Let the function g(a, b) = f(a) + f(b).

Let the function g(a, b) = f(a) + f(b).

For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?

A: x +3
B: x^2
C: |x|
D: 1/x
E: x/4

We can use plug-in method to solve this question.

Since \(g(a, b) = f(a) + f(b)\), then:
\(g(a + b, a + b)=f(a+b)+f(a+b)=2f(a+b)\);
\(g(a, a) + g(b, b)=f(a)+f(a)+f(b)+f(b)=2f(a)+2f(b)\).

Thus the question asks: for which function f below will \(2f(a+b)=2f(a)+2f(b)\) --> \(f(a+b)=f(a)+f(b)\)?

Say a=-1 and b=1, then the question becomes: for which function f below will \(f(0)=f(-1)+f(1)\).

A: x +3
\(f(0)=0+3=3\)
\(f(-1)+f(1)=(-1+3)+(1+3)=6\)
No match.

B: x^2
\(f(0)=0^2=0\)
\(f(-1)+f(1)=(-1)^2+(1)^2=2\)
No match.

C: |x|
\(f(0)=|0|=0\)
\(f(-1)+f(1)=|-1|+|1|=2\)
No match.

D: 1/x
\(f(0)=\frac{1}{0}=undefined\)
\(f(-1)+f(1)=\frac{1}{-1}+\frac{1}{1}=0\)
No match.

E: x/4
\(f(0)=\frac{0}{4}=0\)
\(f(-1)+f(1)=\frac{-1}{4}+\frac{1}{4}=0\)
Match.

Answer: E.

Hope it's clear.

for-which-of-the-following-functions-is-f-a-b-f-b-f-a-124491.html
for-which-of-the-following-functions-f-is-f-x-f-1-x-for-85751.html
functions-problem-need-help-93184.html

Hope it helps.
_________________

The question is basically asking for which function f(a + b) = f(a) + f(b)

A)\(LHS = a + b + 3\)
\(RHS = a + 3 + b + 3 = a + b + 6.\)
Not equal

B)\(LHS = (a + b)^2\)
\(RHS = a^2 + b^2\)
Not equal

C)\(LHS = |a + b|\)
\(RHS = |a| + |b|\)
If a & b are of different polarity not equal.

D)\(LHS = \frac{1}{a + b}\)
\(RHS = \frac{1}{a} + \frac{1}{b}\)
Not equal

E)\(LHS = \frac{a + b}{4}\)
\(RHS = \frac{a}{4} + \frac{b}{4} = \frac{a + b}{4}\).
Equal.

Answer should be E

Bunuel and MacFauz,
your explanations make a lot of sense.
More so, Bunuel, you took time to explain the breakdown.

Thanks.

let a = 1 and b = 2, a+b = 3

A) x+3,
2(a+b+3) = 2(a+3) + 2(b+3)
2(a+b+3) = 2(a+b+3) + 6 [Can never be true]

B) x^2
2(a+b)^2 = 2(a^2)+2(b^2) [Can/Cannot be true]

C)|x|
2|a+b| = 2|a| + 2|b| [Can/Cannot be true]

D)1/x
2/(a+b) = (2/a)+(2/b) = 2(a+b)/ab [Can/Cannot be true]

E)x/4
2*(a+b)/4 = 2(a/4) + 2(b/4) = 2(a+b)/4 [Always true]

Ans 'E'

g(a+b,a+b) = g(a,a)+g(b,b) = f(a) + f(a) + f(b) + f(b)= 2f(a) + 2f(b) = RHS
g(a+b,a+b) = f(a+b)+f(a+b) = 2f(a+b) = LHS

We divide by 2 both RHS and LHS and get

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

I have an issue here: how do I know what to plug in from the answer choices? Since those are expressed in terms of X.

My approach was to take both a and b as the answer choice, but I don´t know if that is correct.

For d)

LHS: f(x/4 + x/4) = x/2
RHS: f(x/4) + f(x/4) = x/2

I'm sure i'm conceptually missing something.

Thanks.

I am not sure what you mean by this last line but I can help you with the various variables.

The options (x+3), x^2 etc are the values of the function f(x)

Option (A) tells you that f(x) = x + 3
So if you want to find f(a) or f(b) or f(a+b), it is quite simple.
If f(x) = x+3, f(a) = a+3
If f(x) = x+3, f(a+b) = a+b+3
etc
Wherever you have x in the expression you put a or a+b or b as the case may be.

In this question, since options give the function f(x), you convert the entire g(x) into f(x).
You get that you need to find the function f(x) such that f(a+b) = f(a) + f(b)

The sum of individual functions of a and b and should be equal to the function of (a+b). We should look for an option where x is in the numerator and there is no addition/subtraction. So the first option I will try is (E)

If f(x) = x/4, f(a) = a/4, f(b) = b/4, f(a+b) = (a+b)/4

f(a) + f(b) = a/4 + b/4 = (a+b)/4

Hence, for option (E), f(a+b) = f(a) + f(b)

Answer (E)

g(a + b, a + b) = g(a, a) + g(b, b)

LHS

g(a + b, a + b)

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

= 2f(a+b)

RHS

g(a, a) + g(b, b)

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

= 2f(a) + 2f(b)

LHS = RHS

2f(a+b) = 2f(a) + 2f(b)

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


\(\frac{a}{4} +\frac{b}{4} = \frac{a+b}{4}\)

Answer =E

here i got confused because the given function "g" had 2 input variables whereas all option choices had single variable. it did not strike me that function g can be completely written in form of f. any tips on this..? how to avoid this or what shd be the line of thinking..?

Yes, the tip is very simple: read the question very carefully.

Note that it says: For which function f below ...

So basically what you are given below (in options) is function f. So obviously it will have a single input.

Now you want certain condition in g to hold. Since you know the equivalency of f and g, convert g to f and you know which condition f should hold.

Hope it makes sense.

I was trying to solve the same problem with values a=2 and b=3, therefore then the question becomes: for which function f below will \(f(5)=f(2)+f(3)\).
Based on this I observed that for option C:|x|,
f(5)=|5|=5 and f(2)+f(3)=|2|+|3|=2+3=5

now I know the OA is E and the functions are equal for option E, but am I missing something in option C, it would be great if someone could shed some light on this.

Even I fully understand how to solve this problem, I don't think I can get the answer in 2 mins.... TOO MUCH CALCULATION

A little bit of forethought will help you solve it quickly.

You want:
g(a + b, a + b) = g(a, a) + g(b, b)

In the expression, you want that when you put a+b, it should be the same as when you put a and b individually.

A: x +3
B: x^2
C: |x|
D: 1/x
E: x/4

Of the given options, x^2 may not be it since (a+b)^2 is not same as a^2 + b^2.
Similarly, |a+b| is not the same as |a| + |b|
In such questions, with x in the denominator, you certainly cannot split 1/(a+b) as 1/a + 1/b.

I will try option (E) first since it has a multiplication (x * 1/4). We know that (a+b)*1/4 = a*(1/4) + b*(1/4)

I'm still a little confused about the commas. I've never seen a function written as such. E.g. g(a+b,a+b)
In this case, do we always assume that the comma becomes an addition of the two?

So if I see g(a-b,a+b), I could assume that would equal to f(a-b)+f(a+b)?

i had the same question two weeks ago, i hope one of the greatest geniuses of all times would reply!

I have the exact same question.
Bunuel, I recall in a previous post involving a similar question, you mentioned substituting with either -1, 0 or 1 wasn't a good idea. So I approached this with \(f(2+3) = f(2) + f(3)\). I realize looking at the examples given here, option C mostly doesn't work, but how could I have avoided this? As in due to the time it takes testing each option, as soon as I found option C satisfied f(2+3), I deemed it prudent to move on. Thanks in advance.
VeritasKarishma

In this question, yes! It's basically giving you the result of putting 2 values, if you put 2 values in the function "g" they will be added.

g(a, b) = f(a) + f(b)
g(a+b, a+b) = f(a+b) + f(a+b) = 2f(a+b)
g(a, a) = f(a) + f(a) = 2f(a)
g(b, b) = 2f(b)

So we want a value for f such that
2f(a+b) = 2f(a) + 2f(b)
i.e. f(a+b) = f(a) + f(b)

Now eliminate some options right away. We know that (a+b)^2 is not a^2 + b^2. We know that |a+b| may not be equal to |a| + |b|. We know that 1/(a+b) is not equal to 1/a + 1/b.
Option (A) gives us a single 3 on LHS but two 3s on RHS.
Only option (E) works.

achanak
You can try plugging in values. When you find that 2, 3 satisfies option (C), you have to check for (D) and (E) too. You will find that (E) works too. After that you will need to take another set of values for just (C) and (E) and see which one works. Since you have an absolute value sign in option (C), assume a negative value.

There is a reason for not assuming -1, 0, 1 usually - They end up validating multiple options. e.g.
x, x^2, |x| are all 0 when you put x = 0.
But when you put x = -2, they are all different.
Same logic goes for 1 and -1 too. Hence, it is a good idea to try other integers.

Ideally, question stem should be written as "For which value of function f(x) below will g(a + b, a + b) = g(a, a) + g(b, b)? instead of "For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?"

Ideally, question stem should be written as "For which value of function f(x) below will g(a + b, a + b) = g(a, a) + g(b, b)?" instead of "For which function f below will g(a + b, a + b) = g(a, a) + g(b, b)?"