浙江省培优联盟高二2024年5月联考数学h试题答案 (更新中)

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浙江省培优联盟高二2024年5月联考数学h试卷答案

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分析(1)先求出函数的导数,通过讨论a的范围,确定函数的单调区间即可
(2)先求出a,b的值,求出函数的解析式,从而求出代数式的最小值即可;
(3)通过讨论①x1,x2,x3都为正数时,②当x1,x2,x3为有一个为负数时的情况,从而证出结论.

解答解:(1)∵f′(x)=$\frac{{ax}^{2}-1}{{bx}^{2}}$,首先x≠0,
∴①当a≤0时,令f′(x)<0,得:ax2-1<0,
∵a≤0,
∴x的单调递减区间为(-∞,0)∪(0,+∞);
②当a>0时,令f′(x)<0,
ax2-1<0,ax2<1,x2<$\frac{1}{a}$,
∵a>0,
∴x的单调递减区间为(-$\frac{1}{\sqrt{a}}$,0)∪(0,$\frac{1}{\sqrt{a}}$),
∴当a≤0,x的单调递减区间为(-∞,0)∪(0,+∞);
a>0,x的单调递减区间为(-$\frac{1}{\sqrt{a}}$,0)∪(0,$\frac{1}{\sqrt{a}}$),
(2)∵对?x>0,都有f(x)>f(1)=2,
∴根据上问分析a不可能≤0,
∴a>0,∴$\frac{1}{\sqrt{a}}$=1,∴a=1,
∵f(1)=$\frac{a+1}{b}$=2,∴b=1,
∴f(x)=$\frac{{x}^{2}+1}{x}$=x+$\frac{1}{x}$,
|[f(x)]3|-|f(x3)|=3x+$\frac{3}{x}$≥2×3=6;
(3)由条件知道x1,x2,x3最多有一个负数,
①当x1,x2,x3都为正数时,由第一问可知:
f(xi)>f($\frac{1}{\sqrt{a}}$)=$\frac{2\sqrt{a}}{b}$,
∴f(x1)+f(x2)+f(x3)≥$\frac{6\sqrt{a}}{b}$>$\frac{2\sqrt{a}}{b}$,
②当x1,x2,x3为有一个为负数时,不妨设x3<0,
∵x2+x3>0,|x3|<$\frac{1}{\sqrt{a}}$,
∴x2>-x3>$\frac{1}{\sqrt{a}}$,
∴f(x2)>f(-x3),
∵f(x)为奇函数,
∴f(x2)+f(x3)>0,
∵f(x1)>$\frac{2\sqrt{a}}{b}$,
∴f(x1)+f(x2)+f(x3)>$\frac{2\sqrt{a}}{b}$.

点评本题考查了导数的应用,考查函数的单调性问题,考查不等式的证明,是一道难题.

浙江省培优联盟高二2024年5月联考数学h

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