Let the function g(a, b) = f(a) + f(b). : GMAT Problem Solving (PS)
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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]

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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.
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink]

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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.

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Re: Let the function g(a, b) = f(a) + f(b). [#permalink]

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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.

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]

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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]

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23 Jul 2013, 01:15
From 100 hardest questions.

Bumping for review and further discussion.
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Re: Let the function g(a, b) = f(a) + f(b). [#permalink]

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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]

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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]

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04 Jun 2014, 20:19
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)

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1858 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Followers: 44 Kudos [?]: 1834 [0], given: 193 Re: Let the function g(a, b) = f(a) + f(b). [#permalink] ### Show Tags 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 _________________ Kindly press "+1 Kudos" to appreciate GMAT Club Legend Joined: 09 Sep 2013 Posts: 12904 Followers: 562 Kudos [?]: 158 [0], given: 0 Re: Let the function g(a, b) = f(a) + f(b). [#permalink] ### Show Tags 22 Jul 2015, 04:29 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Current Student Joined: 14 May 2014 Posts: 47 Schools: Broad '18 (WA) GMAT 1: 700 Q44 V41 GPA: 3.11 Followers: 0 Kudos [?]: 1 [0], given: 39 Re: Let the function g(a, b) = f(a) + f(b). [#permalink] ### Show Tags 23 Jul 2015, 02:21 Bunuel 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 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. here i got confused because the given function "g" had 2 input variables whereas all option choices had single variable. it did not strike me that function g can be completely written in form of f. any tips on this..? how to avoid this or what shd be the line of thinking..? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7076 Location: Pune, India Followers: 2088 Kudos [?]: 13300 [1] , given: 222 Re: Let the function g(a, b) = f(a) + f(b). [#permalink] ### Show Tags 23 Jul 2015, 03:26 1 This post received KUDOS Expert's post riyazgilani wrote: here i got confused because the given function "g" had 2 input variables whereas all option choices had single variable. it did not strike me that function g can be completely written in form of f. any tips on this..? how to avoid this or what shd be the line of thinking..? Yes, the tip is very simple: read the question very carefully. Note that it says: For which function f below ... So basically what you are given below (in options) is function f. So obviously it will have a single input. Now you want certain condition in g to hold. Since you know the equivalency of f and g, convert g to f and you know which condition f should hold. Hope it makes sense. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Let the function g(a, b) = f(a) + f(b). [#permalink]

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Hello from the GMAT Club BumpBot!

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Re: Let the function g(a, b) = f(a) + f(b). [#permalink]

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03 Nov 2016, 22:54
Bunuel 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

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.

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.

I was trying to solve the same problem with values a=2 and b=3, therefore then the question becomes: for which function f below will $$f(5)=f(2)+f(3)$$.
Based on this I observed that for option C:|x|,
f(5)=|5|=5 and f(2)+f(3)=|2|+|3|=2+3=5

now I know the OA is E and the functions are equal for option E, but am I missing something in option C, it would be great if someone could shed some light on this.
Re: Let the function g(a, b) = f(a) + f(b).   [#permalink] 03 Nov 2016, 22:54
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