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If f(x) = 2^x + 2^-2, and x is an integer

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If f(x) = 2^x + 2^-2, and x is an integer [#permalink] New post 06 Dec 2012, 04:07
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Question Stats:

33% (00:00) correct 67% (00:40) wrong based on 9 sessions
If f(x) = 2^(x) + 2^(-2), and x is an integer, is f(x) = √(a + 4)?

(1) -4 < x < 4
(2) 4^(x) + 4^(-x) = a
[Reveal] Spoiler: OA

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Expert Post
Magoosh GMAT Instructor
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Re: If f(x) = 2^x + 2^-2, and x is an integer [#permalink] New post 06 Dec 2012, 15:07
Expert's post
gmatbull wrote:
If f(x) = 2^(x) + 2^(-2), and x is an integer, is f(x) = √(a + 4)?
(1) -4 < x < 4
(2) 4^(x) + 4^(-x) = a

An interesting DS problem. The prompt involves this letter a, and if we don't know the value of this, then we can't determine anything.

For Statement #1, without doing any calculations, we see that not matter what range is established for x, we have no idea what a is, and without some indication of the value of a, we have no way of answering the prompt question. Statement #1, alone and by itself, is insufficient.

Statement #2 is interesting
For example, let's consider x = 0
Then f(0) = 1 + 1/4 = 5/4 = 1.25
Meanwhile, sqrt(4^(x) + 4^(-x) + 4) = sqrt(1 + 1 + 4) = sqrt(6). For that value, they are not equal.

Let's consider x = 1
Then f(1) = 2 + 1/4 = 9/4 = 2.25
Meanwhile, sqrt(4^(x) + 4^(-x) + 4) = sqrt(4 + 1/4 + 4) = sqrt(33/4) = [sqrt(33)]/2. For that value they are not equal.

Let's set the two expressions equal and see what happens:
2^(x) + 2^(-2) = sqrt(4^(x) + 4^(-x) + 4)
square both sides
(2^x)^2 + 2*(2^x)*(2^(-2)) + (2^(-2))^2 = 4^(x) + 4^(-x) + 4
4^(x) +(1/2)*(2^x) + 1/16 = 4^(x)+ 4^(-x) + 4
(1/2)*(2^x) + 1/16 = 4^(-x) + 4
(1/2)*(2^x) - 4^(-x) = 4 - 1/16
(1/2)*(2^x) - 4^(-x) = 63/16

There are really no straightforward algebraic ways to simplify this. Let g(x) = sqrt(4^(x) + 4^(-x) + 4). Looking at a graph of f(x) & g(x), we see that the functions appear to approach each other asymptotically as x gets large in the positive direction.
Attachment:
graph of f(x) & g(x).JPG
graph of f(x) & g(x).JPG [ 75.08 KiB | Viewed 555 times ]

In fact, the two functions actually cross at x = 2.9831285117658, and asymptotically approach each other thereafter
Given that x has to be an integer, the two expressions will never be equal.

Therefore, despite what the above post says is the OA, it seems to me the answer must be B. It's really not clear to me that there's an efficient way to solve this without substantial recourse to technology.

Notice that if the function in the prompt were f(x) = 2^(x) + 2^(-x) instead of 2^(x) + 2^(-2), and the expression in the prompt were f(x) = √(a + 2) instead of f(x) = √(a + 4), then the expressions would always be equal, for all real numbers.

Does all this make sense?

Mike :-)
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Last edited by mikemcgarry on 07 Dec 2012, 11:01, edited 1 time in total.
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Re: If f(x) = 2^x + 2^-2, and x is an integer [#permalink] New post 07 Dec 2012, 01:31
mikemcgarry wrote:
gmatbull wrote:
If f(x) = 2^(x) + 2^(-2), and x is an integer, is f(x) = √(a + 4)?
(1) -4 < x < 4
(2) 4^(x) + 4^(-x) = a

An interesting DS problem. The prompt involves this letter a, and if we don't know the value of this, then we can't determine anything.

For Statement #1, without doing any calculations, we see that not matter what range is established for x, we have no idea what a is, and without some indication of the value of a, we have no way of answering the prompt question. Statement #1, alone and by itself, is insufficient.

Statement #2 is interesting
For example, let's consider x = 0
Then f(0) = 1 + 1 = 2
Meanwhile, sqrt(4^(x) + 4^(-x) + 4) = sqrt(1 + 1 + 4) = sqrt(6). For that value, they are not equal.

Let's consider x = 1
Then f(1) = 2 + 1/2 = 2.5
Meanwhile, sqrt(4^(x) + 4^(-x) + 4) = sqrt(4 + 1/4 + 4) = sqrt(33/4) = [sqrt(33)]/2. For that value they are not equal.

Let's set the two expressions equal and see what happens:
2^(x) + 2^(-2) = sqrt(4^(x) + 4^(-x) + 4)
square both sides
(2^x)^2 + 2*(2^x)*(2^(-x)) + (2^(-x))^2 = 4^(x) + 4^(-x) + 4
4^(x) + 2 + 4^(-x) = 4^(x) + 4^(-x) + 4
2 = 4
This is a contradiction. They only thing we did was set the two expressions equal, so if that leads to a contradiction, it must mean that the two expressions can never be equal. Both expressions are defined for all real numbers, and for absolutely no value are these two expressions ever equal.

That provides a very clear and definitive answer to the prompt question. This statement, by itself, is unambiguously sufficient.

Therefore, despite what the above post says is the OA, it seems to me the answer must be B.

Notice that if the expression in the prompt were f(x) = √(a + 2) instead of f(x) = √(a + 4), then the expressions would always be equal, for all real numbers.

Does all this make sense?

Mike :-)


You made a mistake in (2) when squaring the LHS up. The question reads "2^(-2)", not 2^(-x) as in your equation.
Expert Post
Magoosh GMAT Instructor
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Re: If f(x) = 2^x + 2^-2, and x is an integer [#permalink] New post 07 Dec 2012, 11:05
Expert's post
catennacio wrote:
You made a mistake in (2) when squaring the LHS up. The question reads "2^(-2)", not 2^(-x) as in your equation.

Actually, I just misread that and treated the wrong function throughout my whole solution. I just corrected this in my response. Thanks for pointing this out.
Mike :-)
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Magoosh Test Prep

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Re: If f(x) = 2^x + 2^-2, and x is an integer   [#permalink] 07 Dec 2012, 11:05
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If f(x) = 2^x + 2^-2, and x is an integer

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