|
|
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.
25 Mar 2020, 07:05
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
69% (02:45) correct
31%
(02:40)
wrong
based on 603
sessions
History
Date
Time
Result
Not Attempted Yet
a is the maximum integer and b is the minimum integer satisfying \(\sqrt{(x-1)^2}+\sqrt{(x+2)^2} < 5\) What is the value of \(a^2+b^2\)?
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
A. \(2\)
B. \(3\)
C. \(4\)
D. \(5\)
E. \(6\)
27 Mar 2020, 02:06
Kudos
Bookmarks
=>
\(\sqrt{(x-1)^2}+\sqrt{(x+2)^2} = |x - 1| + |x - 2|.\)
Case 1: \(x ≥ 1\)
\(|x - 1| + |x + 2| < 5\)
=>\( (x - 1) + (x + 2) < 5,\) since \(x – 1 ≥ 0\) and \(x + 2 ≥ 0\)
=> \(2x - 1 < 5\)
=> \(2x < 4\)
=> \(x < 2\)
Thus, we have \(1 ≤ x < 2.\)
Case 2: \(-2 ≤ x < 1\)
\(|x - 1| + |x + 2| < 5\)
=> \(-(x - 1) + (x + 2) < 5\), since \(x – 1 < 0\) and \(x + 2 ≥ 0\)
=> \(-x + 1 + x + 2 < 5\)
=> \(3 < 5\), which is always valid.
Thus, we have \(-2 ≤ x < 1.\)
Case 3: \(x < -2\)
\(|x - 1| + |x + 2| < 5\)
=> \(-(x - 1) - (x + 2) < 5\), since \(x – 1 < 0\) and \(x + 2 < 0\)
=> \(-2x - 1 < 5\)
=> \(-2x < -6\)
=> \(x > -3\)
Thus, we have \(-3 < x < -2.\)
When we combine those solutions, we have \(-3 < x < 2.\)
However, \(x\) is an integer. \(a\), the maximum possible integer of \(x\) is \(1\), and \(b\), the minimum possible integer is \(-2.\)
Then \(a^2 + b^2 = 1^2 + (-2)^2 = 1 + 4 = 5.\)
Therefore, the answer is D.
Answer: D
\(\sqrt{(x-1)^2}+\sqrt{(x+2)^2} = |x - 1| + |x - 2|.\)
Case 1: \(x ≥ 1\)
\(|x - 1| + |x + 2| < 5\)
=>\( (x - 1) + (x + 2) < 5,\) since \(x – 1 ≥ 0\) and \(x + 2 ≥ 0\)
=> \(2x - 1 < 5\)
=> \(2x < 4\)
=> \(x < 2\)
Thus, we have \(1 ≤ x < 2.\)
Case 2: \(-2 ≤ x < 1\)
\(|x - 1| + |x + 2| < 5\)
=> \(-(x - 1) + (x + 2) < 5\), since \(x – 1 < 0\) and \(x + 2 ≥ 0\)
=> \(-x + 1 + x + 2 < 5\)
=> \(3 < 5\), which is always valid.
Thus, we have \(-2 ≤ x < 1.\)
Case 3: \(x < -2\)
\(|x - 1| + |x + 2| < 5\)
=> \(-(x - 1) - (x + 2) < 5\), since \(x – 1 < 0\) and \(x + 2 < 0\)
=> \(-2x - 1 < 5\)
=> \(-2x < -6\)
=> \(x > -3\)
Thus, we have \(-3 < x < -2.\)
When we combine those solutions, we have \(-3 < x < 2.\)
However, \(x\) is an integer. \(a\), the maximum possible integer of \(x\) is \(1\), and \(b\), the minimum possible integer is \(-2.\)
Then \(a^2 + b^2 = 1^2 + (-2)^2 = 1 + 4 = 5.\)
Therefore, the answer is D.
Answer: D
General Discussion
ArunSharma12
Joined: 25 Oct 2015
Last visit: 20 Jul 2022
Posts: 512
Given Kudos: 74
Location: India
Schools: IIML-IPMX '22 (A)
GMAT 1: 650 Q48 V31
GMAT 2: 720 Q49 V38 (Online)
GPA: 4
Products:
25 Mar 2020, 10:39
Kudos
Bookmarks
MathRevolution
First I tried to do it the harder way, which is squaring both sides and solving for x. It is time consuming and does not guarantees that one would get the correct values for a and b.
In my second attempt, looking at options we can see that a,b can not be any integer \(\ge{3}\).
that means, a and b will fit this equation \(-2\le b<a\le{2}\)
checking for maximum value, for a =2
\(\sqrt{(2-1)^2}+\sqrt{(2+2)^2} = 5\); this does not satisfy the condition, so a =1.
checking for minimum value, for b = -2
\(\sqrt{(-2-1)^2}+\sqrt{(-2+2)^2} = 3\); this satisfies the condition, so b =-2.
\(a^2+b^2 = (1)^2+(-2)^2\) = 5
Ans: D
