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
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..
\(|2x+3|^2≤12^2......4x^2+9+12x\leq{144}......4x^2+12x-135\leq{0}.....(4x^2+30x-18x-135)=2x(2x+15)-9(2x+15)=(2x-9)(2x+15)\leq{0}\)
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