2023-2024学年陕西省高一期中考试质量监测(方块包菱形)数学.考卷答案

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2023-2024学年陕西省高一期中考试质量监测(方块包菱形)数学.考卷答案试卷答案

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分析（1）根据定义可得f（x）+f（-x）=2，进而求出m值；
（2）根据定义可得g（x）+g（-x）=2，得出g（x）=2-g（-x），设x＜0时，则-x＞0，求出g（x）即可；
（3）恒有g（x）＜f（t）成立，则g（x）=-x²+ax+1＜f（t）_min=3，求出a的范围．

解答解：（1）因为函数f（x）的图象关于点（0，1）对称，
∴f（x）+f（-x）=2，
即$\frac{{{x^2}+mx+m}}{x}+\frac{{{x^2}-mx+m}}{-x}=2$，
所以2m=2，
∴m=1．
（2）因为函数g（x）在（-∞，0）∪（0，+∞）上的图象关于点（0，1）对称，
则g（x）+g（-x）=2，
∴g（x）=2-g（-x），
∴当x＜0时，则-x＞0，
∴g（-x）=x²-ax+1，
∴g（x）=2-g（-x）=-x²+ax+1；
（3）由（1）知，$f（t）=\frac{{{t^2}+t+1}}{t}=t+\frac{1}{t}+1（t＞0）$，
∴f（t）_min=3，
又当x＜0时，g（x）=-x²+ax+1
∴g（x）=-x²+ax+1＜3，
∴ax＜2+x²又x＜0，
∴$a＞\frac{2}{x}+x$，
∴$a＞-2\sqrt{2}$．

点评考查了新定义类型的做题方法和恒成立问题的转化．要紧扣定义．