1686

已知函数 ${\displaystyle f(x)=\frac{{\mathrm{e}}^x}{\cos x}-\tan x}$, ${\displaystyle x\in (0,\frac{\pi }{2})}$.

1. 证明: ${\displaystyle f(x)>2\tan x}$;
2. 已知数列 ${\displaystyle \{{\theta }_n\}}$ 满足: ${\displaystyle {\theta }_n\in (0,\frac{\pi }{2})}$, ${\displaystyle \tan {\theta }_{n+1}=f({\theta }_n)}$, ${\displaystyle n\in {\mathbb{N}}^{\ast }}$. 记 ${\displaystyle \lambda ={\mathrm{e}}^{\frac{\pi }{2}}-1}$.
   1. 证明: ${\displaystyle \lambda \tan {\theta }_n-\tan {\theta }_{n+1}<\lambda +1}$;
   2. 是否存在小于 ${\displaystyle \lambda }$ 的实数 ${\displaystyle a}$, 使得 ${\displaystyle \tan {\theta }_n<a^n}$ 对任意的正整数 ${\displaystyle n}$ 成立? 若存在, 求 ${\displaystyle a}$ 的取值范围; 若不存在, 请说明理由.

1. 证明 原不等式等价于 ${\displaystyle {\mathrm{e}}^x>3\sin x}$. 记 ${\displaystyle g(x)=\frac{{\mathrm{e}}^x}{3\sin x}}$, 则 ${\displaystyle g^{\prime }(x)=\frac{{\mathrm{e}}^x(\sin x-\cos x)}{3{\sin }^{2}x}}$, 容易知道 ${\displaystyle g(x)}$ 的最小值为 ${\displaystyle g\left(\frac{\pi }{4}\right)}$. 这样本题变成了 ${\displaystyle {\mathrm{e}}^{\frac{\pi }{4}}}$ 和 ${\displaystyle \frac{3\sqrt{2}}{2}}$ 的大小比较, 也即 ${\displaystyle {\mathrm{e}}^{\pi },\frac{81}{4}}$ 的大小比较. 注意到 $${\mathrm{e}}^{\pi }-\frac{81}{4}={\int }_0^{\pi }\frac{{(1-\sin x)}^2(18045-17030\sin x){\mathrm{e}}^x{\sin }^{2}x}{2064}\mathrm{d}x>0,$$ 所以得证! (出题人你要不还是给个 ${\displaystyle {\mathrm{e}}^{\pi }}$ 吧)
2. 1. 证明 经过一段代数变形, 原不等式等价于 $$\begin{array}{rl}& ({\mathrm{e}}^{\frac{\pi }{2}}-1)\tan {\theta }_n-(\frac{{\mathrm{e}}^{{\theta }_n}}{\cos {\theta }_n}-\tan {\theta }_n)<{\mathrm{e}}^{\frac{\pi }{2}}\\ ⟺& {\mathrm{e}}^{\frac{\pi }{2}}\tan {\theta }_n-\frac{{\mathrm{e}}^{{\theta }_n}}{\cos {\theta }_n}<{\mathrm{e}}^{\frac{\pi }{2}}\\ ⟺& \sin {\theta }_n-\cos {\theta }_n-{\mathrm{e}}^{{\theta }_n-\frac{\pi }{2}}<0.\end{array}$$ 记 ${\displaystyle h(x)=\sin x-\cos x-{\mathrm{e}}^{x-\frac{\pi }{2}}}$, ${\displaystyle x\in (0,\frac{\pi }{2})}$. 则 $$h^{\prime }(x)=\sin x+\cos x-{\mathrm{e}}^{x-\frac{\pi }{2}}=\underset{>1}{\underbrace{\sqrt{2}\sin (x+\frac{\pi }{4})}}-\underset{<1}{\underbrace{{\mathrm{e}}^{x-\frac{\pi }{2}}}}>0,$$ 所以 ${\displaystyle h({\theta }_n)<h\left(\frac{\pi }{2}\right)=0}$. 得证!
   2. 解 2.1 的结论式等价于 $$\begin{array}{}\text{(*)}& \tan {\theta }_{n+1}-\frac{\lambda +1}{\lambda -1}>\lambda (\tan {\theta }_n-\frac{\lambda +1}{\lambda -1}).\end{array}$$ 由 1, ${\displaystyle \tan {\theta }_{n+1}>2\tan {\theta }_n}$. 而 ${\displaystyle \tan {\theta }_1>0}$, 所以 ${\displaystyle \mathrm{\exists }N:\tan {\theta }_N>\frac{\lambda +1}{\lambda -1}}$. 这样对 (*) 累乘: $$\tan {\theta }_n-\frac{\lambda +1}{\lambda -1}>{\lambda }^{n-N}(\tan {\theta }_N-\frac{\lambda +1}{\lambda -1})\stackrel{\mathrm{\Delta }}{=}{\lambda }^n{b}_N.$$ 如果 ${\displaystyle a^n>\tan {\theta }_n}$ 恒成立, 则 $$a^n>\tan {\theta }_n>\frac{\lambda +1}{\lambda -1}+{\lambda }^n{b}_N>{\lambda }^n{b}_N\Rightarrow {\left(\frac{a}{\lambda }\right)}^n>b_N.$$ 但 ${\displaystyle a<\lambda }$, 所以 ${\displaystyle {\left(\frac{a}{\lambda }\right)}^n\to 0}$, 产生了矛盾. 所以这样的 ${\displaystyle a}$ 不存在.