Which of the following expressions is defined for all : Quant Question Archive [LOCKED]
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# Which of the following expressions is defined for all

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Which of the following expressions is defined for all [#permalink]

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06 Sep 2008, 09:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Which of the following expressions is defined for all integer values of z, such that (z^2) <9?
I. (z+1)/z
II. (z-4)/(z^2-4z+4)
III. 18/(z^2-4z-5)

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?
Director
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06 Sep 2008, 09:40
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. (z-4)/(z^2-4z+4)
III. 18/(z^2-4z-5)

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^2-4z+4 = 0
(z-2)^2 = 0
z = 2

III. The expression is undefined when z^2-4z-5 = 0
(z-5)(z+1) = 0
z = 5 or -1
Director
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06 Sep 2008, 09:41
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. (z-4)/(z^2-4z+4)
III. 18/(z^2-4z-5)

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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06 Sep 2008, 09:59
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. (z-4)/(z^2-4z+4)
III. 18/(z^2-4z-5)

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

(z-3) < 0 & (z+3 ) > 0 means z<3 and z > -3 --> -3 < z < 3

or

(z-3) > 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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06 Sep 2008, 10:16
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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06 Sep 2008, 11:08
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
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07 Sep 2008, 22:52
Very good question. Initially, I thought something is missing in the question..
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08 Sep 2008, 13:46
A

divide by 0 is never defined !
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08 Sep 2008, 14:39

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.
Re: Math Question   [#permalink] 08 Sep 2008, 14:39
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