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Difficulty:
45%
(medium)
Question Stats:
65% (01:40) correct
35%
(01:43)
wrong
based on 673
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History
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If x is not equal to zero, is |x| < 1 ?
(1) x^2 < 1
(2) |x| < 1/x
(1) x^2 < 1
(2) |x| < 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
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
01 Oct 2012, 06:05
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If x is not equal to zero, is |x| < 1 ?
Is \(|x| < 1\)? --> is \(-1<x<1\) (\(x\neq{0}\)).
(1) x^2 < 1 --> -1<x<1. Directly answers the question. Sufficient.
(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.
Answer: D.
Is \(|x| < 1\)? --> is \(-1<x<1\) (\(x\neq{0}\)).
(1) x^2 < 1 --> -1<x<1. Directly answers the question. Sufficient.
(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.
Answer: D.
30 Sep 2012, 06:34
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carcass
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.
