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question: x id non zero, is \(| x | < 1\) ??? now to be < 1 x must be negative, so the question become " is x negative ???

1) for a square to be < 1 the must be for example \(\frac{- 1}{2}\) ----> x is negative . suff

2) for x to be minor of a certain number that is the reciprocal i.e. : \(-2 < - 1/2\) --------> x is negative. suff

is correct or I should rethink the entire part regarding inequalities ?? because this statement should not take you more that 50 seconds to solve. otherwise you get in trouble with this exam.

Re: If x is not equal to zero, is | x | < 1 ?? [#permalink]

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30 Sep 2012, 05:34

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carcass wrote:

If x is not equal to zero, is \(| x | < 1\) ??

1) \(x^2\) \(< 1\)

2) \(| x |\) < \(\frac{1}{x}\)

I would like to know if I do in the right manner

question: x id non zero, is \(| x | < 1\) ??? now to be < 1 x must be negative, so the question become " is x negative ???

1) for a square to be < 1 the must be for example \(\frac{- 1}{2}\) ----> x is negative . suff

2) for x to be minor of a certain number that is the reciprocal i.e. : \(-2 < - 1/2\) --------> x is negative. suff

is correct or I should rethink the entire part regarding inequalities ?? because this statement should not take you more that 50 seconds to solve. otherwise you get in trouble with this exam.

Thanks

The condition \(x\) non-zero was given just because statement (2) has \(x\) in the denominator.

x id non zero, is \(| x | < 1\) ??? Not necessarily. \(x\) can be greater than \(1\) or less than \(-1\), in which case, definitely \(|x|\) is not less than \(1.\)

now to be < 1 x must be negative, so the question become " is x negative ??? NO \(x\) can be between \(0\) and \(1\).

(1) \(x^2\) \(< 1\), take square root from both sides and obtain \(|x|<1\), because \(\sqrt{x^2}=|x|\). Sufficient.

(2) Since \(|x|<\frac{1}{x}\), it follows that \(x>0\), because \(|x|>0\). \(x\) cannot be negative!!! Then, the given inequality becomes \(x<\frac{1}{x}\), or \(x^2<1\) (we can multiply both sides by \(x\), which is positive) and we are again in the situation from (1) above. Sufficient.

Answer D.
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PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If x is not equal to zero, is | x | < 1 ?? [#permalink]

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30 Sep 2012, 06:58

Perfect Explanation, Evajager, @Carcass this is a very common mistake which I am also prone to make. Thanks @Evajager - Your explanation for {B}

Quote:

(2) Since , it follows that , because . cannot be negative!!! Then, the given inequality becomes , or (we can multiply both sides by , which is positive) and we are again in the situation from (1) above. Sufficient.

I remember skipping a Question just like this in one of my exams because I gt stuck!!! in solving - | x | < 1/x

Thanks guys you rock!!!
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Re: If x is not equal to zero, is | x | < 1 ?? [#permalink]

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30 Sep 2012, 12:29

EvaJager wrote:

carcass wrote:

If x is not equal to zero, is \(| x | < 1\) ??

1) \(x^2\) \(< 1\)

2) \(| x |\) < \(\frac{1}{x}\)

I would like to know if I do in the right manner

question: x id non zero, is \(| x | < 1\) ??? now to be < 1 x must be negative, so the question become " is x negative ???

1) for a square to be < 1 the must be for example \(\frac{- 1}{2}\) ----> x is negative . suff

2) for x to be minor of a certain number that is the reciprocal i.e. : \(-2 < - 1/2\) --------> x is negative. suff

is correct or I should rethink the entire part regarding inequalities ?? because this statement should not take you more that 50 seconds to solve. otherwise you get in trouble with this exam.

Thanks

The condition \(x\) non-zero was given just because statement (2) has \(x\) in the denominator.

x id non zero, is \(| x | < 1\) ??? Not necessarily. \(x\) can be greater than \(1\) or less than \(-1\), in which case, definitely \(|x|\) is not less than \(1.\)

now to be < 1 x must be negative, so the question become " is x negative ??? NO \(x\) can be between \(0\) and \(1\).

(1) \(x^2\) \(< 1\), take square root from both sides and obtain \(|x|<1\), because \(\sqrt{x^2}=|x|\). Sufficient.

(2) Since \(|x|<\frac{1}{x}\), it follows that \(x>0\), because \(|x|>0\). \(x\) cannot be negative!!! Then, the given inequality becomes \(x<\frac{1}{x}\), or \(x^2<1\) (we can multiply both sides by \(x\), which is positive) and we are again in the situation from (1) above. Sufficient.

Answer D.

Correction: in (2) and we are again in the situation from (1) above. - not exactly, but similar as we have \(x^2<1\) but in addition \(x>0.\) In (1) \(x\) could also be negative. The conclusion is still correct, as now \(|x|=x<1.\)
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

question: x id non zero, is \(| x | < 1\) ??? now to be < 1 x must be negative, so the question become " is x negative ???

1) for a square to be < 1 the must be for example \(\frac{- 1}{2}\) ----> x is negative . suff

2) for x to be minor of a certain number that is the reciprocal i.e. : \(-2 < - 1/2\) --------> x is negative. suff

is correct or I should rethink the entire part regarding inequalities ?? because this statement should not take you more that 50 seconds to solve. otherwise you get in trouble with this exam.

Thanks

I would suggest you to keep 4 ranges in mind when dealing with squares, reciprocals etc.

You can do the question logically or by plugging in numbers.

Ques: If x is not equal to zero, is \(| x | < 1\) ?? Logically, | x | implies distance from 0 on the number line. So the question becomes "Is x a distance of less than 1 away from 0?" i.e. "Is -1 < x < 1?" given x is not 0.

Statement 1: \(x^2\) \(< 1\) Square of a number will be less than 1 only if the absolute value of the number is less than 1. This means \(| x | < 1\). The number needn't be negative. If it is positive, it should be less than 1. If it negative, it should be greater than -1. Or, do you remember how to solve inequalities using the wave? \(x^2 < 1\) is \(x^2 - 1 < 0\) which is \((x - 1)(x + 1) < 0\). This implies -1 < x < 1 but x is not 0. Or plug in numbers from the given 4 ranges. YOu will see that x must lie in -1 < x < 1. Sufficient.

Statement 2: \(| x | < \frac{1}{x}\) First of all, x cannot be negative since | x | is never negative. Since | x | is less than 1/x, 1/x must be positive. Also, x cannot be greater than 1 since then, | x | will be greater than 1/x. Or plug in numbers from the given 4 ranges. You will see that x must lie in 0 < x < 1. Sufficient.

Re: If x is not equal to zero, is | x | < 1 ?? [#permalink]

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01 Oct 2012, 03:24

VeritasPrepKarishma wrote:

carcass wrote:

If x is not equal to zero, is \(| x | < 1\) ??

1) \(x^2\) \(< 1\)

2) \(| x |\) < \(\frac{1}{x}\)

I would like to know if I do in the right manner

question: x id non zero, is \(| x | < 1\) ??? now to be < 1 x must be negative, so the question become " is x negative ???

1) for a square to be < 1 the must be for example \(\frac{- 1}{2}\) ----> x is negative . suff

2) for x to be minor of a certain number that is the reciprocal i.e. : \(-2 < - 1/2\) --------> x is negative. suff

is correct or I should rethink the entire part regarding inequalities ?? because this statement should not take you more that 50 seconds to solve. otherwise you get in trouble with this exam.

Thanks

I would suggest you to keep 4 ranges in mind when dealing with squares, reciprocals etc.

You can do the question logically or by plugging in numbers.

Ques: If x is not equal to zero, is \(| x | < 1\) ?? Logically, | x | implies distance from 0 on the number line. So the question becomes "Is x a distance of less than 1 away from 0?" i.e. "Is -1 < x < 1?" given x is not 0.

Statement 1: \(x^2\) \(< 1\) Square of a number will be less than 1 only if the absolute value of the number is less than 1. This means \(| x | < 1\). The number needn't be negative. If it is positive, it should be less than 1. If it negative, it should be greater than -1. Or, do you remember how to solve inequalities using the wave? \(x^2 < 1\) is \(x^2 - 1 < 0\) which is \((x - 1)(x + 1) < 0\). This implies -1 < x < 1 but x is not 0. Or plug in numbers from the given 4 ranges. YOu will see that x must lie in -1 < x < 1. Sufficient.

Statement 2: \(| x | < \frac{1}{x}\) First of all, x cannot be negative since | x | is never negative. Since | x | is less than 1/x, 1/x must be positive. Also, x cannot be greater than 1 since then, | x | will be greater than 1/x. Or plug in numbers from the given 4 ranges. You will see that x must lie in 0 < x < 1. Sufficient.

Answer (D)

This is the same reasoning that I followed in the first instance with some errors but the path was correct. Specifically the stimulus evaluation : \(| x | < 1\) ------ in this scenario the first one is \(x < 1\) AND \(-x < 1\) so \(x > -1\) ------> \(- 1 < x < 1\)

(2) |x| < 1/x. Since the left hand side of the inequity (|x|) is an absolute value, which cannot be negative (actually in this case we know that its positive as we are given that \(x\neq{0}\)) then the right hand side (1/x) must also be positive, which means that \(x>0\), so \(|x|=x\).

Hence, \(|x| < \frac{1}{x}\) becomes \(x<\frac{1}{x}\). Since we know that \(x>0\) we can safely cross-multiply: \(x^2<1\). The same inequality as in (1). Sufficient.

Re: If x is not equal to zero, is |x| < 1 ? [#permalink]

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If x is not equal to zero, is |x| < 1 ? [#permalink]

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20 Nov 2016, 14:48

The general rule of thumb with DS + ineq questions is if you can avoid plug numbers DO IT!

If x is not equal to zero, is |x| < 1 ?

Unraveling the question we have case1 where x < 1 OR case 2 where x > -1

REMEMBER that with ineq. DS questions A suffiecient asnswer is the answer that verifies case1 OR case2 OR BOTH

(1) x^2 < 1 "ROOTING" both sides (i.e. raised both sides in 1/2 power) we have sqrt(x^2) = sqrt(1) = 1 hence |x| < 1 which is SUFFICIENT as verifies case1 and case2 (effectivly if you solve it you get exactly the 2 cases from the question stem). Is important to know that "ROOTING" is allowed here because BOTH sides of the enequality are NOT NEGATIVE.

(2) |x| < 1/x Solving the ineq. above gives two cases:

caseA: x < 1 / x and caseB x > - 1/x

CaseA: Since x>0 we can multiply both sides and get x^2 < 1. Rooting both sides again (as shown in (1) above) you get |x| < 1 this verifies again case1 and case2 from the question stem. At this point YOU DONT need to go and check caseB.

D is the correct answer

gmatclubot

If x is not equal to zero, is |x| < 1 ?
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