0?

Thanks in advance.

Posted from my mobile device
Moderator:
Bunuel
Math Expert
111019 posts