已知函數(shù)f(x)=ax+lnx(a∈R).
(Ⅰ)若a=2,求曲線y=f(x)在x=1處切線的斜率;
(Ⅱ)求f(x)的單調(diào)區(qū)間;
(Ⅲ)設(shè)g(x)=x2-2x+2,若對(duì)任意x1∈(0,+∞),均存在x2∈[0,1],使得f(x1)<g(x2),求a的取值范圍.
【答案】
分析:(Ⅰ)把a(bǔ)的值代入f(x)中,求出f(x)的導(dǎo)函數(shù),把x=1代入導(dǎo)函數(shù)中求出的導(dǎo)函數(shù)值即為切線的斜率;
(Ⅱ)求出f(x)的導(dǎo)函數(shù),分a大于等于0和a小于0兩種情況討論導(dǎo)函數(shù)的正負(fù),進(jìn)而得到函數(shù)的單調(diào)區(qū)間;
(Ⅲ)對(duì)任意x
1∈(0,+∞),均存在x
2∈[0,1],使得f(x
1)<g(x
2),等價(jià)于f(x)
max<g(x)
max,分別求出相應(yīng)的最大值,即可求得實(shí)數(shù)a的取值范圍.
解答:解:(Ⅰ)由已知
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,則f'(1)=2+1=3.
故曲線y=f(x)在x=1處切線的斜率為3;
(Ⅱ)
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.
①當(dāng)a≥0時(shí),由于x>0,故ax+1>0,f'(x)>0
所以,f(x)的單調(diào)遞增區(qū)間為(0,+∞).
②當(dāng)a<0時(shí),由f'(x)=0,得
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.
在區(qū)間
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上,f'(x)>0,在區(qū)間
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上f'(x)<0,
所以,函數(shù)f(x)的單調(diào)遞增區(qū)間為
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,單調(diào)遞減區(qū)間為
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;
(Ⅲ)由已知,轉(zhuǎn)化為f(x)
max<g(x)
max,
因?yàn)間(x)=x
2-2x+2=(x-1)
2+1,x∈[0,1],
所以g(x)
max=2…(9分)
由(Ⅱ)知,當(dāng)a≥0時(shí),f(x)在(0,+∞)上單調(diào)遞增,值域?yàn)镽,故不符合題意.
當(dāng)a<0時(shí),f(x)在(0,-
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)上單調(diào)遞增,在(-
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,+∞)上單調(diào)遞減,
故f(x)的極大值即為最大值,f(-
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)=-1+ln(-
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)=-1-ln(-a),
所以2>-1-ln(-a),解得a<-
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.
點(diǎn)評(píng):此題考查學(xué)生會(huì)利用導(dǎo)數(shù)求曲線上過某點(diǎn)切線方程的斜率,會(huì)利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,掌握不等式恒成立時(shí)所滿足的條件,是一道中檔題.