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，\(n - n{\left( {1 - \dfrac{x}{n}} \right)^n}{{\rm{e}}^x} \leqslant {x^2}\)．
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．