Quod Erat Demonstrandum

2012/09/25

答家軒

Filed under: Pure Mathematics — johnmayhk @ 7:07 下午

Question

For positive numbers a,b, prove that

\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2}=0

has at least one root in (-1,1).

Solution

Let

f(x)=\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2}=\frac{a}{(x+1)(x^2+x-1)}+\frac{b}{(x-1)(x^2+x+2)}

For x\rightarrow 1^{-},

(x-1)(x^2+x+2) \rightarrow 0^{-}

\frac{b}{(x-1)(x^2+x+2)} \rightarrow -\infty

also,

\frac{a}{(x+1)(x^2+x-1)} \rightarrow \frac{a}{2};

hence f(x) \rightarrow -\infty as x \rightarrow 1^{-}.

For x\rightarrow (\frac{-1+\sqrt{5}}{2})^{+},

(x+1)(x^2+x-1) \rightarrow 0^{+}

\frac{a}{(x+1)(x^2+x-1)} \rightarrow +\infty

also,

\frac{b}{(x-1)(x^2+x+2)} is a finite negative number;

hence f(x) \rightarrow +\infty as x \rightarrow (\frac{-1+\sqrt{5}}{2})^{+}.

And, f(x) is continuous on (\frac{-1+\sqrt{5}}{2},1);

thus, f(x)=0 has at least one root in (\frac{-1+\sqrt{5}}{2},1) \subset (-1,1).

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