《2025年学霸高考黑题数学人教版》

典型例题（浙江高考）
已知$1 ＜ a\leq2$，函数$f(x)=e^{x}-x - a$，其中$e = 2.71828·s$为自然对数的底数。
(1) 证明：函数$y = f(x)$在$(0,+\infty)$上有唯一零点。
(2) 记$x_{0}$为函数$y = f(x)$在$(0,+\infty)$上的零点，证明：
①$\sqrt{a - 1}\leq x_{0}\leq\sqrt{2(a - 1)}$；
②$x_{0}f(e^{x_{0}})\geq(e - 1)(a - 1)a$。

答案： 证明：
(1)方法一：$\because f'(x)=e^{x}-1,x＞0,\therefore e^{x}＞1,\therefore f'(x)＞0$，
$\therefore f(x)$在$(0,+\infty)$上单调递增$\therefore1＜a\leq2,\therefore f(2)=e^{2}-2-a\geq e^{2}-4＞0$，
$f(0)=1-a＜0,\therefore$由零点存在定理理得$f(x)$在$(0,+\infty)$上有唯一零点.
方法二：函数$y=f(x)$在$(0,+\infty)$内有唯一零点等价于方程$e^{x}-x=a$在
$(0,+\infty)$内有唯一实根，又等价于直线$y=a$与曲线$g(x)=e^{x}-x(x＞0)$
只有1个交点.由$g'(x)=e^{x}-1＞0$在$(0,+\infty)$内恒成立，$\therefore g(x)$在
$(0,+\infty)$内单调递增，故$g(x)＞g(0)=1$因此，当$1＜a\leq2$，直线$y=a$与
曲线$g(x)=e^{x}-x(x＞0)$只有1个交点，即$f(x)$在$(0,+\infty)$上有唯一零点.
(2)①$\because f(x_{0})=0,\therefore e^{x_{0}}-x_{0}-a=0,\Leftrightarrow\sqrt{a}-1\leq x_{0}\leq\sqrt{2(a-1)}\Leftrightarrow e^{x_{0}}-x_{0}-1\leq x_{0}^{2}\leq2(e^{x_{0}}-1)$.令$g(x)=e^{x}-x-1-\frac{x^{2}}{2}(0＜x＜2)$，$h(x)=e^{x}-x-1-\frac{x^{2}}{2}$.
一方面：$h'(x)=e^{x}-1-x=h_{1}(x)$，$h_{1}'(x)=e^{x}-1＞0,\therefore h'(x)＞h'(0)=0$，
$\therefore h(x)$在$(0,2)$上单调递增，$\therefore h(x)＞h(0)=0,\therefore e^{x}-x-1-\frac{x^{2}}{2}＞0$，
$\therefore2(e^{x}-1)＞x^{2}$.
另一方面：$\because1＜a\leq2,\therefore0＜a-1\leq1,\therefore$当$x_{0}\geq1$时，$\sqrt{a}-1\leq x_{0}$成立，因
此只需证明当$0＜x＜1$时，$g(x)=e^{x}-x-1-x^{2}\leq0$.
$\because g'(x)=e^{x}-1-2x=g_{1}(x)$，$g_{1}'(x)=e^{x}-2=0\Rightarrow x=\ln2$，当$x\in(0,\ln2)$
时，$g_{1}'(x)＜0$，当$x\in(\ln2,1)$时，$g_{1}'(x)＞0$，$\therefore g'(x)$在
$(0,1)$上$g'(x)＜max\{g'(0),g'(1)\}\because g'(0)=0,g'(1)=e-3＜0,\therefore g'(x)＜0,\therefore g(x)$在
$(0,1)$上单调递减，$\therefore g(x)＜g(0)=0,\therefore e^{x}-x-1＜x^{2}$.
综上，$\therefore e^{x_{0}}-x_{0}-1\leq x_{0}^{2}\leq2(e^{x_{0}}-1),\therefore\sqrt{a}-1\leq x_{0}\leq\sqrt{2(a-1)}$.
②令$t(x_{0})=x_{0}f(e^{x_{0}})=x_{0}f(x_{0}+a)=x_{0}[(e^{x}-1)x_{0}+a(e^{x}-2)]$
$\because t'(x_{0})=2(e^{x}-1)x_{0}+a(e^{x}-2)＞0,\sqrt{a}-1\leq x_{0}\leq\sqrt{2(a-1)}$，
$\therefore t(x_{0})\geq t(\sqrt{a}-1)=\sqrt{a}-1[(e^{a}-1)\sqrt{a}-1+a(e^{x}-2)]=(e^{a}-1)(a-1)+a\sqrt{a}-1(e^{a}-2).\because1＜a\leq2,\therefore e^{a}＞(e-1)a＞1,\therefore t(x_{0})\geq(e-1)(a-1)+2\sqrt{a}-1(e^{a}-2).\because1＜a\leq2,\therefore$只需证明$2(a-1)\sqrt{a}-1(e^{a}-2)\geq(e-1)(a-1)^{2}$，即只需证明$4(e^{a}-2)^{2}\geq(e-1)^{2}(a-1)$.
令$s(a)=4(e^{a}-2)^{2}-(e-1)^{2}(a-1)(1＜a\leq2)$，则$s'(a)=8e^{a}(e^{a}-2)-(e-1)^{2}8e^{a}-2-(e-1)^{2}＞0,\therefore s(a)＞s(1)=4(e-2)^{2}＞0$，即
$4(e^{a}-2)^{2}\geq(e-1)^{2}(a-1)$成立，因此$x_{0}f(e^{x_{0}})\geq(e-1)(a-1)a$.
方法二：$x_{0}f(e^{x_{0}})\geq(e-1)(a-1)a\Leftrightarrow x_{0}f(x_{0}+a)\geq(e-1)(a-1)a$.设
$h(x_{0})=x_{0}f(x_{0}+a)$，由①知，$\sqrt{a}-1\leq x_{0}\leq\sqrt{2(a-1)}$，则由$e^{x_{0}+a}\geq1+x_{0}+a$得$h'(x_{0})=(x_{0}+1)e^{0+x_{0}a}-2(x_{0}+a)+(x_{0}+1)(x_{0}+a+1)-2(x_{0}+a)＞0$，
从而只要证$h(\sqrt{a}-1)-(e-1)(a-1)a\geq0$.
上式左边$=e^{\sqrt{a}-1+a}-2a+\sqrt{a}-1(a-ae-1)=a(e-2)+(a-1)\sqrt{a}-1(a-ae-1)=a(e-2)+(a-1)\sqrt{a}-1(a-ae-1)$，使用不等式$e^{x}\geq x+1,e\geq ex$
可得$e^{\sqrt{a}-1+a}-2a+\sqrt{a}-1(a-ae-1)\geq(1+\sqrt{a}-1)ae-2a+\sqrt{a}-1(a-ae-1)=a(e-2)+(a-1)\sqrt{a}-1(a-ae-1)=a(e-2)+(a-1)\sqrt{a}-1(a-ae-1)＞0$，因此$x_{0}f(e^{x_{0}})\geq(e-1)(a-1)a$.
answer:证明：
(1)方法一：$\because f'(x)=e^{x}-1,x＞0,\therefore e^{x}＞1,\therefore f'(x)＞0$，
$\therefore f(x)$在$(0,+\infty)$上单调递增$\therefore1＜a\leq2,\therefore f(2)=e^{2}-2-a\geq e^{2}-4＞0$，
$f(0)=1-a＜0,\therefore$由零点存在定理理得$f(x)$在$(0,+\infty)$上有唯一零点.
方法二：函数$y=f(x)$在$(0,+\infty)$内有唯一零点等价于方程$e^{x}-x=a$在
$(0,+\infty)$内有唯一实根，又等价于直线$y=a$与曲线$g(x)=e^{x}-x(x＞0)$
只有1个交点.由$g'(x)=e^{x}-1＞0$在$(0,+\infty)$内恒成立，$\therefore g(x)$在
$(0,+\infty)$内单调递增，故$g(x)＞g(0)=1$因此，当$1＜a\leq2$，直线$y=a$与
曲线$g(x)=e^{x}-x(x＞0)$只有1个交点，即$f(x)$在$(0,+\infty)$上有唯一零点.
(2)①$\because f(x_{0})=0,\therefore e^{x_{0}}-x_{0}-a=0,\Leftrightarrow\sqrt{a}-1\leq x_{0}\leq\sqrt{2(a-1)}\Leftrightarrow e^{x_{0}}-x_{0}-1\leq x_{0}^{2}\leq2(e^{x_{0}}-1)$.令$g(x)=e^{x}-x-1-\frac{x^{2}}{2}(0＜x＜2)$，$h(x)=e^{x}-x-1-\frac{x^{2}}{2}$.
一方面：$h'(x)=e^{x}-1-x=h_{1}(x)$，$h_{1}'(x)=e^{x}-1＞0,\therefore h'(x)＞h'(0)=0$，
$\therefore h(x)$在$(0,2)$上单调递增，$\therefore h(x)＞h(0)=0,\therefore e^{x}-x-1-\frac{x^{2}}{2}＞0$，
$\therefore2(e^{x}-1)＞x^{2}$.
另一方面：$\because1＜a\leq2,\therefore0＜a-1\leq1,\therefore$当$x_{0}\geq1$时，$\sqrt{a}-1\leq x_{0}$成立，因
此只需证明当$0＜x＜1$时，$g(x)=e^{x}-x-1-x^{2}\leq0$.
$\because g'(x)=e^{x}-1-2x=g_{1}(x)$，$g_{1}'(x)=e^{x}-2=0\Rightarrow x=\ln2$，当$x\in(0,\ln2)$
时，$g_{1}'(x)＜0$，当$x\in(\ln2,1)$时，$g_{1}'(x)＞0$，$\therefore g'(x)$在
$(0,1)$上$g'(x)＜max\{g'(0),g'(1)\}\because g'(0)=0,g'(1)=e-3＜0,\therefore g'(x)＜0,\therefore g(x)$在
$(0,1)$上单调递减，$\therefore g(x)＜g(0)=0,\therefore e^{x}-x-1＜x^{2}$.
综上，$\therefore e^{x_{0}}-x_{0}-1\leq x_{0}^{2}\leq2(e^{x_{0}}-1),\therefore\sqrt{a}-1\leq x_{0}\leq\sqrt{2(a-1)}$.
②令$t(x_{0})=x_{0}f(e^{x_{0}})=x_{0}f(x_{0}+a)=x_{0}[(e^{x}-1)x_{0}+a(e^{x}-2)]$
$\because t'(x_{0})=2(e^{x}-1)x_{0}+a(e^{x}-2)＞0,\sqrt{a}-1\leq x_{0}\leq\sqrt{2(a-1)}$，
$\therefore t(x_{0})\geq t(\sqrt{a}-1)=\sqrt{a}-1[(e^{a}-1)\sqrt{a}-1+a(e^{x}-2)]=(e^{a}-1)(a-1)+a\sqrt{a}-1(e^{a}-2).\because1＜a\leq2,\therefore e^{a}＞(e-1)a＞1,\therefore t(x_{0})\geq(e-1)(a-1)+2\sqrt{a}-1(e^{a}-2).\because1＜a\leq2,\therefore$只需证明$2(a-1)\sqrt{a}-1(e^{a}-2)\geq(e-1)(a-1)^{2}$，即只需证明$4(e^{a}-2)^{2}\geq(e-1)^{2}(a-1)$.
令$s(a)=4(e^{a}-2)^{2}-(e-1)^{2}(a-1)(1＜a\leq2)$，则$s'(a)=8e^{a}(e^{a}-2)-(e-1)^{2}8e^{a}-2-(e-1)^{2}＞0,\therefore s(a)＞s(1)=4(e-2)^{2}＞0$，即
$4(e^{a}-2)^{2}\geq(e-1)^{2}(a-1)$成立，因此$x_{0}f(e^{x_{0}})\geq(e-1)(a-1)a$.
方法二：$x_{0}f(e^{x_{0}})\geq(e-1)(a-1)a\Leftrightarrow x_{0}f(x_{0}+a)\geq(e-1)(a-1)a$.设
$h(x_{0})=x_{0}f(x_{0}+a)$，由①知，$\sqrt{a}-1\leq x_{0}\leq\sqrt{2(a-1)}$，则由$e^{x_{0}+a}\geq1+x_{0}+a$得$h'(x_{0})=(x_{0}+1)e^{0+x_{0}a}-2(x_{0}+a)+(x_{0}+1)(x_{0}+a+1)-2(x_{0}+a)＞0$，
从而只要证$h(\sqrt{a}-1)-(e-1)(a-1)a\geq0$.
上式左边$=e^{\sqrt{a}-1+a}-2a+\sqrt{a}-1(a-ae-1)=a(e-2)+(a-1)\sqrt{a}-1(a-ae-1)=a(e-2)+(a-1)\sqrt{a}-1(a-ae-1)$，使用不等式$e^{x}\geq x+1,e\geq ex$
可得$e^{\sqrt{a}-1+a}-2a+\sqrt{a}-1(a-ae-1)\geq(1+\sqrt{a}-1)ae-2a+\sqrt{a}-1(a-ae-1)=a(e-2)+(a-1)\sqrt{a}-1(a-ae-1)=a(e-2)+(a-1)\sqrt{a}-1(a-ae-1)＞0$，因此$x_{0}f(e^{x_{0}})\geq(e-1)(a-1)a$.

变式训练 1.（2025·湖南长沙雅礼中学模拟）
设函数$f(x)=e^{x}-a\ln x$，$a\in R$。
(1) 当$a = e$时，证明：$f(x)＞0$。
(2) 当$a ＞ 0$时，证明：$f(x)\geq a(2-\ln a)$。

答案： 证明：
(1)当$a=e$时，$f(x)=e^{x}-e\ln x$，定义域为$(0,+\infty)$，
$f'(x)=e^{x}-\frac{e}{x}$.记$g(x)=f'(x)=e^{x}-\frac{e}{x}$，则$g'(x)=e^{x}+\frac{e}{x^{2}}＞0,x\in(0,+\infty)$，所以$f'(x)$在$(0,+\infty)$上单调递增.又$f'(1)=0$，所以当
$x\in(0,1)$时，$f'(x)＜0,f(x)$单调递减，当$x\in(1,+\infty)$时，$f'(x)＞0$，
$f(x)$单调递增，所以$f(x)\geq f(1)=e^{1}-e\ln1=e＞0$，故$f(x)＞0$.
(2)记$g(x)=\ln x+x+1,x＞0$，则$g'(x)=\frac{1}{x}-1=\frac{1-x}{x}$，所以$g(x)$在$(0,1)$上单调递增，在$(1,+\infty)$上单调递减，即$\ln x\leq x-1$.又$a＞0$，所以$f(x)=e^{x}-a\ln x\geq e^{x}-a(x-\1)$，
记$h(x)=e^{x}-a(x-1)$，则$h'(x)=e^{x}-a＞1-a$.
当$0＜a\leq1$时，$h(x)$单调递增，所以$h(x)＞h(0)=1+a$，记$\varphi(a)=1+a-a(2-\ln a)=a\ln a+a+1$，则$\varphi'(a)=\ln a\leq0$，所以$\varphi(a)$在$(0,1]$上单调
递减，所以$\varphi(a)\geq\varphi(1)=0$，即$1+a\geq a(2-\ln a)$，所以$h(x)＞a(2-\ln a)$.
当$a＞1$时，$h(x)$在$(0,\ln a)$上单调递减，在$(\ln a,+\infty)$上单调递增，所
以$h(x)\geq h(\ln a)=a-a(\ln a-1)=a(2-\ln a)$.
综上，$h(x)\geq a(2-\ln a)$，所以$f(x)\geq h(x)\geq a(2-\ln a)$.