2022年高考数学试题解答

(2022年北京高考)已知函数$f(x)=e^x\ln (1+x)$.

(I)求曲线$y=f(x)$在点$(0,f(0))$处的切线方程;

(II)设$g(x)=f'(x)$,讨论函数$g(x)$在$[0,+\infty)$上的单调性;

(II)证明:对任意的$s,t\in (0,+\infty)$,有$f(s+t)>f(s)+f(t)$.

解.(I)由于$\displaystyle f'\left( x \right) =e^x\left[ \ln \left( 1+x \right) +\frac{1}{1+x} \right]$,则$f'(0)=1$.又$f(0)=0$,故曲线$y=f(x)$在点$(0,f(0))$处的切线方程为$y=x$.

(II)因为$\displaystyle g(x)=f'(x)=e^x\left[ \ln \left( 1+x \right) +\frac{1}{1+x} \right]$,则
$$
g'\left( x \right) =e^x\left[ \ln \left( 1+x \right) +\frac{2}{1+x}-\frac{1}{\left( 1+x \right) ^2} \right].
$$
令
$$
h\left( x \right) =\ln \left( 1+x \right) +\frac{2}{1+x}-\frac{1}{\left( 1+x \right) ^2},
$$
则
$$
h'\left( x \right) =\frac{1}{1+x}-\frac{2}{\left( 1+x \right) ^2}+\frac{2}{\left( 1+x \right) ^3}=\frac{x^2+1}{\left( 1+x \right) ^3}>0.
$$
故$h(x)$在$[0,+\infty)$上单调递增,则$h(x)\geqslant h(0)=1>0$,

则$g'\left( x \right)>0$,故函数$g(x)$在$[0,+\infty)$上单调递增.

(II)令
$$F\left( s \right) =f\left( s+t \right) -f\left( s \right) -f\left( t \right),
$$
由函数$g(x)=f'(x)$在$[0,+\infty)$上单调递增可知
$$
F'\left( s \right) =f'\left( s+t \right) -f'\left( s \right) >0
$$
故$F(s)>F(0)=0$,即$f(s+t)>f(s)+f(t)$.

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(2022年新高考二卷)已知函数$f(x)=xe^{ax}-e^x$.

(1)当$a=1$时,讨论$f(x)$的单调性;

(2)当$x>0$时, $f(x)<-1$,求实数$a$的取值范围;

(3)设$n\in \mathbb{N}^\ast$,证明:
$$
\frac{1}{\sqrt{1^2+1}}+\frac{1}{\sqrt{2^2+2}}+\cdots +\frac{1}{\sqrt{n^2+n}}>\ln \left( n+1 \right).
$$
解. (1)当$a=1$时, $f(x)=xe^{x}-e^x=(x-1)e^x$,
则$f'(x)=xe^x$.故$f(x)$在$(-\infty,0)$上单调递减,在$(0,+\infty)$上单调递增.

 

(2)先证明$xe^{x/2}-e^x<-1$.

这等价于
$$
x-e^{x/2}<-e^{-x/2},\quad e^{x/2}-e^{-x/2}-x>0
$$
令$g\left( x \right) =e^{x/2}-e^{-x/2}-x$,
则$\displaystyle g'\left( x \right) =\frac{1}{2}\left( e^{x/2}+e^{-x/2} \right) -1>\frac{1}{2}\cdot 2\sqrt{e^{x/2}\cdot e^{-x/2}}-1=0$.故$g(x)>g(0)=0$.

 

若$\displaystyle a\leqslant \frac{1}{2}$,则$xe^{ax}-e^x< xe^{x/2}-e^x< -1$.

若$a>\frac{1}{2}$且$a\neq 1$,则$xe^{ax}-e^x<-1\Leftrightarrow e^{\left( 1-a \right) x}-e^{-ax}-x>0$.

令$h\left( x \right) =e^{\left( 1-a \right) x}-e^{-ax}-x$,则
$$
h'\left( x \right) =\left( 1-a \right) e^{\left( 1-a \right) x}+ae^{-ax}-1,
$$
而
$$
h''\left( x \right) =\left( 1-a \right) ^2e^{\left( 1-a \right) x}-a^2e^{-ax}=e^{-ax}\left[ \left( 1-a \right) ^2e^x-a^2 \right].
$$
则当$0<x<\ln \frac{a^2}{\left( 1-a \right) ^2}$时, $h''(x)<0$,
此时$h'(x)<h'(0)=0$,则$h(x)<h(0)=0$,矛盾.

若$a=1$,则$f(x)=(x-1)e^x$,显然$f(2)=e^2>-1$,矛盾.

综上所述, $\displaystyle a\leqslant \frac{1}{2}$.

解法二.令$h\left( x \right) =xe^{ax}-e^x+1$,则
$h'\left( x \right) =\left( 1+ax \right) e^{ax}-e^x$,
而
$$
h'\left( x \right) =\left( 1+ax \right) e^{ax}-e^x=e^{ax}\left( 1+ax-e^{\left( 1-a \right) x} \right),
$$
令$F\left( x \right) =1+ax-e^{\left( 1-a \right) x}$,
则$F'\left( x \right) =a-\left( 1-a \right) e^{\left( 1-a \right) x}$.

若$\displaystyle a\leqslant \frac{1}{2}$,
$$
F'\left( x \right) = a-\left( 1-a \right) e^{\left( 1-a \right) x}< a-\left( 1-a \right) =2a-1\leqslant 0,
$$
此时$F(x)< F(0) =0,h'(x)<0$,则$h(x) < h(0) = 0$.

若$\frac{1}{2}< a < 1$,当$\displaystyle 0 < x < \frac{1}{1-a}\ln \frac{a}{1-a}$时$F'\left( x \right)>0$,此时$F(x)>F(0)=0,h'(x)>0$,则$h(x)>h(0)=0$,矛盾.

若$a\geqslant 1$,则$f\left( x \right) =xe^{ax}-e^x\geqslant xe^x-e^x=\left( x-1 \right) e^x$,故$f\left( 2 \right) \geqslant e^2>-1$,矛盾.

 

综上所述, $\displaystyle a\leqslant \frac{1}{2}$.

(3)只需证明
$$
\frac{1}{\sqrt{n^2+n}}>\ln \left( n+1 \right) -\ln n,
$$
之后累加即可.即证
$$
\ln \left( 1+\frac{1}{n} \right) <\frac{\frac{1}{n}}{\sqrt{1+\frac{1}{n}}}.
$$

利用(2)中的结论,当$\displaystyle a=\frac{1}{2}$时,有
$$
xe^{x/2}-e^x<-1,\quad x < e^{x/2}-e^{-x/2}.
$$
将$x$换成$\ln (1+x)$,便有
$$
\ln \left( 1+x \right) <\sqrt{1+x}-\frac{1}{\sqrt{1+x}}=\frac{x}{\sqrt{1+x}}.
$$
再取$x=\frac{1}{n}$可得
$$
\ln \left( 1+\frac{1}{n} \right) <\frac{\frac{1}{n}}{\sqrt{1+\frac{1}{n}}}=\frac{1}{\sqrt{n^2+n}}.
$$
于是
$$\begin{aligned}
&\frac{1}{\sqrt{1^2+1}}+\frac{1}{\sqrt{2^2+2}}+\cdots +\frac{1}{\sqrt{n^2+n}}
\\
&>\left( \ln 2-\ln 1 \right) +\left( \ln 3-\ln 2 \right) +\cdots +\ln \left( n+1 \right) -\ln n
\\
&=\ln \left( n+1 \right).
\end{aligned}$$

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(2022年高考甲卷)已知函数$\displaystyle f(x)=\frac{e^x}{x}-\ln x+x-a$.

(1)若$f(x)\geqslant 0$,求$a$的取值范围;

(2)证明:若$f(x)$有两个零点$x_1,x_2$,则$x_1x_2<1$.

解. (1)首先注意到
$$
f\left( x \right) =\frac{e^x}{x}-\ln x+x-a=e^{x-\ln x}-\ln x+x-a,
$$
令$u=x-\ln x\geqslant 1$,要使$f(x)\geqslant 0$,只需$g(u)=e^u+u-a\geqslant 0$,则$e+1-a\geqslant 0$,即$a\leqslant e+1$.

(2)由$f(x)=0$可知$g(u)=0$,则$a>e+1$.

设$g(u_0)=0$,则$x-\ln x=u_0>1$有两个不同零点$x_1,x_2$,则
$x_1-\ln x_1=u_0,x_2-\ln x_2=u_0$.于是$x_1-x_2=\ln x_1-\ln x_2$.

利用对数平均不等式
$$\sqrt{x_1x_2}<\frac{x_1-x_2}{\ln x_1-\ln x_2}=1,$$
于是$x_1x_2<1$.

---

(2022年新高考一卷)设$\displaystyle a=0.1e^{0.1},b=\frac{1}{9},c=-\ln 0.9$,则( )

(A) $a< b< c$

(B) $c< b< a$

(C) $c< a< b$

(D) $a< c< b$

在$\displaystyle\frac{x}{1+x}<\ln (1+x)$中取$x=\frac{1}{9}$可得$\displaystyle 0.1<\ln \frac{10}{9}$,则$\displaystyle e^{0.1}<\frac{10}{9},0.1e^{0.1}<\frac{1}{9}$,故$a< b$.

或者在$\ln x< x-1$中取$x=0.9$亦可.

考虑
$$
a-c=0.1e^{0.1}+\ln 0.9=0.1e^{0.1}+\ln \left( 1-0.1 \right),
$$
令$f\left( x \right) =xe^x+\ln \left( 1-x \right)$,则
$$
f'\left( x \right) =\left( x+1 \right) e^x+\frac{1}{x-1}=\frac{\left( 1-x^2 \right) e^x-1}{1-x},
$$
令$g\left( x \right) =\left( 1-x^2 \right) e^x-1$,则$g'\left( x \right) =\left( 1-2x-x^2 \right) e^x$,

当$0\leqslant x\leqslant 0.1$时, $1-2x-x^2>0$,则$g'\left( x \right)>0$,此时$g(x)\geqslant g(0)=0$,即$f'(x)\geqslant 0$.

因此$f\left( x \right)$在区间$\left[0,0.1\right]$上递增,故$f\left( 0.1 \right) >f\left( 0 \right) =0$,故$a-c>0,a>c$.

综上所述, $b>a>c$,答案为C选项.

(2022年新高考一卷)
已知函数$f(x)=e^x-ax$和$g(x)=ax-\ln x$有相同的最小值.

(1)求$a$;

(2)证明:存在直线$y=b$,其与两条曲线$y=f(x)$和$y=g(x)$共有三个不同的交点,并且从左到右的三个交点的横坐标成等差数列.

解. (1)若$a\leqslant 0$,则$f(x)=e^x-ax$单调递增,无最小值.

若$a>0$, $\displaystyle f'(x)=e^x-a,g'(x)=a-\frac{1}{x}$,则
$$
\left[ f\left( x \right) \right] _{\min}=f\left( \ln a \right) =a-a\ln a=\left[ g\left( x \right) \right] _{\min}=g\left( \frac{1}{a} \right) =1+\ln a,
$$
即
$$
\left( a+1 \right) \ln a-a+1=0,\quad \ln a-\frac{a-1}{a+1}=0.
$$
令$\displaystyle h(a)=\ln a-\frac{a-1}{a+1}$,
则$\displaystyle h'(a)=\frac{1}{a}-\frac{2}{(a+1)^2}=\frac{a^2+1}{a(a+1)^2}>0$.

故$h(a)$在$(0,+\infty)$上单调递增,且$h(1)=0$,故$\displaystyle h(a)=\ln a-\frac{a-1}{a+1}=0$有唯一解$a=1$.

(2)要使直线$y=b$与两条曲线$y=f(x)$和$y=g(x)$共有三个不同的交点,只需
$e^x-x=x-\ln x=b$,此时$e^x+\ln x-2x=0$.

令$F(x)=e^x+\ln x-2x$,则$\displaystyle F'(x)=e^x+\frac{1}{x}-2\geqslant x+1+\frac{1}{x}-2=x+\frac{1}{x}-1\geqslant 2-1=1>0$,则$F(x)$在$(0,+\infty)$上单调递增,又$\displaystyle F(1)=e-2>0,F\left( \frac{1}{2} \right) =\sqrt{e}-\ln 2-1 < 0$,或者利用
$$
F\left( e^{-4} \right) = e^{e^{-4}}-4-\frac{2}{e^4}< e- 4-\frac{2}{e^4} < 0.
$$

故$F(x)$在$(0,+\infty)$上有唯一零点,记为$x_0$,令$b=e^{x_0}-x_0$即可.此时,直线$y=b$与两条曲线$y=f(x)$和$y=g(x)$共有三个不同的交点$x_1,x_0,x_2$且$x_1< x_0< x_2$.

下面证明: $2x_0=x_1+x_2$,即证$x_2=2x_0-x_1$.

利用$e^{x_1}-x_1=b$, $e^{x_0}-x_0=b=x_0-\ln x_0$和$x_2-\ln x_2=b$.

由于$2x_0-x_1>x_0$,只需证明
$$
\left( 2x_0-x_1 \right) -\ln \left( 2x_0-x_1 \right) =b.
$$
则$2x_0-\ln \left( 2x_0-x_1 \right) =x_1+b=e^{x_1},e^{x_0}+\ln x_0-\ln \left( 2x_0-x_1 \right) =e^{x_1}$,
故等价于证明
$$
e^{x_0}-e^{x_1}=x_0-x_1=\ln \frac{2x_0-x_1}{x_0},e^{x_0-x_1}=\frac{x_0-b}{x_1-b}
=\frac{2x_0-x_1}{x_0},
$$
即$\left( x_1-x_0 \right) \left( x_1-x_0+b \right) =0$.又$x_1-x_0>0$,只需证明$x_1-x_0+b=0$,即$x_0=x_1+b=e^{x_1}$.

事实上,
$$
g\left( e^{x_1} \right) =e^{x_1}-\ln e^{x_1}=e^{x_1}-x_1=b=g\left( x_0 \right),
$$
又$x_2>1>e^{x_1}>0>x_1$, $g(x)$在$(0,1)$上单调递减,故$e^{x_1}=x_0=x_1+b$,得证.

解法二. (同构法)利用$e^{x_1}-x_1=b$, $e^{x_0}-x_0=b=x_0-\ln x_0$和$x_2-\ln x_2=b$可得$\ln x_0+e^{x_0}=2x_0$.

再由$e^{x_1}-x_1=b=x_0-\ln x_0=e^{\ln x_0}-\ln x_0$, $e^{x_0}-x_0=b=x_2-\ln x_2=e^{\ln x_2}-\ln x_2$.由$g(x)=e^x-x$在$(-\infty,0)$上递减,在$(0,+\infty)$上递增以及$x_1,\ln x_0<0$,而$x_0,\ln x_2>0$可知$x_1=\ln x_0,x_0=\ln x_2$.

于是$x_1+x_2=\ln x_0+e^{x_0}=2x_0$.

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(2022年全国乙卷)已知函数$f(x)=\ln (1+x)+axe^{-x}$.

(1)当$a=1$时,求曲线$y=f(x)$在点$(0,f(0))$处的切线方程;

(2)若$f(x)$在区间$(-1,0),(0,+\infty)$各恰有一个零点,求$a$的取值范围.

解. (1)当$a=1$时, $f(x)=\ln (1+x)+xe^{-x}$,则
$$
f'\left( x \right) =\frac{1}{1+x}+\left( 1-x \right) e^{-x}.
$$
故$f(0)=0,f'(0)=2$,则曲线$y=f(x)$在点$(0,f(0))$处的切线方程为$y=2x$.

 

(2) 解法一. (参变分离)当$x\neq 0$时,由$f(x)=\ln (1+x)+axe^{-x}=0$可得
$$
a=-\frac{e^x\ln \left( 1+x \right)}{x}.
$$
令$g\left( x \right) =-\frac{e^x\ln \left( 1+x \right)}{x}$,则
$$
g'\left( x \right) =-\frac{e^x\left[ \left( x^2-1 \right) \ln \left( 1+x \right) +x \right]}{x^2\left( 1+x \right)}.
$$
令$h\left( x \right) =\left( x^2-1 \right) \ln \left( 1+x \right) +x$,则
$$
h'\left( x \right) =2x\ln \left( 1+x \right) +x=x\left( 2\ln \left( 1+x \right) +1 \right).
$$
于是$h(x)$在$\displaystyle\left( -1,\frac{1}{\sqrt{e}}-1 \right)$和$\left(0,+\infty\right)$上单调递增,在$\displaystyle\left(\frac{1}{\sqrt{e}}-1,0\right)$上单调递减,又$h(0)=0$,则存在$\displaystyle x_0\in \left( -1,\frac{1}{\sqrt{e}}-1 \right)$,使得$h(x_0)=0$.此时,当$-1< x< x_0$时, $h(x)< 0$,则$g'(x)> 0$;当$x\geqslant x_0$时, $h(x)\geqslant 0$,则$g'(x)\leqslant 0$.

于是$g(x)$在$\left( -1,x_0\right)$上单调递增,在$\left(x_0,0\right)$和$\left(0,+\infty\right)$上单调递减.

注意到
$$
\lim _{x\rightarrow 0}g\left( x \right) =\lim _{x\rightarrow 0}\left( -\frac{e^x\ln \left( 1+x \right)}{x} \right) =-1,
$$
要使$f(x)$在区间$(-1,0),(0,+\infty)$各恰有一个零点,只需$g(x)$的图象与水平直线$y=a$在区间$(-1,0),(0,+\infty)$各有一个交点,于是$a<-1$.

(解法二.) 由于
$$
f'\left( x \right) =\frac{1}{1+x}+a\left( 1-x \right) e^{-x}=\frac{e^x+a\left( 1-x^2 \right)}{\left( 1+x \right) e^x},
$$

(i)若$a\geqslant 0$,则当$x>0$时, $f(x)=\ln (1+x)+axe^{-x}>0$,
与$f(x)$在区间$(0,+\infty)$恰有一个零点矛盾.

(ii)若$-1\leqslant a<0$,当$0<x\leqslant 1$时,
$$
e^x+a\left( 1-x^2 \right) \geqslant e^x-\left( 1-x^2 \right) =e^x+x^2-1>0,
$$
此时$f'(x)>0$;

当$x\geqslant 1$时,
$$
e^x+a\left( 1-x^2 \right) \geqslant e^x >0,
$$
此时$f'(x)>0$也成立.

因此当$x>0$时恒有$f'(x)>0$,故$f(x)$在区间$(0,+\infty)$上单调递增,
则$f(x)\geqslant f(0)=0$,与$f(x)$在区间$(0,+\infty)$恰有一个零点也矛盾.

(iii)若$a<-1$,令$g(x)=e^x+a\left( 1-x^2 \right)$,
则$g'(x)=e^x-2ax$在$(-1,+\infty)$上单调递增,
且$\displaystyle g'(-1)=\frac{1}{e}+2a<\frac{1}{e}-2<0,g'(0)=1>0$,则存在$x_0\in (-1,0)$,使得$g'(x_0)=0$.当$x\in (-1,x_0)$时, $g'(x)<0$, $g(x)$单调递减;当$x\in (x_0,+\infty)$时, $g'(x)>0$, $g(x)$单调递增.

由于$\displaystyle g(-1)=\frac{1}{e}>0,g(0)=1+a<0,g(1)=e>0$,则存在$x_1\in (-1,0),x_2\in (0,1)$,使得$g(x_1)=g(x_2)=0$.当$x\in (-1,x_1)$和$(x_2,+\infty)$时, $g(x)>0,f'(x)>0$, $g(x)$单调递增;当$x\in (x_1,x_2)$时, $g(x)<0,f'(x)<0$, $f(x)$单调递减.

由于$-1< x_1< 0< x_2$且$f(0)=0$,则$f(x_1)>0,f(x_2)<0$,故$f(x)$在区间$(-1,x_1),(x_2,+\infty)$各恰有一个零点,即$f(x)$在区间$(-1,0),(0,+\infty)$各恰有一个零点.

综上所述, $a$的取值范围为$a<-1$.

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(2022年浙江高考)已知数列$\{a_n\}$满足$\displaystyle a_1=1,a_{n+1}=a_n-\frac{1}{3}a_n^2\ (n\in \mathbb{N}^\ast)$,则( )

(A) $\displaystyle 2<100a_{100}<\frac{5}{2}$

(B) $\displaystyle\frac{5}{2}<100a_{100}<3$

(C) $\displaystyle 3<100a_{100}<\frac{7}{2}$

(D) $\displaystyle\frac{7}{2}<100a_{100}<4$

解. 由$\displaystyle a_{n+1}-a_n=-\frac{1}{3}a_n^2<0$可知数列$\{a_n\}$单调递减, 则$a_n\leqslant 1$.再由$\displaystyle a_{n+1}=a_n\left( 1-\frac{1}{3}a_n \right) >0$可知$0< a_n< 1$.

于是
$$
\frac{1}{a_{n+1}}=\frac{1}{a_n\left( 1-\frac{1}{3}a_n \right)}=\frac{3}{a_n\left( 3-a_n \right)}=\frac{1}{a_n}+\frac{1}{3-a_n},
$$
故
$$
\frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{1}{3-a_n}>\frac{1}{3}.
$$
递推得
$$\begin{aligned}
\frac{1}{a_n} &=\left( \frac{1}{a_n}-\frac{1}{a_{n-1}} \right) +\left( \frac{1}{a_{n-1}}-\frac{1}{a_{n-2}} \right) +\cdots +\left( \frac{1}{a_2}-\frac{1}{a_1} \right) +\frac{1}{a_1}
\\
&>\frac{n-1}{3}+1=\frac{n+2}{3}.
\end{aligned}$$
则
$$
\frac{1}{a_{100}}>\frac{102}{3}=34.
$$

 

进而由$\displaystyle a_n<\frac{3}{n+2}$可得
$$
\frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{1}{3-a_n}<\frac{1}{3-\frac{3}{n+2}}=\frac{1}{3}\left( 1+\frac{1}{n+1} \right),
$$
故
$$\begin{aligned}
\frac{1}{a_n} &=\left( \frac{1}{a_n}-\frac{1}{a_{n-1}} \right) +\left( \frac{1}{a_{n-1}}-\frac{1}{a_{n-2}} \right) +\cdots +\left( \frac{1}{a_2}-\frac{1}{a_1} \right) +\frac{1}{a_1}
\\
&<\frac{n-1}{3}+\frac{1}{3}\left( \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right) +1
\\
&=\frac{n+2}{3}+\frac{1}{3}\left( \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n} \right).
\end{aligned}$$
因此
$$\begin{aligned}
\frac{1}{a_{100}} &<\frac{102}{3}+\frac{1}{3}\left( \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{100} \right)
\\
& =34+\frac{1}{3}\left( \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{7} \right) +\frac{1}{3}\left( \frac{1}{8}+\frac{1}{9}+\cdots +\frac{1}{100} \right)
\\
&<34+\frac{1}{3}\times \frac{1}{2}\times 6+\frac{1}{3}\times \frac{1}{8}\times 93<40.
\end{aligned}$$
于是
$$
\frac{5}{2}<100a_{100}<\frac{100}{34}<3.
$$
答案为B选项.

 

(2022年浙江高考)设函数$\displaystyle f(x)=\frac{e}{2x}+\ln x,x>0$.

(I)求函数$f(x)$的单调区间;

(II)若$y=f(x)$在互不相同的三点$(x_1,f(x_1)),(x_2,f(x_2)),(x_3,f(x_3))$的切线均过点$(a,b)$.

(i)若$a>e$,证明: $\displaystyle 0 < b-f\left( a \right) <\frac{1}{2}\left( \frac{a}{e}-1 \right)$;

(ii)若$0< a < e$且$x_1< x_2< x_3$,证明: $\displaystyle\frac{2}{e}+\frac{e-a}{6e^2}<\frac{1}{x_1}+\frac{1}{x_3}
<\frac{2}{a}-\frac{e-a}{6e^2}$.

注: $e=2.71828\cdots$是自然对数的底数.

证. (I)因为
$$
f'\left( x \right) =-\frac{e}{2x^2}+\frac{1}{x}=\frac{2x-e}{2x^2}
$$
函数$f(x)$的单调递减区间为$\displaystyle\left(0,\frac{e}{2}\right)$,单调递增区间为$\displaystyle\left[\frac{e}{2},+\infty\right)$.

 

(II)函数$f(x)$在点$(x_i,f(x_i))\ (i=1,2,3)$处的切线方程为
$$
y-\left( \frac{e}{2x_i}+\ln x_i \right) =\left( -\frac{e}{2x_{i}^{2}}+\frac{1}{x_i} \right) \left( x-x_i \right),
$$
即
$$
y=\left( -\frac{e}{2x_{i}^{2}}+\frac{1}{x_i} \right) x+\frac{e}{x_i}+\ln x_i-1.
$$
由于这三点处的切线均经过$(a,b)$,则有
$$
b=\left( -\frac{e}{2x_{i}^{2}}+\frac{1}{x_i} \right) a+\frac{e}{x_i}+\ln x_i-1.
$$
这表明关于$x$的方程
$$
\left( -\frac{e}{2x^2}+\frac{1}{x} \right) a+\frac{e}{x}+\ln x-1-b=0
$$
有三个不同正根$x_1,x_2,x_3$.

(i)若$a>e$,令
$$
g\left( x \right) =\left( -\frac{e}{2x^2}+\frac{1}{x} \right) a+\frac{e}{x}+\ln x-1-b,
$$
则
$$
g'\left( x \right) =\frac{\left( x-e \right) \left( x-a \right)}{x^3},
$$
故$g(x)$在$(0,e]$和$[a,+\infty)$上单调递增,在$(e,a)$上单调递减.

要使$g(x)=0$有三个不同正根,只需保证极大值$\displaystyle g\left( e \right) =\frac{a}{2e}+1-b>0$
和极小值$\displaystyle g\left( a \right) =\frac{e}{2a}+\ln a-b<0$,即
$$
0< b-f\left( a \right) =b-\frac{e}{2a}-\ln a<\frac{1}{2}\left( \frac{a}{e}-\frac{e}{a} \right) +1-\ln a.
$$
要证
$$
\frac{1}{2}\left( \frac{a}{e}-\frac{e}{a} \right) +1-\ln a<\frac{1}{2}\left( \frac{a}{e}-1 \right),$$
只需证
$$
\ln a+\frac{e}{2a}-\frac{3}{2}>0,\quad a>e.
$$

令$\displaystyle h\left( a \right) =\ln a+\frac{e}{2a}-\frac{3}{2}$,则$\displaystyle h'\left( a \right) =\frac{1}{a}-\frac{e}{2a^2}=\frac{2a-e}{2a^2}>0$,
故$h(a)$在$(e,+\infty)$上递增.故$h\left( a \right) >h\left( e \right) =0$,得证.

 

(ii)若$0< a< e$,则$g(x)$在$(0,a]$和$[e,+\infty)$上单调递增,在$(a,e)$上单调递减.类似地由极大值$\displaystyle g\left( a \right) =\frac{e}{2a}+\ln a-b>0$和极小值$\displaystyle g\left( e \right) =\frac{a}{2e}+1-b<0$可得
$$
\frac{a}{2e}+1< b <\frac{e}{2a}+\ln a.
$$

令$\displaystyle t_i=\frac{1}{x_i}\ (i=1,2,3)$,由$0< x_1< a< x_2< e< x_3$可知$\displaystyle t_3<\frac{1}{e}< t_2< \frac{1}{a}< t_1$.

要证
$$\frac{2}{e}+\frac{e-a}{6e^2}<\frac{1}{x_1}+\frac{1}{x_3}
<\frac{2}{a}-\frac{e-a}{6e^2},$$
只需证
$$\frac{2}{e}+\frac{e-a}{6e^2}< t_1+t_3
<\frac{2}{a}-\frac{e-a}{6e^2}.$$
等价于证明
$$
\left( t_1+t_3-\frac{2}{e}-\frac{e-a}{6e^2} \right) \left( t_1+t_3-\frac{2}{a}+\frac{e-a}{6e^2} \right) <0,
$$
即
$$
\left( t_1+t_3 \right) ^2-\left( \frac{2}{e}+\frac{2}{a} \right) \left( t_1+t_3 \right) +\left( \frac{2}{e}+\frac{e-a}{6e^2} \right) \left( \frac{2}{a}-\frac{e-a}{6e^2} \right) <0.
$$

 

此时$t_i\ (i=1,2,3)$均为方程
$$
-\frac{ae}{2}t^2+\left( a+e \right) t-\ln t-1-b=0
$$
的三个根.

将
$$
\begin{cases}
-\frac{ae}{2}t_{1}^{2}+\left( a+e \right) t_1-\ln t_1-1-b=0\\
-\frac{ae}{2}t_{3}^{2}+\left( a+e \right) t_3-\ln t_3-1-b=0\\
\end{cases}
$$
两式相减整理得
$$
t_1+t_3=\frac{2}{e}+\frac{2}{a}-\frac{2}{ae}\cdot \frac{\ln t_1-\ln t_3}{t_1-t_3}.
$$

只需证
$$
-\frac{2}{ae}\cdot \frac{\ln t_1-\ln t_3}{t_1-t_3}\left( t_1+t_3 \right) +\left( \frac{2}{e}+\frac{e-a}{6e^2} \right) \left( \frac{2}{a}-\frac{e-a}{6e^2} \right) <0,
$$
即证
$$
2\ln \frac{t_1}{t_3}\cdot \frac{\frac{t_1}{t_3}+1}{\frac{t_1}{t_3}-1}>\left( \frac{2a}{e}+\frac{e-a}{6e^2}a \right) \left( \frac{2e}{a}-\frac{e-a}{6e} \right).
$$
注意到$\displaystyle u=\frac{t_1}{t_3}>\frac{e}{a}>1$且函数$\displaystyle G\left( u \right) =2\ln u\cdot \frac{u+1}{u-1}$单调递增,因为
$$
G'\left( u \right) =\frac{2}{\left( u-1 \right) ^2}\left( u-\frac{1}{u}-2\ln u \right) >0,\quad u>1.
$$
这里$\displaystyle u-\frac{1}{u}-2\ln u>0,u>1$.

故$\displaystyle G\left( u \right)>G\left( \frac{e}{a} \right)$,只需证明
$$
G\left( \frac{e}{a} \right) >\left( \frac{2a}{e}+\frac{e-a}{6e^2}a \right) \left( \frac{2e}{a}-\frac{e-a}{6e} \right).
$$

记$\displaystyle w=\frac{e}{a}>1$,等价于证明
$$
2\ln w\cdot \frac{w+1}{w-1}>\left( \frac{2}{w}+\frac{w-1}{6w^2} \right) \left( 2w-\frac{w-1}{6w} \right) =\frac{\left( 13w-1 \right) \left( 12w^2-w+1 \right)}{36w^3}.
$$
令
$$
H\left( w \right) =2\ln w-\frac{\left( 13w-1 \right) \left( 12w^2-w+1 \right) \left( w-1 \right)}{36w^3\left( w+1 \right)},
$$
则
$$
H'\left( w \right) =\frac{\left( w-1 \right) ^2\left( 72w^3-49w^2-20w+3 \right)}{36w^4\left( w+1 \right) ^2}>0,
$$
这里$72w^3-49w^2-20w+3=49w^2\left( w-1 \right) +20w\left( w^2-1 \right) +3w^3+3>0$.

故$H(w)$在$(1,+\infty)$上单调递增,则$H(w)>H(1)=0$,证毕.

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(2022年天津高考)
已知$f(x)=e^x-a\sin x,g(x)=b\sqrt{x}$.

(1)求函数$y=f(x)$在$(0,f(0))$处的切线方程;

(2)若$y=f(x)$和$y=g(x)$有公共点.

i)当$a=0$时,求$b$的取值范围;

ii)求证: $a^2+b^2>e$.

解. (1)因为$f(0)=1$,而$f'(x)=e^x-a\cos x,f'(0)=1-a$,则函数$y=f(x)$在$(0,f(0))$处的切线方程为$y=(1-a)x+1$.

(2)

i)当$a=0$时, $f(x)=e^x$,要使$y=f(x)$和$y=g(x)$有公共点,只需保证方程$f(x)=g(x)$有解,即$e^x=b\sqrt{x}$有解.

(参变分离)显然$x>0$,则$\displaystyle b=\frac{e^x}{\sqrt{x}}$.令$\displaystyle h(x)=\frac{e^x}{\sqrt{x}}$,则$\displaystyle h'\left( x \right) =\frac{e^x\left( 2x-1 \right)}{2x^{3/2}}$.于是$h(x)$在$\displaystyle\left(0,\frac{1}{2}\right)$上递减,在$\displaystyle\left[\frac{1}{2},+\infty\right)$上递增且$\displaystyle h\left( \frac{1}{2} \right) =\sqrt{2e}$.

由于$y=h(x)$的图象与水平直线$y=b$有交点,则$b$的取值范围为$\displaystyle\left[ \sqrt{2e},+\infty\right)$.

ii)要使$y=f(x)$和$y=g(x)$有公共点,只需保证方程$f(x)=g(x)$有解,即$e^x-a\sin x=b\sqrt{x}$有正数解$t$,则$a\sin t+b\sqrt{t}=e^t$.

由柯西不等式可得
$$
\left( a^2+b^2 \right) \left( \sin ^2t+t \right) \geqslant \left( a\sin t+b\sqrt{t} \right) ^2=e^{2t}.
$$
则
$$
a^2+b^2\geqslant \frac{e^{2t}}{\sin ^2t+t}\geqslant \frac{e^{2t}}{t^2+t}.
$$
下面证明$\displaystyle\frac{e^{2t}}{t^2+t}>e$,即$e^{2t}>et^2+et$.

令$F(x)=e^{2t}-et^2-et$,则$F'(x)=2e^{2t}-2et-e,F''(x)=4e^{2t}-2e$,
则$F'(x)$在$\displaystyle\left( 0,\frac{1}{2}\ln \frac{e}{2} \right)$上递减,在$\displaystyle\left[\frac{1}{2}\ln \frac{e}{2},+\infty\right)$上递增.
又$\displaystyle F'\left( 0 \right) =2-e<0,F'\left( \frac{1}{2} \right) =0,F'\left( 1 \right) =2e^2-3e>0$,则$F(x)$在$\displaystyle\left( 0,\frac{1}{2}\right)$上递减,在$\displaystyle\left[\frac{1}{2},+\infty\right)$上递增.故
$$
F\left( x \right) \geqslant F\left( \frac{1}{2} \right) =\frac{e}{4}>0.
$$
因此
$$
a^2+b^2\geqslant \frac{e^{2t}}{\sin ^2t+t}\geqslant \frac{e^{2t}}{t^2+t}>e.
$$

另证.利用$e^t>t+1$可得$e^{t-1}>t$,则$e^{2t-1}>t^2+t$,故$\displaystyle\frac{e^{2t}}{t^2+t}>e$.