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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