tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:
We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5
how does this become x>-1??
Is \(|x| < 1\)?
Is \(|x| < 1\), means is \(x\) in the range (-1,1) or is \(-1<x<1\) true?
(1) \(|x + 1| = 2|x - 1|\)
Two key points: \(x=-1\) and \(x=1\) (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------A. \(x<-1\) (blue range) --> \(|x + 1| = 2|x - 1|\) becomes: \(-x-1=2(-x+1)\) --> \(x=3\), not OK, as this value is not in the range we are checking (\(x<-1\));
B. \(-1\leq{x}\leq{1}\) (green range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(-x+1)\) --> \(x=\frac{1}{3}\). OK, as this value is in the range we are checking (\(-1\leq{x}\leq{1}\));
C. \(x>1\) (red range) --> \(|x + 1| = 2|x - 1|\) becomes: \(x+1=2(x-1)\) --> \(x=3\). OK, as this value is in the range we are checking (\(x>1\)).
So we got TWO values of \(x\) (two solutions): \(\frac{1}{3}\) and \(3\), first is in the range (-1,1) but second is out of the range. Not sufficient.
(2) \(|x - 3|\neq{0}\)
Just says that \(x\neq{3}\). But we don't know whether \(x\) is in the range (-1,1) or not.
(1)+(2) \(x=\frac{1}{3}\) or \(x=3\) AND \(x\neq{3}\) --> means \(x\) can have only value \(\frac{1}{3}\), which is in the range (-1,1). Sufficient.
Answer: C.
Hope it helps.