Explanation

The question asks which of the functions in the answer choices is such that performing the function on a + b yields the same answer as performing the function to a and b individually and then adding those answers together.

The correct answer should be such that f(a + b) = f(a) + f(b) is true for any values of a and b. Test some numbers, for example a = 2 and b = 3:



Alternatively, use logic—for what kinds of operations are performing the operation on two numbers and then summing results the same as summing the original numbers and then performing the operation?

Multiplication or division would work, but squaring, square-rooting, adding, or subtracting would not. The correct function can contain only multiplication and/or division.

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!

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

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