北大本科06数分期中试题(李伟固命题)

Eufisky posted @ 2015年8月06日 00:44 in 数学分析 with tags 北大 , 1112 阅读

前些年在百度文库(http://wenku.baidu.com/link?url=Y_HDeeYMcyEGuUJWt3fNqC7N08AEqMVfleNVGcv7hC2t9EVO0-MFFHWuqnLYyiDJ4H7ATgg1fOQGN1Lta2RW4Z4pOrm6aZ468WkrTqNqZzG)找到一份李伟固命制的06年北大期中考试题，当时感觉难度稍大，正好结识了Veer大神，便把这份题发给他。事实证明，V神秒得还是很顺利！

1.给定实数 $\lambda_i(1\leq i\leq n)$ ,满足 $\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {k = 1,2,3, \cdots } \right)$ .令 $f\left( x \right) = \prod\limits_{i = 1}^n {\frac{1}{{1 - {\lambda _i}x}}}$ .证明: $f^{(k)}(0)>0,k=1,2,3,\cdots$ .

证明：令 $g\left( x \right) = \ln f\left( x \right) = - \ln \left( {1 - {\lambda _1}x} \right)-\ln \left( {1 - {\lambda _2}x} \right) -\cdots -\ln \left( {1 - {\lambda _n}x} \right)$ $\left( {x \in U\left( {0;\delta } \right)\text{使得对}\forall {\lambda _i},\text{有}1 - {\lambda _i}x > 0} \right)$ ,则

$g\left( x \right) = \left( {\sum\limits_{i = 1}^n {{\lambda _i}} } \right)x + \left( {\frac{1}{2}\sum\limits_{i = 1}^n {\lambda _i^2} } \right){x^2} + \cdots + \left( {\frac{1}{k}\sum\limits_{i = 1}^n {\lambda _i^k} } \right){x^k} + \cdots$ 由函数幂级数展开的唯一性可知 ${g^{\left( k \right)}}\left( 0 \right) = \frac{1}{k}\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {x \in U\left( {0;\delta } \right)} \right)$ .

另一方面 $f\left( x \right) = {e^{g\left( x \right)}}\left( {x \in U\left( {0;\delta } \right)} \right)$ .首先注意到对任意可导函数 $F(x)$ ,有 ${\left( {{e^{F\left( x \right)}}} \right)^\prime } = F'\left( x \right){e^{F\left( x \right)}}$ .其次注意到对可导函数组 $F_1,F_2,\cdots,F_s$ ,有 ${\left( {{F_1}{F_2}{F_3} \cdots {F_s}} \right)^\prime } = {{F'}_1}{F_2}{F_3} \cdots {F_s} + {F_1}{F_2}^\prime {F_3} \cdots {F_s} + \cdots + {F_1}{F_2}{F_3} \cdots {F_s}^\prime$ ,从而归纳可证

${f^{\left( k \right)}}\left( x \right) = {\left( {{e^{g\left( x \right)}}} \right)^{\left( k \right)}} = \left( {\sum\limits_{j \in {N_ + },{k_i} \in {N_ + }} {{g^{\left( {{k_1}} \right)}}\left( x \right){g^{\left( {{k_2}} \right)}}\left( x \right) \cdots {g^{\left( {{k_j}} \right)}}\left( x \right)} } \right){e^{g\left( x \right)}}.$ 由 ${g^{\left( k \right)}}\left( 0 \right) > 0,k=1,2,3,\cdots$ 且 $g(0)=0$ ,所以 ${f^{\left( k \right)}}\left( 0 \right) > 0,k=1,2,3,\cdots$ .

可知 $f\left( x \right) = \prod\limits_{i = 1}^n {\frac{1}{{1 - {\lambda _i}x}}} = \left( {1 + {\lambda _1}x + \lambda _1^2{x^2} + \cdots } \right)\left( {1 + {\lambda _2}x + \lambda _2^2{x^2} + \cdots } \right) \cdots \left( {1 + {\lambda _n}x + \lambda _n^2{x^2} + \cdots } \right).$ 由幂级数的乘积公式归纳可得

${f^{\left( k \right)}}\left( 0 \right) = \sum\limits_{\substack{{k_1} + {k_2} + \cdots + {k_n} = k\\ \left( {{k_1},{k_2}, \cdots ,{k_n}} \right)}} {\lambda _1^{{k_1}}\lambda _2^{{k_2}} \cdots \lambda _n^{{k_n}}} \left( \text{其中}{{k_i} \in N} \right).$ 若能通过 $\sum\limits_{i = 1}^n {\lambda _i^k} > 0\left( {k = 1,2,3, \cdots } \right)$ 得出 $\sum\limits_{\substack{{k_1} + {k_2} + \cdots + {k_n} = k\\ \left( {{k_1},{k_2}, \cdots ,{k_n}} \right)}} {\lambda _1^{{k_1}}\lambda _2^{{k_2}} \cdots \lambda _n^{{k_n}}}>0$ 即可得到证明.

2.令 $D = \left\{ {u = \left( {x,y} \right) \in {\mathbb{R}^2}\left| {\left\| u \right\| = \sqrt {{x^2} + {y^2}} \le \frac{1}{2}} \right.} \right\}$ . $f(u)=f(x,y)$ 是全平面上的连续可微函数满足 $\left\| {\nabla f\left( {0,0} \right)} \right\| = 1,\left\| {\nabla f\left( u \right) - \nabla f\left( v \right)} \right\| \le \left\| {u - v} \right\|$ .那么对于任意的 $u,v\in D$ ,证明函数 $f|_D$ 在 $D$ 中唯一点处达到其最大值.

证明：对 $u\in D$ ,有 $\left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| {u - \left( {0,0} \right)} \right\|$ ,即

$1 - \left\| {\nabla f\left( u \right)} \right\| = \left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( u \right)} \right\| \le \left\| {\nabla f\left( u \right) - \nabla f\left( {0,0} \right)} \right\| \le \left\| u \right\|,$ 亦即 $1 - \left\| u \right\| \le \left\| {\nabla f\left( u \right)} \right\|$ ,则 $\nabla f\left( u \right) \ne \left( {0,0} \right)$ ,所以 $f$ 的最大值只可能在边界上取得.

易知 $f$ 在其边界上的函数可设为 $g\left( t \right) = f\left( {\frac{1}{2}\cos \theta ,\frac{1}{2}\sin \theta } \right),g'\left( \theta \right) = \frac{1}{2}\left( { - \sin \theta {f_x} + \cos \theta {f_y}} \right),\theta \in \left[ {0,2\pi } \right).$ 现假设 $f$ 在其边界上有两点取得最大值,不妨设为 $\theta=\theta_1$ 和 $\theta=\theta_2$ ,记 ${u_1} = \left( {\frac{1}{2}\cos {\theta _1},\frac{1}{2}\sin {\theta _1}} \right),{u_2} = \left( {\frac{1}{2}\cos {\theta _2},\frac{1}{2}\sin {\theta _2}} \right)$ ,则由 $g'\left( {{\theta _1}} \right) = g'\left( {{\theta _2}} \right) = 0$ 可得 $- \sin {\theta _1}{f_x}\left( {{u_1}} \right) + \cos {\theta _1}{f_y}\left( {{u_1}} \right) = 0$ ,即 $\nabla f\left( {{u_1}} \right)$ 与 $\left( { - \sin {\theta _1},\cos {\theta _1}} \right)$ 垂直即可.设 $\nabla f\left( {{u_1}} \right) = {a_1}\left( {\cos {\theta _1},\sin {\theta _1}} \right)$ ,由于 $f$ 在 $u_1$ 处取得最大值,则 $f$ 在 $u_1$ 点沿方向 $(\cos\theta_1,\sin \theta_1)$ 的方向导数需大于等于 $0$ ,所以 $a_1\geq 0$ .

另一方面由 $\left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {\nabla f\left( {{u_1}} \right)} \right\| \le \left\| {\nabla f\left( {0,0} \right) - \nabla f\left( {{u_1}} \right)} \right\|$ 当且仅当 ${\nabla f\left( {{0,0}} \right)}$ 与 ${\nabla f\left( {{u_1}} \right)}$ 异向取等可知

$\left\| {\nabla f\left( {0,0} \right)} \right\| - \left\| {{u_1}} \right\| \le \left\| {\nabla f\left( {{u_1}} \right)} \right\| = \left| {{a_1}} \right|,$

即 $a_1\geq \frac12$ .同理 $\nabla f\left( {{u_2}} \right) = {a_2}\left( {\cos {\theta _2},\sin {\theta _2}} \right),a_2\geq \frac12$ .由于不等式取等需要与 ${\nabla f\left( {0,0} \right)}$ 异向,故 $a_1\geq\frac12,a_2\geq\frac12$ 中有一个是严格的.不妨设为 $a_1>\frac12$ ,再设 $\overrightarrow {{r_1}} = \left( {\cos {\theta _1},\sin {\theta _1}} \right),\overrightarrow {{r_2}} = \left( {\cos {\theta _2},\sin {\theta _2}} \right)$ ,则

$\begin{align*}&{\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\|^2} - {\left\| {{u_1} - {u_2}} \right\|^2}\\= &{\left\| {{a_1}\overrightarrow {{r_1}} - {a_2}\overrightarrow {{r_2}} } \right\|^2} - {\left\| {\frac{1}{2}\overrightarrow {{r_1}} - \frac{1}{2}\overrightarrow {{r_2}} } \right\|^2} = a_1^2 + a_2^2 - 2{a_1}{a_2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} - \left( {\frac{1}{4} - \frac{1}{2}\overrightarrow {{r_1}} \overrightarrow {{r_2}} + \frac{1}{4}} \right)\\= &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right)\overrightarrow {{r_1}} \overrightarrow {{r_2}} \left( \text{由于}{\overrightarrow {{r_1}}\text{与} \overrightarrow {{r_2}} \text{不同向},\text{所以}\overrightarrow {{r_1}} \overrightarrow {{r_2}} < 1,\text{且}2{a_1}{a_2} - \frac{1}{2} > 0} \right)\\> &a_1^2 + a_2^2 - \frac{1}{2} - \left( {2{a_1}{a_2} - \frac{1}{2}} \right) = {\left( {{a_1} - {a_2}} \right)^2} \ge 0,\end{align*}$

因此 ${\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\|^2} - {\left\| {{u_1} - {u_2}} \right\|^2} > 0$ ,即 $\left\| {\nabla f\left( {{u_1}} \right) - \nabla f\left( {{u_2}} \right)} \right\| > \left\| {{u_1} - {u_2}} \right\|$ ,矛盾,从而 $f$ 在边界上只有一点取得最大值,即函数 $f|_D$ 在 $D$ 中唯一点处达到其最大值.

3.讨论级数 $\sum\limits_{n = 1}^{ + \infty } {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}},\alpha\in \mathbb{R}$ 的收敛性.

解：(1)当 $\alpha\leq0$ 时,我们可知 $\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}$ 发散,从而级数 $\sum\limits_{n = 1}^{ + \infty } {{{\left( { - 1} \right)}^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}}}$ 发散.

(2)当 $\alpha>0$ 时,由于 $\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^{n - 1}}\frac{{\sin \left( {\ln n} \right)}}{{{n^\alpha }}} = 0$ ,我们讨论其前 $2n$ 项和数列的收敛性即可,也就是级数 $\sum\limits_{n = 1}^{ + \infty } {\left[ {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$ 的收敛性.

而

$\begin{align*}&\left| {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\= &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)\left( {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right) + \frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right) - \sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left| {\sin \left( {\ln \left( {2k - 1} \right)} \right)} \right|\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\frac{{2\cos \left[ {\frac{{\ln \left( {2k - 1} \right) + \ln \left( {2k} \right)}}{2}} \right]\sin \left[ {\frac{{\ln \left( {2k - 1} \right) - \ln \left( {2k} \right)}}{2}} \right]}}{{{{\left( {2k} \right)}^\alpha }}}} \right|\\\le &\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right] + \left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha}}}.\end{align*}$

已知 $\left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha }}} \sim \frac{1}{{{{\left( {2k} \right)}^{\alpha + 1}}}}$ ,从而 $\sum\limits_{k = 1}^{ + \infty } {\left| {\ln \left( {1 - \frac{1}{{2k}}} \right)} \right|\frac{1}{{{{\left( {2k} \right)}^\alpha }}}}$ 收敛.又 $\sum\limits_{k = 1}^{ + \infty } {\left[ {\frac{1}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{1}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$ 是收敛的,故 $\sum\limits_{n = 1}^{ + \infty } {\left| {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right|}$ 收敛,因此 $\sum\limits_{n = 1}^{ + \infty } {\left[ {\frac{{\sin \left( {\ln \left( {2k - 1} \right)} \right)}}{{{{\left( {2k - 1} \right)}^\alpha }}} - \frac{{\sin \left( {\ln \left( {2k} \right)} \right)}}{{{{\left( {2k} \right)}^\alpha }}}} \right]}$ 亦收敛.

综上所述, 当 $\alpha\leq 0$ 时,级数发散;当 $\alpha>0$ 时级数收敛.

3.求 $\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\}.$

解：首先, $\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^1 {\frac{{\cos kx}}{k}} = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \cos x = 0,\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^2 {\frac{{\cos kx}}{k}} = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \left( {\cos x + \frac{{\cos 2x}}{2}} \right) = - \frac{1}{2}$ .

现记 ${f_n}\left( x \right) = \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}}$ ,则 $\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_1}\left( x \right) = 0 \ge - \frac{1}{2},\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_1}\left( x \right) = - \frac{1}{2} \ge - \frac{1}{2}$ .

现假设 $n=k-1(k\geq2,k\in\mathbb{N}_+)$ 时,有 $\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_n}\left( x \right) = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_{k - 1}}\left( x \right) \ge - \frac{1}{2}$ 成立;

当 $n=k$ 时,注意到 ${f_n}^\prime \left( x \right) = \sum\limits_{k = 1}^n {\left( { - \sin kx} \right)} = \sum\limits_{i = 1}^k {\left( { - \sin ix} \right)} = \sum\limits_{i = 1}^k {\frac{{\cos \left( {i + \frac{1}{2}} \right)x - \cos \left( {i - \frac{1}{2}} \right)x}}{{2\sin \frac{x}{2}}}} ,$

则 ${f_k}^\prime \left( x \right) = \frac{{\cos \left( {k + \frac{1}{2}} \right)x - \cos \frac{x}{2}}}{{2\sin \frac{x}{2}}} = - \frac{{\sin \frac{{k + 1}}{2}x\sin \frac{k}{2}x}}{{\sin \frac{x}{2}}},x \in \left[ {0,\frac{\pi }{2}} \right].$

由 $f_k(x)$ 的连续性可知 $\min f_k(x)$ 的点只可能在端点或稳定点,从而令 $f'_k(x)=0$ ,则 $x = \frac{{2j\pi }}{{k + 1}},\frac{{2j\pi }}{k},j \in \mathbb{Z}$ 且 $\frac{{2j\pi }}{{k + 1}},\frac{{2j\pi }}{k} \in \left( {0,\frac{\pi }{2}} \right)$ ,而

${f_k}\left( {\frac{{2j\pi }}{{k + 1}}} \right) = \sum\limits_{i = 1}^n {\frac{{\cos i\left( {\frac{{2j\pi }}{{k + 1}}} \right)}}{i}} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2kj\pi }}{{k + 1}}}}{k} = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k}.$

由于 $\frac{{2j\pi }}{{k + 1}} \in \left[ {0,\frac{\pi }{2}} \right]$ ,所以 $\frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k} > 0$ ,则 ${f_k}\left( {\frac{{2j\pi }}{{k + 1}}} \right) = {f_{k - 1}}\left( {\frac{{2j\pi }}{{k + 1}}} \right) + \frac{{\cos \frac{{2j\pi }}{{k + 1}}}}{k} \ge - \frac{1}{2}.$

而 ${f_k}\left( {\frac{{2j\pi }}{k}} \right) = \sum\limits_{i = 1}^n {\frac{{\cos i\left( {\frac{{2j\pi }}{k}} \right)}}{i}} = {f_{k - 1}}\left( {\frac{{2j\pi }}{k}} \right) + \frac{1}{k} \ge - \frac{1}{2}$ ,又

${f_k}\left( 0 \right) = \sum\limits_{i = 1}^k {\frac{1}{i}} > - \frac{1}{2},{f_k}\left( {\frac{\pi }{2}} \right) = \sum\limits_{i = 1}^k {\frac{{\cos i\frac{\pi }{2}}}{i}} = \left\{ \begin{array}{l}\frac{1}{2}\left[ { - 1 + \left({\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s + 1\\\frac{1}{2}\left[ { - 1 + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \cdots + \left( {\frac{1}{{s - 1}} - \frac{1}{s}} \right)} \right],k = 2s\end{array} \right.$ ,从而 ${f_k}\left( {\frac{\pi }{2}} \right) \ge - \frac{1}{2}$ .

综上归纳可知 $\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} {f_n}\left( x \right) = \mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} \ge - \frac{1}{2}$ ,又 $\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^2 {\frac{{\cos kx}}{k}} = - \frac{1}{2}$ ,因此

$\mathop {\inf }\limits_{n \ge 1} \left\{ {\mathop {\min }\limits_{x \in \left[ {0,\frac{\pi }{2}} \right]} \sum\limits_{k = 1}^n {\frac{{\cos kx}}{k}} } \right\} = - \frac{1}{2}.$

4.函数 $f(x)$ 在 $[0,1]$ 上二次可导, $f(0)=2,f'(0)=-2,f(1)=1$ .证明存在 $c\in (0,1)$ ,使得 $f(c)f'(c)+f''(c)=0$ .

证明：令 $F\left( x \right) = \frac{1}{2}{f^2}\left( x \right) + f'\left( x \right)$ ,则 $F(0)=\frac12\times 2^2-2=0$ .现假设不存在 $\xi\in (0,1]$ 使得 $F(\xi)=0$ ,则 $F$ 在 $(0,1]$ 上不变号.倘若 $\forall x\in (0,1]$ 都有 $F(x)<0$ ,即 $\frac{1}{2}{f^2}\left( x \right) + f'\left( x \right) < 0$ ,则 $f'\left( x \right) < - \frac{1}{2}{f^2}\left( x \right) \le 0$ ,则 $f$ 在 $[0,1]$ 上单调递减,由 $f(0)=2,f(1)=1$ 可知 $1\leq f(x)\leq 2,f(x)\neq 0$ ,从而 $\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}} < - \frac{1}{2}$ .由 $f,f'$ 的连续性可知 $\int_0^1 {\frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}}dx} < \int_0^1 {\left( { - \frac{1}{2}} \right)dx}$ ,即 $\left. { - \frac{1}{{f\left( x \right)}}} \right|_0^1 < - \frac{1}{2}$ ,得到 $-\frac12<-frac12$ ,矛盾.

倘若 $\forall x\in (0,1]$ 都有 $F(x)>0$ ,即 $\frac{1}{2}{f^2}\left( x \right) + f'\left( x \right) > 0$ ,由于 $f(1)=1\neq 0$ ,则 $\forall x\in U^-(1,\delta)$ ,有 $\int_x^1 {\frac{{f'\left( t \right)}}{{{f^2}\left( t \right)1}}dt} > \int_x^1 {\left( { - \frac{1}{2}} \right)dt}$ ,即 $\left. { - \frac{1}{{f\left( t \right)}}} \right|_x^1 > \frac{{x - 1}}{2}$ ,从而 $\frac{1}{{f\left( x \right)}} > \frac{{x + 1}}{2}$ .

当 $x\in U^-(1,\delta)$ 时,有 $\frac{1}{{f\left( x \right)}} > \frac{{x + 1}}{2} > 0\left( {\delta \le 1} \right)$ ,故 $f(x)$ 在 $[0,1]$ 上不为 $0$ ,则 $\frac{1}{{f\left( 0 \right)}} > \frac{{0 + 1}}{2}$ ,则 $\frac12>\frac12$ ,矛盾.

因此存在 $\xi\in (0,1]$ 使得 $F(\xi)=0$ ,从而 $F(0)=F(\xi)=0$ ,由罗尔定理可知 $\exists c\in (0,1)$ ,使得

$F'\left( c \right) = f\left( c \right)f'\left( c \right) + f''\left( c \right) = 0.$

5. $A$ 和 $B$ 是自然数 $\mathbb{N}$ 的两个无穷子集,满足 $A\cap B=\text{空集},A\cup B=\mathbb{N}$ ,对于任意的自然数 $c>0$ ,是否存在两个递增的数列 $\{a_n\},\{b_n\},\{a_n\}\in A,\{b_n\}\in B$ ,使得 $\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c$ .

证明：对 $\forall l\geq s\in \mathbb{N}_+$ .记 $s\sim l=\{s,s+1,\cdots,l\}$ ,则若对 $\forall N\in \mathbb{N}_+$ , $\exists n_2\geq n_1\geq N\geq c^2$ 有 $x_{n_1}\sim x_{n_2}\subset B$ .令 $x_{n_1}=v_1,x_{n_2}=u_1$ ,则 $v_1\sim u_1\subset B$ .若 $cv_1\sim cu_1+c[\sqrt{u_1}+1]\not\subset B$ ,则 $\exists a\in A,b\in B$ ,使得 $|a-cb|\leq c$ 或 $|a-cb|\leq \sqrt{b}+1$ .

若 $cv_1\sim cu_1+c[\sqrt{u_1}+1]\subset B$ ,令 $u_2=u_1+[\sqrt{u_1}+1]$ ,则 $cv_1\sim cu_2\subset B$ .以此类推,若始终成立 $c^kv_1 \sim c^k u_{k+1}\subset B$ ,易知 $u_{k+1}=u_k+[\sqrt{u_k}+1]\geq u_k+\sqrt{u_1}$ ,即 $u_{k+1}-u_k\geq \sqrt{u_1}$ ,所以 $u_k$ 有趋于无穷大的趋势,所以存在某个 $k_0\in \mathbb{N}_+$ ,使得 $\frac{u_{k_0}}{v_1}>c$ .而 $c^{k_0}v_1\sim c^{k_0} u_{k_0+1}\subset B$ .令 $x_{n_{k_0}}\in B$ 满足 $c^{k_0}v_1\sim c^{k_0}u_{k_0+1}\subset c^{k_0}v_1\sim x_{n_{k_0}}\subset B,x_{n_{k_0}}+1\in A$ ,令 $a_1=x_{n_{k_0}}+1$ ,则 $c^{k_0}v_1<\frac{a_1}{c}\leq x_{n_{k_0}+1}$ ,则 $\exists b_1\in B$ ,使得 $\left| {\frac{{{a_1}}}{c} - {b_1}} \right| \le 1$ ,即 $\left| {{a_1} - c{b_1}} \right| \le c$ .

若不始终成立 $c^kv_1\sim c^ku_{k+1}$ ,则 $\exists b_1\in B,a_1\in A$ 有 $|a_1-cb_1|\leq c$ 或 $|a_1-cb_1|\leq \sqrt{b_1}+1$ ,即 $|a_1-cb_1|\leq \sqrt{a_1}+1$ .接着再取 $n_1\geq n_2\geq N$ ,有 $x_{n_2}\geq x_{n_1}>a_1,b_1$ 且 $x_{n_1}\sim x_{n_2}\subset B$ ,则 $\exists a_2\in A,b_2\in B$ 有 $|a_2-cb_2|\leq \sqrt{b_2}+1,\cdots$ ,故存在递增数列 $\{a_n\}\subset A,\{b_n\}\subset B$ ,有 $|a_n-cb_n|\leq \sqrt{b_n}+1$ ,则 $\left| {\frac{{{a_n}}}{{{b_n}}} - c} \right| \le \frac{1}{{\sqrt {{b_n}} }} + \frac{1}{{{b_n}}}$ ,所以 $\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = c$ .