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m11 q24

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m11 q24 [#permalink] New post 08 Feb 2009, 17:33
Function F(x) is defined as follows:
if x is positive or 0 then F(x) = 1 - x;
if x is negative then F(x) = x - 1.

Which of the following is true about F(x)?

I. F(|x|) = |F(x)|

II. F(x) = F(-x)

III. |F(x + 1)| = |x|

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) not I, II, or III

[Reveal] Spoiler: OA
C

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Re: m11 q24 [#permalink] New post 15 Feb 2009, 14:38
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ConkergMat wrote:
Function F(x) is defined as follows:
if x is positive or 0 then F(x) = 1 - x;
if x is negative then F(x) = x - 1.

Which of the following is true about F(x)?

I. F(|x|) = |F(x)|
II. F(x) = F(-x)
III. |F(x + 1)| = |x|


I. F(|x|) = |F(x)| :- not possible.

First, F(|x|) could be -ve, 0 and +ve but |F(x)| is always 0 or +ve.

If |x| is greater than 1, F(|x|) = 1 - lxl is -ve fraction or > 1.
If |x| is smaller than 1, F(|x|) = 1 - |x| is +ve fraction.
If |x| is 0, F(|x|) is 1.

But |F(x)| is always 0 or +ve. So not true.


II. F(x) = F(-x):- not possible.
If x is +ve, -x is -ve and vice versa.

If x is 0, F(x) = F(-x) = 1
If x is +ve, F(x) = 1 - x, which could be +ve if x is smaller than 1 or -ve if x is > 1, and F(-x) = -x-1, which could also be -ve if x is >-1 or +ve if x is <-1.
Similarly, if x is negative, then F(x) and F(-x) will have +ve or -ve values.


They are not equal. not suff.



III. |F(x + 1)| = |x|:- is possible.

|F(x + 1)| is always |x| whether (x+1) is +ve or -ve.

If (x + 1) is +ve, |F(x + 1)| = |1- (x + 1)| = |x|
If (x + 1) is -ve, |F(x + 1)| = |(x + 1)-1| = |x|

III is true.
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Re: m11 q24 [#permalink] New post 18 Feb 2009, 07:53
Thanks. What a tough question though (to get in 2 min).
Maybe its one to guess and move on :)
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Re: m11 q24 [#permalink] New post 24 Feb 2010, 06:43
I just draw the graph.
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Re: m11 q24 [#permalink] New post 24 Feb 2010, 07:30
flyingbunny wrote:
I just draw the graph.


How do you graph this, what does it look like?
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Re: m11 q24 [#permalink] New post 24 Feb 2010, 11:46
GMAT TIGER's explanation is right but you are not supposed to write Math major test.
Just take x=2 and x=-2.
Then apply on the rule. Its time consuming not tough. In fact it is easy. It took 2 min for me to solve this one.

It will take 5-6 min to draw a graph for most of the guys.
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Re: m11 q24 [#permalink] New post 24 Feb 2010, 12:11
Answer is clearly C
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Re: m11 q24 [#permalink] New post 24 Feb 2010, 23:43
hmm, i am an engineer and i got this wrong :oops:
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Re: m11 q24 [#permalink] New post 27 Feb 2010, 02:02
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by substituting values only ans choice c is valid
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Re: m11 q24 [#permalink] New post 08 Sep 2010, 02:25
Its C.

f(2) = -1 and f(-2) = -3
Substituting these values in the equations shows that only C hold good.
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Re: m11 q24 [#permalink] New post 01 Mar 2011, 09:18
just by looking at the functions you can tell that 1 & 2 are not valid because they are saying that the value of X is the same, before and after the function which we know is untrue as depending on the value of X it will either be subtracted from 1 or 1 will be added to it.

for answer 3, if you subtract 1 and then add 1 back, you will have your original number in absolute terms. if you subtract your original number from 1 and then add 1 back, you will have your number in absolute terms. therefore answer is C.
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Re: m11 q24 [#permalink] New post 01 Mar 2011, 12:13
C , tough one though
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Re: m11 q24 [#permalink] New post 02 Mar 2011, 10:41
It seems as if it's time consuming and then you begin to panic and want to guess so you can do it in under 2 minutes, but if you do as silasaaa2 suggests and just substitute. Pick a positive and negative number and substitute for each. You WILL find the answer in less than 2 minutes.
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Re: m11 q24 [#permalink] New post 02 Mar 2011, 17:13
F(0) = 1
F(1) = 0
F(-1) = -2

Statement I is false. F(1) != F(-1) a, d gone
Statement II is false. F(1) != F(-1) b gone
Statement III is true. x = 0 = F(1), e gone. C it is.
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Re: m11 q24 [#permalink] New post 02 Mar 2012, 15:06
I just took two values for x and tested them all. It took me about 2 minutes and 30 seconds to get the answer. A bit slow but it worked.
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Re: m11 q24 [#permalink] New post 02 Mar 2012, 20:32
30 seconds to substitute and confirm C.
Do not solve these type of questions unless of course this is your 37th question and you have 15 mns spare time. Even then, I would prefer not to solve, but substitute for this question :)
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Re: m11 q24 [#permalink] New post 26 Apr 2012, 05:13
ConkergMat wrote:
Function F(x) is defined as follows:
if x is positive or 0 then F(x) = 1 - x;
if x is negative then F(x) = x - 1.

Which of the following is true about F(x)?

I. F(|x|) = |F(x)|

II. F(x) = F(-x)

III. |F(x + 1)| = |x|

(A) I only
(B) II only
(C) III only
(D) I and III only
(E) not I, II, or III

[Reveal] Spoiler: OA
C

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followed substitution method and the answer is C
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Re: m11 q24 [#permalink] New post 05 Mar 2013, 21:56
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F(x) = 1 - x, for x greater than equal to zero.....condition 1
F(x) = x -1 , for x less than zero.....condition 2

I- F(|x|) = 1- |x| and |F(x)| = | 1 - x |.....condition 1
and F(|x|) = |x| -1 and |F(x)| = | x - 1 |.....condition 2

F(|x|) not equal to F(|x|)

II-F(-x) = 1- (-x) = 1+ x.....condition 1
and F(-x) = (-x) -1 = -( x + 1).....condition 2

F(x) not equal toF(-x)


III-|F(x +1)| = |1- (x+1)| = |x|.....condition 1
and |F(x +1)| = |(x + 1) - 1| = |x|.....condition 2

Hence, III is correct answer and C is the correct choice
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Re: m11 q24 [#permalink] New post 06 Mar 2013, 08:31
vipinktyagi wrote:
F(x) = 1 - x, for x greater than equal to zero.....condition 1
F(x) = x -1 , for x less than zero.....condition 2

I- F(|x|) = 1- |x| and |F(x)| = | 1 - x |.....condition 1
and F(|x|) = |x| -1 and |F(x)| = | x - 1 |.....condition 2

F(|x|) not equal to F(|x|)

II-F(-x) = 1- (-x) = 1+ x.....condition 1
and F(-x) = (-x) -1 = -( x + 1).....condition 2

F(x) not equal toF(-x)


III-|F(x +1)| = |1- (x+1)| = |x|.....condition 1
and |F(x +1)| = |(x + 1) - 1| = |x|.....condition 2

Hence, III is correct answer and C is the correct choice


This question doesn't look like hardest category question....indeed good detailed explanation...
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Re: m11 q24 [#permalink] New post 06 Mar 2013, 08:32
This is not a level 700-800 question...
Re: m11 q24   [#permalink] 06 Mar 2013, 08:32
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