If 0<x<2, which of the following must be greatest?

[ghag.kamlesh]

Hello,

Can anyone please help me to understand the following doubt on the QT question that appeared in GMATPrep ?

My doubt is should I consider here "x" as a real integer only ?

Because the question does not say so. And that is why if I consider 'x' as decimal, then answer is different, and if 'x' is integer as per solution given, then answer is different.

Attached here the screenshot of the question from GMAT Prep.

http://s22.postimg.org/tp2xznptt/GMATPREP.jpg

I hope someone will share his or her views on this.

Thanks in advance.
Kamlesh

[RonPurewal]

nobody is saying that x must be an integer.

all they're doing in the answer key is using the "plug in your own numbers" method. (they don't show this method in their answer keys very often -- even though it works on a huge number of the problems -- so, admittedly, it's a bit of a surprise to see them doing that.)
the point is that the problem has to work out the same way for EVERY number between 0 and 2, or else the problem would be unsolvable. i.e., if the correct answer were, say, C for some values of x and D for others, then there would not be any correct answer to the overall problem.

they're just using 1 because 1 is, by a mile, the easiest number within the interval 0 < x < 2.
if you did this problem with 1.1 or 0.3 or √2 or pi/2 instead of 1, it would not work out any differently.

[RonPurewal]

by the way -- DO NOT post problems as image files, unless it's actually necessary to do so -- e.g., if there's a diagram in the problem.
from now on, if you unnecessarily post a problem as an image file, the thread will be deleted.

the issue, of course, is that problems posted as pictures are not searchable, and so this thread becomes useless for users who are searching for the problem at hand.

thanks.

--

for searchers: the problem is this:

If 0 < x < 2, which of the following must be greatest?
(A) x
(B) x^2
(C) 2x
(D) x + 2
(E) 2^x

[RonPurewal]

if I consider 'x' as decimal, then answer is different, and if 'x' is integer as per solution given, then answer is different.

this is false. if this were true, then the problem would not be a valid problem, as there would be more than one "correct" answer.

i'm curious -- what decimal value did you think gave an answer besides (D)?
whatever that value was, you must have made some error in your arithmetic, because (x + 2) is still the biggest of these expressions.

--

if you want a proof, here's a proof. (NOTE: if you are actually going through this kind of mental gymnastics when you're taking the test, then you should really get on methods like "plug in your own numbers".)

* (A) x is clearly smaller than (C) 2x. eliminate (A).

* (B) x^2 is x times x.
since x is less than 2, x times x must be less than 2 times x.
so, (B) is smaller than (C). eliminate (B).

* (C) 2x is x + x.
(D) is x + 2. again, we know that x is less than 2, so (D) must be bigger. eliminate (C).

* (E) is not as easy to prove; I don't know if there is an algebraic way to prove this (without using things that are beyond what the gmat actually tests).
if you know what the graphs look like, then you'll know that (D) x + 2 is a straight line, while (E) 2^x is curved upward.
since both (D) and (E) are equal to 4 when x = 2, it follows that the straight line is on top for smaller values. (try drawing the line and the curve if you don't see why this is so.)
other than that, you pretty much just have to plug in numbers and see what happens.
if x is anything between 0 and 1, then 2^x is less than 2, but x + 2 is more than 2.
for the other values, just try enough to see what the deal is: e.g., if x = 1.5, then 2^x is 2√2 = approximately 2.8, while x + 2 is substantially bigger (3.5).

again, all of this is unnecessary.
THE POINT of this problem, especially choice (E) (which can't be easily resolved with textbook methods), is for you to REALIZE that you only need to plug in ONE VALUE.

[ghag.kamlesh]

Hello Ron,

Thank you for your explanation. Now, I understand very clearly that what mistake I was doing.
Also, I will keep it in mind about the rules for posting the questions on the forum.
Thanks once again.

[tim]

:)