2014年华约自主招生试题

1、$5$个正整数，任意$4$个的和构成集合$\left\{ {44,45,46,47} \right\}$，求这$5$个整数．

2、甲乙两人采用五局三胜制比赛，单局甲获胜的概率为$p$且$p > \dfrac 12$，甲最终获胜的概率为$q$，当$p$为何值时$q - p$最大？

3、已知$f(x) = \dfrac {\sqrt 2 }2 \left( {\cos x - \sin x} \right)\sin \left( x + \dfrac{\pi }{4}\right) - 2a\sin x + b$的最大值为$1$，最小值为$- 4$，求$a,b$的值．

4、已知$f\left( x \right)$的反函数为${f^{ - 1}}\left( x \right)$，$g\left( x \right)$的反函数为${g^{ - 1}}\left( x \right)$．
（1）求证：$f\left( {g\left( x \right)} \right)$的反函数为${g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right)$；
（2）若$f\left( {g\left( x \right)} \right)$为奇函数，求证${g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right)$也为奇函数．

5、已知椭圆$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$（$a > b > 0$），圆${x^2} + {y^2} = {b^2}$，过椭圆上的动点$M$作圆的两条切线，切点分别为$P$、$Q$，直线$PQ$与坐标轴的交点为$E$、$F$，求$\triangle EOF$面积的最小值．

6、已知数列$\left\{ {{a_n}} \right\}$满足${a_{n + 1}} = n{p^n} + q{a_n},a_1=0$．

（1）若$q = 1$，求$\left\{ {{a_n}} \right\}$的通项公式；

（2）若$\left| p \right| < 1$，$\left| q \right| < 1$，求证：数列$\left\{ {{a_n}} \right\}$有界．

7、求证：当$x \leqslant n$时，$n - n{\left( {1 - \dfrac{x}{n}} \right)^n}{{\rm{e}}^x} \leqslant {x^2}$．

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参考答案

1、设这$5$个整数分别为${a_1} , {a_2} , {a_3} , {a_4} , {a_5}$，则
它们任意$4$个的和（共有$5$个）之和为$4\left( {{a_1} + {a_2} + {a_3} + {a_4} + {a_5}} \right)$必为$4$的倍数．

考虑到 $$44 \equiv 0\left( {\bmod 4} \right),\\45 \equiv 1\left( {\bmod 4} \right),\\46 \equiv 2\left( {\bmod 4} \right),\\47 \equiv 3\left( {\bmod 4} \right),$$ 于是这$5$个整数任意$4$个的和为 $$44 , 45 , 46 , 46 , 47.$$ 因此 $${a_1} + {a_2} + {a_3} + {a_4} + {a_5} = \dfrac{1}{4}\left( {44 + 45 + 46 + 46 + 47} \right) = 57.$$ 这$5$个整数分别为 $$57 - 44 , 57 - 45 , 57 - 46 , 57 - 46 , 57 - 47$$ 即 $$10 , 11 , 11 , 12 , 13.$$

2、根据题意 $$\begin{split}q &= {\rm C}_2^2{p^2} \cdot p + {\rm C}_3^2{p^2}\left( {1 - p} \right) \cdot p + {\rm C}_4^2{p^2}{\left( {1 - p} \right)^2} \cdot p\\&= {p^3} + 3{p^3}\left( {1 - p} \right) + 6{p^3}{\left( {1 - p} \right)^2}\\&= 6{p^5} - 15{p^4} + 10{p^3},\end{split}$$ 于是 $$q - p = 6{p^5} - 15{p^4} + 10{p^3} - p.$$ 因此 $$\begin{split}{\left( {6{p^5} - 15{p^4} + 10{p^3} - p} \right)^\prime } &= 30{p^4} - 60{p^3} + 30{p^2} - 1 \\&= 30{p^2}{\left( {p - 1} \right)^2} - 1,\end{split}$$ 由$p\left( {p - 1} \right) = - \sqrt {\dfrac{1}{{30}}}$，解得 $$p = \dfrac{{1 + \sqrt {1 - 4\sqrt {\dfrac{1}{{30}}} } }}{2}$$ 为使得$q-p$最大的$p$的值．

3、根据题意 $$\begin{split}f\left( x \right) &= \dfrac{{\sqrt 2 }}{2}\left( {\cos x - \sin x} \right)\left( {\dfrac{{\sqrt 2 }}{2}\sin x + \dfrac{{\sqrt 2 }}{2}\cos x} \right) - 2a\sin x + b\\&= \dfrac{1}{2}\left( {1 - 2{{\sin }^2}x} \right) - 2a\sin x + b\\&= - {\sin ^2}x - 2a\sin x + b + \dfrac{1}{2},\end{split}$$ 令$y = f\left( x \right)$，$t = \sin x$，$t \in \left[ { - 1 , 1} \right]$，则 $$y = - {t^2} - 2at + b + \dfrac{1}{2},$$ 考虑到对称轴为$t = - a$，于是 $$\begin{cases} - a < - 1 \\ {\left. y \right|_{t = - 1}} = 1 \\ {\left. y \right|_{t = 1}} = - 4 \\ \end{cases}\lor\begin{cases} - 1 \leqslant - a < 0 \\ {\left. y \right|_{t = - a}} = 1 \\ {\left. y \right|_{t = 1}} = - 4 \\ \end{cases}\lor\begin{cases} 0 \leqslant - a \leqslant 1 \\ {\left. y \right|_{t= -a}} = 1 \\ {\left. y \right|_{t = - 1}} = - 4 \\ \end{cases}\lor\begin{cases} - a > 1 \\ {\left. y \right|_{t = 1}} = 1 \\ {\left. y\right|_{t = - 1}} = - 4 \\ \end{cases}$$
解得$a = \pm \dfrac{5}{4}$，$b = - 1$．

4、（1）设$g\left( a \right) = b$，$f\left( b \right) = c$，则 $$f\left( {g\left( a \right)} \right) = c,a = {g^{ - 1}}\left( b \right),b = {f^{ - 1}}\left( c \right),$$ 于是$a = {g^{ - 1}}\left( {{f^{ - 1}}\left( c \right)} \right)$．因此$f\left( {g\left( x \right)} \right)$的反函数为${g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right)$．

（2）用分析法，${g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right)$为奇函数只需要
$$\begin{split}&\qquad \forall x , {g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right) + {g^{ - 1}}\left( {{f^{ - 1}}\left( { - x} \right)} \right) = 0\\&\Leftarrow \forall x , {g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right) = - {g^{ - 1}}\left( {{f^{ - 1}}\left( { - x} \right)} \right)\\&\Leftarrow \forall x , f\left( {g\left( {{g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right)} \right)} \right) = f\left( {g\left( { - {g^{ - 1}}\left( {{f^{ - 1}}\left( { - x} \right)} \right)} \right)} \right)\\&\Leftarrow \forall x , f\left( {g\left( {{g^{ - 1}}\left( {{f^{ - 1}}\left( x \right)} \right)} \right)} \right) = - f\left( {g\left( {{g^{ - 1}}\left( {{f^{ - 1}}\left( { - x} \right)} \right)} \right)} \right)\\&\Leftarrow \forall x , x = - \left( { - x} \right).\end{split}$$

5、设$M\left( {a\cos \theta , b\sin \theta } \right)$，则 $$PQ:xa\cos \theta + yb\sin \theta = {b^2},$$ 即 $$\dfrac{x}{{\dfrac{{{b^2}}}{{a\cos \theta }}}} + \dfrac{y}{{\dfrac{b}{{\sin \theta }}}} = 1,$$ 于是 $${S_{\triangle EOF}} = \left| {\dfrac{1}{2} \cdot \dfrac{{{b^2}}}{{a\cos \theta }} \cdot \dfrac{b}{{\sin \theta }}} \right| = \dfrac{{{b^3}}}{a} \cdot \dfrac{1}{{\left| {\sin 2\theta } \right|}}\geqslant \dfrac{{{b^3}}}{a},$$ 等号当且仅当$\theta = \dfrac{{\rm \pi}}{4}$时取得，因此$\triangle EOF$面积的最小值为$\dfrac{{{b^3}}}{a}$．

6、（1）若$p=1$，则$a_n=\dfrac{n(n-1)}{2}$；

若$p\ne 1$，则$a_n=\dfrac{p-p^n}{\left(1-p\right)^2}-\dfrac{(n-1)p^n}{1-p}$．

（2）由${a_{n + 1}} = n{p^n} + q{a_n},a_1=0$，得 $$\dfrac{a_{n+1}}{q^{n+1}}=\dfrac{a_n}{q^n}+\dfrac{1}{q}\cdot n\cdot \left(\dfrac{p}{q}\right)^n,$$ 所以 $$\dfrac{a_{n+1}}{q^{n+1}}=\dfrac{1}{q}\left[\left(\dfrac{p}{q}\right)+2\left(\dfrac{p}{q}\right)^2+3\left(\dfrac{p}{q}\right)^3+\cdots+n\left(\dfrac{p}{q}\right)^n\right],$$ 故 $$a_{n+1}=pq^{n-1}+2p^2q^{n-2}+\cdots+np^n.$$ 而当$\left| p \right| < 1$时，数列$\left\{np^n\right\}$有界，即存在常数$M>0$，使得$\left|np^n\right|<M$恒成立，所以 $$\begin{split}\left|a_{n+1}\right|&\leqslant \left|p\right|\cdot\left|q\right|^{n-1}+2\left|p\right|^2\cdot\left|q\right|^{n-2}+\cdots+n\left|p\right|^{n}\\&<M\left(1+\left|q\right|+\left|q\right|^2+\cdots+\left|q\right|^{n-1}\right)\\&<\dfrac{M}{1-\left|q\right|},\end{split}$$ 故数列$\left\{a_n\right\}$有界．

7、法一
令$f\left( x \right) = {x^2} + n{\left( {1 - \dfrac{x}{n}} \right)^n}{{\rm e}^x}$，则只需要证明当$x \leqslant n$时，$f\left( x \right) \geqslant n$．而
$$\begin{split}f'\left( x \right) &= 2x + {{\rm e}^x}\left[ {n{{\left( {1 - \dfrac{x}{n}} \right)}^n} + {n^2}{{\left( {1 - \dfrac{x}{n}} \right)}^{n - 1}}\left( { - \dfrac{1}{n}} \right)} \right]\\&= x\left[ {2 - {{\rm e}^x}{{\left( {1 - \dfrac{x}{n}} \right)}^{n - 1}}} \right].\end{split}$$

情形1    当$n = 1$时，有 $$f\left( x \right) = {x^2} + \left( {1 - x} \right){{\rm e}^x},$$ 于是 $$f'\left( x \right) = x\left( {2 - {{\rm e}^x}} \right),$$ 可得$f\left( x \right)$在$\left( { - \infty , 0} \right)$上单调递减，在$(0,\ln 2)$上单调递增，在$(\ln 2,1)$上单调递减．因此$f\left( x \right)$的极小值为$f\left( 0 \right) = 1$，以及$f\left( 1 \right) = 1$，原命题得证．

情形2    当$n \geqslant 2$时，令$g\left( x \right) = {{\rm e}^x}{\left( {1 - \dfrac{x}{n}} \right)^{n - 1}}$，则 $$g'\left( x \right) = {{\rm e}^x} \cdot \dfrac{{1 - x}}{n} \cdot {\left( {1 - \dfrac{x}{n}} \right)^{n - 2}},$$ 于是当$x = 1$时，$g\left( x \right)$取得最大值${\rm e}{\left( {1 - \dfrac{1}{n}} \right)^{n - 1}}$．

由于$\ln \left( {1 - \dfrac{1}{n}} \right) < - \dfrac{1}{n}$，因此${\rm e} < {\left( {1 - \dfrac{1}{n}} \right)^{ - n}}$．于是 $${\rm e}{\left( {1 - \dfrac{1}{n}} \right)^{n - 1}} < {\left( {1 - \dfrac{1}{n}} \right)^{ - 1}} = \dfrac{n}{{n - 1}} \leqslant 2,$$ 因此 $$2 - {{\rm e}^x}{\left( {1 - \dfrac{x}{n}} \right)^{n - 1}} > 0,$$ 从而$f\left( x \right)$的最小值为$f\left( 0 \right) = n$，原命题得证．

法二

由于 $$\ln \left( {1 + \dfrac{x}{n}} \right) < \dfrac{x}{n},$$ 于是 $${{\rm e}^x} > {\left( {1 + \dfrac{x}{n}} \right)^n},$$ 从而
$$\begin{split}n - n{\left( {1 - \dfrac{x}{n}} \right)^n}{{\rm e}^x} &\leqslant n - n{\left( {1 - \dfrac{x}{n}} \right)^n}{\left( {1 + \dfrac{x}{n}} \right)^n}\\&= n - n{\left( {1 - \dfrac{{{x^2}}}{{{n^2}}}} \right)^n}\\&\leqslant n - n\left( {1 - \dfrac{{{x^2}}}{{{n^2}}} \cdot n} \right)\\&= {x^2}.\end{split}$$ 其中倒数第二步用到了伯努利不等式．

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