Question

 已知函數(shù)f (x) =

（1）若函數(shù)y = f (x)存在零點(diǎn)a² + 1，且直線y = x – 1與函數(shù)y = f (x)的圖象相切，求a的值．       

（2）當(dāng)b = 1時(shí)，討論f (x)的單調(diào)性．

 

 

 

 

 

 

 

 

 

 

 

Answer 1

【答案】

 【解析】（1）由f (a² + 1) = 0得+ ln1= 0，∴= 0即b = 0．設(shè)直線y = x–1與曲線y = f (x)切點(diǎn)為P(x₀，y₀)，則，

∴x₀ – a² = 1，∴x₀ – 1 = 0，x₀ = 1，∴a = 0．……5分

（2）b = 1時(shí)，f (x) = ，f (x)定義域?yàn)?a²，+∞)．       

f′(x) = ，設(shè)h (x) = x² – x + a²，二次方程h (x) = 0對(duì)應(yīng)的判別式= 1 – 4a²．……6分

①當(dāng)＜0即a＞或a＜–時(shí)，對(duì)一切x＞a²，都有f′(x)＞0，此時(shí)f (x)是(a²，+∞)上的單調(diào)遞增函數(shù)．……7分

②當(dāng)=0即a =±時(shí)，僅對(duì)x = 有f′(x) = 0，對(duì)于其余的x＞a²都有f′(x)＞0．此時(shí)f (x)是(a²，+∞)上的單調(diào)遞增函數(shù)．……8分       

③當(dāng)＞0即–＜a＜時(shí)，方程h (x) = 0有兩個(gè)不同實(shí)根x₁ = ，a²＜x₁＜x₂，f (x)，f′(x)隨x的變化如下表．

x	(a2，x1)	x1	(x1，x2)	x2	(x2，+∞)
f′(x)	+	0	–	0	+
f (x)	↑	極大值	↓	極小值	↑

此時(shí)y=f (x)在上單調(diào)遞增，在上單調(diào)遞減．在上單調(diào)遞增．  ……13分