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

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

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Updated on: 30 Nov 2012, 01:41
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Question Stats:

39% (02:44) correct 61% (02:14) wrong based on 634 sessions

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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

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Originally posted by gmatbull on 29 Nov 2012, 18:17.
Last edited by Bunuel on 30 Nov 2012, 01:41, edited 1 time in total.
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Joined: 02 Sep 2009
Posts: 53067
Re: Let the function g(a, b) = f(a) + f(b).  [#permalink]

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30 Nov 2012, 02:02
23
22
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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29 Nov 2012, 21:55
14
5
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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##### General Discussion
Director
Joined: 21 Dec 2009
Posts: 512
Concentration: Entrepreneurship, Finance
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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Joined: 29 May 2013
Posts: 21
Concentration: Marketing, Entrepreneurship
Schools: Molson '17 (M$) GMAT Date: 08-20-2014 GPA: 3.34 Re: Let the function g(a, b) = f(a) + f(b). [#permalink] ### Show Tags 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' Intern Joined: 13 Dec 2013 Posts: 36 GMAT 1: 620 Q42 V33 Re: Let the function g(a, b) = f(a) + f(b). [#permalink] ### Show Tags 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. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8895 Location: Pune, India Re: Let the function g(a, b) = f(a) + f(b). [#permalink] ### Show Tags 04 Jun 2014, 20:19 1 1 1 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) _________________ Karishma Veritas Prep GMAT Instructor Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options > SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1820 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: Let the function g(a, b) = f(a) + f(b). [#permalink] ### Show Tags 04 Jun 2014, 22:11 1 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 Intern Joined: 14 May 2014 Posts: 41 Schools: Broad '18 (WA$)
GMAT 1: 700 Q44 V41
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Re: Let the function g(a, b) = f(a) + f(b).  [#permalink]

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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.

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

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23 Jul 2015, 03:26
1
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.
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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.
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Re: Let the function g(a, b) = f(a) + f(b).  [#permalink]

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11 Oct 2017, 00:09
Even I fully understand how to solve this problem, I don't think I can get the answer in 2 mins.... TOO MUCH CALCULATION
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Re: Let the function g(a, b) = f(a) + f(b).  [#permalink]

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11 Oct 2017, 02:43
3
patrickpeng wrote:
Even I fully understand how to solve this problem, I don't think I can get the answer in 2 mins.... TOO MUCH CALCULATION

You want:
g(a + b, a + b) = g(a, a) + g(b, b)

In the expression, you want that when you put a+b, it should be the same as when you put a and b individually.

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

Of the given options, x^2 may not be it since (a+b)^2 is not same as a^2 + b^2.
Similarly, |a+b| is not the same as |a| + |b|
In such questions, with x in the denominator, you certainly cannot split 1/(a+b) as 1/a + 1/b.

I will try option (E) first since it has a multiplication (x * 1/4). We know that (a+b)*1/4 = a*(1/4) + b*(1/4)
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Re: Let the function g(a, b) = f(a) + f(b).  [#permalink]

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15 Oct 2018, 05:26
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Re: Let the function g(a, b) = f(a) + f(b).   [#permalink] 15 Oct 2018, 05:26
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