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If x is an integer and |2x+3|≤12
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18 Oct 2018, 10:46
18 Oct 2018, 10:46
2
1
3
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00:00
A
B
C
D
E
Question Stats:
43% (01:40) correct
56% (01:35) wrong based on 44 sessions
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If x is an integer and |2x+3|≤12 , then which of the following must be true?
x< -9
x< -8
x> -8
x< 4
x> 4
x< -9
x< -8
x> -8
x< 4
x> 4
Re: If x is an integer and |2x+3|≤12
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18 Oct 2018, 17:39
18 Oct 2018, 17:39
2
Expert Reply
Sawant91 wrote:
If x is an integer and |2x+3|≤12 , then which of the following must be true?
x< -9
x< -8
x> -8
x< 4
x> 4
x< -9
x< -8
x> -8
x< 4
x> 4
Hi
One way is to just substitute few values...
1) put x as 0..
|2x+3|≤12.......|2∗0+3|≤12...yes
So 0 is one value and is in the region of choices C and D
2) put x as 4
4*2+3≤12.....11≤12....yes
So 4 is also a value but is not in D, do eliminate D
Ans is C
Second way is to open the modulus..
A) square both sides..
|2x+3|2≤122......4x2+9+12x≤144......4x2+12x−135≤0.....(4x2+30x−18x−135)=2x(2x+15)−9(2x+15)=(2x−9)(2x+15)≤0
Two case
x<9/2 and x>-15/2 or X<5 and X>-8
Or X>9/2 and X<-15/2... No overlap, so not possible
So C
B) open the point through critical point
|2x+3|≤12
(i) 2x+3≤12.....2x≤9....x≤9/2 or x≤4 that is x<5
(ii) -(2x+3)≤12.....-2x-3≤12.....-2x≤15......-15/2≤x...x>-8
C
You can also check the following post on absolute modulus...
https://gre.myprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html
