How to solve these FUNCTIONS?

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How to solve these FUNCTIONS?

by himu » Wed Feb 20, 2013 8:22 pm
Let the function g(a, b) = f(a) + f(b).

For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?


X+3

X^2

|X|

1/X

X/4

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by [email protected] » Wed Feb 20, 2013 8:31 pm
himu wrote:Let the function g(a, b) = f(a) + f(b).

For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?
g(a + b, a + b) = f(a + b) + f(a + b) = 2*f(a + b)
g(a, a) = f(a) + f(a) = 2*f(a)
g(b, b) = f(b) + f(b) = 2*f(b)

Hence, we need to find a function which satisfies the condition 2*f(a + b) = 2*f(a) + 2*f(b), i.e. f(a + b) = f(a) + f(b)

This relation will satisfied only by linear functions.
Hence, only possible options are (A) f(x) = x + 3 or (C) f(x) = |x| or (E) f(x) = x/4

For option (A) ---> f(a + b) = (a + b + 3) ≠ f(a) + f(b) = (a + 3) + (b + 3) = (a + b + 6)
For option (C) ---> f(a + b) = |a + b| may not be equal to f(a) + f(b) = |a| + |b|
For option (E) ---> f(a + b) = (a + b)/4 = f(a) + f(b) = a/4 + b/4 = (a + b)/4

The correct answer is E.
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by GMATGuruNY » Thu Feb 21, 2013 3:44 am
Let the function g(a, b) = f(a) + f(b).

For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?

A: x+3

B: x^2

C: |x|

D: 1/x

E: x/4



OA : x/4
We can plug in values.
Since one of the answer choices is |x|, we should plug in at least one negative value.
Let a=-2 and b=3.
Then a+b = -2+3 = 1.

The question becomes:
For which function f below will g(1, 1) = g(-2, -2) + g(3, 3)?

Since g(a,b) = f(a) + f(b):
g(1,1) = f(1) + f(1) = 2f(1)
g(-2,-2) = f(-2) + f(-2) = 2f(-2)
g(3,3) = f(3) + f(3) = 2f(3)

Substituting these relationships into g(1,1) = g(-2,-2) + g(3,3), we get:
2f(1) = 2f(-2) + 2f(3)
f(1) = f(-2) + f(3).

Thus, the question now becomes:
For which function f below will f(1) = f(-2) + f(3)?

A: x+3
f(1) = 1+3 = 4.
f(-2) + f(3) = (-2+3) + (3+3) = 10.
Doesn't work. Eliminate A.

B: x²
f(1) = 1² = 1.
f(-2) + f(3) = (-2)² + 3² = 13.
Doesn't work. Eliminate B.

C: |x|
f(1) = |1| = 1.
f(-2) + f(3) = |-2| + |3| = 5.
Doesn't work. Eliminate C.

D: 1/x
f(1) = 1/1 = 1.
f(-2) + f(3) = -1/2 + 1/3 = -1/6.
Doesn't work. Eliminate D.

The correct answer is E.

E: x/4
f(1) = 1/4
f(-2) + f(3) = -2/4 + 3/4 = 1/4.
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by pemdas » Thu Feb 21, 2013 3:50 pm
first of all, this q. is well beyond GMAT :)
secondly, the main trick here is that all q. choices contain only one variable (it doesn't matter what it states a or b or a+b, we speak about the number of variables), but the target function, For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)? contains two variables in one function of the left-hand-side and right-hand-side. This is why we must simplify this function to contain one variable on both sides. We will do it with the help of given function g(a, b) = f(a) + f(b)

Just before moving, try to understand some more ...
g(a, b) = f(a) + f(b) What does it tell us?
It tells that the function of two variables a and b varies as the sum of one-variable functions

g(a + b, a + b) = g(a, a) + g(b, b) What does it tell us?
It tells us that two-variable function of two identical variables (a+b) and (a+b) varies as the sum of two two-variable functions

The last part in the preceding sentence is important, as we see that the function of two variables, g(a + b, a + b), behaves the way as the first function g(a, b) = f(a) + f(b) does.

So, g(a,a)=f(a)+f(a) and g(b,b)=f(b)+f(b) OR g(a,a)=2f(a) and g(b,b)=2f(b). The same way, g(a+b, a+b)=f(a+b)+f(a+b)=2f(a+b)

We already know g(a + b, a + b)= g(a, a) + g(b, b) which is equivalent to 2f(a+b)=2f(a)+2f(b) OR f(a+b)=f(a)+f(b). This means we have simplified two-variable function and set it as one-variable function actually. We can begin checking with the variable x

a) X+3 and f(a+b)=f(a)+f(b). Let f(a+b) be f(x) then f(a+b)=a+b+3, because f(x)=x+3.
We check further, Let f(a) be f(x), then f(a)=a+3 and the same with f(b)=b+3. So, f(a+b)=f(a)+f(b) is equivalent to a+b+3=a+3+b+3. Obviously, this is false, since a+b+3=!a+b+6

b) X^2. Let f(a+b) be f(x) then f(a+b)=(a+b)^2 and with the same logic f(a)=a^2 and f(b)=b^2. We check whether f(a+b)=f(a)+f(b), (a+b)^2=a^2+b^2? No, this is false

c) |X|. Let f(a+b) be f(x) then f(a+b)=|a+b|, accordingly, f(a)=|a| and f(b)=|b|. Is |a+b|=|a|+|b|? No, false

d) 1/X. Let f(a+b) be f(x) then f(a+b)=1/(a+b), consequently, f(a)=1/a and f(b)=1/b. We check if 1/(a+b)=1/a +1/b? No, false

e) X/4. Let f(a+b) be f(x) then f(a+b)=(a+b)/4, also f(a)=a/4 and f(b)=b/4. We may check if (a+b)/4=a/4+b/4. Yes, indeed.

p.s. I have decided to give detailed explanation for this q. because the solution by Anurag was in PhD style and the solution by Mitch was as usually intuitive, plug and check.

himu wrote:Let the function g(a, b) = f(a) + f(b).

For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?


X+3

X^2

|X|

1/X

X/4
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