设正数$x,y$满足$xy=1$,求$m=\dfrac{x+y}{[x]\cdot[y]+[x]+[y]+1}$的取值范围.
正确答案是$\left\{\dfrac 12\right\}\cup\left[\dfrac 56,\dfrac 54\right)$.
分析与解 不妨设$x\geqslant 1$,当$x=1$时,$m=\dfrac 12$;
当$1<x< 2$时,有\[m=\dfrac 12\left(x+\dfrac 1x\right),\]取值范围是$\left(1,\dfrac 54\right)$;
当$x\geqslant 2$时,设$k\leqslant x<k+1$,$k\in\mathbb N^*$且$k\geqslant 2$,则有\[m=\dfrac{x+\dfrac 1x}{k+1},\]考虑到对勾函数的单调性,$m$取值范围是$$\left[\dfrac{k+\dfrac 1k}{k+1},\dfrac{k+1+\dfrac{1}{k+1}}{k+1}\right),$$即$\left[\dfrac{k^2+1}{k^2+k},\dfrac{k^2+2k+2}{k^2+2k+1}\right)$.
注意到一方面当$k\geqslant 3$时,$\dfrac{k^2+1}{k^2+k}$随$k$单调递增,且当$k=2$与$k=3$时,均有$\dfrac{k^2+1}{k^2+k}=\dfrac 56$;
另一方面,$\dfrac{k^2+2k+2}{k^2+2k+1}$随$k$单调递减,于是\[\bigcup_{k=2}^{\infty}\left[\dfrac{k^2+1}{k^2+k},\dfrac{k^2+2k+2}{k^2+2k+1}\right)=\left[\dfrac 56,\dfrac {10}{9}\right).\]
综上所述,所求$m$的取值范围是$\left\{\dfrac 12\right\}\cup\left[\dfrac 56,\dfrac 54\right)$.