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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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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, 02:41, 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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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]

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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

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

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New post 30 Nov 2012, 03:52
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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New post 02 May 2014, 00:08
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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New post 04 Jun 2014, 04:18
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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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]

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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]

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New post 23 Jul 2015, 03:21
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..?

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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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New post 03 Nov 2016, 23:54
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.

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.


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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New post 11 Oct 2017, 01:09
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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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] 11 Oct 2017, 03:43
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