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Which of the following expressions is defined for all [#permalink]
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06 Sep 2008, 10:31
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Which of the following expressions is defined for all integer values of z, such that (z^2) <9? I. (z+1)/z II. (z4)/(z^24z+4) III. 18/(z^24z5)
a) None b) I only c) II only d) III only e) II and III
*** What confuses me is z<3 and z<3. Am I taking the right approach?



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Re: Math Question [#permalink]
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06 Sep 2008, 10:40
HVD1975 wrote: Which of the following expressions is defined for all integer values of z, such that (z^2) <9? I. (z+1)/z II. (z4)/(z^24z+4) III. 18/(z^24z5)
a) None b) I only c) II only d) III only e) II and III
*** What confuses me is z<3 and z<3. Am I taking the right approach? I get A The question is asking which expressions, I, II, and/or III, are defined. Each of these expressions are undefined when their denominators are 0. The prompt gives the restriction that z^2 < 9, which means 3 < z < 3 I. The expression is undefined when z = 0. II. The expression is undefined when z^24z+4 = 0 (z2)^2 = 0 z = 2 III. The expression is undefined when z^24z5 = 0 (z5)(z+1) = 0 z = 5 or 1



Director
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Re: Math Question [#permalink]
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06 Sep 2008, 10:41
HVD1975 wrote: Which of the following expressions is defined for all integer values of z, such that (z^2) <9? I. (z+1)/z II. (z4)/(z^24z+4) III. 18/(z^24z5)
a) None b) I only c) II only d) III only e) II and III
*** What confuses me is z<3 and z<3. Am I taking the right approach? E if z = 0, equation 1 is undefined



VP
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Re: Math Question [#permalink]
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06 Sep 2008, 10:59
HVD1975 wrote: Which of the following expressions is defined for all integer values of z, such that (z^2) <9? I. (z+1)/z II. (z4)/(z^24z+4) III. 18/(z^24z5)
a) None b) I only c) II only d) III only e) II and III
*** What confuses me is z<3 and z<3. Am I taking the right approach? z ^ 2 < 9 is a quadratic inequality z ^2 9 < 0 (z3) < 0 & (z+3 ) > 0 means z<3 and z > 3 > 3 < z < 3 or (z3) > 0 & (z+3) <0 z> 3 or z < 3 means z is between [ infinity, 3] and [3, Infinity] I am guessing that we ignore the second one because this graph is a parabola and we are only concerned with the points at which it cuts the x axis. Hence we ignore the second set. The rest of the solution is already explained.



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Re: Math Question [#permalink]
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06 Sep 2008, 11:16
But why 3<z<3 and not z<3 and z<3
1. If z<3 then z could be 2,1,0,1,2,3,4,5, etc 2. If z<3 then z could be 4,5,6, etc
Shouldn't z<3 cancels the other integers less than 3 (2,1,0,etc)?



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Re: Math Question [#permalink]
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06 Sep 2008, 12:08
HVD1975 wrote: But why 3<z<3 and not z<3 and z<3
1. If z<3 then z could be 2,1,0,1,2,3,4,5, etc 2. If z<3 then z could be 4,5,6, etc
Shouldn't z<3 cancels the other integers less than 3 (2,1,0,etc)? The prompt says z^2 < 9 If z < 3, then z^2 > 9.



SVP
Joined: 17 Jun 2008
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Re: Math Question [#permalink]
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07 Sep 2008, 23:52
Very good question. Initially, I thought something is missing in the question..



Senior Manager
Joined: 31 Jul 2008
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Re: Math Question [#permalink]
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08 Sep 2008, 14:46
A
divide by 0 is never defined !



Manager
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Re: Math Question [#permalink]
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08 Sep 2008, 15:39
Answer should be e.
x^2 < 9 ... only 2,1,0,1,2 statisfy this condition.
and 0 does not satisfy the first option. so that is not defined for all x^2<9
2 and 3 are good. e.










