2024年新高考2卷精选试题解答

**(2024年新高考2卷19题)**
已知双曲线$C:x^2-y^2=m\ (m>0)$,点$P_1(5,4)$在$C$上, $k$为常数, $0< k<1$.按照如下方式依次构造点$P_n\ \left(n=2,3,\cdots\right)$:过$P_{n-1}$作斜率为$k$的直线与$C$的左支交于点$Q_{n-1}$,令$P_n$为$Q_{n-1}$关于$y$轴的 对称点,记$P_{n}$的坐标为$\left(x_{n},y_{n}\right)$.

(1)若 $k=\frac 12$,求 $x_{2},y_{2}$;

(2)证明:数列$\left\{x_{n}-y_{n}\right\}$是公比为$\displaystyle\frac{1+k}{1-k}$的等比数列;

(3)设$S_n$为$\triangle P_nP_{n+1}P_{n+2}$的面积,证明:对任意正整数$n$, $S_n=S_{n+1}$.

**解.** 将点$P_1(5,4)$代入$C$的方程可得$m=9$.

(1)若 $k=\frac 12$,则直线$P_1Q_1$的方程为
$y=\frac{1}{2}x+\frac{3}{2}$.

联立双曲线$C$的方程$x^2-y^2=9$可得$x^2-2x-15=0$,
即$(x+3)(x-5)=0$,故$Q_1$为$(3,0)$, $P_2$为$(-3,0)$,
即$x_2=-3,y_2=0$.

(2)由$P_{n-1}\left(x_{n-1},y_{n-1}\right)$可得直线$P_{n-1}Q_{n-1}$的方程为
$y=kx+y_{n-1}-kx_{n-1}$.
联立双曲线方程$x^2-y^2=9$可得
$$
\left( 1-k^2 \right) x^2-2k\left( y_{n-1}-kx_{n-1} \right) x-\left( y_{n-1}-kx_{n-1} \right) ^2-9=0.
$$
由韦达定理可知$\displaystyle -x_n+x_{n-1}=\frac{2k\left( y_{n-1}-kx_{n-1} \right)}{1-k^2}$,
则$\displaystyle x_n=x_{n-1}-\frac{2k\left( y_{n-1}-kx_{n-1} \right)}{1-k^2}
=\frac{1+k^2}{1-k^2}x_{n-1}-\frac{2k}{1-k^2}y_{n-1}$.

于是$y_n=-kx_n+y_{n-1}-kx_{n-1}=-\frac{2k}{1-k^2}x_{n-1}
+\frac{1+k^2}{1-k^2}y_{n-1}$.

因此
$$
x_n-y_n=\frac{\left( 1+k \right) ^2}{1-k^2}x_{n-1}-\frac{(1+k)^2}{1-k^2}y_{n-1}
=\frac{1+k}{1-k}\left( x_{n-1}-y_{n-1} \right),
$$
故数列$\left\{x_{n}-y_{n}\right\}$是公比为$\displaystyle\frac{1+k}{1-k}$的等比数列.

(3)因为$\triangle P_{n+1}P_{n+2}P_{n+3}$和$\triangle P_nP_{n+1}P_{n+2}$都以边$P_{n+1}P_{n+2}$为底,要证$S_n=S_{n+1}$,只需证明$P_n$到直线$P_{n+1}P_{n+2}$的距离等于$P_{n+3}$到直线$P_{n+1}P_{n+2}$的距离,也等价于直线$P_{n+1}P_{n+2}$平行于直线$P_nP_{n+3}$.

直线$P_{n+1}P_{n+2}$的斜率为$\displaystyle k_1=\frac{y_{n+2}-y_{n+1}}{x_{n+2}-x_{n+1}}$,
直线$P_nP_{n+3}$的斜率为$\displaystyle k_2=\frac{y_ {n+3}-y_{n}}{x_{n+3}-x_{n}}$.

要证$\displaystyle\frac{y_ {n+2}-y_{n+1}}{x_{n+2}-x_{n+1}}=\frac{y_ {n+3}-y_{n}}{x_{n+3}-x_{n}}$,即证$\displaystyle 1-\frac{y_{n+2}-y_{n+1}}{x_{n+2}-x_{n+1}}
=1-\frac{y_{n+3}-y_n}{x_{n+3}-x_n}$,
等价于证明
$$
\frac{\left( x_{n+2}-y_{n+2} \right) -\left( x_{n+1}-y_{n+1} \right)}{x_{n+2}-x_{n+1}}=\frac{\left( x_{n+3}-y_{n+3} \right) -\left( x_n-y_n \right)}{x_{n+3}-x_n}.
$$

记$\displaystyle t=\frac{1+k}{1-k}$,由(2)可知数列$\left\{x_{n}-y_{n}\right\}$是公比为$t$的等比数列,
则等价于证明$\displaystyle\frac{t^2-t}{x_{n+2}-x_{n+1}}=\frac{t^3-1}{x_{n+3}-x_n}$,
只需证明$\displaystyle\frac{x_{n+3}-x_n}{x_{n+2}-x_{n+1}}=\frac{t^3-1}{t^2-t}$.

事实上,由$x_n-y_n=\left(x_1-y_1\right) t^{n-1}=t^{n-1}$和
$\displaystyle x_n+y_n=\frac{9}{x_n-y_n}=\frac{9}{t^{n-1}}$
可知$\displaystyle x_n=\frac{1}{2}\left( t^{n-1}+\frac{9}{t^{n-1}} \right)$.

于是
$$\begin{aligned}
\frac{x_{n+3}-x_n}{x_{n+2}-x_{n+1}} &=\frac{\frac{1}{2}\left( t^{n+2}+\frac{9}{t^{n+2}} \right) -\frac{1}{2}\left( t^{n-1}+\frac{9}{t^{n-1}} \right)}{\frac{1}{2}\left( t^{n+1}+\frac{9}{t^{n+1}} \right) -\frac{1}{2}\left( t^n+\frac{9}{t^n} \right)}\\
&=\frac{\left( t^{n+2}-t^{n-1} \right) \left( 1-\frac{9}{t^{2n+1}} \right)}{\left( t^{n+1}-t^n \right) \left( 1-\frac{9}{t^{2n+1}} \right)}=\frac{t^3-1}{t^2-t}.
\end{aligned}$$
得证.

故直线$P_{n+1}P_{n+2}$平行于直线$P_nP_{n+3}$,则对任意正整数$n$, $S_n=S_{n+1}$.

 

**(2024年甲卷高考)**
已知$f(x)=(1-ax)\ln (1+x)-x$.

(1)若$a=-2$时,求$f(x)$的极值;

(2)当$x\geqslant 0$时, $f(x)\geqslant 0$,求$a$的取值范围.

**解.** 函数的定义域为$(-1,+\infty)$.

(1)当$a=-2$时, $f\left( x \right) =\left( 1+2x \right) \ln \left( 1+x \right) -x$.则$\displaystyle f\left( x \right) =2\ln \left( 1+x \right) +\frac{x}{x+1}$.

令$\displaystyle g\left( x \right) =2\ln \left( 1+x \right) +\frac{x}{x+1}$,则$\displaystyle g'\left( x \right) =\frac{2x+3}{(1+x)^2}>0$,故$g(x)$在$(-1,+\infty)$上单调递增.又$g(0)=0$,则当$x\in (-1,0)$时,有$f'(x)=g(x)<0$;
当$x\in [0,+\infty)$时,有$f'(x)=g(x)\geqslant 0$.

故$g(x)$在$(-1,0)$上单调递减, $g(x)$在$[0,+\infty)$上单调递增,则$f(x)$的极小值为$f(0)=0$.

(2) **解法一:** 由于$\displaystyle f'\left( x \right) =-a\ln \left( 1+x \right) -\frac{\left( a+1 \right) x}{x+1}$,
令$\displaystyle h(x)=-a\ln \left( 1+x \right) -\frac{\left( a+1 \right) x}{x+1}$,则$\displaystyle h'\left( x \right) =-\frac{ax+2a+1}{\left( 1+x \right) ^2}$.

若$\displaystyle a\leqslant -\frac{1}{2}$,则当$x\geqslant 0$时,
$ax+2a+1\leqslant 2a+1\leqslant 0$,于是$h'(x)\geqslant 0$,又$h(0)=0$,故$f'\left( x \right) =h(x)\geqslant h(0)=0$,因此$f(x)\geqslant f(0)=0$,符合题意;

若$\displaystyle -\frac{1}{2}< a<0$,则当$\displaystyle 0< x<-2-\frac{1}{a}$时,
$ax+2a+1>0$,于是$h'(x)<0$,又$h(0)=0$,故$f'\left( x \right) =h(x)< h(0)=0$,
因此$f(x)< f(0)=0$,矛盾;

若$a\geqslant 0$,则当$x>0$时,
$ax+2a+1 >0$,于是$h'(x)<0$,又$h(0)=0$,故$f'\left( x \right) =h(x)< h(0)=0$,
因此$f(x)< f(0)=0$,也矛盾.

综上所述, $a$的取值范围是$\displaystyle\left( -\infty ,-\frac{1}{2} \right]$.

**解法二:** 若$a=0$,则$f(1)=\ln 2-1<0$,矛盾.

若$a>0$,则$\displaystyle f\left( \frac{1}{a} \right) =-\frac{1}{a}<0$,也矛盾;

若$a<0$,则$1-ax>0$,当$x\geqslant 0$时, $f(x)\geqslant 0$等价于$\displaystyle\ln \left( 1+x \right) -\frac{x}{1-ax}\geqslant 0$.

令$\displaystyle m(x)=\ln \left( 1+x \right) -\frac{x}{1-ax}$,
则$\displaystyle m'(x)=\frac{a^2x\left( x-\frac{2a+1}{a^2} \right)}{\left( 1+x \right) \left( 1-ax \right) ^2}$.

若$\displaystyle a\leqslant -\frac{1}{2}$,则$2a+1\leqslant 0$, $m'(x)\geqslant 0$,故$m(x)\geqslant m(0)=0$,符合题意;

若$\displaystyle -\frac{1}{2}< a<0$,则$2a+1> 0$,当$\displaystyle 0< x<\frac{2a+1}{a^2}$时, $m'(x)< 0$,此时$m(x)< m(0)=0$,矛盾.

综上所述, $a$的取值范围是$\displaystyle\left( -\infty ,-\frac{1}{2} \right]$.

**注:** 事实上,由泰勒公式可知
$$
f\left( x \right) =\left( 1-ax \right) \ln \left( 1+x \right) -x=\left( -a-\frac{1}{2} \right) x^2+\left( \frac{a}{2}+\frac{1}{3} \right) x^3+o\left( x^3 \right),
$$
则$\displaystyle a\leqslant -\frac{1}{2}$.