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Senior Manager  Joined: 21 Dec 2009
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Let the function g(a, b) = f(a) + f(b).  [#permalink]

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Question Stats: 39% (02:42) correct 61% (02:14) wrong based on 721 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, 19:17.
Last edited by Bunuel on 30 Nov 2012, 02:41, edited 1 time in total.
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Joined: 02 Sep 2009
Posts: 58435
Re: Let the function g(a, b) = f(a) + f(b).  [#permalink]

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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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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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Senior Manager  Joined: 21 Dec 2009
Posts: 457
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Re: Let the function g(a, b) = f(a) + f(b).  [#permalink]

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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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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 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: 35 GMAT 1: 620 Q42 V33 Re: Let the function g(a, b) = f(a) + f(b). [#permalink] Show Tags 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 V Joined: 16 Oct 2010 Posts: 9706 Location: Pune, India Re: Let the function g(a, b) = f(a) + f(b). [#permalink] Show Tags 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: 1749 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 1 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: 39 Schools: Broad '18 (WA$)
GMAT 1: 700 Q44 V41 GPA: 3.11
Re: Let the function g(a, b) = f(a) + f(b).  [#permalink]

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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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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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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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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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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

A little bit of forethought will help you solve it quickly.

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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_________________ Re: Let the function g(a, b) = f(a) + f(b).   [#permalink] 15 Oct 2018, 06:26
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