2023新教材數(shù)學(xué)高考第一輪專題練習(xí)--專題四導(dǎo)數(shù)的應(yīng)用專題檢測(cè)題組

1、2023新高考數(shù)學(xué)第一輪專題練習(xí)4.2導(dǎo)數(shù)的應(yīng)用一、選擇題1.(2022屆四川綿陽質(zhì)檢,3)函數(shù)f(x)=x3-3x2+1是減函數(shù)的區(qū)間為()A.(2,+)B.(-,2)C.(-,0)D.(0,2)答案D由f (x)=3x2-6x0,得0x2,函數(shù)f(x)=x3-3x2+1是減函數(shù)的區(qū)間為(0,2).故選D.2.(2022屆河南許昌聯(lián)考,11)若f(x)=13x3-ax2的單調(diào)減區(qū)間是(-4,0),則a的值是()A.-2B.2C.-4D.4答案Af (x)=x2-2ax,依題意得,f (-4)=16+8a=0,解得a=-2.檢驗(yàn):當(dāng)a=-2時(shí),f (x)=x2+4x=x(x+4),令f (x)

2、0,得-4x0,即f(x)的單調(diào)減區(qū)間為(-4,0).故a=-2滿足題意,故選A.3.(2021河南平頂山月考,7)設(shè)函數(shù)f(x)在R上可導(dǎo),其導(dǎo)函數(shù)為f (x),且函數(shù)f(x)在x=-3處取得極大值,則函數(shù)y=xf (x)的圖象可能是()答案D函數(shù)f(x)在x=-3處取得極大值,當(dāng)x0;當(dāng)x=-3時(shí),f (x)=0;當(dāng)x-3時(shí),f (x)0,當(dāng)x-3時(shí),xf (x)0;當(dāng)x=-3時(shí),xf (x)=0;當(dāng)-3x0;當(dāng)x=0時(shí),xf (x)=0;當(dāng)x0時(shí),xf (x)f(c)f(d)B. f(b)f(a)f(c)C. f(c)f(b)f(a)D. f(c)f(b)f(d)答案C顯然f(x)在(a

3、,c)上遞增,在(c,d)上遞減,而abc,故f(a)f(b)0且有f(2)=1e,則f(x)1ex的解集為()A.-,12B.12,+C.(-,2)D.(2,+)答案D構(gòu)造g(x)=ex2f(x),則g(x)=12ex2f(x)+ex2f (x),因?yàn)?2f(x)+f (x)0,所以g(x)0,所以g(x)是R上的增函數(shù),且g(2)=ef(2)=1,不等式f(x)1ex可化為ex2f(x)1,即g(x)g(2),所以x(2,+),故選D.6.(2022屆河南段考三,6)函數(shù)f(x)=xe2x-x2-x-14的極大值為()A.-12B.-12eC.0D.-14答案B函數(shù)f(x)的定義域?yàn)镽,

4、f (x)=(2x+1)(e2x-1),令f (x)=0,得x=0或x=-12.當(dāng)x0時(shí), f (x)0, f(x)單調(diào)遞增;當(dāng)-12x0時(shí), f (x)0),f (2)=0,2a-2=0,可得a=1.f(x)=12x2-x-2ln x, f (x)=x-2x-1=(x+1)(x-2)x(x0).當(dāng)0x2時(shí), f (x)2時(shí), f (x)0, f(x)單調(diào)遞增,f(x)有極小值f(2)=-2ln 2,無極大值.f(x)有最小值-2ln 2,無最大值.故選A.8.(2021北京市八一學(xué)校10月月考,7)已知函數(shù)f(x)=x3+mln x在區(qū)間1,2上不是單調(diào)函數(shù),則m的取值范圍是()A.(-,-

5、3)B.-24,-3C.(-24,-3)D.(-24,+)答案C由題意得f (x)=3x2+mx=3x3+mx(x0),當(dāng)m0時(shí), f (x)0,所以f(x)在(0,+)上單調(diào)遞增,不滿足題意.當(dāng)m0,解得x3-m3,令f (x)0,解得0x3-m3,所以f(x)在0,3-m3上單調(diào)遞減,在3-m3,+上單調(diào)遞增,要使得函數(shù)f(x)=x3+mln x在區(qū)間1,2上不是單調(diào)函數(shù),則有13-m32,解得-24m0,設(shè)a=f(0),b=f12,c=f(3),則a,b,c的大小關(guān)系為()A.cabB.cbaC.cabD.bc0,所以f (x)f12=b.又f(x)=f(2-x),所以c=f(3)=f(

6、-1),所以c=f(-1)f(0)=a,所以cab.故選A.10.(2021東北三省四市聯(lián)考,12)已知函數(shù)f(x)=ex-3,g(x)=12+lnx2,若f(m)=g(n)成立,則m-n的最大值為()A.1-ln 2B.ln 2C.2ln 2D.ln 2-1答案A不妨設(shè)f(m)=g(n)=t,em-3=12+lnn2=t(t0),m-3=ln t,lnn2=t-12,即m=3+ln t,n=2et-12,故m-n=3+ln t-2et-12(t0),令h(t)=3+ln t-2et-12(t0),h(t)=1t-2et-12(t0),h(t)=-1t2-2et-1212時(shí),h(t)0,當(dāng)0t

7、0,即當(dāng)t=12時(shí),h(t)取得極大值同時(shí)也是最大值,此時(shí)h12=3+ln12-2=1-ln 2,即m-n的最大值為1-ln 2,故選A.解題關(guān)鍵令f(m)=g(n)=t得到m,n,利用消元法將m-n轉(zhuǎn)化為關(guān)于t的函數(shù),構(gòu)造函數(shù),求函數(shù)的導(dǎo)數(shù),利用導(dǎo)數(shù)研究函數(shù)的最值即可得到結(jié)論.11.(2021河南平頂山模擬,11)已知f(x)為定義在R上的偶函數(shù),其導(dǎo)函數(shù)為f (x),對(duì)于任意的x0,2,總有f (x)cos x+f(x)sin x0成立,則下列不等式成立的是()A.3f(0)2f6B.f42f3C.3f-3f6D.3f-40成立,所以當(dāng)x0,2時(shí),F(x)0,所以F(x)在0,2上是增函數(shù)

8、,F(0)F6,F4F3,即f(0)cos0f6cos 6,f4cos 4f3cos 3,所以3f(0)2f6,f4F6,即f-3cos 3f6cos 6,所以3f-3f6,故C正確,同理3f-42f6,故D錯(cuò)誤.故選C.思路分析構(gòu)造函數(shù)F(x)=f(x)cosx,對(duì)其求導(dǎo),根據(jù)題中條件,得到F(x)在0,2上是增函數(shù),可判斷A,B錯(cuò)誤;再由f(x)與y=cos x均為偶函數(shù),可得F(x)為偶函數(shù),進(jìn)而可判斷C正確,D錯(cuò)誤.12.(2022屆昆明月考,12)若函數(shù)f(x)=ex-ax2(aR)在(0,+)有兩個(gè)不同的零點(diǎn),則實(shí)數(shù)a的取值范圍是()A.e22,+B.e2,+C.e4,+D.e24

9、,+答案D函數(shù)f(x)=ex-ax2(aR)有兩個(gè)不同的零點(diǎn),即方程ex-ax2=0有兩個(gè)不同根,即a=exx2有兩個(gè)不同根,令g(x)=exx2(x0),則g(x)=ex(x-2)x3,當(dāng)x(0,2)時(shí),g(x)0,當(dāng)x=2時(shí),g(x)min=g(2)=e24.實(shí)數(shù)a的取值范圍是e24,+.故選D.二、解答題13.(2022屆安徽蚌埠質(zhì)檢,21)已知函數(shù)f(x)=ex,g(x)=1x.(1)求函數(shù)y=f(x)g(x)的極值;(2)求證:f(x)+g(x)f(x)g(x)1.解析(1)記F(x)=f(x)g(x)=exx,定義域?yàn)?-,0)(0,+),則F(x)=xex-exx2=(x-1)e

10、xx2,令F(x)=0,解得x=1,列表如下:x(-,0)(0,1)1(1,+)F(x)-0+F(x)單調(diào)遞減單調(diào)遞減極小值單調(diào)遞增結(jié)合表格可知函數(shù)y=f(x)g(x)的極小值為F(1)=e,無極大值.(2)證明:f(x)+g(x)f(x)g(x)=ex+1xexx=xex+1ex=x+1ex.構(gòu)造函數(shù)G(x)=x+1ex,定義域?yàn)?-,0)(0,+),G(x)=1-1ex.當(dāng)x0時(shí),G(x)0;當(dāng)x0時(shí),G(x)G(0)=1,即f(x)+g(x)f(x)g(x)1.14.(2022屆合肥聯(lián)考(二),21)已知函數(shù)f(x)=ex(x2-2x+m),mR.(1)討論f(x)的單調(diào)性;(2)當(dāng)x(

11、0,+)時(shí),不等式f(x)exln x恒成立,求m的取值范圍.解析(1)f (x)=ex(x2+m-2),當(dāng)m2時(shí),f (x)0恒成立,f(x)在R上單調(diào)遞增;當(dāng)m0),g(x)=1-2x2+2xx,令g(x)=0x=1+32(負(fù)值舍去),當(dāng)x0,1+32時(shí),g(x)0,g(x)單調(diào)遞增,當(dāng)x1+32,+時(shí),g(x)0,g(x)單調(diào)遞減,則g(x)max=g1+32=ln 1+32-1+322+21+32=ln 1+32+32,所以mln 1+32+32,即m的取值范圍是ln 1+32+32,+.15.(2021河南尖子生診斷性考試,21)已知函數(shù)f(x)=ex-ax2(其中e為自然對(duì)數(shù)的底數(shù)

12、,a為常數(shù)).(1)若f(x)在(0,+)上有極小值0,求實(shí)數(shù)a的值;(2)若f(x)在(0,+)上有極大值M,求證:Ma.解析(1)f (x)=ex-2ax.設(shè)f(x0)=0(x0(0,+),則f (x0)=0.由ex0-ax02=0,ex0-2ax0=0,解得x0=2,a=e24.經(jīng)檢驗(yàn),a=e24滿足f(x)在(0,+)上有極小值,且極小值為0.故a=e24.(2)證明:設(shè)f(x)在(0,+)上的極大值點(diǎn)為x1,則f (x1)=0,即ex1-2ax1=0,則有a=ex12x1.此時(shí)M=f(x1)=ex1-ax12.故M-a=ex1-ax12-a=ex1-a(x12+1)=ex1-(x12

13、+1)ex12x1=ex11-12x1+1x10(當(dāng)且僅當(dāng)x1=1時(shí)取等號(hào)).而當(dāng)x1=1時(shí),a=e2,f (x)=ex-ex,f (x)=ex-e,x(0,1)時(shí),f (x)0.則f (x)在(0,1)上單調(diào)遞減,在(1,+)上單調(diào)遞增,且f (1)=0.則f (x)f (1)=0,故f(x)在(0,+)上單調(diào)遞增,此時(shí)f(x)在(0,+)上無極值.與已知條件矛盾,故x11,則M-a0,即M-12x2+1;(3)當(dāng)x0時(shí),若曲線y=f(x)在曲線y=ax2+1的上方,求實(shí)數(shù)a的取值范圍.解析(1)f (x)=-xex.令f (x)=0,解得x=0.隨x的變化, f (x)和f(x)的變化情況

14、如下:x(-,0)0(0,+)f (x)+0-f(x)極大值故函數(shù)f(x)在x=0處取得極大值, f(0)=1,無極小值.(2)證明:令g(x)=f(x)+12x2-1=x+1ex+12x2-1(x0),則g(x)=-xex+x=x1-1ex=xex-1ex.由x0得ex-10,于是g(x)0,故函數(shù)g(x)是(0,+)上的增函數(shù).所以g(x)g(0)=0,即f(x)-12x2+1.(3)當(dāng)a-12時(shí),由(2)知f(x)-12x2+1ax2+1,滿足題意.令h(x)=f(x)-ax2-1=x+1ex-ax2-1,則h(x)=-xex-2ax=-x1ex+2a.當(dāng)-12a0時(shí),若x0,ln-12

15、a,則h(x)0,h(x)在0,ln-12a上是減函數(shù).所以h(x)h(0)=0,不合題意.當(dāng)a0時(shí),h(x)0,h(x)在(0,+)上是減函數(shù),所以h(x)h(0)=0,不合題意.綜上所述,實(shí)數(shù)a的取值范圍是-,-12.解后反思(1)先求導(dǎo),根據(jù)f (x)和f(x)隨x變化的情況表,求得極值;(2)令g(x)=f(x)+12x2-1,求導(dǎo),根據(jù)函數(shù)g(x)的單調(diào)性和最值證明;(3)分a-12,-12a0及a0三種情況討論,求得實(shí)數(shù)a的取值范圍.17.(2022屆北京師大附中10月月考,20)已知函數(shù)f(x)=lnx-1x-ax(aR).(1)若曲線y=f(x)在點(diǎn)(1, f(1)處的切線與x

16、軸平行.(i)求a的值;(ii)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若1a2,求證:f(x)0,令g(x)=2-ln x-2x2(x0),則g(x)=-1x-4x0,所以g(x)在(0,+)上單調(diào)遞減,且g(1)=2-ln 1-2=0,所以當(dāng)0x0,即f (x)0;當(dāng)x1時(shí),g(x)0,即f (x)0.所以f(x)的單調(diào)增區(qū)間為(0,1),單調(diào)減區(qū)間為(1,+).(2)證明:當(dāng)1a2時(shí),要證明f(x)=lnx-1x-ax0),即證ax2-x+1-lnxx0,等價(jià)于證明ax2-x+1-ln x0.令h(x)=ax2-x+1-ln x(x0),則h(x)=2ax-1-1x=2ax2-x-1x,由h(x

17、)=0得2ax2-x-1=0,易知方程2ax2-x-1=0有一正一負(fù)兩個(gè)根,因?yàn)閤0,所以負(fù)根不符合題意,舍去.令其正根為x0,則2ax02-x0-1=0,當(dāng)x(0,x0)時(shí)h(x)0,所以h(x)在(0,x0)上單調(diào)遞減,在(x0,+)上單調(diào)遞增,所以h(x)min=h(x0)=ax02-x0+1-ln x0=1+x02-x0+1-ln x0=3-x02-ln x0,因?yàn)閔(1)=2a-20,h12=a-30,所以12x00,ln x00,所以h(x)0,即f(x)0,令h(x)=ax2-x+1-ln x,利用導(dǎo)數(shù)判斷h(x)的單調(diào)性,證明h(x)min0即可.18.(2022屆太原模擬,2

18、1)已知函數(shù)f(x)=ax+ln x.(1)若曲線y=f(x)在點(diǎn)(m,2)(m0)處的切線方程為y=-x+3,求f(x)的單調(diào)區(qū)間;(2)若方程f(x)-1=0在x1e,e上有兩個(gè)實(shí)數(shù)根,求實(shí)數(shù)a的取值范圍.解析(1)f (x)=-ax2+1x.由題意可得2=-m+3,且-am2+1m=-1,解得a=2,m=1,所以f(x)=2x+ln x,則f (x)=-2x2+1x=x-2x2,當(dāng)x2時(shí),f (x)0,函數(shù)f(x)單調(diào)遞增,當(dāng)0x2時(shí),f (x)0,函數(shù)f(x)單調(diào)遞減,所以f(x)的單調(diào)遞增區(qū)間為(2,+),單調(diào)遞減區(qū)間為(0,2).(2)方程f(x)-1=0在x1e,e上有兩個(gè)實(shí)數(shù)根

19、,即方程a=x(1-ln x)在x1e,e上有兩個(gè)實(shí)數(shù)根,令h(x)=x(1-ln x),則h(x)=1-ln x-1=-ln x,當(dāng)1ex0,h(x)單調(diào)遞增;當(dāng)1xe時(shí),h(x)0,h(x)單調(diào)遞減,所以h(x)max=h(1)=1,又h1e=2e,h(e)=0,所以2ea0時(shí),函數(shù)g(x)=exf(x)+tex-x+1在R上有唯一零點(diǎn),求t的值. 解析(1)當(dāng)t=-4時(shí),f(x)=-4ex-1ex-2,則f (x)=-4ex+1ex=(1+2ex)(1-2ex)ex,令f (x)=0,得x=-ln 2,f(x)的單調(diào)遞增區(qū)間是(-,-ln 2),單調(diào)遞減區(qū)間是(-ln 2,+),f(x)的極大值是f(-ln 2)=-412-2-2=-6,無極小值.(2)當(dāng)t0時(shí),g(x)=exf(x)+tex-x+1=te2x+(t-2)ex-x,則g(x)=2te2x+(t-2)ex-1=(tex-1)(2ex+1),令g(x)=0,得x=-ln t,g(x)在(-,-ln t)上單調(diào)遞減,在(-ln t,+)上單調(diào)遞增,g(x)的極小值是g(-ln t),只要g(-ln t)=0,即可滿足函數(shù)g(x)在R上有唯一零點(diǎn),g(-ln t)=ln t-1t+1=0,令F(t)=ln t-1t+1(t0),則F(t)=1t+1t20,F(t)在(0,+)上單調(diào)遞增,F(1)=0,