設(shè)m為實(shí)數(shù),函數(shù)f(x)=lnx+mx.
(Ⅰ)求函數(shù)f(x)的單調(diào)區(qū)間;
(Ⅱ)當(dāng)m=e時,直線y=ax+b2是曲線y=f(x)的切線,求a+b的最小值;
(Ⅲ)若方程f(x)=(m+1)x+n-2(n∈R)有兩個實(shí)數(shù)根x1,x2(x1<x2),證明:2x1+x2>e.
(注:e=2.71828…是自然對數(shù)的底數(shù))
y
=
ax
+
b
2
【答案】(Ⅰ)當(dāng)m≥0時,函數(shù)f(x)在(0,+∞)上單調(diào)遞增;當(dāng)m<0時,函數(shù)f(x)在上單調(diào)遞增,在上單調(diào)遞減.
(Ⅱ)e-2ln2;
(Ⅲ)證明過程見解答.
(
0
,-
1
m
)
(
-
1
m
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+
∞
)
(Ⅱ)e-2ln2;
(Ⅲ)證明過程見解答.
【解答】
【點(diǎn)評】
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發(fā)布:2024/6/27 10:35:59組卷:229引用:7難度:0.3
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