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).
Explanation:
f(x + xy) = f(x) + f(xy)
Substituting y = 1 ⇒ f(x + x) = f(x) + f(x) ⇒ f(2x) = 2f(x)
Substituting y = 2 ⇒ f(x + 2x) = f(x) + f(2x) ⇒ f(3x) = f(x) + 2f(x) ⇒ f(3x) = 3f(x)
Hence, we can say, f(ax) = a × f(x) ...(1)
Substituting a = 2020 and x = 1 ⇒ f(2020) = 2020 × f(1) ⇒ 1 = 2020 × f(1) ⇒ f(1) = 1/2020
Now, putting x = 1 and a = 2021 in (1) we get, f(2021) = 2021 × f(1) = 2021/2020
Hence, option (a).
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