Re: If y=|x+1|/x and x!=0, is xy>0?

@Zarrolou,
Regret for not responding within time.


If y=\frac{|x+1|}{x} and x\neq{}0, is xy>0 ?

A)x^2+2x+1>0

B)y\neq{0}


First lets simplify the inequality

\frac{|x+1|}{x} –y = 0 ----------> \frac{|x+1|-xy}{x} = 0 -----> we know x\neq{}0, then |x+1|-xy must be zero. Hence |x+1| - xy = 0 --------> |x+1| = xy


We are asked whether xy>0 --------> Whether |x+1| > 0 ? --------> We know the expression within modules can either be zero or greater than zero. For xy to be greater than zero |x+1| has to be greater than zero.
|x+1| will be zero only when x\neq{-1} and for any other value of x, |x+1| will always be greater than zero
So the question can be rephrased as whether x\neq{-1}


Statement 1) x^2 + 2x + 1 > 0

This is an quadratic inequality.
Rule :- For any quadratic inequation ax^2 + bx + c > 0, if b^2 – 4ac = 0 and a > 0 then the inequality holds true outside the interval of roots
In our case b^2 – 4ac = 4 – 4 = 0 and a > 0 so x^2 + 2X + 1 > 0 will hold true for all values beyond the Root(s) of equation (Towards any direction - Positive or Negative)
x^2 + 2x + 1 = 0 --------> x(x+1) +1(x + 1) = 0 ---------> (x+1)(x+1) = 0 ----------> x=Root = -1
So x^2 + 2x + 1 > 0 will hold true for any of x except for -1
This reveals that x\neq{-1} and xy>0 ----------------> Sufficient


Statement 2) y\neq{0}

From the question stem we know x\neq{}0
As per Statement 2, y\neq{0}-------------->That means both X and Y are not zero.

|x+1| = xy
xy can be either Positive or Negative
|x+1| can be Zero or Positive
Combining both these inferences we can conclude that XY must be Positive. Sufficient

Answer = D

Regards,

Narenn