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

For x = 1, try the options given

(A) x = 1
(B) \(x^2 = 1^2 = 1\)
(C) 2x = 2
(D) x + 2 = 1 + 2 = 3
(E) \(2^x = 2^1 = 2\)

Answer = D

Hi,

The best way to solve any MUST-BE/COULD –BE question in GMAT is plugging in,

Just plug values based out of the range given in the question,

Let’s pick x= 1

(A) x --> 1

(B) x^2 -->1

(C) 2x -->2

(D) x + 2 -->3

(E) 2^x -->2

So answer has to be D.

Here we are little lucky, because we plugged in value, we got the answer.

But sometime we might have to Plug more than once to get to a final answer.

Some students might think, we have plugged in only one value, what if we try other values and we might get more options,

Just remember this in GMAT, MUST BE means “Always”. So here all the other four options were not greater than D for x = 1. So we can eliminate.

So answer is D.

Hope this helps.
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A graphical solution is the best in this case.


At \(x = 0\) , try to compare values of all functions, only two functions have non zero values \(x+2 = 2, 2^x = 1\).
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why can it not be 1.5? are they looking for integers specifically

x can be any number satisfying 0 < x < 2, including fractions. For any x from that range options D is the greatest.
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0 < x < 2
x > 0 , x < 2
x = 1
D

If x has to be less than 2 then how option D is the answer ? since D says x=2.

Option D is x + 2, not x = 2.
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The key word here is MUST.
That is, the correct answer must yield the biggest value for ALL values of x between 0 and 2
So, we need only test ONE of the possible x-values.

Let's test x = 1

We get....
(A) x = 1
(B) x^2 = 1^2 = 1
(C) 2x = (2)(1) = 2
(D) x + 2 = 1 + 2 = 3
(E) 2^x = 2^1 = 2

Since D yields the biggest value, then the correct answer MUST be D

Cheers,
Brent

Use X as 1.5 (3/2) to see which option has the highest answer.

Option D clearly has the highest answer with 3.5.

Smash D

When x = 1 (note that 0 < 1 < 2), then the choices offered become 1, 12 = 1, 2(1) = 2, 1 + 2 = 3, and 21 = 2. The greatest of these values is 3, which corresponds to x + 2

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