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Bunuel
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 14025 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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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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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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mbaprep2016
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

Yes , for X=-2 seems a valid number to plug in, this will make f(x+1) undefined .
(x+1)/(x+2) = (-2+1)/(-2+2) will become undefined.
But question itself is not bounded by x is not equal to 2, but rather x is not equal to 1.

Can someone please clarify.
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Bunuel
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.

I solved it fairly quickly so I will mention my approach.
The question is asking must be true. So let's try to make it not true.
II. f(x) > 0 -- If you try n=-2 you get the value of the function as -2. So this if false
III. f(x) ≠ 0 -- Well 0 brings the expression to 0.

Now I was actually trying to evaluate options but then I thought do I have an option which says None of these? No? Hence A.

40 seconds :)

Thank you!
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Solving for A as B and C can easily be eliminated at x=0

lets say f(x+1)=x+1/x+2 and f(x) = x/x+1. Cross multiplying to see if f(x+1) > f(x), we get

LHS = (x+1)(x+1) = x(x+1)+ x+1

RHS = x(x+2) = x(x+1) + x

Hence we see that LHS will also have an extra 1, doesn't matter if x is positive or negative.

PS: please replace x with n.
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