(1)設(shè)a>0,討論y=f(x)的單調(diào)性;
(2)若對任意x∈(0,1)恒有f(x)>1,求a的取值范圍.
解:
(1)f(x)的定義域?yàn)?-∞,1)∪(1,+∞).對f(x)求導(dǎo)數(shù),得f′(x)=. ①當(dāng)a=2時(shí),f′(x)=,f′(x)在(-∞,0),(0,1)和(1,+∞)上均大于0,且f(x)在x=0處連續(xù),所以f(x)在(-∞,1),(1,+∞)上為增函數(shù). ②當(dāng)0<a<2時(shí),f′(x)>0,f(x)在(-∞,1),(1,+∞)上為增函數(shù). ③當(dāng)a>2時(shí),0<<1,令f′(x)=0,解得x1=-,x2=. 當(dāng)x變化時(shí),f′(x)和f(x)的變化情況如下表:
|