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设 \(g(x)=\dfrac{x}{x^2+1}+\dfrac{kx}{x^2+k^2},\,k>0\) 则取 \(k=\dfrac{a}{b}\) 时有 \(f(x)=g\left(\dfrac{x}{b}\right)\),接下来只需讨论 \(g(x)\) 的情况 不妨设 \(k>1\),\(k<1\) 情况同理,\(k=1\) 情况平凡略。 \(g(x)\) 为奇函数,只讨论正半边,设 \(x=\tan \theta,\,\theta \in \left[0,\,\dfrac{\pi}{2}\right)\),则 \(g(\theta)=\sin \theta \cos \theta \left(\dfrac{k+\sin^2 \theta+k^2 \cos^2 \theta}{\sin^2 \theta+k^2 \cos^2 \theta}\right)\) 再设 \(m=(1-k^2)\sin^2 \theta + k^2 \in \left(1,\,k^2\right]\) 则 \[ \begin{aligned} g(m)&=\dfrac{\sqrt{-m^2+(k^2+1)m-k^2}}{k^2-1}\left(\dfrac{m+k}{m}\right)\\ &=\dfrac{1}{k^2-1}\sqrt{\left[k^2+1-\left(m+\dfrac{k^2}{m}\right)\right]\left[2k+\left(m+\dfrac{k^2}{m}\right)\right]} \end{aligned} \] 记 \(n=m+\dfrac{k^2}{m}\in [2k,\,k^2+1]\),里面是个二次函数 \(h(n)=(k^2+1-n)(2k+n)\),对称轴是 \(\dfrac{(k-1)^2}{2}
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Sat, 10 Jun 2023 09:04 GMT