Question

已知函數(shù)f(x)=ln(e^x+1)—ax.

(Ⅰ)設(shè)a＞0,討論f(x)的單調(diào)性;

(Ⅱ)當(dāng)a=9時(shí),若△ABC的三個(gè)頂點(diǎn)A、B、C都在函數(shù)y=f(x)的圖像上，且橫坐標(biāo)成等差數(shù)列，求證：△ABC為鈍角三角形.

Answer 1

解：(Ⅰ)由已知f′(x)=

當(dāng)a≥1時(shí),f′(x)＜0,y=f(x)在R單調(diào)遞減；

當(dāng)0＜a＜1時(shí),解f′(x)＞0得(1-a)(ex+1)＞1即e^x＞-1+  ∴x＞ln

∴當(dāng)0＜a＜1時(shí),y=f(x)在(1n,+∞)內(nèi)單調(diào)遞增；在(-∞,In)內(nèi)單調(diào)遞減

(Ⅱ)當(dāng)a=9時(shí),f(x)=ln(ex+1)-9x在(-∞,+∞)上單調(diào)遞減

設(shè)A(x₁,f(x₁)),B(x₂,f(x₂)),C(x₃,f(x₃))不妨設(shè)x₁＜x₂＜x₃

∴=(x₁-x₂，f(x₁)-f(x₂))，=(x₃-x₂，f(x₃)-f(x₂))

又∵·=(x₁-x₂)(x₃-x₂)+(f(x₁)-f(x₂))(f(x₃)-f(x₂))

又由f(x)的單調(diào)性知：

x₁-x₂＜0，x₃-x₂＞0，f(x₁)- f(x₂)＞0，f(x₃)-f(x₂)＜0

∴＜0   ∴cos∠ABC=＜0

∴△BAC為鈍角三角形