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

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12 Dec 2011, 06:52

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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?

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..
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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.

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.
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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.
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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]

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09 Dec 2014, 11:41

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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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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.

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.

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