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!

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