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

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Let the function g(a, b) = f(a) + f(b). [#permalink] New post 29 Nov 2012, 18:17
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 30 Nov 2012, 01:41, edited 1 time in total.
OA added.
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink] New post 29 Nov 2012, 21:55
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gmatbull 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)?

A: x +3
B: x^2
C: |x|
D: 1/x
E: x/4

OA to be posted later.


The question is basically asking for which function f(a + b) = f(a) + f(b)

A)\(LHS = a + b + 3\)
\(RHS = a + 3 + b + 3 = a + b + 6.\)
Not equal

B)\(LHS = (a + b)^2\)
\(RHS = a^2 + b^2\)
Not equal

C)\(LHS = |a + b|\)
\(RHS = |a| + |b|\)
If a & b are of different polarity not equal.

D)\(LHS = \frac{1}{a + b}\)
\(RHS = \frac{1}{a} + \frac{1}{b}\)
Not equal

E)\(LHS = \frac{a + b}{4}\)
\(RHS = \frac{a}{4} + \frac{b}{4} = \frac{a + b}{4}\).
Equal.

Answer should be E
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink] New post 30 Nov 2012, 02:02
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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

We can use plug-in method to solve this question.

Since \(g(a, b) = f(a) + f(b)\), then:
\(g(a + b, a + b)=f(a+b)+f(a+b)=2f(a+b)\);
\(g(a, a) + g(b, b)=f(a)+f(a)+f(b)+f(b)=2f(a)+2f(b)\).

Thus the question asks: for which function f below will \(2f(a+b)=2f(a)+2f(b)\) --> \(f(a+b)=f(a)+f(b)\)?

Say a=-1 and b=1, then the question becomes: for which function f below will \(f(0)=f(-1)+f(1)\).

A: x +3
\(f(0)=0+3=3\)
\(f(-1)+f(1)=(-1+3)+(1+3)=6\)
No match.

B: x^2
\(f(0)=0^2=0\)
\(f(-1)+f(1)=(-1)^2+(1)^2=2\)
No match.

C: |x|
\(f(0)=|0|=0\)
\(f(-1)+f(1)=|-1|+|1|=2\)
No match.

D: 1/x
\(f(0)=\frac{1}{0}=undefined\)
\(f(-1)+f(1)=\frac{1}{-1}+\frac{1}{1}=0\)
No match.

E: x/4
\(f(0)=\frac{0}{4}=0\)
\(f(-1)+f(1)=\frac{-1}{4}+\frac{1}{4}=0\)
Match.

Answer: E.

Hope it's clear.

for-which-of-the-following-functions-is-f-a-b-f-b-f-a-124491.html
for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html
functions-problem-need-help-93184.html

Hope it helps.
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink] New post 30 Nov 2012, 02:52
Bunuel and MacFauz,
your explanations make a lot of sense.
More so, Bunuel, you took time to explain the breakdown.

Thanks.
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink] New post 23 Jul 2013, 01:15
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink] New post 01 May 2014, 23:08
Bunuel wrote:
From 100 hardest questions.

Bumping for review and further discussion.




let a = 1 and b = 2, a+b = 3

A) x+3,
2(a+b+3) = 2(a+3) + 2(b+3)
2(a+b+3) = 2(a+b+3) + 6 [Can never be true]

B) x^2
2(a+b)^2 = 2(a^2)+2(b^2) [Can/Cannot be true]

C)|x|
2|a+b| = 2|a| + 2|b| [Can/Cannot be true]

D)1/x
2/(a+b) = (2/a)+(2/b) = 2(a+b)/ab [Can/Cannot be true]

E)x/4
2*(a+b)/4 = 2(a/4) + 2(b/4) = 2(a+b)/4 [Always true]

Ans 'E'
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink] New post 04 Jun 2014, 03:18
g(a+b,a+b) = g(a,a)+g(b,b) = f(a) + f(a) + f(b) + f(b)= 2f(a) + 2f(b) = RHS
g(a+b,a+b) = f(a+b)+f(a+b) = 2f(a+b) = LHS

We divide by 2 both RHS and LHS and get

f(a+b)=f(a) + f(b)

I have an issue here: how do I know what to plug in from the answer choices? Since those are expressed in terms of X.

My approach was to take both a and b as the answer choice, but I don´t know if that is correct.

For d)

LHS: f(x/4 + x/4) = x/2
RHS: f(x/4) + f(x/4) = x/2

I'm sure i'm conceptually missing something.

Thanks.
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink] New post 04 Jun 2014, 20:19
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Enael wrote:
g(a+b,a+b) = g(a,a)+g(b,b) = f(a) + f(a) + f(b) + f(b)= 2f(a) + 2f(b) = RHS
g(a+b,a+b) = f(a+b)+f(a+b) = 2f(a+b) = LHS

We divide by 2 both RHS and LHS and get

f(a+b)=f(a) + f(b)

I have an issue here: how do I know what to plug in from the answer choices? Since those are expressed in terms of X.

My approach was to take both a and b as the answer choice, but I don´t know if that is correct.



I am not sure what you mean by this last line but I can help you with the various variables.

The options (x+3), x^2 etc are the values of the function f(x)

Option (A) tells you that f(x) = x + 3
So if you want to find f(a) or f(b) or f(a+b), it is quite simple.
If f(x) = x+3, f(a) = a+3
If f(x) = x+3, f(a+b) = a+b+3
etc
Wherever you have x in the expression you put a or a+b or b as the case may be.

In this question, since options give the function f(x), you convert the entire g(x) into f(x).
You get that you need to find the function f(x) such that f(a+b) = f(a) + f(b)

The sum of individual functions of a and b and should be equal to the function of (a+b). We should look for an option where x is in the numerator and there is no addition/subtraction. So the first option I will try is (E)

If f(x) = x/4, f(a) = a/4, f(b) = b/4, f(a+b) = (a+b)/4

f(a) + f(b) = a/4 + b/4 = (a+b)/4

Hence, for option (E), f(a+b) = f(a) + f(b)

Answer (E)
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink] New post 04 Jun 2014, 22:11
g(a + b, a + b) = g(a, a) + g(b, b)

LHS

g(a + b, a + b)

= f(a+b) + f(a+b)

= 2f(a+b)

RHS

g(a, a) + g(b, b)

= f(a) + f(a) + f(b) + f(b)

= 2f(a) + 2f(b)

LHS = RHS

2f(a+b) = 2f(a) + 2f(b)

f(a+b) = f(a) + f(b)


\(\frac{a}{4} +\frac{b}{4} = \frac{a+b}{4}\)

Answer =E
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Re: Let the function g(a, b) = f(a) + f(b).   [#permalink] 04 Jun 2014, 22:11
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