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Re: For which of the following functions is f(a + b) = f(a) + f(
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24 Jul 2020, 07:22
Carcass wrote:
For which of the following functions is f(a + b) = f(a) + f(b) for all positive numbers a and b?
A. \(f(x)=x^2\)
B. \(f(x)= x+1\)
C. \(f(x) = \sqrt{x}\)
D. \(f(x)=\frac{2}{x}\)
E. \(f(x) = -3x\)
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
Re: For which of the following functions is f(a + b) = f(a) + f(
[#permalink]
19 Aug 2022, 10:00
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)