Question

已知函數(shù)f(x)=ln(1+x²)+ax(a≤0)．

(Ⅰ)若f(x)在x=0處取得極值，求a的值；

(Ⅱ)討論f(x)的單調(diào)性；

(Ⅲ)證明：(1+)(1+)…(1+)(n∈N₊，e為自然對(duì)數(shù)的底數(shù))．

Answer 1

(Ⅰ)∵f′(x)=,∵x=0使f(x)的一個(gè)極值點(diǎn)則f′(0)=0a=0,驗(yàn)證知a=0符合條件 

(Ⅱ)∵f′(x)=

i)若a=0時(shí), ∵f′(x)   f′(x)＜0x＜0

∴f(x)在(0,+∞)單調(diào)遞增,在(-∞,0)單調(diào)遞減 

ii)若時(shí),f′(x)≤0時(shí),x∈R恒成立,

∴f(x)在R上單調(diào)遞減

iii)若-1＜a＜0時(shí),由f′(x)＞0ax²+2x+a＞0

再令f′(x)＜0,可得x＞

∴f(x)在單調(diào)遞增,

在(-∞,)和(,+∞)上單調(diào)遞減

綜上所述,若a≤-1時(shí),f(x)在(-∞,+∞)上單調(diào)遞減;若-1＜a＜0時(shí),f(x)在(,)單調(diào)遞減

在上單調(diào)遞減,上單調(diào)遞減

若a=0時(shí),f(x)在(0,+∞)單調(diào)遞增,在(-∞,0)單調(diào)遞減 

(Ⅲ)由(Ⅱ)知,當(dāng)a=-1時(shí),f(x)在(-∞,+∞)上單調(diào)遞減

當(dāng)x∈(0,+∞)時(shí),由f(x)＜f(0)=0  ∴l(xiāng)n(1+x²)＜x 

∴l(xiāng)n［(1+)(1+)…(1+)］

=(1+)+ln(1+)+…+ln(1+)＜

∴(1+)(1+)…(1+)＜,命題得證.