练习题集[90]基础练习

1.$f(x),g(x)$定义在$\mathbf R$上,下列说法正确的是(  )
A.若$f(f(x))=f(x)$,则$f(x)=x$
B.若$f(f(x))=x$,则$f(x)=x$
C.若$f(g(x))=x$,则$g(f(x))=x$
D.若$f(g(x))=x$,则$g(x_1)=g(x_2)$等价于$x_1=x_2$

2.设$A,B$为抛物线$y^2=2px$($p>0$)上不同的两点,$O$为坐标原点,则$\left|\overrightarrow{OA}+\overrightarrow{OB}\right|^2-\left|\overrightarrow{AB}\right|^2$的最小值为______.

3.已知$A,B$分别为椭圆$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$($a>b>0$)的右顶点和上顶点,直线$y=kx$($k>0$)与椭圆交于$C,D$两点.若四边形$ACBD$的面积的最大值为$2c^2$,则椭圆的离心率为______.

4.已知实数$x,y,z$满足$\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1$,则$\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}$的值是______.

5.已知数列$\{x_n\}$满足$$x_{n+1}=\left(\dfrac 2{n^2}+\dfrac 3n+1\right)x_n+n+1,n\in\mathbf N^*,$$且$x_1=3$,求数列$\{x_n\}$的通项公式.

6.已知正整数数列$\{a_n\}$满足$a_1=2$,$a_{n+1}=a_n^2-a_n+1$($n\in\mathbf N^*$),求证:数列$\{a_n\}$中的任意两项都互质.

7.已知$a,b>0$,$n\in\mathbf N^*$,求证:$\dfrac{1}{a+b}+\dfrac{1}{a+2b}+\cdots +\dfrac{1}{a+nb}<\dfrac{n}{\sqrt{\left(a+\dfrac 12b\right)\left(a+\dfrac{n+1}2b\right)}}$.


参考答案

1.D.

选项A的反例:$f(x)=|x|$;
选项B的反例:$f(x)=-x$;
选项C的反例:$f(x)=\begin{cases} \tan x,&x\neq \dfrac{\pi}2+k\pi,\\1,&x=\dfrac{\pi}2+k\pi,\end{cases} $其中$k\in\mathbf Z$,$g(x)=\arctan x$.

2.$-4p^2$.

由极化恒等式,所求代数式即$4\overrightarrow {OA}\cdot \overrightarrow{OB}$.设$A(2pm^2,2pm),B(2pn^2,2pn)$,则$$4\overrightarrow{OA}\cdot\overrightarrow{OB}=16p^2\left[\left(mn+\dfrac 12\right)^2-\dfrac 14\right]\geqslant -4p^2.$$当且仅当$mn=-\dfrac 12$时取等号.

3.$\dfrac{\sqrt 2}2$.

仿射为圆,可得$\left(\dfrac 12\cdot \sqrt 2a\cdot 2a\right)\cdot \dfrac ba=2c^2$.

4.$0$.

为了出现所求的代数式,在题中条件左右两边同乘以$(x+y+z)$,有$$(x+y+z)\cdot\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=x+y+z,$$于是所求代数式的值为$$\sum_{cyc}\left[(x+y)\cdot \dfrac{z}{x+y}\right]-(x+y+z)=0.$$

5.根据题意,有$$x_{n+1}=\dfrac{(n+1)(n+2)}{n^2}x_n+n+1,$$于是$$\dfrac{x_{n+1}}{(n+1)^2(n+2)}=\dfrac{x_n}{n^2(n+1)}+\dfrac{1}{(n+1)(n+2)},$$进而可得$$\dfrac{x_{n+1}}{(n+1)^2(n+2)}+\dfrac{1}{n+2}=\dfrac{x_n}{n^2(n+1)}+\dfrac{1}{n+1},$$因此$$\dfrac{x_n}{n^2(n+1)}+\dfrac{1}{n+1}=\dfrac{x_{n-1}}{(n-1)^2\cdot n}+\dfrac{1}{n}=\cdots =\dfrac{x_1}{2}+\dfrac 12=2,$$所以$x_n=n^2(2n+1),n\in\mathbf N^*$.

6.根据题意,有$$a_{n+1}-1=a_n\left(a_n-1\right)=a_n\cdots a_1,$$于是对任意$p,q\in\mathbf N^*$且$p<q$,都有$ a_p \mid \left( a_q-1\right)$,因此$\left(a_p,a_q\right)=1$,因此数列$\{a_n\}$中任意两项都互质.

7.柯西加裂项 根据题意,有\[\begin{split} LHS&<\sqrt n\cdot \sqrt{\dfrac{1}{(a+b)^2}+\dfrac{1}{(a+2b)^2}+\cdots +\dfrac{1}{(a+nb)^2}}\\&<\sqrt n \cdot \sqrt{\dfrac{1}{\left(a+\dfrac 12b\right)\left(a+\dfrac 32b\right)}+\dfrac{1}{\left(a+\dfrac 32b\right)\left(a+\dfrac 52b\right)}+\cdots +\dfrac{1}{\left(a+\dfrac{n-1}2b\right)\left(a+\dfrac{n+1}2\right)}}\\&=\sqrt n\cdot \sqrt{\dfrac 1b\left(\dfrac 1{a+\dfrac 12b}-\dfrac{1}{a+\dfrac 32b}+\dfrac{1}{a+\dfrac 32b}-\dfrac 1{a+\dfrac 52b}+\cdots +\dfrac{1}{a+\dfrac{2n-1}2b}-\dfrac{1}{a+\dfrac{2n+1}2b}\right)}\\&=RHS,\end{split}\]因此原不等式得证.

A-L-G不等式 根据题意,有\[\begin{split} LHS&=\sum_{i=1}^n\dfrac{2}{\left(a+\dfrac{2i+1}2b\right)+\left(a+\dfrac{2i-1}2b\right)}\\&<\sum_{i=1}^n\dfrac{\ln\left(a+\dfrac{2i+1}2b\right)-\ln\left(a+\dfrac{2i-1}2b\right)}{b}\\&=\dfrac{\ln\left(a+\dfrac {2n+1}2b\right)-\ln\left(a+\dfrac 1b\right)}{b}\\&=n\cdot \dfrac{\ln\left(a+\dfrac {2n+1}2b\right)-\ln\left(a+\dfrac 1b\right)}{\left(a+\dfrac{2n+1}2b\right)-\left(a+\dfrac 12b\right)}\\&<\dfrac{n}{\sqrt{\left(a+\dfrac 12b\right)\left(a+\dfrac{2n+1}2b\right)}}=RHS,\end{split}\]因此原不等式得证.

此条目发表在练习题集分类目录。将固定链接加入收藏夹。

发表评论