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Suppose that a continuous 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 \(f(2021)\).

Question and options

Suppose that a continuous real-valued function f(x)f(x) of real numbers satisfies f(x+xy)=f(x)+f(xy)f(x + xy) = f(x) + f(xy) for all real x,y,x, y, and that f(2020)=1f(2020) = 1. Compute f(2021)f(2021).
  1. 20212020\frac{2021}{2020}
  2. 20202019\frac{2020}{2019}
  3. 1
  4. 20202021\frac{2020}{2021}

Answer and explanation

Correct answer: A - \(\frac{2021}{2020}\)
Put u=x, v=xy (any u not 0, any v): f(u+v)=f(u)+f(v); x=0 gives f(0)=0. A continuous additive function is linear, f(x)=kx. f(2020)=1 gives k=1/2020, so f(2021)=2021/2020.

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