For which of the following functions f(x) is f(a + b) = f(a) ?

I did not really understand what the question is asking, can someone explain the question?
Thank you in advance

I explained very similar question in detail HERE.

Also, check below topics:

13. Functions

• Theory
  How to Interpret Unfamiliar Symbols
  Functions on GMAT
  A Closer Look at GMAT Function Questions
  
  • Questions
    Operations/functions defining algebraic/arithmetic expressions
    Rounding Functions Problems
    Various Functions Problems
    DS Questions
    PS Questions

For more check below:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread

Hope it helps.

In my opinion, it is very basic fundamental question. I don't know whether we really need number to put in and check. If student's basic math fundamentals are clear, Student should be able to solve it visually.

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) = 5x

f(3) = 5(3) = 15

f(1) = 5(1) = 5

f(2) = 5(2) = 10

Since 15 equals 5 + 10 = 15, choice B can be correct.

C) f(x) = 2x + 1

f(3) = 2(3) + 1 = 7

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

f(2) = 2(2) + 1 = 5

Since 7 does not equal 5 + 3, choice C is not correct.

D) f(x) = √x

f(3) = √3

f(1) = √1 = 1

f(2) = √2

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

E) f(x) = x - 2

f(3) = 3 - 2 = 1

f(1) = 1 - 2 = -1

f(2) = 2 - 2 = 0

Since 1 does not equal -1 + 0, choice E is not correct.

Since every answer choice except B is eliminated, the correct answer is B.

Answer: B

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) \(f(x) = 5x\)
=> POSSIBLE: Satisfies all the conditions above. In Test Situation we can mark and move on. But I am solving the problem to complete the solution.

(C) \(f(x) = 2x + 1\)
=> Eliminate : Doesn't Satisfy Point 1 above. It has a constant added. (+1)

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

(E) \(f(x) = x - 2\)
=> Eliminate : Doesn't Satisfy Point 1 above. It has a constant subtracted. (-2)

So, Answer will be B.

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) \(f(x) = 5x\)

=> f(a+b) = \(5*(a+b)\) = 5a + 5b
f(a) + f(b) = \(5a\) + \(5b\) = 5a + 5b
=> f(a+b) = f(a) + f(b) => TRUE. In Test Situation we can mark and move on. But I am solving the problem to complete the solution.

(C) \(f(x) = 2x + 1\)

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

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

=> 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

(E) \(f(x) = x - 2\)

=> f(a+b) = \(a+b - 2\)
=> f(a) + f(b) = \(a - 2\) + \(b - 2\) = \(a+b - 4\) ≠ \(a+b - 2\)
=> f(a+b) ≠ f(a) + f(b) => FALSE

So, Answer will be B
Hope it helps!

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