2024年新高考1卷精选试题解答

**(2024年新高考1卷18题)**
已知函数$f(x)=\ln\frac x{2-x}+ax+b(x-1)^{3}$.

(1)若$b=0$,且$f'(x)\geqslant 0$,求$a$的最小值;

(2)证明:曲线$y=f(x)$是中心对称图形;

(3)若$f(x)>-2$当且仅当$1< x<2$,求$b$的取值范围.

**解.** 函数$f(x)$的定义域为$(0,2)$.

(1)若$b=0$,则$f\left( x \right) =\ln \frac{x}{2-x}+ax=\ln x-\ln \left( 2-x \right) +ax$,
故$f'\left( x \right) =\frac{1}{x}-\frac{-1}{2-x}+a=\frac{2}{x\left( 2-x \right)}+a$.

由$f'(x)\geqslant 0$可知$a\geqslant -\frac{2}{x\left( 2-x \right)}$.
由基本不等式可得$x\left( 2-x \right) \leqslant \left[ \frac{x+\left( 2-x \right)}{2} \right]^2=1$,
则$-\frac{2}{x\left( 2-x \right)}\leqslant -2$,当且仅当$x=1$时取等号成立.故$a\geqslant -2$.故$a$的最小值为$-2$.

(2)令$F\left( x \right) =f\left( x+1 \right) -a=\ln \frac{1+x}{1-x}+ax+bx^3$,则$F\left( -x \right) =\ln \frac{1-x}{1+x}-ax-bx^3=-F\left( x \right)$,
故$F(x)$是奇函数,它的图象关于$(0,0)$对称.从而$y=f\left( x \right)$关于$(1,a)$对称.故曲线$y=f(x)$是中心对称图形.

(3)由函数$f(x)$的连续性可知$f(1)=a=-2$.

否则,若$a<-2$,则$f(1)=a<-2$.当$x\to 2$时, $f(x)\to +\infty$,存在$x_0\in (1,2)$,使得$f(x_0)=-2$,
则$f(x)>-2$当且仅当$x_0< x<2$,与题设矛盾;

若$a>-2$,则$f(1)=a>-2$.故$x=1$时$f(x)> -2$也成立,与题设矛盾.

 

于是$f\left( x \right) =\ln \frac{x}{2-x}-2x+b\left( x-1 \right) ^3$.
令$u=x-1$,则$u\in (0,1)$, $g\left( u \right) =f\left( u+1 \right)+2=\ln \frac{1+u}{1-u}-2u+bu^3$,
则$g(u)>0$当且仅当$0< u<1$.

**解法一:** 由$g'\left( u \right) =u^2\left( 3b+\frac{2}{1-u^2} \right)$

若$b\geqslant -\frac{2}{3}$,由$3b+\frac{2}{1-u^2}>3b+2\geqslant 0$可知$g'(u)>0$,故$g(u)$在$(0,1)$上单调递增,则$g(u)>g(0)=0$,符合题意.

若$b<-\frac{2}{3}$,则当$0< u<\sqrt{1+\frac{2}{3b}}$时,$g'(u)<0$,故$g(u)$在$\left( 0,\sqrt{1+\frac{2}{3b}} \right)$
上单调递减,此时$g(u)< g(0)=0$,矛盾.

综上所述, $b\geqslant -\frac{2}{3}$.

 

**解法二:** 利用参变分离可知, $b>\frac{2u-\ln \frac{1+u}{1-u}}{u^3}$.

令$G\left( u \right) =\frac{2u-\ln \frac{1+u}{1-u}}{u^3},0< u<1$.则
$$G'\left( u \right) =\frac{1}{u^4}\left[ \frac{-4u^3+6u}{u^2-1}+3\ln \frac{1+u}{1-u} \right].$$

令$h\left( u \right) =\frac{-4u^3+6u}{u^2-1}+3\ln \frac{1+u}{1-u},0< u<1$,则$h'\left( u \right) =-\frac{4u^4}{\left( u^2-1 \right) ^2}<0$,
故$h(u)$在$(0,1)$上单调递减,则$h(u)< h(0)=0$.

故$G'(u)<0$,函数$G(u)$在$(0,1)$上单调递减.

由洛必达法则可知$\lim_{u\to 0}G(u)=-\frac{2}{3}$.

因此$b\geqslant -\frac{2}{3}$.

**分析:** 事实上,由泰勒展开
$$g\left( u \right) =\ln \frac{1+u}{1-u}-2u+bu^3=\left( b+\frac{2}{3} \right) u^3+\frac{2}{5}u^5+o\left( u^5 \right)$$
可知, $g(u)>0$当且仅当$0< u<1$,必须有$b+\frac{2}{3}\geqslant 0$,则$b\geqslant -\frac{2} {3}$.

 

**(2024年新高考1卷19题)**
设$m$为正整数,数列$a_1,a_2,\cdots,a_{4m+2}$是公差不为$0$的等差数列,若从中删去两项$a_i$和$a_j\ \left(i< j\right)$后剩余的$4m$项可被平均分为$m$组,且每组的$4$个数都能构成等差数列,则称数列$a_1,a_2,\cdots,a_{4m+2}$是$(i,j)-$可分数列.

(1)写出所有的$(i,j)$, $1\leqslant i < j\leqslant 6$,使得数列$a_1,a_2,\cdots,a_6$是$(i,j)-$可分数列;

(2)当$m\geqslant 3$时,证明:数列$a_1,a_2,\cdots,a_{4m+2}$是$(2,13)-$可分数列;

(3)从$1,2,\cdots,4m+2$中一次任取两个数$i$和 $j\ \left ( i< j\right )$,记数列$a_1,a_2,\cdots,a_{4m+2}$
是$(i,j)-$可分数列的概率为$P_m$,证明: $P_m>\frac18$.

**解.** (1) $(1,2),(1,6),(5,6)$;

(2)将数列$a_1,a_2,\cdots,a_{4m+2}$剔除$a_2$和$a_{13}$后,分成如下$m$组:
$$\begin{array}{lll} a_1,a_4,a_7,a_{10};& a_3,a_6,a_9,a_{12};& \\ a_5,a_8,a_{11},a_{14};& a_{15},a_{16},a_{17},a_{18};& \qquad\qquad\cdots\\ a_{4k-1},a_{4k},a_{4k+1},a_{4k+2};& \qquad\qquad\cdots& a_{4m-1},a_{4m},a_{4m+1},a_{4m+2}.\\ \end{array}$$

则以上各组的四个数依次成等差数列.故数列$a_1,a_2,\cdots,a_{4m+2}$是$(2,13)-$可分数列;

(3)下面证明满足$a_1,a_2,\cdots,a_{4m+2}$
是$(i,j)-$可分数列的$(i,j)$的个数$x_m\geqslant m^2+m+1$.

由(1)可知$x_1=3$.

记数列$a_1,a_2,\cdots,a_{4m+2}$为$A_m$, 数列$a_1,a_2,\cdots,a_{4m+6}$为$A_{m+1}$.

若$A_m$是$(i,j)-$可分数列,则$A_{m+1}$也是$(i,j)-$可分数列,这样的$(i,j)$共有$x_m$组;

 

当$0\leqslant n\leqslant m+1$且$n\in \mathbf{N}^\ast$时,将$A_{m+1}$剔除掉$a_{4n+1},a_{4m+6}$两项后剩余的$4m+4$项中连续$4$项依次进行分组可知, $A_{m+1}$是$(4n+1,4m+6)-$可分数列,这样的$(i,j)$共有$m+2$组;

当$0\leqslant n\leqslant m-1$且$n\in \mathbf{N}^\ast$时,记$k=m-n+1$.先将$a_1,a_2,\cdots,a_{4n}$这$4n$项中连续$4$项依次进行分组,再按照如下分组
$$\begin{array}{llll} a_{4n+1},&a_{4n+k+1},&a_{4n+2k+1},&a_{4n+3k+1},\\ a_{4n+3},&a_{4n+k+3},& a_{4n+2k+3},& a_{4n+3k+3},\\ a_{4n+4},& a_{4n+k+4},&a_{4n+2k+4},& a_{4n+3k+4},\\ \quad\vdots & \qquad\vdots &\qquad\vdots & \qquad\vdots \\ a_{4n+k},& a_{4n+2k},& a_{4n+3k},& a_{4n+4k},\\ a_{4n+k+2}, &a_{4n+2k+2},& a_{4n+3k+2},& a_{4n+4k+2}. \end{array}$$
可知, $A_{m+1}$是$(4n+2,4m+5)-$可分数列,这样的$(i,j)$共有$m$组.

因此$x_{m+1}\geqslant x_m+m+2+m=x_m+2m+2$,故
$$x_m=\sum_{k=2}^m{\left( x_k-x_{k-1} \right)}+x_1\geqslant\sum_{k=2}^m{2k}+3=m^2+m+1,$$
于是
$$P_m=\frac{x_m}{C_{4m+2}^{2}}\geqslant \frac{m^2+m+1}{\left( 2m+1 \right) \left( 4m+1 \right)}>\frac{1}{8}.$$

**注:** 实际上,可以证明$x_m=m^2+m+1$.

 

**(2024年天津高考20题)** 设$f(x)=x\ln x$. (1)求$f(x)$在点$(1,f(1))$处的切线方程; (2)若$f(x)\geqslant a(x-\sqrt{x})$对$x\in(0,+\infty)$都成立,求$a$的取值范围; (3)当$x_1,x_2\in (0,1)$时,求证: $|f(x_1)-f(x_2)|\leqslant |x_1-x_2|^{1/2}$. **解.** 函数$f(x)$的定义域为$(0,+\infty)$. (1)由于$f'(x)=\ln x+1$, $f(1)=0$, $f'(1)=1$,故所求切线方程为$y=x-1$. (2)由$f(x)\geqslant a(x-\sqrt{x})$可得$\ln x\geqslant a\left( 1-\frac{1}{\sqrt{x}} \right)$. 令$g\left( x \right) =\ln x-a\left( 1-\frac{1}{\sqrt{x}} \right)$, 则$g'\left( x \right) =\frac{1}{x}-\frac{a}{2x\sqrt{x}} =\frac{\sqrt{x}-\frac{a}{2}}{x\sqrt{x}}$. 若$a\leqslant 0$,则$g(x)$在$(0,+\infty)$上单调递增, 而$g(1)=0$,故当$0< x<1$时, $g(x)<0$,矛盾. 若$a>0$,则$g(x)$在$\left( 0,\frac{a^2}{4} \right)$ 上单调递减,在$\left[\frac{a^2}{4},+\infty\right)$ 上单调递增.若$\frac{a^2}{4}\neq 1$, 则必有$g\left( \frac{a^2}{4} \right) < g\left( 1 \right) =0$,矛盾.故$\frac{a^2}{4}=1$,解得$a=2$. (3) **解法一:** 不妨设$0\leqslant x_1\leqslant x_2\leqslant 1$, 要证$|f(x_1)-f(x_2)|\leqslant |x_1-x_2|^{1/2}$, 只需证 $$-\sqrt{x_2-x_1}\leqslant x_1\ln x_1-x_2\ln x_2\leqslant \sqrt{x_2-x_1}.$$ 一方面,由$\left( \sqrt{x_2-x_1}+\sqrt{x_1} \right) ^2=x_2+2\sqrt{\left( x_2-x_1 \right) x_1}\geqslant x_2$可知$\sqrt{x_2-x_1}\ge \sqrt{x_2}-\sqrt{x_1}$. 下面证明: $x_1\ln x_1-x_2\ln x_2\leqslant \sqrt{x_2}-\sqrt{x_1}$, 这等价于证明$x_1\ln x_1+\sqrt{x_1}\leqslant x_2\ln x_2+\sqrt{x_2}$. 令$F\left( x \right) =x\ln x+\sqrt{x}$,则$F'\left( x \right) =\ln x+1+\frac{1}{2\sqrt{x}}$. 而$F''\left( x \right) =\frac{1}{x}-\frac{1}{4x\sqrt{x}} =\frac{4\sqrt{x}-1}{4x\sqrt{x}}$, 故$F'(x)$在$\left(0,\frac{1}{16}\right)$上递减,在$\left(\frac{1}{16},1\right)$上递增. 在不等式$\ln x<\frac{1}{2}\left( x-\frac{1}{x} \right)\ (x>1)$中令$x=2$可知$\ln 2<\frac{3}{4}$. 则$F'\left( x \right) \geqslant F'\left( \frac{1}{16} \right) =3-4\ln 2>0$,故$F(x)$在$\left(0,1\right)$上递增, 于是$x_1\ln x_1+\sqrt{x_1}\leqslant x_2\ln x_2+\sqrt{x_2}$. 从而有$x_1\ln x_1-x_2\ln x_2\leqslant \sqrt{x_2}-\sqrt{x_1}\leqslant \sqrt{x_2-x_1}$. 另一方面,由$x_2-x_1\leqslant 1$可知$x_2-x_1=\sqrt{x_2-x_1}\cdot \sqrt{x_2-x_1}\leqslant \sqrt{x_2-x_1}$,则$-\sqrt{x_2-x_1}\leqslant x_1-x_2$. 下面证明: $x_1\ln x_1-x_2\ln x_2\geqslant x_1-x_2$, 这等价于证明$x_1\ln x_1-x_1\geqslant x_2\ln x_2-x_2$. 令$G\left( x \right) =x\ln x-x$,则$G'\left( x \right) =\ln x<0$.故$G(x)$在$\left(0,1\right)$上递减, 于是$x_1\ln x_1-x_1\geqslant x_2\ln x_2-x_2$. 从而有$x_1\ln x_1-x_2\ln x_2\geqslant x_1-x_2\geqslant -\sqrt{x_2-x_1}$. **解法二:** 不妨设$0\leqslant x_1\leqslant x_2\leqslant 1$,由柯西不等式可得 $$\begin{aligned} \left| x_1\ln x_1-x_2\ln x_2 \right|^2 &=\left| \int_{x_1}^{x_2}{\left( 1+\ln x \right) dx} \right|^2\leqslant \int_{x_1}^{x_2}{\left( 1+\ln x \right) ^2dx}\int_{x_1}^{x_2}{1^2dx}\\ &\leqslant \int_0^1{\left( 1+\ln x \right) ^2dx}\int_{x_1}^{x_2}{dx}=\int_{x_1}^{x_2}{dx}=\left| x_1-x_2 \right|, \end{aligned}$$ 于是$|f(x_1)-f(x_2)|\leqslant |x_1-x_2|^{1/2}$. 这里利用了分部积分公式以及 $$\begin{aligned} \int_0^1{\left( 1+\ln x \right) ^2dx} &=\left[ x\left( 1+\ln x \right) ^2 \right] _{0}^{1}-\int_0^1{2\left( 1+\ln x \right) dx}\\ &=1-\left[ 2x\left( 1+\ln x \right) \right] _{0}^{1}+\int_0^1{2dx}=1. \end{aligned}$$