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If the function f(n) is defined as f(n) = n/(n + 1), for all integer

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If the function f(n) is defined as f(n) = n/(n + 1), for all integer  [#permalink]

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New post 09 Jun 2015, 02:43
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If the function f(n) is defined as f(n) = n/(n + 1), for all integer values of n such that n ≠ –1, which of the following must be true?

I. f(x + 1) > f (x)
II. f(x) > 0
III. f(x) ≠ 0

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) II and III only


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If the function f(n) is defined as f(n) = n/(n + 1), for all integer  [#permalink]

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New post Updated on: 10 Jun 2015, 06:22
2
Bunuel wrote:
If the function f(n) is defined as f(n) = n/(n + 1), for all integer values of n such that n ≠ –1, which of the following must be true?

I. f(x + 1) > f (x)
II. f(x) > 0
III. f(x) ≠ 0

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) II and III only


Kudos for a correct solution.


Checking I. f(x + 1) > f (x)

f(x + 1) =(x+1)/(x+2) and f (x)= (x)/(x+1)

@x=1, f(x+1) = 2/3, f(x) = 1/2 i.e. f(x + 1) > f (x)
@x=2, f(x+1) = 3/4, f(x) = 2/3 i.e. f(x + 1) > f (x)
@x=-3, f(x+1) = 2, f(x) = 3/2 i.e. f(x + 1) > f (x)
@x=-4, f(x+1) = 3/2, f(x) = 4/3 i.e. f(x + 1) > f (x)
@x=-10, f(x+1) = 9/8, f(x) = 10/9 i.e. f(x + 1) > f (x)
@x=-100, f(x+1) = 99/98, f(x) = 100/99 i.e. f(x + 1) > f (x)

CONCEPT: if a/b = c/d where a>b
then, (a+x)/(b+x) < c/d where x > 0


i.e. TRUE

Checking II. f(x) > 0
Checked with above working already
FALSE

Checking III. f(x) ≠ 0
@x=0, f(x)=0
i.e. FALSE

Answer: Option
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Originally posted by GMATinsight on 09 Jun 2015, 06:42.
Last edited by GMATinsight on 10 Jun 2015, 06:22, edited 1 time in total.
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Re: If the function f(n) is defined as f(n) = n/(n + 1), for all integer  [#permalink]

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New post 10 Jun 2015, 02:02
1
If n = 0, then \(f(n)=\frac{n}{n+1}=\frac{0}{1}=0\), which negates II. and III. Therefore the answer has to be A.
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Re: If the function f(n) is defined as f(n) = n/(n + 1), for all integer  [#permalink]

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New post 15 Jun 2015, 01:15
Bunuel wrote:
If the function f(n) is defined as f(n) = n/(n + 1), for all integer values of n such that n ≠ –1, which of the following must be true?

I. f(x + 1) > f (x)
II. f(x) > 0
III. f(x) ≠ 0

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) II and III only


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Set up a chart to test a representative set of possible x values, including positives, negatives, and zero:
Attachment:
2015-06-15_1214.png
2015-06-15_1214.png [ 81.8 KiB | Viewed 2672 times ]

I. ALWAYS TRUE: In all cases where f(x) and f(x + 1) were defined, f(x + 1) > f(x).

II. USUALLY TRUE: However, f(x) can equal 0 when x = 0.

III. USUALLY TRUE: However, f(x) can equal 0 when x = 0.

Therefore, only statement I must be true.

The correct answer is A.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If the function f(n) is defined as f(n) = n/(n + 1), for all integer  [#permalink]

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New post 23 Oct 2016, 02:55
A faster way:

try some values for f(n)
f(-2) = 2
f(0) = 0
f(1) = 1/2

you see directly that II and III are wrong.
f(x)>0, wrong f(0)=0
f(x) ≠ 0, wrong f(0)=0

So, I only, since you don't have the answer choice "None".
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Re: If the function f(n) is defined as f(n) = n/(n + 1), for all integer  [#permalink]

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New post 23 Oct 2016, 22:06
i have a doubt please correct me where m i wrong
in A

f(x+1) > f(x), so the equation become
x+1/x+2 > x/x+1
let x =1
2/3 > 1/2 answer is yes
x= -3
-3+1/-3+2 > -3/-3+1
-2/-1 > -3/-2 i.e. 2> 3/2 yes

what x=-2 solution is not defined, how we deal with this case
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Re: If the function f(n) is defined as f(n) = n/(n + 1), for all integer  [#permalink]

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New post 20 Apr 2018, 00:01
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Re: If the function f(n) is defined as f(n) = n/(n + 1), for all integer &nbs [#permalink] 20 Apr 2018, 00:01
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