Question

(理)設(shè)函數(shù)f(x)=e^x-e^-x.

(1)證明f(x)的導(dǎo)數(shù)f′(x)≥2;

(2)若對(duì)所有x≥0都有f(x)≥ax,求a的取值范圍.

(文)設(shè)函數(shù)f(x)=2x³+3ax²+3bx+8c在x=1及x=2時(shí)取得極值.

(1)求a、b的值;

(2)若對(duì)任意的x∈［0,3］,都有f(x)＜c²成立,求c的取值范圍.

Answer 1

解：(理)(1)f(x)的導(dǎo)數(shù)f′(x)=e^x+e^-x.由于e^x+e^-x≥=2,

故f′(x)≥2.(當(dāng)且僅當(dāng)x=0時(shí),等號(hào)成立)

(2)令g(x)=f(x)-ax,則g′(x)=f′(x)-a=e^x+e^-x-a.

①若a≤2,當(dāng)x＞0時(shí),g′(x)=e^x+e^-x-a＞2-a≥0,故g(x)在(0,+∞)上為增函數(shù).所以,x≥0時(shí),g(x)≥g(0),即f(x)≥ax.

②若a＞2,方程g′(x)=0的正根為x₁=ln,此時(shí),若x∈(0,x₁),則g′(x)＜0,故g(x)在該區(qū)間為減函數(shù).所以,x∈(0,x₁)時(shí),g(x)＜g(0)=0,即f(x)＜ax,與題設(shè)f(x)≥ax相矛盾.綜上,滿足條件的a的取值范圍是(-∞,2].

(文)(1)f′(x)=6x²+6ax+3b.因?yàn)楹瘮?shù)f(x)在x=1及x=2取得極值,則有f′(1)=0,f′(2)=0.

即解得a=-3,b=4.

(2)由(1)可知,f(x)=2x³-9x²+12x+8c,f′(x)=6x²-18x+12=6(x-1)(x-2).當(dāng)x∈(0,1)時(shí),f′(x)＞0;當(dāng)x∈(1,2)時(shí),f′(x)＜0;當(dāng)x∈(2,3)時(shí),f′(x)＞0.所以,當(dāng)x=1時(shí),f(x)取得極大值f(1)=5+8c.又f(0)=8c,f(3)=9+8c，則當(dāng)x∈[0,3]時(shí),f(x)的最大值為f(3)=9+8c.因?yàn)閷?duì)于任意的x∈[0,3],有f(x)＜c²恒成立,所以9+8c＜c².解得c＜-1或c＞9,因此c的取值范圍為(-∞,-1)∪(9,+∞).