《2025年学霸高考黑题数学人教版》

变式训练1. (2025·广东东莞模拟)已知函数$f(x)=x - a\ln x+\frac{1 + a}{x}(a\in\mathbf{R})$.
(1)求$f(x)$的单调区间；
(2)若在$[1,e](e = 2.71828·s)$上存在一点$x_0$，使得$f(x_0)\leq0$成立，求$a$的取值范围.

答案： 变式训练1. 解：
(1)由题意知$f^{\prime}(x)=1-\frac{a}{x}-\frac{1 + a}{x^{2}}=\frac{x^{2}-ax-(1 + a)}{x^{2}}=\frac{(x + 1)[x-(1 + a)]}{x^{2}}$, 当$1 + a\leq0$,即$a\leq - 1$时,在$(0,+\infty)$上$f^{\prime}(x)＞0$,所以函数$f(x)$在$(0,+\infty)$上单调递增; 当$1 + a＞0$,即$a＞-1$时,在$(a + 1,+\infty)$上$f^{\prime}(x)＞0$,在$(0,a + 1)$上$f^{\prime}(x)＜0$,所以函数$f(x)$在$(a + 1,+\infty)$上单调递增,在$(0,a + 1)$上单调递减.
(2)若在$[1,e]$上存在一点$x_{0}$,使得$f(x_{0})\leq0$成立,即$f(x)_{\min}\leq0$,$x\in[1,e]$. ①由
(1)可知,当$a\leq - 1$时,函数$f(x)$在$[1,e]$上单调递增,$f(x)_{\min}=f(1)=1 + 1 + a\geq0$,即$a\geq - 2$; ②当$a＞-1$时,函数$f(x)$在$(0,a + 1)$上单调递减,在$(a + 1,+\infty)$上单调递增. (i)当$1 + a\geq e$,即$a\geq e - 1$时,函数$f(x)$在$[1,e]$上单调递减,$f(x)_{\min}=f(e)=e^{2}+1 - a\geq\frac{e^{2}+1}{e}$,因为$\frac{e^{2}+1}{e}＞e - 1$,所以$a\geq\frac{e^{2}+1}{e - 1}$ (ii)当$1 + a\leq1$,即$-1＜a\leq0$时,函数$f(x)$在$[1,e]$上单调递增,$f(x)_{\min}=f(1)=1 + 1 + a\leq0$,即$a\leq - 2$(舍). (iii)当$1＜1 + a＜e$,即$0＜a＜e - 1$时,函数$f(x)$在$[1,a + 1]$上单调递减,在$[a + 1,e]$上单调递增,$f(x)_{\min}=f(1 + a)=a + 2 - a\ln(1 + a)$,此时$0＜\ln(1 + a)＜1$,则$0＜a\ln(1 + a)＜a$,所以$2 + a - a\ln(1 + a)＞2$,即$f(1 + a)＞0$,所以$f(1 + a)\leq0$无解. 综上,实数$a$的取值范围是$a\geq\frac{e^{2}+1}{e - 1}$或$a\leq - 2$.

典型例题2. (2025·陕西安康三模)已知函数$f(x)=3x^2 - 8\sin(x + \varphi)$，其中$|\varphi|\leq\pi$.
(1)若$f(x)$是偶函数，求$\varphi$；
(2)当$\varphi = 0$时，讨论$f(x)$在$[0,+\infty)$上的零点个数；
(3)已知$\varphi\in(-\frac{\pi}{2},0]$，若$\forall x\geq0$，$f(x)\geq0$，求$\varphi$的取值范围.

答案： 典型例题2. 解：
(1)因为函数$f(x)$是偶函数,所以$f(-x)=f(x)$,即$3(-x)^{2}-8\sin(-x+\varphi)=3x^{2}-8\sin(x+\varphi)$,$\sin(-x+\varphi)=\sin x\cos\varphi+\cos x\sin\varphi$,$\sin(-x)\cos\varphi+\cos(-x)\sin\varphi=\sin x\cos\varphi+\cos x\sin\varphi$,所以$2\sin x\cos\varphi = 0$,所以$\cos\varphi = 0$,又$\vert\varphi\vert\leq\pi$,所以$\varphi=\pm\frac{\pi}{2}$
(2)当$\varphi = 0$时,$f(x)=3x^{2}-8\sin x$,$f(0)=0$,可得$f^{\prime}(x)=6x - 8\cos x$,则$h(x)=6x - 8\cos x$,则$h^{\prime}(x)=6 + 8\sin x$ ①当$x\geq\pi$时$f^{\prime}(x)=6x - 8\cos x＞0$,所以$f(x)\geq3\pi^{2}-8＞0$; ②当$0＜x＜\pi$时,$h^{\prime}(x)=6 + 8\sin x＞0$,所以$f^{\prime}(x)$在$(0,\pi)$上单调递增,又$f^{\prime}(0)=-8,f^{\prime}(\pi)=6\pi + 8＞0$,所以存在$x_{1}\in(0,\pi)$,使得$f^{\prime}(x_{1})=0$,当$x\in(0,x_{1})$时$f^{\prime}(x)＜0$,所以$f(x)$在$(0,x_{1})$上单调递减,当$x\in(x_{1},\pi)$时$f^{\prime}(x)＞0$,所以$f(x)$在$(x_{1},\pi)$上单调递增,而$f(0)=0,f(x_{1})＜0,f(\pi)＞0$,所以$f(x)$在$(x_{1},\pi)$上存在一个零点. 综上,函数$f(x)$在$[0,+\infty)$上有两个零点.
(3)当$\varphi\in\left(-\frac{\pi}{2},0\right]$时, ①当$x\geq\frac{2\pi}{3}$时,$f^{\prime}(x)=6x - 8\cos(x+\varphi)＞0$,所以$f(x)\geq\frac{4\pi^{2}}{3}-8＞0$; ②当$0\leq x＜\frac{2\pi}{3},\varphi\in\left(-\frac{\pi}{2},0\right]$时,若$\sin(x+\varphi)\leq0$,则$f(x)\geq0$成立,只需考虑$\sin(x+\varphi)＞0$,又$g(x)=f^{\prime}(x)$,此时$g^{\prime}(x)=6 + 8\sin(x+\varphi)＞0$,$f^{\prime}(x)$单调递增. 又$f^{\prime}(0)=-8\cos\varphi＜0,f^{\prime}\left(\frac{\pi}{2}\right)=3\pi + 8\sin\varphi＞0$,所以存在$x_{0}\in\left(0,\frac{\pi}{2}\right)$,使得$f^{\prime}(x_{0})=6x_{0}-8\cos(x_{0}+\varphi)=0$,可得$\cos(x_{0}+\varphi)=\frac{3x_{0}}{4}$,若$x\in(0,x_{0})$,则$f^{\prime}(x)＜0,f(x)$在$(0,x_{0})$上单调递减,若$x\in\left(x_{0},\frac{\pi}{2}\right)$,则$f^{\prime}(x)＞0,f(x)$在$\left(x_{0},\frac{\pi}{2}\right)$上单调递增,所以$f(x)\geq f(x_{0})=3x_{0}^{2}-8\sin(x_{0}+\varphi)=3x_{0}^{2}-8\sqrt{1-\frac{9x_{0}^{2}}{16}}\geq0$,解得$x_{0}\geq\frac{2\sqrt{3}}{3}$ 此时$\cos(x_{0}+\varphi)=\frac{3x_{0}}{4}-\frac{\sqrt{3}}{2}$,所以$x_{0}+\varphi\leq\frac{\pi}{6}$,从而$\varphi\leq\frac{\pi}{6}-x_{0}\leq\frac{\pi}{6}-\frac{2\sqrt{3}}{3}$,所以$\varphi$的取值范围为$\left(\frac{\pi}{2},\frac{\pi}{6}-\frac{2\sqrt{3}}{3}\right]$

变式训练2. (2025·山东济宁二模)已知函数$f(x)=xe^x - a$，$a\in\mathbf{R}$.
(1)讨论$f(x)$零点的个数；
(2)若$|f(x)|＞ax(\ln x + 1)$，求实数$a$的取值范围.

答案： 变式训练2. 解：
(1)由题意知$f^{\prime}(x)=(x + 1)e^{x}$,故$f(x)$在$(-\infty,-1)$上单调递减,在$(-1,+\infty)$上单调递增,$f(x)_{\min}=f(-1)=-\frac{1}{e}-a$, ①当$-\frac{1}{e}-a＞0$,即$a＜-\frac{1}{e}$时,$f(x)_{\min}＞0$,所以$f(x)$无零点; ②当$-\frac{1}{e}-a=0$,即$a=-\frac{1}{e}$时,$f(x)_{\min}=0$,所以$f(x)$只有1个零点; ③当$-\frac{1}{e}-a＜0$,即$a＞-\frac{1}{e}$时,$f(x)_{\min}＜0$. 又$f(a)=ae^{a}-a=a(e^{a}-1)$,若$a\geq0$,则$e^{a}\geq1,f(a)\geq0$,若$a＜0$,则$e^{a}＜1,f(a)＜0$,所以$f(a)\geq0$恒成立,故存在$x_{1}\in(-1,a]$使得$f(x_{1})=0$. 因为当$x＜-1$时,$xe^{x}＜0$,$f(x)=xe^{x}-a＜0$,$f(x)$在$(-\infty,-1)$没有零点,因此$f(x)$只有1个零点; 当$-\frac{1}{e}＜a＜0$时,$f\left(\frac{1}{a}\right)=\frac{e^{\frac{1}{a}}}{a}-a$,令$t=\frac{1}{a}\in(-\infty,-e)$,则$f\left(\frac{1}{a}\right)=\frac{1}{a}e^{\frac{1}{a}}-a=te^{t}-\frac{1}{t}=\frac{t^{2}e^{t}-1}{t}$,令$g(t)=t^{2}e^{t}-1,t\in(-\infty,-e)$,则$g^{\prime}(t)=t(t + 2)e^{t}＞0$,所以$g(t)$在$(-\infty,-e)$上单调递增,所以$g(t)＜g(-e)=e^{-e + 2}-1＜0$,所以$f\left(\frac{1}{a}\right)=\frac{g(t)}{t}＞0$,故存在$x_{2}\in\left(\frac{1}{a},1\right)$使得$f(x_{2})=0$,因此$f(x)$有2个零点. 综上,当$a＜-\frac{1}{e}$时,$f(x)$无零点,当$a=-\frac{1}{e}$或$a\geq0$时,$f(x)$有1个零点,当$-\frac{1}{e}＜a＜0$时,$f(x)$有2个零点.
(2)①当$a\leq0$时,由$x＞0$,得$f(x)＞0$,所以$\vert f(x)\vert＞ax(\ln x + 1)$等价于$xe^{x}-a＞ax(\ln x + 1)$对$x\in(0,+\infty)$恒成立,即$e^{x}＞a\left(\ln x+\frac{1}{x}+1\right)$对$x\in(0,+\infty)$恒成立, 令$h(x)=\ln x+\frac{1}{x}+1,x＞0$,则$h^{\prime}(x)=\frac{x - 1}{x^{2}}$,当$x\in(0,1)$时,$h^{\prime}(x)＜0$,当$x\in(1,+\infty)$时,$h^{\prime}(x)＞0$,所以$h(x)$在$(0,1)$内单调递减,在$(1,+\infty)$内单调递增,所以$h(x)\geq h(1)=2$.又$e^{x}＞0$,所以$e^{x}＞a\left(\ln x+\frac{1}{x}+1\right)$对$x\in(0,+\infty)$恒成立,所以$a\leq0$时成立; ②当$a＞0$时, (i)若$x\in\left(0,\frac{1}{e}\right)$,则$ax(\ln x + 1)＜0$,显然成立. (ii)若$x\in\left[\frac{1}{e},+\infty\right)$,$\vert f(x)\vert＞ax(\ln x + 1)$等价于$xe^{x}-a＞ax(\ln x + 1)$或$xe^{x}-a＜-ax(\ln x + 1)$,即$\frac{e^{x}}{a}＞\ln x+\frac{1}{x}+1$或$-\frac{e^{x}}{a}＜\ln x+\frac{1}{x}-1$, 对于$-\frac{e^{x}}{a}＜\ln x+\frac{1}{x}-1$,取$x = 1$,得$-\frac{e}{a}＜0$,与$a＞0$矛盾,故不成立; 对于$\frac{e^{x}}{a}＞\ln x+\frac{1}{x}+1$,即$\frac{\ln x+\frac{1}{x}+1}{e^{x}}$对$x\in\left[\frac{1}{e},+\infty\right)$恒成立, 令$t(x)=\frac{\ln x+\frac{1}{x}+1}{e^{x}},x\in\left[\frac{1}{e},+\infty\right)$,则$t^{\prime}(x)=\frac{x^{2}-1}{x^{2}}e^{-x}＜0$, 所以$t(x)$在$\left[\frac{1}{e},+\infty\right)$内单调递减,所以$t(x)\leq t\left(\frac{1}{e}\right)=e^{1-\frac{1}{e}}$,所以$0＜a＜e^{1-\frac{1}{e}}$. 综上,实数$a$的取值范围是$\left(-\infty,e^{1-\frac{1}{e}}\right)$.