2021新高考數(shù)學(xué)二輪總復(fù)習(xí)專題突破練7熱點(diǎn)小專題二導(dǎo)數(shù)的應(yīng)用含解析

1、專題突破練7熱點(diǎn)小專題二、導(dǎo)數(shù)的應(yīng)用一、單項(xiàng)選擇題1.設(shè)曲線y=ax-ln(x+1)在點(diǎn)(0,0)處的切線方程為y=2x,則a=()a.0b.1c.2d.32.(2020天津河北區(qū)線上測試,6)已知函數(shù)f(x)=3x+2cos x,若a=f(32),b=f(2),c=f(log27),則a,b,c的大小關(guān)系是()a.abcb.cabc.bacd.bca3.(2020河南開封三模,文9,理7)已知函數(shù)f(x)=x(x-c)2在x=2處取極大值,則c=()a.-2或-6b.2或6c.2d.64.(2020山東德州二模,8)已知函數(shù)f(x)的定義域?yàn)閞,且f(x)+13ex的解集為()a.(1,+)

2、b.(-,1)c.(0,+)d.(-,0)5.已知函數(shù)f(x)=aex-x2-(2a+1)x,若函數(shù)f(x)在區(qū)間(0,ln 2)上有極值,則實(shí)數(shù)a的取值范圍是()a.(-,-1)b.(-1,0)c.(-2,-1)d.(-,0)(0,1)6.(2020山東濟(jì)南一模,8)已知直線y=ax+b(b0)與曲線y=x3有且只有兩個公共點(diǎn)a(x1,y1),b(x2,y2),其中x1ax-2ex在x(0,+)上恒成立,則實(shí)數(shù)a的取值范圍是()a.a2b.a2c.a0d.0a28.(2020江西名校大聯(lián)考,理12)已知函數(shù)f(x)=-13x3+x2,xm,x-m,xm,若存在實(shí)數(shù)a,使得函數(shù)g(x)=f(x

3、)-a恰好有4個零點(diǎn),則實(shí)數(shù)m的取值范圍是()a.(0,2)b.(2,+)c.(0,3)d.(3,+)二、多項(xiàng)選擇題9.函數(shù)y=f(x)的導(dǎo)函數(shù)y=f(x)的圖象如圖所示,以下命題錯誤的是()a.-3是函數(shù)y=f(x)的極值點(diǎn)b.-1是函數(shù)y=f(x)的最小值點(diǎn)c.y=f(x)在區(qū)間(-3,1)上單調(diào)遞增d.y=f(x)在x=0處切線的斜率小于零10.(2020山東聊城二模,10)下列關(guān)于函數(shù)f(x)=x3-3x2+2x的敘述正確的是()a.函數(shù)f(x)有三個零點(diǎn)b.點(diǎn)(1,0)是函數(shù)f(x)圖象的對稱中心c.函數(shù)f(x)的極大值點(diǎn)為x=1-33d.存在實(shí)數(shù)a,使得函數(shù)g(x)=f(x)2+a

4、f(x)在r上為增函數(shù)11.(2020山東濰坊臨朐模擬,12)已知函數(shù)f(x)=xln x+x2,x0是函數(shù)f(x)的極值點(diǎn),以下結(jié)論中正確的是()a.0x01ec.f(x0)+2x0012.(2020山師大附中月考,12)設(shè)函數(shù)f(x)=|lnx|,x0,ex(x+1),x0,若方程f(x)2-af(x)+116=0有六個不等的實(shí)數(shù)根,則實(shí)數(shù)a可能的取值是()a.12b.23c.1d.2三、填空題13.(2020全國,文15)設(shè)函數(shù)f(x)=exx+a.若f(1)=e4,則a=.14.(2020全國,文15)曲線y=ln x+x+1的一條切線的斜率為2,則該切線的方程為.15.(2020山東

5、淄博4月模擬,16)已知函數(shù)f(x)=2sin x+sin 2x,則f(x)的最小值是.16.已知函數(shù)f(x)=log2x,g(x)=x+a-x(a0),若對x1x|g(x)=x+a-x,x24,16,使g(x1)=f(x2)成立,則實(shí)數(shù)a的取值范圍是.專題突破練7熱點(diǎn)小專題二、導(dǎo)數(shù)的應(yīng)用1.d解析y=ax-ln(x+1),y=a-1x+1.y|x=0=a-1=2,得a=3.2.d解析因?yàn)楹瘮?shù)f(x)=3x+2cosx,所以導(dǎo)數(shù)函數(shù)f(x)=3-2sinx,可得f(x)=3-2sinx0在r上恒成立,所以f(x)在r上為增函數(shù),又因?yàn)?=log24log27332,所以bca,故選d.3.d解

6、析f(x)=(x-c)2+2x(x-c),f(2)=(2-c)2+22(2-c)=0,解得c=6或c=2.驗(yàn)證知當(dāng)c=2時,函數(shù)在x=2處有極小值,舍去,當(dāng)c=6時滿足題意,故c=6.4.c解析令g(x)=f(x)+1ex,f(x)+10,故g(x)在r上單調(diào)遞增,且g(0)=3,由f(x)+13ex,可得f(x)+1ex3,即g(x)g(0),所以x0,故選c.5.a解析f(x)=aex-2x-(2a+1),令g(x)=aex-2x-(2a+1).由函數(shù)f(x)在區(qū)間(0,ln2)上有極值g(x)在區(qū)間(0,ln2)上單調(diào)且存在零點(diǎn).所以g(0)g(ln2)=(a-2a-1)(2a-2ln2

7、-2a-1)0,即a+10,解得a0)與曲線y=x3有且只有兩個公共點(diǎn),即為b=x3-ax有兩個根,即函數(shù)y=x3-ax與y=b恰有兩個交點(diǎn),作出兩個函數(shù)的圖象,可知x1是極大值點(diǎn)時滿足題意.y=3x2-a,3x12=a.又b=x13-ax1=x23-ax2,x13-x23=a(x1-x2),(x1-x2)(x12+x1x2+x22)=a(x1-x2).x1ax-2ex在x(0,+)上恒成立,等價于f(x+1)f(ex)在(0,+)上恒成立,因?yàn)閤(0,+)時,1x+11時,f(x)0恒成立,即當(dāng)x1時,ax2恒成立,a2x,所以a2,故選a.8.b解析g(x)=f(x)-a的零點(diǎn)個數(shù)等價于直

8、線y=a與函數(shù)f(x)圖象的交點(diǎn)個數(shù).令y=-13x3+x2,y=-x2+2x,當(dāng)x2時,y0,當(dāng)0x0.所以函數(shù)y=-13x3+x2在(-,0),(2,+)上單調(diào)遞減,在(0,2)上單調(diào)遞增,畫出函數(shù)f(x)的大致圖象如圖所示,由圖可知當(dāng)m2時,存在直線y=a與函數(shù)f(x)圖象的交點(diǎn)為4個;當(dāng)0m2時,直線y=a與函數(shù)f(x)圖象的交點(diǎn)至多為3個;當(dāng)m0時,直線y=a與函數(shù)f(x)圖象的交點(diǎn)至多為2個.所以m的取值范圍為(2,+).故選b.9.bd解析根據(jù)導(dǎo)函數(shù)的圖象可知,當(dāng)x(-,-3)時,f(x)0),f(x)=lnx+1+2x.x0是函數(shù)f(x)的極值點(diǎn),f(x0)=0,即lnx0+1+2x0=0,f(x)在(0,+)上單調(diào)遞增,且f1e=2e0,x0,f(x)-,0x00,即選項(xiàng)d正確,選項(xiàng)c不正確.故選ad.12.bc解析當(dāng)x0時,f(x)=ex(x+1),則f(x)=ex(x+1)+ex=ex(x+2).由f(x)0得,x+20,即x0得,x+20,即-2x0,此時f(x)為增函數(shù),即當(dāng)x=-2時,f(x)取得極小值f(-2)=-1e2,作出f(x)的圖象如圖:由圖象可知當(dāng)00,g(1)0,0,0a20,1