# 2018年考研数二第18题解析：导数、单调性

## 解析

$$f^{‘}(x) = 1 – 2 \ln x \cdot \frac{1}{x} + 2k \cdot \frac{1}{x} \Rightarrow$$

$$f^{‘}(x) = \frac{1}{x}(x – 2 \ln x + 2k).$$

$$g^{‘}(x) = 1 – \frac{2}{x}.$$

$$g^{‘}(x) = 1 – \frac{2}{x} = 0 \Rightarrow$$

$$x = 2.$$

$$g(2) = 2 – 2 \ln 2 + 2k \geqslant 2 – 2 ln 2 + 2(\ln 2 – 1) \Rightarrow$$

$$g(2) \geqslant 0 \Rightarrow$$

$$x – 2 \ln x + 2k > 0 \quad (x \neq 0).$$

$$f^{‘}(x) = \frac{1}{x}(x – 2 \ln x + 2k) > 0 \quad (x \neq 0).$$

$$f(1) = 1 – 0 + 0 -1 = 0.$$

• 当 $0 < x < 1$ 时，有：

$$\left\{\begin{matrix} x – 1 < 0;\\ f(x) = x – \ln^{2} x + 2k \ln x – 1 < 0 \end{matrix}\right. \Rightarrow$$

$$(x-1)(x – \ln^{2} x + 2k \ln x – 1) > 0.$$

• 当 $x > 1$ 时，有：

$$\left\{\begin{matrix} x – 1 > 0;\\ f(x) = x – \ln^{2} x + 2k \ln x – 1 > 0 \end{matrix}\right. \Rightarrow$$

$$(x-1)(x – \ln^{2} x + 2k \ln x – 1) > 0.$$

• 当 $x = 1$ 时，有：

$$\left\{\begin{matrix} x – 1 = 0;\\ f(x) = x – \ln^{2} x + 2k \ln x – 1 = 0 \end{matrix}\right. \Rightarrow$$

$$(x-1)(x – \ln^{2} x + 2k \ln x – 1) = 0.$$