Question 11
Suppose that a real-valued function f(x) of real numbers satisfies f(x + xy) = f(x) + f(xy) for all real x, y, and that f(2020) = 1. Compute f(2021)
Solution
We are given a real-valued function f(x) of real numbers that satisfies f(x + xy) = f(x) + f(xy) for all real x, y.
f(2020) = 1.
We need to find f(2021)
f(x + xy) = f(x) + f(xy)
We know that when f(x+y) = f(x) + f(y)
Then, f(x) = kx
So, here, assuming xy = w
f(x+w) = f(x) + f(w) = kx
We know that f(2020) = k*2000 = 1
k = $$\dfrac{1}{2000}$$
So, f(2021) = k*2021 = $$\dfrac{2021}{2020}$$
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