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If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?
(1) \(x^2+2x+1>0\)
(2) \(y\neq{0}\)
My own question, as always any feedback is appreciated
Click here for the OE
(1) \(x^2+2x+1>0\)
(2) \(y\neq{0}\)
My own question, as always any feedback is appreciated
Click here for the OE
Updated on: 14 Aug 2022, 11:00
Originally posted by BrushMyQuant on 16 May 2013, 11:47.
Last edited by BrushMyQuant on 14 Aug 2022, 11:00, edited 3 times in total.
Last edited by BrushMyQuant on 14 Aug 2022, 11:00, edited 3 times in total.
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Question can be written as |x+1| = xy
so, to prove that xy> 0 we essentially need to prove that |x|1| > 0
STAT1
x^2 + 2x + 1 > 0
means,
(x+1) ^ 2 >0
taking square root on both the sides does not change the inequality
so, we have
|x+1| > 0
which means that xy > 0
So, SUFFICIENT
STAT2
y != 0
we know that
|x+1| = xy => xy is non negative. so, it is either 0 or positive
since we know that both x and y are not equal to zero so, xy > 0
So, SUFFICIENT
So, Answer will be D
Hope it helps!
Watch the following video to learn Basics of Inequalities
Check out these posts to understand Absolute Values and Inequalities in Detail
How to Solve: Inequalities (Basic)
How to Solve: Inequality Problems - Algebra and Sine Wave/Wavy Method
How to Solve: Absolute Value (Basics)
How to Solve: Absolute Value Problems
How to Solve: Absolute Value + Inequality Problems
so, to prove that xy> 0 we essentially need to prove that |x|1| > 0
STAT1
x^2 + 2x + 1 > 0
means,
(x+1) ^ 2 >0
taking square root on both the sides does not change the inequality
so, we have
|x+1| > 0
which means that xy > 0
So, SUFFICIENT
STAT2
y != 0
we know that
|x+1| = xy => xy is non negative. so, it is either 0 or positive
since we know that both x and y are not equal to zero so, xy > 0
So, SUFFICIENT
So, Answer will be D
Hope it helps!
Watch the following video to learn Basics of Inequalities
Check out these posts to understand Absolute Values and Inequalities in Detail
How to Solve: Inequalities (Basic)
How to Solve: Inequality Problems - Algebra and Sine Wave/Wavy Method
How to Solve: Absolute Value (Basics)
How to Solve: Absolute Value Problems
How to Solve: Absolute Value + Inequality Problems
16 May 2013, 15:41
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Bookmarks
@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=-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 nonzero.
|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
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=-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 nonzero.
|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
Wow! This question is crazier than I anticipated. 
