Question

(理)已知函數(shù)f(x)=(m∈R,e=2.718 28…是自然對(duì)數(shù)的底數(shù)).

(1)求函數(shù)f(x)的極值;

(2)當(dāng)x＞0時(shí),設(shè)f(x)的反函數(shù)為f^-1(x),對(duì)0＜p＜q,試比較f(q-p)、f^-1(q-p)及f^-1(q)-f^-1(p)的大小.

(文)已知函數(shù)f(x)=x³+bx²+cx+d(b、c、d∈R且都為常數(shù))的導(dǎo)函數(shù)為f′(x)=3x²+4x,且f(1)=7,設(shè)F(x)=f(x)-ax²(a∈R).

(1)當(dāng)a＜2時(shí),求F(x)的極小值;

(2)若對(duì)任意的x∈［0,+∞),都有F(x)≥0成立,求a的取值范圍并證明不等式a²-13a+39≥.

Answer 1

答案：(理)解:(1)∵當(dāng)x＞0時(shí),f(x)=e^x-1在(0,+∞)上單調(diào)遞增,且f(x)=e^x-1＞0;

當(dāng)x≤0時(shí),f(x)=x³+mx²,此時(shí)f′(x)=x²+2mx=x(x+2m).

①若m=0時(shí),f′(x)=x²≥0,則f(x)=x³在(-∞,0］上單調(diào)遞增,且f(x)=x³≤0.又f(0)=0,可知函數(shù)f(x)在R上單調(diào)遞增,無(wú)極值.

②若m＜0,令f′(x)=x(x+2m)＞0x＜0或x＞-2m(舍去).函數(shù)f(x)=x³+mx²在(-∞,0］上單調(diào)遞增,同理,函數(shù)f(x)在R上單調(diào)遞增,無(wú)極值.

③若m＞0,令f′(x)=x(x+2m)＞0x＞0或x＜-2m.函數(shù)f(x)=x³+mx²在(-∞,-2m］上單調(diào)遞增,在(-2m,0］上單調(diào)遞減.此時(shí)函數(shù)f(x)在x=-2m處取得極大值:f(-2m)=m³+4m³=m³＞0;又f(x)在(0,+∞)上單調(diào)遞增,故在x=0處取得極小值:f(0)=0.綜上可知,當(dāng)m＞0時(shí),f(x)的極大值為m³,極小值為0;當(dāng)m≤0時(shí),f(x)無(wú)極值.

(2)當(dāng)x＞0時(shí),設(shè)y=f(x)=e^x-1y+1=e^xx=ln(y+1).∴f^-1(x)=ln(x+1)(x＞0).①比較f(q-p)與f^-1(q-p)的大小.記g(x)=f(x)-f^-1(x)=e^x-ln(x+1)-1(x＞0).∵g′(x)=e^x-在(0,+∞)上是單調(diào)遞增函數(shù),∴g′(x)＞g′(0)=e⁰-=0恒成立.∴函數(shù)g(x)在(0,+∞)上單調(diào)遞增.∴g(x)＞g(0)=e⁰-ln(0+1)-1=0.當(dāng)0＜p＜q時(shí),有q-p＞0,∴g(q-p)=e^q-p-ln(q-p+1)-1＞0.∴e^q-p-1＞ln(q-p+1),

即f(q-p)＞f^-1(q-p).①9分

②比較f^-1(q-p)與f^-1(q)-f^-1(p)的大小.ln(q-p+1)-［ln(q+1)-ln(p+1)］=ln(q-p+1)-ln(q+1)+ln(p+1)

=ln=ln

=ln=ln［+1］.

∵0＜p＜q,∴+1＞1,故ln［+1］＞0.∴l(xiāng)n(q-p+1)＞ln(q+1)-ln(p+1),

即f^-1(q-p)＞f^-1(q)-f^-1(p).②

(注:也可用分析法或考查函數(shù)h(x)=ln(x-p+1)-ln(x+1)+ln(p+1),x∈(p,+∞).

求導(dǎo)可知h(x)在(p,+∞)上單調(diào)遞增.∴h(x)＞h(p)恒成立,而h(p)=0.∴h(x)＞0在x∈(p,+∞)上恒成立.∵q∈(p,+∞),∴h(q)＞0恒成立)∴由①②可知,當(dāng)0＜p＜q時(shí),有f(q-p)＞f^-1(q-p)＞f^-1(q)-f^-1(p).12分

(文)解：(1)∵f′(x)=3x²+4x=3x²+2bx+c,∴b=2,c=0.∴f(x)=x³+2x²+d.

又∵f(1)=7,∴d=4.∴f(x)=x³+2x²+4.2分

∴F(x)=f(x)-ax²=x³+2x²+4-ax²=x³+(2-a)x²+4.則F′(x)=3x²+2(2-a)x.

由F′(x)=0求得x₁=,x₂=0.∵a＜2,∴x₁＜x₂.

當(dāng)x變化時(shí),F′(x)、F(x)的變化情況如下:

X	(-∞,)		(,0)	0	(0,+∞)
F′(x)	+	0	-	0	+
F(x)	增函數(shù)	極大值	減函數(shù)	極小值	增函數(shù)

∴當(dāng)x=0時(shí),F(x)取得極小值4.

(2)由(1)知F(x)=x³+(2-a)x²+4.∵F(x)≥0在［0,+∞)上恒成立當(dāng)x∈［0,+∞)時(shí),F(x)_min≥0.

①若2-a＞0,即a＜2時(shí),由(1)可知F(x)_min=F(0)=4＞0,符合題意;

②若2-a≤0,即a≥2時(shí),由F′(x)=0求得x₁=,x₂=0,且x₁＞x₂.

∴當(dāng)x∈［0,+∞)時(shí),F(x)_min=F()≥0,即()³-(a-2)()²+4≥0，解不等式得2≤a≤5.綜上所述,應(yīng)有a≤5.

要證不等式a²-13a+39≥,只需證6-a+≥-(a-6)²+3.∵a≤5,∴6-a+≥2,

-(a-6)²+3≤2.(當(dāng)a=5時(shí),“=”成立)故a²-13a+39≥成立(當(dāng)a=5時(shí)，“=”成立).