Bunuel wrote:
Official Solution:
First of all: is \(|x| \gt 1\) means is \(x \lt -1\) \((-2, -3, -4, ...)\) or is \(x \gt 1\) \((2, 3, 4, ...)\), so for YES answer \(x\) can be any integer but -1, 0, and 1.
(1)\((1 - 2x)(1 + x) \lt 0\). Rewrite as \((2x - 1)(x + 1) \gt 0\) (so that the coefficient of \(x^2\) is positive after expanding): roots are \(x=-1\) and \(x=\frac{1}{2}\). \(\gt\)" sign means that the given inequality holds true for: \(x \lt -1\)and \(x \gt \frac{1}{2}\). \(x\) could still equal 1, so not sufficient.
(2) \((1 - x)(1 + 2x) \lt 0\). Rewrite as \((x - 1)(2x + 1) \gt 0\): roots are \(x=-\frac{1}{2}\) and \(x=1\). "\(\gt\)" sign means that the given inequality holds true for: \(x \lt - \frac{1}{2}\) and \(x \gt 1\). \(x\) could still equal -1, so not sufficient.
(1)+(2) Intersection of the ranges from (1) and (2) is \(x \lt -1\) and \(x \gt 1\). Sufficient.
Answer: C
Hi Experts! I'm confused:
In statement (1) when you get X<-1 and X>1/2, doesn't this show that |X| is not >1?
When you plot the quadratic equation (2X-1)(X+1)>0 ---> 2X^2+X-1 >0, you get the range X<-1 and X>1/2, thus X obviously has a value at 1 -- meaning that X is not greater than 1. Wouldn't this make (1) sufficient?
Similarly in statement (2), the range is X<-1/2 and X>1, thus again X obviously has a value at -1 -- meaning that X is not less than -1. Again wouldn't this make (2) sufficient?
Thank you!