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Manager  Joined: 27 Feb 2010
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For which of the following functions is f(a + b) = f(a) + f(b) for all  [#permalink]

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For which of the following functions is f(a + b) = f(a) + f(b) for all positive numbers a and b?

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

B. $$f(x)= x+1$$

C. $$f(x) = \sqrt{x}$$

D. $$f(x)=\frac{2}{x}$$

E. $$f(x) = -3x$$
Math Expert V
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Responding to a PM.

For which of the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?
A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Similar questions to practice:
function-85751.html
functions-problem-need-help-93184.html
let-the-function-g-a-b-f-a-f-b-143311.html

Hope it helps.
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Re: For which of the following functions is f(a+b)=f(b)+f(a)  [#permalink]

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enigma123 wrote:
For which of the following functions is f(a+b)=f(b)+f(a) for all positive numbers a and b?

A. f(x)=x^2
B. f(x)=x+1
C. f(x)=√x
D. f(x)=2/x
E. f(x)=-3x

One approach is to plug in numbers. Let's let a = 1 and b = 1

So, the question becomes, "Which of the following functions are such that f(1+1) = f(1) + f(1)?"
In other words, for which function does f(2) = f(1) + f(1)?

A) If f(x)=x², does f(2) = f(1) + f(1)?
Plug in to get: 2² = 1² + 1²? (No, doesn't work)
So, it is not the case that f(2) = f(1) + f(1), when f(x)=x²

B) If f(x)=x+1, does f(2) = f(1) + f(1)?
Plug in to get: 2+1 = 1+1 + 1+1? (No, doesn't work)
So, it is not the case that f(2) = f(1) + f(1)
.
.
.
A, B, C and D do not work.
So, at this point, we can conclude that E must be the correct answer.
Let's check E anyway (for "fun")

E) If f(x)=-3x, does f(2) = f(1) + f(1)?
Plugging in 2 and 1 we get: (-3)(2) = (-3)(1) + (-3)(1)
Yes, it works

Cheers,
Brent
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Question asks you to check which of the provided options satisfy the equality "f(a+b) = f(a) + f(b)"
if you apply f(x) = -3x to f(a+b) then
f(a+b) = -3(a+b) = -3a -3b
f(a) = -3a
f(b) = -3b
f(a) + f(b) = -3a -3b = f(a+b)
Hope I am clear.
##### General Discussion
Math Expert V
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For which of the following functions is f(a + b) = f(a) + f(b) for all  [#permalink]

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For which of the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?

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

B. $$f(x)= x+1$$

C. $$f(x) = \sqrt{x}$$

D. $$f(x)=\frac{2}{x}$$

E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Similar questions:
http://gmatclub.com/forum/for-which-of- ... 24491.html
http://gmatclub.com/forum/for-which-of- ... 85751.html
http://gmatclub.com/forum/let-the-funct ... 43311.html

Hope it helps.
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Re: For which of the following functions is f(a + b) = f(a) + f(b) for all  [#permalink]

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Thank you Bunuel. This is really a good explaination. I did some samples based on your explaination and i am confident that I can handle these kind or problems.
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Re: For which of the following functions is f(a + b) = f(a) + f(b) for all  [#permalink]

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Bunuel wrote:
hardnstrong wrote:
yes can anyone explain how to proceed with function problems .....this is one of my weakest area

For which of the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Hope it helps.

Thanks bunuel
You have solved my problem here. i think i can handle these questions now
+1
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Re: For which of the following functions is f(a + b) = f(a) + f(b) for all  [#permalink]

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Bunuel wrote:
hardnstrong wrote:
yes can anyone explain how to proceed with function problems .....this is one of my weakest area

For which of the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Hope it helps.

thanks a ton for the explanation. There are not many problems on functions in samples and I am glad now i know how to go about for questions ike this.
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Re: For which of the following functions is f(a + b) = f(a) + f(b) for all  [#permalink]

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For these kind of questions where you have to test each choice, ALWAYS start with E.

E. f(x) = -3x
f(a+b) = -3(a+b) = -3a -3b
f(a) = -3a ; f(b) = -3b
f(a) + f(b)= -3a -3b = f(a+b)

On test day - stop.

Remember, just substitute for x with whatever is in the brackets in f ( )
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Re: For which of the following functions is f(a + b) = f(a) + f(b) for all  [#permalink]

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I agree with AbhayPrasanna.

If you want an explanation of why it doesn't work for the other functions, look at the type of each function.

$$x^2$$ is a quadratic function, so $$f(a+b) = (a+b)^2$$ and $$f(a) + f(b) = a^2 + b^2$$ - Wrong

$$x+1$$ has a constant variable in it, though its linear. So $$f(a+b) = a+b+1$$ and $$f(a) + f(b) = (a+1)+(b+1) = a+b+2$$ - Wrong

$$\sqrt{x}$$ is a root function. $$f(a+b) = \sqrt{a+b}$$ and $$f(a)+f(b) = \sqrt{a}+\sqrt{b}$$ - Wrong

$$2/x$$ is a fraction type function. So $$f(a+b) = 2/(a+b)$$ and $$f(a)+f(b) = 2/a + 2/b = 2(a+b)/ab$$ - Wrong

$$-3x$$ is a linear function without constants. So this must be the answer. But to check:

$$f(a+b) = -3*(a+b)$$ and $$f(a) + f(b) = -3*a + (-3*b) = -3* (a+b)$$ - Right

This kind of elimination is not necessary on the exam. However, it's good to know why the other options don't work.
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Bunuel wrote:
hardnstrong wrote:
yes can anyone explain how to proceed with function problems .....this is one of my weakest area

For which of the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Hope it helps.

Thanks for really fleshing out the algebra on this problem Bunuel. The problems seem fairly easy once you understand how to work functions properly.
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f(a+b) has to be equal to f(a) + f(b)

A. f(x)=x^2
B. f(x)=x+1
C. f(x)=√x
D. f(x)=2/x
E. f(x)=-3x

A) f(x) = x^2
f(a) = a^2; f(b)=b^2
f(a+b) = (a+b)^2 = a^2+b^2 +2ab
f(a) +f(b) = a^2 + b^2 <> f(a+b)

so on and so forth...each option will lead to the same result except for E

Moreover, just by observing it can be found out the square roots, sqaures and x in the denominator will not be correct answers hence just try with E and you can find out..
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enigma123 wrote:
For which of the following functions is f(a+b)=f(b)+f(a) for all positive numbers a and b?
A. f(x)=x^2
B. f(x)=x+1
C. f(x)=√x
D. f(x)=2/x
E. f(x)=-3x

Can someone please tell me how to solve this? As OA is not provided I started by substituing the value of f(x) in the answer choice but got stuck. Can someone please help?

You can save time by using an intuitive method. Look for the expression that satisfies the distributive property i.e. x * (y + z) = (x * y) + (x * z)

When you put (a+b), it should give you individual functions in a and b which means that you will get two separate, comparable terms in a and b.
Squares, roots, addition and division by the variable does not satisfy the distributive property.
Multiplication does. So check for option (E) first.

One rule of thumb - in such questions, try the options which have multiplication/addition first. These two operators have various properties which make such relations possible.
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BigUp wrote:
Hello GmatClub,

For which of the following functions is f(a+b) = f(a)+f(b) for all positive numbers a and b?

f(x)=x^2
f(x)=x+1
f(x)=sqrt(x)
f(x)=2/x
f(x)=-3x

First, a remark: $$a$$ and $$b$$ are considered positive because the function in C, the square root is not defined for negative numbers and the function in D is not defined for $$x=0$$ ($$x$$ being in the denominator). The other functions are defined for any real number.

For each function, we translate the given equality and check whether is holds for any positive $$a$$ and $$b$$. If the equality holds for any $$a$$ and $$b$$, we should get an identity, which means the same expression on both sides of the equal sign.
For $$f(a+b)$$ we take the expression of any of the given functions and replace $$x$$ by $$(a+b)$$.

(A) $$(a+b)^2=a^2+b^2$$ or $$a^2+2ab+b^2=a^2+b^2$$. Necessarily $$ab=0$$, which cannot hold, $$a$$ an $$b$$ being positive. NO
(B) $$a+b+1=a+1+b+1$$ gives $$1=2$$, impossible. NO
(C) $$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$ NO (check for example $$a=b=1$$)
(D) $$\frac{2}{a+b}=\frac{2}{b}+\frac{2}{b}$$ NO (again, check for $$a=b=1$$)
(E) $$-3(a+b)=-3a+(-3b)$$ or $$-3a-3b=-3a-3b$$ YES!!!

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Best to answer this with tiny-winie numbers such as a=1, b=1 and a+b=1...

A. (1) + (1) = 4 OUT!
B. (1+1) + (1+1) = 3 OUT!
C. 1 + 1 = $$\sqrt{2}$$ OUT!
D. 2/1 + 2/1 = 2/2 OUT!
E. -3(1) -3(1) = -3(2) BINGO!

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Re: For which of the following functions is f(a + b) = f(a) + f(b) for all  [#permalink]

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Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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Well, i did not start with picking no's or keeping the no's as a and b,

But just worked in the following way :-
if, f(a+b) = f(a) + f(b)
Then,
since our answers are in this format,

f(x+x) = f(x) + f(x) should also hold true, for all positive no's a and b (we havent been told that it has to be distinct, so this should be perfectly valid)

Hence, you need to evaluate f(2x) here, and check weather it is equal to 2f(x)

This seems to be the case only for E, hence, the answer.
Quick and easy.
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zz0vlb wrote:
For which of the following functions is f(a+b)=f(a)+f(b) for all positive numbers a and b?

A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

f(a+b) = f(a) + f(b) holds true for pure multiplication with any number

D: $$\frac{2}{x} = 2 * x^{-1}$$ ......... power function.... discard

E: Only Multiplication.... This will be the answer

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

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sagarag wrote:
For which of the following functions is F(a+b) = f(a) + f(b) for all positives numbers a and b?

A) f(x) = x^2
B) f(x) = x+1
C) f(x) = x^1/2
D) f(x) = 2/x
e) f(x) = -3x

hi sagar..
you can eliminate the choices by looking at the choices..
the answer cannot be a variable added or subtracted with a constant.. since that value will get added/subtracted twice on right side... B is out
it cannot be a variable multiplied with another variable or with self.. A and C out..
the answer can be a variable multiplied or divided by a constant...
D is out as a constant is divided by variable.. E follows the above rule
ans E..
you can also find answer by testing values ..
take E for example..
since f(x) = -3x, f(a) = -3a f(b) = -3b.. f(a+b)=-3(a+b)=-3a+(-3b)=f(a)+f(b)... E is the ans
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Re: For which of the following functions is f(a+b)=f(b)+f(a)  [#permalink]

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Bunuel wrote:
Responding to a PM.

For which of the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?
A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Similar questions to practice:
function-85751.html
functions-problem-need-help-93184.html
let-the-function-g-a-b-f-a-f-b-143311.html

Hope it helps.

I always get confused with these kind of questions and I like the method to pick numbers to check whether answer choices are equal to the main statement.
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Location: Pune, India
Re: For which of the following functions is f(a+b)=f(b)+f(a)  [#permalink]

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3
pepo wrote:
Bunuel wrote:
Responding to a PM.

For which of the following functions is f(a+b)= f(a)+f(b) for all positive numbers a and b?
A. $$f(x)=x^2$$
B. $$f(x)= x+1$$
C. $$f(x) = \sqrt{x}$$
D. $$f(x)=\frac{2}{x}$$
E. $$f(x) = -3x$$

A. $$f(a+b)=(a+b)^2=a^2+2ab+b^2\neq{f(a)+f(b)}=a^2+b^2$$

B. $$f(a+b)=(a+b)+1\neq{f(a)+f(b)}=a+1+b+1$$

C. $$f(a+b)=\sqrt{a+b}\neq{f(a)+f(b)}=\sqrt{a}+\sqrt{b}$$.

D. $$f(a+b)=\frac{2}{a+b}\neq{f(a)+f(b)}=\frac{2}{a}+\frac{2}{b}$$.

E. $$f(a+b)=-3(a+b)=-3a-3b=f(a)+f(b)=-3a-3b$$. Correct.

OR, as f(a+b)= f(a)+f(b) must be true for all positive numbers a and b, then you can randomly pick particular values of a and b and check for them:

For example: $$a=2$$ and $$b=3$$

A. $$f(a + b) = f(5) = 5^2 = 25\neq{f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13}$$

B. $$f(a + b) = f(5) = 5 + 1 = 6\neq{f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7}$$

C. $$f(a + b)=f(5)=\sqrt{5}\neq{f(a) + f(b)=f(2)+f(3)=\sqrt{2}+\sqrt{3}}$$

D. $$f(a + b)=f(5)=\frac{2}{5}\neq{f(a)+f(b)=f(2)+f(3)=\frac{2}{2}+\frac{2}{3}=\frac{5}{3}}$$

E. $$f(a + b)=f(5)=-3*(5) =-15=f(a)+f(b)=f(2)+f(3)=-3*(2)-3*(3)=-15$$. Correct.

It might happen that for some choices of a and b other options may be "correct" as well. If this happens just pick some other numbers and check again these "correct" options only.

Similar questions to practice:
function-85751.html
functions-problem-need-help-93184.html
let-the-function-g-a-b-f-a-f-b-143311.html

Hope it helps.

I always get confused with these kind of questions and I like the method to pick numbers to check whether answer choices are equal to the main statement.

Here are a couple of posts on functions. They could help you.

http://www.veritasprep.com/blog/2015/03 ... s-on-gmat/
http://www.veritasprep.com/blog/2015/03 ... questions/
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Karishma
Veritas Prep GMAT Instructor Re: For which of the following functions is f(a+b)=f(b)+f(a)   [#permalink] 15 Jan 2016, 08:59

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# For which of the following functions is f(a + b) = f(a) + f(b) for all  