# 对题目的等价转化往往就是解题的突破口

## 二、解析

$$\tan x = ax$$

$$a = \frac{\tan x}{x}$$

$$f(x) = \frac{\tan x}{x}$$

\begin{aligned} f ^ { \prime } ( x ) \\ \\ & = \left( \frac{\tan x}{x} \right)^{\prime} \\ \\ & = \frac { x \sec ^ { 2 } x – \tan x } { x ^ { 2 } } \\ \\ & = \frac { x \frac{1}{\cos ^{2} x} – \frac{\sin x}{\cos x} } { x ^ { 2 } } \\ \\ & = \frac { x \frac{1}{\cos ^{2} x} – \frac{ \cos x \sin x}{\cos ^{2} x} } { x ^ { 2 } } \\ \\ & = \textcolor{springgreen}{\frac{x – \cos x \sin x}{x^{2} \cos^{2} x}} \\ \\ & = \frac { \textcolor{orangered}{2} \textcolor{springgreen}{x} – \textcolor{orangered}{2} \textcolor{springgreen}{\sin x \cos x} } { \textcolor{orangered}{2} \textcolor{springgreen}{x ^ { 2 } \cos ^ { 2 } x} } \\ \\ & = \frac { 2 x – \sin 2 x } { 2 x ^ { 2 } \cos ^ { 2 } x } \end{aligned}

$$\begin{cases} x > 0 \\ 2 x > \sin 2 x \\ \cos x > 0 \end{cases}$$

$$f(x) = \frac { 2 x – \sin 2 x } { 2 x ^ { 2 } \cos ^ { 2 } x } > 0$$

$$\lim _ { x \rightarrow 0 ^ { + } } f ( x ) = 1$$

$$f \left( 0 ^ { + } \right) < f ( x ) < f \left( \frac { \pi } { 4 } \right) \Rightarrow$$

$$1 < f ( x ) < \frac { 4 } { \pi }$$

$$\textcolor{springgreen}{ 1 < a < \frac { 4 } { \pi } }$$