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

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.

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

The correct answer is E

Cheers,
Brent

Thanks for really fleshing out the algebra on this problem Bunuel. The problems seem fairly easy once you understand how to work functions properly.

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..

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.

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!

Answer: E

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.

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/03 ... s-on-gmat/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/03 ... questions/

Attached is a visual that should help. Alternate method: plug in a=1 and b=1 to make the math easier.

hi

A: (a + b) ^ 2 is not equal to a^ + b ^2, discard
B: a + b + 1 is not equl to b + 1 + a + 1, discard
C: square rootover (a + b) is not equal to rootover a + rootover b , discard
D: 2/ a+b is not equal to 2/a + 2/b
E: -3 (a + b) is equal to -3a -3b, so this is the correct choice

thanks

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


I substituted x in a and b -> f(a+b)= f(a) + f(b) --> f(x+x)= f(x) + f(x) --> f(2x)= 2 f(x)
so I have to find the option that satisfies this equation.
a) f(2x)= 4x^2 vs 2f(x)=2x^2 ( Not equal)
b) f(2x)=2x+1 vs 2f(x)= 2x+2 ( Not equal)
c) f(2x)= sqrt 2*sqrt x vs 2f(x)= 2 * sqrt x (Not equal)
d) f(2x)=1/x vs 2f(x)= 4/x (Not equal)
e) f(2x)=-6x vs 2f(x)= -6x (equal)
Therefore e) is the answer.

We need to determine when f(a + b) = f(a) + f(b). Before we evaluate each answer choice it may be easier to use numerical values for a and b. If we let a = 1 and b = 2, our new function looks like:

f(1 + 2) = f(1) + f(2)

f(3) = f(1) + f(2)

So we must determine when the output of f(3) equals the sum of the outputs of f(1) and f(2).

Let’s now evaluate each answer choice.

A) f(x) = x^2

f(3) = 3^2 = 9

f(1) = 1^2 = 1

f(2) = 2^2 = 4

Since 9 does not equal 1 + 4, choice A is not correct.

B) f(x) = x + 1

f(3) = 3 + 1 = 4

f(1) = 1 + 1 = 2

f(2) = 2 + 1 = 3

Since 4 does not equal 2 + 3, choice B is not correct.

C) f(x) = √x

f(3) = √3

f(1) = √1 = 1

f(2) = √2

Since √3 does not equal 1 + √2, choice C is not correct.

D) f(x) = 2/x

f(3) = 3/2

f(1) = 2/1 = 2

f(2) = 2/2 = 1

Since 3/2 does not equal 2 + 1, choice D is not correct.

Since we have eliminated all the other answer choices, we know the answer is E. However, let’s show as an exercise that answer choice E satisfies the given property for our choice of numbers:

E) f(x) = -3x

f(3) = -3(3) = -9

f(1) = -3(1) = -3

f(2) = -3(2) = -6

Since -9 equals -3 + (-6), choice E is correct.

Answer: E

Solution:

Use the options to answer the question

A. f(a+b)=(a+b)^2=a^2+2ab+b^2 and this is NOT equal to f(a)+f(b) which is a^2+b^2 (Eliminate)

B. f(a+b)=(a+b)+1 and this is NOT equal to f(a)+f(b) which =a+1+b+1=a+b+2 (Eliminate)

C. f(a+b)=√(a+b) and this is NOT equal to f(a)+f(b) which is = √a+√b (Eliminate)

D. f(a+b)=2/a+b and this is NOT equal to f(a)+f(b) which is =2/a+2/b (Eliminate)


E. f(a+b)=−3(a+b)=−3a−3b and this is EQUAL to f(a)+f(b)=−3a−3b. (option e)

Devmitra Sen
GMAT SME

Video solution from Quant Reasoning starts at 8:40
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Given that f(a + b) = f(a) + f(b) and we need to find which of the following can be the value of f(x) which satisfies this.

Let's solve the problem using two methods

Method 1: Logic (Eliminate Option Choices)

f(a+b) = f(a) + f(b)

Now, this can be true only when

1. We don't have any constant term added or subtracted from any term of x. As if we have one then on Left Hand Side(LHS) that constant term will be added or subtracted only once, but on Right Hand Side(RHS) it will be added or subtracted twice.
2. We don't have x in the denominator (in general) as we wont be able to match the LHS and RHS then.
3. We don't have any power of x ≠ 1 in the numerator. As otherwise (in general) we wont be able to match the LHS and RHS.
4. We have a term of x in the numerator with power of 1 with any positive or negative constant multiplied with it. Ex 2x, -3x, etc

Using above logic we can eliminate the answer choices

(A) \(f(x)=x^2\)
=> Eliminate : Doesn't Satisfy Point 3 above. Power of x is 2.

(B) \(f(x)= x+1\)
=> Eliminate : Doesn't Satisfy Point 1 above. It has a constant added (+1)

(C) \(\sqrt{x}\)
=> Eliminate : Doesn't Satisfy Point 3 above. Power of x is \(\frac{1}{2}\)

(D) \(f(x)=\frac{2}{x}\)
=> Eliminate : Doesn't Satisfy Point 2 above. x is in denominator

(E) \(-3x\)
=> POSSIBLE: Satisfies all the conditions above.

So, Answer will be E.

Method 2: Algebra (taking all option choices)

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

To find f(a+b) we need to compare what is inside the bracket in f(a+b) and f(x)
=> We need to substitute x with a+b in \(f(x)=x^2\) to get the value of f(a+b)

=> f(a+b) = \((a+b)^2\) = \(a^2 + 2ab + b^2\)
f(a) = \(a^2\) and f(b) = \(b^2\)
=> f(a) + f(b) = \(a^2\) + \(b^2\) = \(a^2 + b^2\) ≠ \(a^2 + 2ab + b^2\)
=> f(a+b) ≠ f(a) + f(b) => FALSE

(B) \(f(x)= x+1\)

=> f(a+b) = \(a+b + 1\)
=> f(a) + f(b) = \(a + 1\) + \(b + 1\) = \(a+b + 2\) ≠ \(a+b + 1\)
=> f(a+b) ≠ f(a) + f(b) => FALSE

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

=> f(a+b) = \(\sqrt{a + b}\)
f(a) + f(b) = \(\sqrt{a}\) + \(\sqrt{b}\) ≠ \(\sqrt{a + b}\)
=> f(a+b) ≠ f(a) + f(b) => FALSE

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

=> f(a+b) = \(\frac{2}{a+b}\)
f(a) + f(b) = \(\frac{2}{a}\) + \(\frac{2}{b}\) = \(\frac{2a + 2b }{ ab}\) ≠ \(\frac{2}{a+b}\)
=> f(a+b) ≠ f(a) + f(b) => FALSE

(E) \(f(x) = -3x\)

=> f(a+b) = \(-3*(a+b)\) = -3a - 3b
f(a) + f(b) = \(-3a\) + \(-3b\) = -3a - 3b
=> f(a+b) = f(a) + f(b) => TRUE

So, Answer will be E
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

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