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江西省2024年初中学业水平考试冲刺(一)1数学.考卷答案试卷答案
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分析令f（x）=$\frac{1}{2}$x2-2lnx，利用导数法求出函数的最值，进而可得关于x的方程$\frac{1}{2}$x2-2lnx=m在区间[$\frac{1}{e}$，e]上有实数解时，m的范围．
解答解：令f（x）=$\frac{1}{2}$x2-2lnx，
则f′（x）=x-$\frac{2}{x}$=$\frac{{x}^{2}-2}{x}$，
当x∈[$\frac{1}{e}$，$\sqrt{2}$）时，f′（x）＜0，f（x）为减函数，x∈[$\sqrt{2}$，e]时，f′（x）＞0，f（x）为增函数，
故当x=$\sqrt{2}$时，f（x）取最小值1-ln2，
又由x=$\frac{1}{e}$时，f（x）=$\frac{1}{2{e}^{2}}+2$，x=2时，f（x）=$\frac{{e}^{2}}{2}-2$，
故x=$\frac{1}{e}$时，f（x）取最大值$\frac{1}{2{e}^{2}}+2$，
若x的方程$\frac{1}{2}$x2-2lnx=m在区间[$\frac{1}{e}$，e]上有实数解，
m∈[1-ln2，$\frac{1}{2{e}^{2}}+2$]
点评本题考查的知识点是函数的零点的判定定理，函数的值域，难度中档．
