jayoptimist wrote:
Is x positive?
(1) 1/(x+1) < 1
(2) (x-1) is a perfect square.
What is wrong with the following line of thought for condition 1
1/(x+1) < 1
1 < (x+1)
1 -1 < x+1-1
0 < x Hence x>0.
We can not multiple \(\frac{1}{x+1}< 1\) by \(x+1\) since we don't know whether this expression (\(x+1\)) is positive or negative, for the first case we should keep the same sign but for the second case we should flip the sign (when multiplying by negative number we should flip the sign of the inequity).
Is x positive?(1) 1/(x+1) < 1. If \(x=10\) then the answer is YES, but if \(x=-10\) then the answer is NO. Not sufficient.
Or if you want to solve this inequality then: \(\frac{1}{x+1}< 1\) --> \(1-\frac{1}{x+1}>0\) --> \(\frac{x}{x+1}>0\) --> \(x>0\) or \(x<-1\), hence \(x\) could be positive as well as negative. Not sufficient.
(2) (x-1) is a perfect square. Given: \(x-1=\{perfect \ square\}\) --> \(x=\{perfect \ square\}+1\). Now, since perfect square is more than or equal to zero, then \(x=\{perfect \ square\}+1=non-negative+1=positive\). Sufficient.
Answer: B.
Hope it's clear.
"\frac{x}{x+1}>0 --> x>0 or x<-1, hence x could be positive as well as negative. Not sufficient."