vdadwal wrote:
If x is an integer, is |x|>1.
(1) (1-2x)(1+x) < 0
(2) (1-x)(1+2x) < 0
Can somebody please explain this question?
Thanks
Vikram
This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach":
x2-4x-94661.html#p731476If x is an integer, is |x| > 1?First of all: is \(|x| > 1\) means is \(x<-1\) (-2, -3, -4, ...) or \(x>1\) (2, 3, 4, ...), so for YES answer \(x\) can be any integer but -1, 0, and 1.
(1) (1 - 2x)(1 + x) < 0 --> rewrite as \((2x-1)(x+1)>0\) (so that the coefficient of x^2 to be positive after expanding): roots are \(x=-1\) and \(x=\frac{1}{2}\) --> "\(>\)" sign means that the given inequality holds true for: \(x<-1\) and \(x>\frac{1}{2}\). \(x\) could still equal to 1, so not sufficient.
(2) (1 - x)(1 + 2x) < 0 --> rewrite as \((x-1)(2x+1)>0\): roots are \(x=-\frac{1}{2}\) and \(x=1\) --> "\(>\)" sign means that the given inequality holds true for: \(x<-\frac{1}{2}\) and \(x>1\). \(x\) could still equal to -1, so not sufficient.
(1)+(2) Intersection of the ranges from (1) and (2) is \(x<-1\) and \(x>1\). Sufficient.
Answer: C.
This question is also discussed here:
m14-72785.htmlSolving inequalities:
x2-4x-94661.html#p731476 (
check this one first)
inequalities-trick-91482.htmldata-suff-inequalities-109078.htmlrange-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535everything-is-less-than-zero-108884.html?hilit=extreme#p868863Hope it helps.
I have a question regarding the above solution let's say in statement 1, when you solve the inequality why do you say that x<-1 AND x >1/2
why is this an AND condition ....why not OR? If this were a quadratic equation x (can be) = 1/2 OR -1 OR both