练习题集[92]基础练习

1．已知点$P$是椭圆$\dfrac{x^2}{16}+\dfrac{y^2}{8}=1$上的一个动点，$F_1,F_2,O$分别为椭圆的左焦点、右焦点和中心，过$F_1$作$\angle F_1PF_2$的角平分线的垂线，垂足为$M$，则$OM$的取值范围是_______．

2．已知函数$f(x)=ax^2+(2a-1)x-3$($a\ne 0$)在区间$\left[-\dfrac 32,2\right]$上的最大值是$1$，则实数$a$的值是_______．

3．已知数列$\{a_n\}$满足$a_1=3$，$a_{n+1}=\dfrac{2n-1}{2n}a_n+\dfrac 1{2n}+3$，求数列$\{a_n\}$的通项公式．

4．设$a>0$，$x_1>0$，且$x_{n+1}=\dfrac 14\left(3x_n+\dfrac a{x_n^3}\right)$，$n\in\mathbb N^*$，则数列$\{x_n\}$是否收敛？

5．设方程$x^6+x^4+x^3+x^2+1=0$的所有虚部为正数的复根的乘积为$z$，求$z$的值(写成三角形式)．

6．已知$f(x)=\dfrac{x^2\tan x}{\tan x-x}$，求证：$f(x)$在$\left(0,\dfrac{\pi}2\right)$上单调递减．

7．已知$n\in\mathbb N^*$，求证：$\ln \left(2^2+1\right)+\ln \left(3^2+1\right)+\cdots +\ln \left(n^2+1\right)<1+2\ln n!$．

参考答案

1．$\left[0,2\sqrt 2\right)$．

作$F_1$关于$\angle F_1PF_2$的角平分线的对称点$F_1'$，则$OM=\dfrac 12F_2F_1'=\dfrac 12|PF_1-PF_2|$，取值范围是$\left(0,2\sqrt 2\right)$．

2．$\dfrac 34$或$-\dfrac{3+2\sqrt 2}2$．
根据题意，下列三个等式$$f\left(-\dfrac 32\right)=1,f(2)=1,\dfrac{-12a-(2a-1)^2}{4a}=1,$$中必然有一个成立，于是$a\in\left\{-\dfrac{10}3,\dfrac 34,\dfrac{-3\pm 2\sqrt 2}2\right\}$．经验证，可得实数$a$的值为$\dfrac 34$或$-\dfrac{3+\sqrt 2}2$．

3．根据题意，有\[2na_{n+1}=(2n-1)a_n+1+6n,\]即\[2n(a_{n+1}-1)=(2n-1)(a_n-1)+6n,\]也即\[2n\left[a_{n+1}-1-2(n+1)\right]=(2n-1)\left(a_n-1-2n\right),\]由于$a_1=3$，于是\[a_{n+1}-1-2(n+1)=0,\]从而$a_n=2n+1,n\in\mathbb N^*$为所求．

4．易得$x_n>0$，$n\in\mathbb N^*$，于是由均值不等式可得$$x_{n+1}=\dfrac 14\left(x_n+x_n+x_n+\dfrac a{x_n^3}\right)\geqslant \sqrt[4]{a},$$归纳易得$x_n\geqslant \sqrt[4]a$，$n\in\mathbb N^*$，$n\geqslant 2$．又$$\dfrac{x_{n+1}}{x_n}=\dfrac 14\left(3+\dfrac{a}{x_n^4}\right)\leqslant 1,$$于是数列$\{x_n\}$单调递减有下界，因此数列$\{x_n\}$收敛．容易计算得数列$\{x_n\}$收敛于$\sqrt[4]a$．

5．观察次数，可得方程等价于$$x^3+x+1+\dfrac 1x+\dfrac{1}{x^3}=0,$$也即$$\left(x+\dfrac 1x\right)^3-2\left(x+\dfrac 1x\right)+1=0,$$也即$$\left(x+\dfrac 1x-1\right)\left[\left(x+\dfrac 1x\right)^2+\left(x+\dfrac 1x\right)-1\right]=0,$$也即$$\left(x^2-x+1\right)\left(x^4+x^3+x^2+x+1\right)=0,$$因此该方程的所有复数根为方程$$\left(x^3+1\right)\left(x^5-1\right)=0$$的所有虚根，为$$\cos\dfrac{\pi}3+{\rm i}\sin\dfrac{\pi}3,\cos\dfrac{5\pi}3+{\rm i}\sin\dfrac{5\pi}3,\cos\dfrac{2\pi}5+{\rm i}\sin\dfrac{2\pi}5,\cos\dfrac{4\pi}5+{\rm i}\sin\dfrac{4\pi}5,\cos\dfrac{6\pi}5+{\rm i}\sin\dfrac{6\pi}5,\cos\dfrac{8\pi}5+{\rm i}\sin\dfrac{8\pi}5,$$进而可得$$z=\cos\left(\dfrac{\pi}3+\dfrac{2\pi}5+\dfrac{4\pi}5\right)+{\rm i}\sin\left(\dfrac{\pi}3+\dfrac{2\pi}5+\dfrac{4\pi}5\right)=\cos\dfrac{23\pi}{15}+{\rm i}\sin\dfrac{23\pi}{15}.$$

6．根据题意，有$$f(x)=\dfrac{x^2\sin x}{\sin x-x\cos x},x\in\left(0,\dfrac{\pi}2\right),$$于是$$f'(x)=-x\cdot \dfrac{x^2+x\sin x\cos x-2\sin^2x}{(\sin x-x\cos x)^2}=-x\cdot \dfrac{2x^2+x\sin 2x+2\cos 2x-2}{2(\sin x-x\cos x)^2}.$$设$\varphi(x)=2x^2+x\sin 2x+2\cos 2x-2$，则$$\varphi'(x)=4x+2x\cos 2x-3\sin 2x,$$进而$$\varphi''(x)=4-4x\sin 2x-4\cos 2x=8\sin ^2x-8x\sin x\cos x=8\sin x\cos x(\tan x-x)>0,$$以下略．

7．欲证明$$\ln\left(2^2+1\right)+\ln\left(3^2+1\right)+\cdots+\ln\left(n^2+1\right)<1+2\ln n!,$$加强到$$\ln\left(2^2+1\right)+\ln\left(3^2+1\right)+\cdots+\ln\left(n^2+1\right)<1+2\ln n!-\dfrac 1{n+\dfrac 12}.$$只需要证明$$\ln \left(n^2+1\right)<2\ln n +\dfrac{1}{\left(n-\dfrac 12\right)\left(n+\dfrac 12\right)},$$即$$\ln\left(1+\dfrac{1}{n^2}\right)<\dfrac{1}{\left(n-\dfrac 12\right)\left(n+\dfrac 12\right)}.$$而我们熟知当$x>0$时，有$$\ln(1+x)<x,$$于是$$\ln\left(1+\dfrac{1}{n^2}\right)<\dfrac 1{n^2}<\dfrac 1{n^2-\dfrac 14},$$命题得证．