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Among the following functions, for which one is g(x − y)

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by guerrero » Sat Aug 10, 2013 9:22 am
Among the following functions, for which one is g(x − y) = g(x) − g(y) for all values of x and y in the domain of y?


(A)g(a)= a^{4}
(B) g(a)= a-7
(C)g(a)=sqrt{a+3}
(D) g(a)={4}/{a-4}
(E) g(a)=7a

OAE
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by Matt@VeritasPrep » Sat Aug 10, 2013 10:38 am
This is a popular question: I think it must've been on the real GMAT at some point, because I've seen about eight different versions of it from eight different GMAT prep sources.

The trick here is that the function isn't defined: all we know that is that whatever it is, it makes g(x-y) = g(x) - g(y). The answer choices are all possible definitions, so we have to take each one and see if it satisfies the equality we were given.

To illustrate one that works and one that doesn't, consider answers B and E.

If g(a) = a - 7, then

g(x-y) = (x-y) - 7
g(x) = x - 7
g(y) = y - 7
g(x) - g(y) = (x - 7) - (y - 7) = x - y

Does g(x-y) = g(x) - g(y)? Only if (x-y) - 7 = x - y. But that's impossible! So this function doesn't work.

As for g(a) = 7a
g(x-y) = 7(x-y)
g(x) = 7x
g(y) = 7y
g(x) - g(y) = 7x - 7y

Does g(x-y) = g(x) - g(y)? Only if 7(x-y) = 7x - 7y. That's always true, so this is our function.

A neat way of noticing this:

If g(x-y) = g(x) - g(y), we have a distributive function. Pretend that you had k instead instead of g. Is k * (x-y) = k*x - k*y? Of course! It's tricky to see when the function is there, but simply multiplying by a constant will satisfy this equality: if g(x) = multiplication by k, we're fine.

The only answer that has multiplication by a constant is E, so that's it.
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by GMATGuruNY » Sat Aug 10, 2013 1:07 pm
guerrero wrote:Among the following functions, for which one is g(x − y) = g(x) − g(y) for all values of x and y in the domain of y?


(A)g(a) = a^{4}
(B) g(a)= a-7
(C)g(a)=sqrt{a+3}
(D) g(a)={4}/{a-4}
(E) g(a)=7a

OAE
An alternate approach is plug in values.
Let x=3 and y=2.
Then g(x-y) = g(3-2) = g(1).
The question becomes:

For which of the following functions does g(1) = g(3) - g(2)?

A: g(a) = a�
g(1) = 1� = 1.
g(3) = 3� = 81.
g(2) = 2� = 16.
g(3) - g(2) = 81-16 = 65.
Since g(1) ≠ g(3) - g(2), eliminate A.

B: g(a) = a-7
g(1) = 1-7 = -6.
g(3) = 3-7 = -4.
g(2) = 2-7 = -5.
g(3) - g(2) = -4 - (-5) = 1.
Since g(1) ≠ g(3) - g(2), eliminate B.

C: g(a) = √(a+3)
g(1) = √(1+3) = 2.
g(3) = √(3+3) = √6.
g(2) = √(2+3) = √5.
g(3) - g(2) = √6 - √5.
Since g(1) ≠ g(3) - g(2), eliminate C.

(D) g(a) = 4/(a-4)
g(1) = 4/(1-4) = -4/3.
g(3) = 4/(3-4) = -4.
g(2) = 4/(2-4) = -2.
g(3) - g(2) = -4 - (-2) = -2.
Since g(1) ≠ g(3) - g(2), eliminate D.

The correct answer is E.

E: g(a) = 7a
g(1) = 7*1 = 7.
g(3) = 7*3 = 21.
g(2) = 7*2 = 14.
g(3) - g(2) = 21-14 = 7.
Success!
g(1) = g(3) - g(2).
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