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# For which of the following functions f(x) is f(a + b) = f(a) ?

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Math Expert
Joined: 02 Sep 2009
Posts: 49993
For which of the following functions f(x) is f(a + b) = f(a) ?  [#permalink]

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31 Jul 2018, 00:31
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15% (low)

Question Stats:

82% (00:59) correct 18% (01:06) wrong based on 49 sessions

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For which of the following functions f(x) is $$f(a + b) = f(a) + f(b)$$?

(A) $$f(x) = x^2$$

(B) $$f(x) = 5x$$

(C) $$f(x) = 2x + 1$$

(D) $$f(x) =\sqrt{x}$$

(E) $$f(x) = x - 2$$

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Posts: 713
Re: For which of the following functions f(x) is f(a + b) = f(a) ?  [#permalink]

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31 Jul 2018, 01:32
Bunuel wrote:
For which of the following functions f(x) is $$f(a + b) = f(a) + f(b)$$?

(A) $$f(x) = x^2$$

(B) $$f(x) = 5x$$

(C) $$f(x) = 2x + 1$$

(D) $$f(x) =\sqrt{x}$$

(E) $$f(x) = x - 2$$

Instead of trying to work with the equations, we'll pick easy numbers to eliminate incorrect answers.
This is an Alternative approach.

Say a = 1 and b = 1, so a+b = 2.
Then (A) gives f(2) = 4 but f(1) + f(1) = 2. NO!
(B) gives f(2) = 10 and f(1)+f(1) = 10. Maybe let's check the others.
(C) gives f(2) = 5 but f(1) + f(1) = 6. NO!
(D) gives f(2) = $$\sqrt{2}$$ while f(1)+f(1) = 2. NO
(E) gives f(2) = 0 and f(1) + f(1) = -2. NO

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Joined: 09 Jul 2018
Posts: 9
Re: For which of the following functions f(x) is f(a + b) = f(a) ?  [#permalink]

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31 Jul 2018, 17:43
I did not really understand what the question is asking, can someone explain the question?
Math Expert
Joined: 02 Sep 2009
Posts: 49993
Re: For which of the following functions f(x) is f(a + b) = f(a) ?  [#permalink]

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01 Aug 2018, 00:50
1
Sandy56 wrote:
I did not really understand what the question is asking, can someone explain the question?

I explained very similar question in detail HERE.

Also, check below topics:

13. Functions

For more check below:
ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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Joined: 17 Mar 2014
Posts: 383
For which of the following functions f(x) is f(a + b) = f(a) ?  [#permalink]

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Updated on: 01 Aug 2018, 20:50
2
Sandy56 wrote:
I did not really understand what the question is asking, can someone explain the question?

Question asks which of below functions can behave like$$f(a+b)=f(a)+f(b)$$.

Now what does$$f(a+b)$$ mean ? It can be any function. when we reading question stem we don't know its value. We need to find out of available options.

Now what does$$f(a+b)=f(a)+f(b)$$ mean ? it means, for example, f(5)=5 and f(6)=6 then f(5+6)= 5+6= f(5)+(6).

Now check available options one by one.

(A)$$f(x)=x^2$$ => $$f(a)=a^2$$ , $$f(b)=b^2$$ and $$f(a+b)=(a+b)^2$$$$= a^2+2ab+b^2$$ which is not equal to $$f(a)+(b)$$

(B)$$f(x)=5x$$ =>$$f(a)=5a$$, $$f(b)=5b$$ and $$f(a+b)=5(a+b)= 5a+5b= f(a) +f(b)$$ Whoohoo..here you go. This is the answer

(C) $$f(x)=2x+1$$

(D)$$f(x)=\sqrt{x}$$

(E) $$f(x)=x−2$$

Pls find similar question to practice: https://gmatclub.com/forum/for-which-of ... 24491.html

Originally posted by ammuseeru on 01 Aug 2018, 20:38.
Last edited by ammuseeru on 01 Aug 2018, 20:50, edited 1 time in total.
Senior Manager
Joined: 17 Mar 2014
Posts: 383
For which of the following functions f(x) is f(a + b) = f(a) ?  [#permalink]

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01 Aug 2018, 20:41
DavidTutorexamPAL wrote:
Bunuel wrote:
For which of the following functions f(x) is $$f(a + b) = f(a) + f(b)$$?

(A) $$f(x) = x^2$$

(B) $$f(x) = 5x$$

(C) $$f(x) = 2x + 1$$

(D) $$f(x) =\sqrt{x}$$

(E) $$f(x) = x - 2$$

Instead of trying to work with the equations, we'll pick easy numbers to eliminate incorrect answers.
This is an Alternative approach.

Say a = 1 and b = 1, so a+b = 2.
Then (A) gives f(2) = 4 but f(1) + f(1) = 2. NO!
(B) gives f(2) = 10 and f(1)+f(1) = 10. Maybe let's check the others.
(C) gives f(2) = 5 but f(1) + f(1) = 6. NO!
(D) gives f(2) = $$\sqrt{2}$$ while f(1)+f(1) = 2. NO
(E) gives f(2) = 0 and f(1) + f(1) = -2. NO

In my opinion, it is very basic fundamental question. I don't know whether we really need number to put in and check. If student's basic math fundamentals are clear, Student should be able to solve it visually.
For which of the following functions f(x) is f(a + b) = f(a) ? &nbs [#permalink] 01 Aug 2018, 20:41
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