Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Philippines

Problem:
Let f:R⟶Rf: \mathbb{R} \longrightarrow \mathbb{R} be a function such that xf(y)=yf(x)x f(y)=y f(x) for all x,y∈Rx, y \in \mathbb{R}. Find the intersection of the graphs of y=f(x)y=f(x) and y=x2+1y=x^{2}+1 if f(1)=−1f(1)=-1.

Solution

Solution:
(ans ϕ=\phi= Null set.
We have that f(x)x=f(y)y=c\frac{f(x)}{x}=\frac{f(y)}{y}=c, a constant ⇒f(x)=cx⇒f(x)=−x\Rightarrow f(x)=c x \Rightarrow f(x)=-x from the given condition. y=f(x)=−xy=f(x)=-x does not intersect the parabola y=x2+1y=x^{2}+1 because x2−x+1=0x^{2}-x+1=0 has no real solutions.)

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