Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

12 Dec 2011, 06:52

4

This post received KUDOS

51

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

15% (low)

Question Stats:

72% (01:55) correct
28% (00:46) wrong based on 806 sessions

HideShow timer Statistics

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?

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.

Answer: E.

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:

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.

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.
_________________

Question asks you to check which of the provided options satisfy the equality "f(a+b) = f(a) + f(b)" Answer is E, 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.

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..
_________________

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.

Answer: E.

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:

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.

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.

Could someone please help me to understand this gmatprep question? Thanks!

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!!!

Answer E.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

Show Tags

09 Dec 2014, 11:41

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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
_________________

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

Show Tags

15 Jan 2016, 07:37

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.

Answer: E.

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:

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.

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.

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...