GRE Prep 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.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If (x + 2)2 ≤ 2 − y, what is the maximum possible value for y?
[#permalink]
02 Feb 2021, 03:09
02 Feb 2021, 03:09
1
1
Expert Reply
6
Bookmarks
00:00
Question Stats:
64% (01:58) correct
35% (02:09) wrong based on 39 sessions
Hide Show timer Statistics
If \((x + 2)^2 ≤ 2 − y\), what is the maximum possible value for \(y\)?
A. -4
B. -2
C. 0
D. 2
E. 4
A. -4
B. -2
C. 0
D. 2
E. 4
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1111
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
If (x + 2)2 2 y, what is the maximum possible value for y?
[#permalink]
03 Feb 2021, 09:38
03 Feb 2021, 09:38
2
\((x + 2)^2 ≤ 2 − y\)
Let's simplify this equation and take y to LHS and everything else to RHS
\(y ≤ 2 - (x + 2)^2 \)
Now, y ≤ 2 - square of a number.
And we know that minimum value of square of a number is 0 as square of a number can never be negative.
So, y ≤ 2 - some non-negative number
=> For maximum y, (x + 2)^2 has to be zero
=> Maximum value of y is 2.
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Inequalities
Let's simplify this equation and take y to LHS and everything else to RHS
\(y ≤ 2 - (x + 2)^2 \)
Now, y ≤ 2 - square of a number.
And we know that minimum value of square of a number is 0 as square of a number can never be negative.
So, y ≤ 2 - some non-negative number
=> For maximum y, (x + 2)^2 has to be zero
=> Maximum value of y is 2.
So, Answer will be D
Hope it helps!
Watch the following video to MASTER Inequalities
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: If (x + 2)2 ≤ 2 − y, what is the maximum possible value for y?
[#permalink]
26 Sep 2021, 06:46
26 Sep 2021, 06:46
1
\(y ≤ 2 - (x + 2)^2\)
If we consider that inequality takes the equal sign and \(y=0\), then \((x + 2)^2 - 2 = 0\) and \(x^2 + 4x +2 = 0\)
There are two roots and two x-intercepts. The middle between two symmetrical intercepts falls within x for the min/max value of the equation
When \(x=-2\), \(y ≤ 2 - (-2 + 2)^2\) and \(y ≤ 2\)
Out of all answer choices only \(D\) is correct
If we consider that inequality takes the equal sign and \(y=0\), then \((x + 2)^2 - 2 = 0\) and \(x^2 + 4x +2 = 0\)
There are two roots and two x-intercepts. The middle between two symmetrical intercepts falls within x for the min/max value of the equation
When \(x=-2\), \(y ≤ 2 - (-2 + 2)^2\) and \(y ≤ 2\)
Out of all answer choices only \(D\) is correct
GeminiHeat wrote:
If \((x + 2)^2 ≤ 2 − y\), what is the maximum possible value for \(y\)?
A. -4
B. -2
C. 0
D. 2
E. 4
A. -4
B. -2
C. 0
D. 2
E. 4
