2022年高考數(shù)學(xué)一輪復(fù)習(xí) 高考大題專項(xiàng)練,2022年高考數(shù)學(xué)一輪復(fù)習(xí),高考大題專項(xiàng)練,2022,年高,數(shù)學(xué),一輪,復(fù)習(xí),高考,專項(xiàng)
高考大題專項(xiàng)練一 高考中的函數(shù)與導(dǎo)數(shù)
一、非選擇題
1.設(shè)函數(shù)f(x)=[ax2-(3a+1)x+3a+2]ex.
(1)若曲線y=f(x)在點(diǎn)(2,f(2))處的切線斜率為0,求a;
(2)若f(x)在x=1處取得極小值,求a的取值范圍.
解:(1)因?yàn)閒(x)=[ax2-(3a+1)x+3a+2]ex,
所以f'(x)=[ax2-(a+1)x+1]ex,f'(2)=(2a-1)e2.
由題設(shè)知f'(2)=0,即(2a-1)e2=0,解得a=12.
(2)由(1)得f'(x)=[ax2-(a+1)x+1]ex=(ax-1)(x-1)ex.
若a>1,則當(dāng)x∈1a,1時(shí),f'(x)<0;
當(dāng)x∈(1,+∞)時(shí),f'(x)>0.
所以f(x)在x=1處取得極小值.
若a≤1,則當(dāng)x∈(0,1)時(shí),ax-1≤x-1<0,
所以f'(x)>0.
所以1不是f(x)的極小值點(diǎn).
綜上可知,a的取值范圍是1,+∞.
2.已知函數(shù)f(x)=ax2+x-1ex.
(1)求曲線y=f(x)在點(diǎn)(0,-1)處的切線方程;
(2)證明:當(dāng)a≥1時(shí),f(x)+e≥0.
答案:(1)解f'(x)=-ax2+(2a-1)x+2ex,f'(0)=2.
因此曲線y=f(x)在點(diǎn)(0,-1)處的切線方程是2x-y-1=0.
(2)證明當(dāng)a≥1時(shí),f(x)+e≥(x2+x-1+ex+1)e-x.
令g(x)=x2+x-1+ex+1,則g'(x)=2x+1+ex+1.
當(dāng)x<-1時(shí),g'(x)<0,g(x)單調(diào)遞減;
當(dāng)x>-1時(shí),g'(x)>0,g(x)單調(diào)遞增;
所以g(x)≥g(-1)=0.
因此f(x)+e≥0.
3.已知函數(shù)f(x)=ln x+12ax2-x-m(m∈Z).
(1)若f(x)是增函數(shù),求a的取值范圍;
(2)若a<0,且f(x)<0恒成立,求m的最小值.
解:(1)f'(x)=1x+ax-1,依題設(shè)可得a≥1x-1x2max,
而1x-1x2=-1x-122+14≤14,當(dāng)x=2時(shí),等號(hào)成立.
所以a的取值范圍是14,+∞.
(2)由(1)可知f'(x)=1x+ax-1=ax2-x+1x,
設(shè)g(x)=ax2-x+1,則g(0)=1>0,g(1)=a<0,
g(x)=ax-12a2+1-14a在區(qū)間(0,+∞)內(nèi)單調(diào)遞減.
因此g(x)在區(qū)間(0,1)內(nèi)有唯一的解x0,使得ax02=x0-1,
而且當(dāng)0
0,當(dāng)x>x0時(shí),f'(x)<0,
所以f(x)≤f(x0)=lnx0+12ax02-x0-m
=lnx0+12(x0-1)-x0-m=lnx0-12x0-12-m.
設(shè)r(x)=lnx-12x-12-m,
則r'(x)=1x-12=2-x2x>0.
所以r(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增.
所以r(x)0,
由f'(x)>0,得02;
由f'(x)<0,得10;
當(dāng)x∈π2,π時(shí),g'(x)<0,
所以g(x)在0,π2單調(diào)遞增,在π2,π單調(diào)遞減.
又g(0)=0,gπ2>0,g(π)=-2,
故g(x)在(0,π)存在唯一零點(diǎn).
所以f'(x)在(0,π)存在唯一零點(diǎn).
(2)解由題設(shè)知f(π)≥aπ,f(π)=0,可得a≤0.
由(1)知,f'(x)在(0,π)只有一個(gè)零點(diǎn),設(shè)為x0,且當(dāng)x∈(0,x0)時(shí),f'(x)>0;當(dāng)x∈(x0,π)時(shí),f'(x)<0,所以f(x)在(0,x0)單調(diào)遞增,在(x0,π)單調(diào)遞減.
又f(0)=0,f(π)=0,所以,當(dāng)x∈[0,π]時(shí),f(x)≥0.
又當(dāng)a≤0,x∈[0,π]時(shí),ax≤0,故f(x)≥ax.
因此,a的取值范圍是(-∞,0].
6.定義在實(shí)數(shù)集上的函數(shù)f(x)=x2+x,g(x)=13x3-2x+m.
(1)求函數(shù)f(x)的圖象在x=1處的切線方程;
(2)若f(x)≥g(x)對(duì)任意的x∈[-4,4]恒成立,求實(shí)數(shù)m的取值范圍.
解:(1)∵f(x)=x2+x,∴當(dāng)x=1時(shí),f(1)=2,
∵f'(x)=2x+1,∴f'(1)=3,
∴所求切線方程為y-2=3(x-1),即3x-y-1=0.
(2)令h(x)=g(x)-f(x)=13x3-x2-3x+m,
則h'(x)=(x-3)(x+1).
∴當(dāng)-40;
當(dāng)-10.
要使f(x)≥g(x)恒成立,即h(x)max≤0,
由上知h(x)的最大值在x=-1或x=4處取得,而h(-1)=m+53,h(4)=m-203,故m+53≤0,即m≤-53,
故實(shí)數(shù)m的取值范圍為-∞,-53.
7.已知函數(shù)f(x)=12ax2-(2a+1)x+2ln x(a∈R).
(1)求f(x)的單調(diào)區(qū)間;
(2)設(shè)g(x)=x2-2x,若對(duì)任意x1∈(0,2],均存在x2∈(0,2],使得f(x1)0).
(1)f'(x)=(ax-1)(x-2)x(x>0).
①當(dāng)a≤0時(shí),x>0,ax-1<0,在區(qū)間(0,2)內(nèi),f'(x)>0,在區(qū)間(2,+∞)內(nèi),f'(x)<0,故f(x)的單調(diào)遞增區(qū)間是(0,2),單調(diào)遞減區(qū)間是(2,+∞).
②當(dāng)02,在區(qū)間(0,2)和1a,+∞內(nèi),f'(x)>0,在區(qū)間2,1a內(nèi),f'(x)<0,故f(x)的單調(diào)遞增區(qū)間是(0,2)和1a,+∞,單調(diào)遞減區(qū)間是2,1a.
③當(dāng)a=12時(shí),f'(x)=(x-2)22x,故f(x)的單調(diào)遞增區(qū)間是(0,+∞).
④當(dāng)a>12時(shí),0<1a<2,在區(qū)間0,1a和(2,+∞)內(nèi),f'(x)>0,在區(qū)間1a,2內(nèi),f'(x)<0,故f(x)的單調(diào)遞增區(qū)間是0,1a和(2,+∞),單調(diào)遞減區(qū)間是1a,2.
(2)對(duì)任意x1∈(0,2],均存在x2∈(0,2],使得f(x1)ln2-1.
故ln2-112時(shí),f(x)在區(qū)間0,1a上單調(diào)遞增,在區(qū)間1a,2上單調(diào)遞減,
故f(x)max=f1a=12a-(2a+1)1a+2ln1a=-12a-2-2lna<0因?yàn)楫?dāng)a>12時(shí),12a+2lna>12a+2lne-1=12a-2>-2.
故a>12時(shí)滿足題意.
綜上,a的取值范圍為(ln2-1,+∞).
8.已知函數(shù)f(x)=x3+kln x(k∈R),f'(x)為f(x)的導(dǎo)函數(shù).
(1)當(dāng)k=6時(shí),
①求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;
②求函數(shù)g(x)=f(x)-f'(x)+9x的單調(diào)區(qū)間和極值;
(2)當(dāng)k≥-3時(shí),求證:對(duì)任意的x1,x2∈[1,+∞),且x1>x2,有f'(x1)+f'(x2)2>f(x1)-f(x2)x1-x2.
答案:(1)解①當(dāng)k=6時(shí),f(x)=x3+6lnx,故f'(x)=3x2+6x.
可得f(1)=1,f'(1)=9,所以曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-1=9(x-1),即y=9x-8.
②依題意,g(x)=x3-3x2+6lnx+3x,x∈(0,+∞).從而可得g'(x)=3x2-6x+6x-3x2,
整理可得g'(x)=3(x-1)3(x+1)x2.
令g'(x)=0,解得x=1.
當(dāng)x變化時(shí),g'(x),g(x)的變化情況如下表:
x
(0,1)
1
(1,+∞)
g'(x)
-
0
+
g(x)
↘
極小值
↗
所以,函數(shù)g(x)的單調(diào)遞減區(qū)間為(0,1),單調(diào)遞增區(qū)間為(1,+∞);g(x)的極小值為g(1)=1,無極大值.
(2)證明由f(x)=x3+klnx,得f'(x)=3x2+kx.
對(duì)任意的x1,x2∈[1,+∞),且x1>x2,令x1x2=t(t>1),
則(x1-x2)[f'(x1)+f'(x2)]-2[f(x1)-f(x2)]
=(x1-x2)3x12+kx1+3x22+kx2-2x13-x23+klnx1x2
=x13-x23-3x12x2+3x1x22+kx1x2-x2x1-2klnx1x2
=x23(t3-3t2+3t-1)+kt-1t-2lnt.①
令h(x)=x-1x-2lnx,x∈[1,+∞).
當(dāng)x>1時(shí),h'(x)=1+1x2-2x=1-1x2>0,
由此可得h(x)在[1,+∞)單調(diào)遞增,
所以當(dāng)t>1時(shí),h(t)>h(1),即t-1t-2lnt>0.
因?yàn)閤2≥1,t3-3t2+3t-1=(t-1)3>0,k≥-3,
所以,x23(t3-3t2+3t-1)+kkt-1t-2lnt≥(t3-3t2+3t-1)-3kt-1t-2lnt=t3-3t2+6lnt+3t-1.②
由(1)②可知,當(dāng)t>1時(shí),g(t)>g(1),即t3-3t2+6lnt+3t>1,故t3-3t2+6lnt+3t-1>0.③
由①②③可得(x1-x2)[f'(x1)+f'(x2)]-2[f(x1)-f(x2)]>0.
所以,當(dāng)k≥-3時(shí),對(duì)任意的x1,x2∈[1,+∞),且x1>x2,有f'(x1)+f'(x2)2>f(x1)-f(x2)x1-x2.
7