老高考新教材適用2025版高考數(shù)學(xué)二輪復(fù)習(xí)專題檢測六函數(shù)與導(dǎo)數(shù)

專題檢測六函數(shù)與導(dǎo)數(shù)一、選擇題:本題共12小題,每小題5分,共60分.在每小題給出的四個選項(xiàng)中,只有一項(xiàng)是符合題目要求的.1.(2022·山東濟(jì)寧一模)定義在R上的奇函數(shù)f(x)滿足f(x-2)=-f(x),則f(2022)=()A.0 B.1 C.-1 D.20222.(2022·浙江·7)已知2a=5,log83=b,則4a-3b=()A.25 B.5 C.259 D.3.(2022·四川瀘州診斷測試)函數(shù)f(x)=0,x=04.(2022·河北石家莊二中模擬)已知函數(shù)f(x)滿足f(x)=f'(2)ex-2-f(0)x+12x2,則f(x)的單調(diào)遞減區(qū)間為(A.(-∞,0) B.(1,+∞) C.(-∞,1) D.(0,+∞)5.(2022·廣東惠州一模)5G技術(shù)的數(shù)學(xué)原理之一便是著名的香農(nóng)公式:C=Wlog21+SN,它表示:在受噪聲干擾的信道中,最大信息傳遞速率C取決于信道帶寬W、信道內(nèi)信號的平均功率S、信道內(nèi)部的高斯噪聲功率N的大小,其中SN叫做信噪比.當(dāng)信噪比較大時(shí),公式中真數(shù)中的1可以忽略不計(jì),按照香農(nóng)公式,若不改變帶寬W,而將信噪比SN從1000提升至5000,則C大約增加了()(附:lg2A.20% B.23% C.28% D.50%6.(2022·四川涼山三模)函數(shù)f(x)=a2x2-sinx,若f(x)在0,π2上有最小值,則實(shí)數(shù)A.(0,+∞) B.(0,1)C.(-∞,0) D.(-1,0)7.(2022·廣東韶關(guān)二模)已知直線l:y=kx(k>0)既是函數(shù)f(x)=x2+1的圖象的切線,同時(shí)也是函數(shù)g(x)=pxx+1+lnx(p∈R)的圖象的切線,則函數(shù)g(x)零點(diǎn)個數(shù)為(A.0 B.1 C.0或1 D.1或28.(2022·全國甲·理12)已知a=3132,b=cos14,c=4sin14A.c>b>a B.b>a>cC.a>b>c D.a>c>b9.(2022·江蘇鹽城模擬)函數(shù)f(x)=12ax2-(a+2)x+2lnx在定義域上單調(diào)遞增的充要條件為(A.a>2 B.a=2 C.a<2 D.a=110.(2022·山東濰坊三模)已知定義域?yàn)镽的函數(shù)f(x)滿足f(1+x)+f(1-x)=0,函數(shù)g(x)=f(x)sinωx(ω>0),若函數(shù)y=g(x+1)為奇函數(shù),則ω的值可以為()A.π4 B.π2 C.π D.11.(2022·廣東佛山三模)已知0<b<a<1,則下列不等式成立的是()A.logab<logba B.logab<1C.alnb<blna D.alna>blnb12.(2022·廣東茂名模擬)設(shè)函數(shù)f(x)=x-ln|A.f(x)為奇函數(shù)B.函數(shù)y=f(x)-1有兩個零點(diǎn)C.函數(shù)y=f(x)+f(2x)的圖象關(guān)于點(diǎn)(0,2)對稱D.過原點(diǎn)與函數(shù)f(x)的圖象相切的直線有且只有一條二、填空題:本題共4小題,每小題5分,共20分.13.(2021·新高考Ⅱ·14)寫出一個同時(shí)具有下列性質(zhì)①②③的函數(shù)f(x):.

①f(x1x2)=f(x1)f(x2);②當(dāng)x∈(0,+∞)時(shí),f'(x)>0;③f'(x)是奇函數(shù).14.(2022·山東煙臺一模)已知f(x)為R上的奇函數(shù),且f(x)+f(2-x)=0,當(dāng)-1<x<0時(shí),f(x)=2x,則f(2+log25)的值為.

15.(2022·北京·14)設(shè)函數(shù)f(x)=-ax+1,x<a,(x-2)2,16.(2022·河北衡水模擬)設(shè)函數(shù)f(x)=x(x+1)(x-2m)的兩個極值點(diǎn)為x1,x2,若f(x1)+f(x2)>0,則實(shí)數(shù)m的取值范圍是.

三、解答題:本題共6小題,共70分.解答應(yīng)寫出文字說明、證明過程或演算步驟.17.(10分)已知函數(shù)f(x)=log32-axx-2((1)求a的值;(2)當(dāng)x∈3,5時(shí),f(x)<log3(x+2k)恒成立,求實(shí)數(shù)k18.(12分)(2022·黑龍江哈爾濱模擬)設(shè)a為實(shí)數(shù),函數(shù)f(x)=x3-3x2+a,g(x)=xlnx.(1)求f(x)的極值;(2)對于?x1∈1,3,?x2∈1e,e,都有f(x1)≥g(x19.(12分)(2022·安徽亳州高三期末)如圖所示,兩村莊A和B相距10km,現(xiàn)計(jì)劃在兩村莊外以AB為直徑的半圓弧AB(異于A,B兩點(diǎn))上選擇一點(diǎn)C建造自來水廠,并沿線段CA和CB鋪設(shè)引水管道.根據(jù)調(diào)研分析,CA段的引水管道造價(jià)為2萬元/km,CB段的引水管道造價(jià)為m萬元/km,設(shè)CA=xkm,鋪設(shè)引水管道的總造價(jià)為y萬元,且已知當(dāng)自來水廠建在半圓弧AB的中點(diǎn)時(shí),y=302.(1)求m的值,并將y表示為x的函數(shù);(2)分析y是否存在最大值,若存在,求出最大值;若不存在,請說明理由.20.(12分)(2020·全國Ⅰ·文20)已知函數(shù)f(x)=ex-a(x+2).(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)若f(x)有兩個零點(diǎn),求a的取值范圍.21.(12分)(2022·全國甲·文20)已知函數(shù)f(x)=x3-x,g(x)=x2+a,曲線y=f(x)在點(diǎn)(x1,f(x1))處的切線也是曲線y=g(x)的切線.(1)若x1=-1,求a;(2)求a的取值范圍.22.(12分)(2022·廣東深圳一模)已知函數(shù)f(x)=2lnx-(a+1)x2-2ax+1(a∈R).(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若函數(shù)f(x)有兩個零點(diǎn)x1,x2.(i)求實(shí)數(shù)a的取值范圍;(ii)求證:x1+x2>21a

專題檢測六函數(shù)與導(dǎo)數(shù)1.A解析:因?yàn)閒(x-2)=-f(x),可得f(x+2)=-f(x),所以f(x+4)=-f(x+2)=f(x),所以f(x)的周期為4.又函數(shù)f(x)是定義在R上的奇函數(shù),所以f(0)=0,所以f(2)=-f(0)=0,f(2022)=f(505×4+2)=f(2)=0.2.C解析:由log83=b,得8b=3,即23b=3,則2a-3b=2a23b=53,所以43.D解析:當(dāng)x≠0時(shí),f(-x)=-x-sin(-x)ln|-所以f(x)是奇函數(shù),排除A,B;易知當(dāng)x=±1時(shí),為臨界點(diǎn).又π6-sinπ6=π6-12>0,lnπ6<0,故fπ4.A解析:由題設(shè)f'(x)=f'(2)ex-2-f(0)+x,則f'(2)=f'(2)-f(0)+2,可得f(0)=2,而f(0)=f'(2)e-2=2,則f'(2)=2e2,所以f(x)=2ex-2x+12x2,即f'(x)=2ex-2+x,則f'(0)=0且f'(x當(dāng)x<0時(shí)f'(x)<0,即f(x)單調(diào)遞減,故f(x)的單調(diào)遞減區(qū)間為(-∞,0).5.B解析:將信噪比SN從1000提升至5000時(shí),C大約增加了Wlog2(1+56.A解析:由題意,函數(shù)f(x)=a2x2-sinx,可得f'(x)=ax-cosx,若a≤0時(shí),當(dāng)x∈0,π2時(shí),可得f'(x)<0,f(x)在0,π2當(dāng)a>0時(shí),令f'(x)=0,即ax-cosx=0,即y=ax與y=cosx的圖象的交點(diǎn)的橫坐標(biāo),畫出函數(shù)y=ax與y=cosx的示意圖象,如圖所示,結(jié)合圖象,可得存在x0∈0,π2,使得f'(x0當(dāng)x∈(0,x0)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈x0,π2時(shí),f'(x)>0,f(x)單調(diào)遞增,此時(shí)函數(shù)f(x)在0,π2上有最小值,符合題意,綜上可得,實(shí)數(shù)a的取值范圍是(0,+∞).7.B解析:設(shè)A(x1,x12+1)是函數(shù)f(x)=x2則k=f'(x1)=2x1,∴x1=k2,①又x12+1=kx1,將①代入②消去x1整理得k2=4,∴k=2(k=-2舍去),設(shè)Bx2,px21+x2+lnx2是函數(shù)g(x)=pxx+1+lnx的切點(diǎn),據(jù)題意g'(x2)=p(1+x2)2+1x2=2,又px21+x2令h(x)=2x2-x+lnx-1(x>0),∴h'(x)=4x-1+1x≥24x·1x-1=3故h(x)=2x2-x+lnx-1(x>0)在定義域上為增函數(shù),又h(1)=0,故x2=1,故g'(1)=1+p4=∴p=4,g(x)=4xx+1+lnx=lnx-4x+1當(dāng)x=1e2時(shí),g1e2<0;當(dāng)x=1時(shí),g(1)由零點(diǎn)存在定理可得,g(x)存在唯一一個x0∈1e2,1,使g(x08.A解析:因?yàn)閍=3132=1-132,構(gòu)造函數(shù)h(x)=1-12x2-cosx,x∈0,π2,則g(x)=h'(x)=-x+sinx,g'(x)=-1+cosx≤0,所以g(x)在0,π2上單調(diào)遞減,g(x)≤g(0)=0,即h'(x)≤0,則h(x)在0,π2上單調(diào)遞減,所以h14=a-b<h(0)=0,即a<b.又cb=4sin19.B解析:由函數(shù)f(x)=12ax2-(a+2)x+2lnx在區(qū)間(0,+∞)上單調(diào)遞增,則f'(x)=ax-(a+2)+2x=a即ax2-(a+2)x+2≥0在區(qū)間(0,+∞)上恒成立,①當(dāng)a=0時(shí),-2x+2≥0?x≤1,不滿足題意;②當(dāng)a<0時(shí),ax2-(a+2)x+2=ax-2a(x-1)≥0,又2a<0,即x-2a(x-1)≤0?x≤1,不滿足題意;③當(dāng)a>0時(shí),ax2-(a+2)x+2=ax-2a(x-1)≥0,又2a>0,ax2-(a+2)x+2≥0在區(qū)間(0,+∞則Δ=(a+2)2-8a=(a-2)2≤0?a=2,綜上,函數(shù)f(x)=12ax2-(a+2)x+2lnx單調(diào)遞增的充要條件為a=210.B解析:因?yàn)閒(1+x)+f(1-x)=0,所以f(x)的圖象關(guān)于點(diǎn)(1,0)對稱,要使g(x+1)=f(x+1)sin[ω(x+1)]為奇函數(shù),因?yàn)閒(x+1)關(guān)于點(diǎn)(0,0)對稱,為奇函數(shù),所以只需使y=sin[ω(x+1)]=sin(ωx+ω)為偶函數(shù)即可,所以ω=π2+kπ,k∈Z故符合題意的為B.11.C解析:選項(xiàng)A:logab-logba=lgb由0<b<a<1,可得lgb<lga<0,則lgblga>0,lgb-lga<0,lgb+lga<0,則(lgb-lga)(lgb選項(xiàng)B:由0<a<1,可得y=logax為(0,+∞)上的減函數(shù),又0<b<a,則logab>logaa=1,故B錯誤;選項(xiàng)C:由0<a<1,可知y=ax為R上的減函數(shù),又b<a,則ab>aa,由a>0,可知y=xa在(0,+∞)上單調(diào)遞增,又b<a,則ba<aa,則ab>ba,又y=lnx為(0,+∞)上的增函數(shù),則lnab>lnba,則alnb<blna,故C正確;選項(xiàng)D:令a=1e,b=1e2alna=1eln1e=-1e,blnb=1e2則alna-blnb=-1e+2e2=2-e12.A解析:f(x)的定義域?yàn)閧x|x≠0},f(-x)=-x-ln|x|-x=由y=f(x)-1=x-ln|x|x-1=-y=f(x)+f(2x)=x-ln|x|x+2x-ln|2x|2g(-x)+g(x)=-x-ln|所以g(x)是奇函數(shù),圖象關(guān)于原點(diǎn)對稱,所以函數(shù)y=f(x)+f(2x)的圖象關(guān)于點(diǎn)(0,2)對稱,C選項(xiàng)正確;對于D選項(xiàng),f(x)圖象上一點(diǎn)(t,f(t)),當(dāng)x>0時(shí),f(x)=1-lnxx,f'(x)=lnx-1x2,則f(t)=1-lntt,f'(t)=lnt-1t2,故過點(diǎn)(t,f將(0,0)代入上式得0-1-lntt=lnt-1t2·(0-t構(gòu)造函數(shù)h(t)=2lnt-t-1(t>0),h'(t)=2t-1=2h(t)在(0,2)上,h'(t)>0,h(t)單調(diào)遞增;在(2,+∞)上,h'(t)<0,h(t)單調(diào)遞減.h(2)=2ln2-3=ln4-lne3<0,所以h(t)<0,即方程2lnt-t-1=0無解,也即當(dāng)x>0時(shí),不存在過原點(diǎn)的切線.當(dāng)x<0時(shí),f(x)=x-ln(-x)x=1-ln(-x同理,設(shè)f(x)圖象上一點(diǎn)(t,f(t)),則f(t)=1-ln(-t)t,f'(t故過點(diǎn)(t,f(t))的切線方程為y-1-ln(-t)t=ln(-將(0,0)代入上式得0-1-ln(-t)t=ln(-t)-1t2·(0構(gòu)造函數(shù)m(t)=2ln(-t)-1-t(t<0),m'(t)=2t-1=2-所以m(t)在(-∞,0)上單調(diào)遞減,m(-1)=2ln1-1-(-1)=0,所以m(t)有唯一零點(diǎn)t=-1,也即當(dāng)x<0時(shí),存在唯一一條過原點(diǎn)的切線.綜上所述,D選項(xiàng)正確.13.f(x)=x2(x∈R),答案不唯一解析:∵f(x)=x2,∴f(x1x2)=(x1x2)2=x12x22=f(x1)f(又f'(x)=2x,∴當(dāng)x∈(0,+∞)時(shí),f'(x)>0,且f'(x)為奇函數(shù),滿足②③.14.-45解析:由題設(shè),f(2-x)=-f(x)=f(-x故f(2+x)=f(x),即f(x)的周期為2,所以f(2+log25)=f2×2+log254=flog254=-f所以f(2+log25)=-2log215.0(第一空答案不唯一)1解析:根據(jù)題意可以用0,2為a的取值的分界點(diǎn),研究函數(shù)f(x)的性質(zhì).當(dāng)a<0時(shí),f(x)=-ax+1,x<a,該函數(shù)的值域?yàn)?-∞,-a2+1),故整個函數(shù)沒有最小值;當(dāng)a=0時(shí),f(x)=-ax+1,x<a,該函數(shù)的值域?yàn)閧1},而函數(shù)f(x)=(x-2)2,x≥a的值域?yàn)閇0,+∞),即存在最小值為0,故a的一個取值可以為0;當(dāng)0<a≤2時(shí),f(x)=-ax+1,x<a,該段函數(shù)的值域?yàn)?-a2+1,+∞),而函數(shù)f(x)=(x-2)2,x≥a的值域?yàn)閇0,+∞),若存在最小值,則需滿足-a2+1≥0,于是結(jié)合0<a≤2可得0<a≤1;當(dāng)a>2時(shí),f(x)=-ax+1,x<a,該段函數(shù)的值域?yàn)?-a2+1,+∞),而函數(shù)f(x)=(x-2)2,x≥a的值域?yàn)閇(a-2)2,+∞),若存在最小值,則滿足-a2+1≥(a-2)2,此時(shí)無解.綜上,a的取值范圍為[0,1],故a的最大值為1.16.(-∞,-1)∪-14,12解析:f(x)=x(x+1)(x-2m)=x3+(1-2m)x2-2mx,f'(x)=3x2+2(1-2m)x-2m,因?yàn)楹瘮?shù)f(x)=x(x+1)(x-2m)的兩個極值點(diǎn)為x1,x所以x1,x2為函數(shù)f'(x)=3x2+2(1-2m)x-2m的兩個零點(diǎn),Δ=4(1-2m)2+24m=16m2+8m+4>0恒成立,x1+x2=4m-23,x1xf(x1)+f(x2)=x13+(1-2m)x12-2mx1+x23+(1-2m)x22-2mx2=x13+x23+(1-2m)(x12+x22)-2m(x1+x2)=(x1+x2)(x12-x1x2+x22)+(1-2m)[(x1+x2)2-2x1x2]-2m(x1+x2)=(x1+x2)[(x1+x2)2-3x1x2]+(1-2m)[(x1+x2)2-2x1x2]-2m(x1+x2)=4m-234m-232+2m+(1因?yàn)閒(x1)+f(x2)>0,所以(2m-1)(4m2+5m+1)<0,則2m-1>0,4m2所以實(shí)數(shù)m的取值范圍是(-∞,-1)∪-14,117.解(1)函數(shù)f(x)=log32-則函數(shù)f(x)=log32-axx-2為奇函數(shù),有f(-x即log32+ax-x-2即log32+ax-x-2·2-ax∴a=-1.(2)由f(x)<log3(x+2k),得log32+xx-2<log3(x+2k)恒成立,即2令g(x)=x+2x-2-x=1+4x-2-x,易知g則g(x)的最大值為g(3)=2.又當(dāng)x∈3,5時(shí),f(x)<log3(x+2即2k>x+2x-2-x在x∈3,∴2k>2,k>1,即實(shí)數(shù)k的取值范圍為(1,+∞).18.解(1)函數(shù)f(x)=x3-3x2+a的定義域?yàn)镽,f'(x)=3x2-6x=3x(x-2),令f'(x)=0,可得x=0或x=2,列表如下:x(-∞,0)0(0,2)2(2,+∞)f'(x)+0-0+f(x)增a減a-4增故函數(shù)f(x)的極大值為f(0)=a,極小值為f(2)=a-4.(2)對于?x1∈1,3,?x2∈1e,e,都有f(x1)≥g(x2),則f由(1)可知,函數(shù)f(x)在[1,2)上單調(diào)遞減,在(2,3]上單調(diào)遞增,故當(dāng)x∈1,3時(shí),f(x)min=f(2)因?yàn)間(x)=xlnx,且x∈1e,e,則g'(x)=1+lnx≥0且g'故函數(shù)g(x)在1e,e上單調(diào)遞增,故g(x)max=g由題意可得a-4≥e,故a≥e+4,故a的取值范圍是[e+4,+∞).19.解(1)因?yàn)锳B為半圓弧的直徑,則∠ACB=90°,則BC=AB由題意可得x>0,所以y=2x+m100-x2當(dāng)點(diǎn)C在AB的中點(diǎn)時(shí),x=52,此時(shí)y=52(2+m)=302,解得m=4,因此y=2x+4100-x2,其中0(2)因?yàn)閥=2x+4100-x2,其中0<x<10,則y'=2因?yàn)楹瘮?shù)f(x)=100-x2-2x在(0,10)上單調(diào)遞減,由f(x)=0可得x=當(dāng)0<x<25時(shí),y'=2(100-x2-2x當(dāng)25<x<10時(shí),y'=2(100-x2-2x故當(dāng)x=25時(shí),函數(shù)y=2x+4100-x2取最大值,即ymax=20.解(1)當(dāng)a=1時(shí),f(x)=ex-x-2,則f'(x)=ex-1.當(dāng)x<0時(shí),f'(x)<0;當(dāng)x>0時(shí),f'(x)>0.所以f(x)在(-∞,0)單調(diào)遞減,在(0,+∞)單調(diào)遞增.(2)f'(x)=ex-a.當(dāng)a≤0時(shí),f'(x)>0,所以f(x)在(-∞,+∞)單調(diào)遞增,故f(x)至多存在1個零點(diǎn),不合題意.當(dāng)a>0時(shí),由f'(x)=0可得x=lna.當(dāng)x∈(-∞,lna)時(shí),f'(x)<0;當(dāng)x∈(lna,+∞)時(shí)f'(x)>0.所以f(x)在(-∞,lna)單調(diào)遞減,在(lna,+∞)單調(diào)遞增,故當(dāng)x=lna時(shí),f(x)取得最小值,最小值為f(lna)=-a(1+lna).①若0<a≤1e,則f(lna)≥0,f(x)在(-∞,+∞)至多存在1個零點(diǎn),不合題意②若a>1e,則f(lna)<0由于f(-2)=e-2>0,所以f(x)在(-∞,lna)存在唯一零點(diǎn).由(1)知,當(dāng)x>2時(shí),ex-x-2>0,所以當(dāng)x>4且x>2ln(2a)時(shí),f(x)=ex2·ex2-a(x+2)>eln(2a)·x2+2-a故f(x)在(lna,+∞)存在唯一零點(diǎn).從而f(x)在(-∞,+∞)有兩個零點(diǎn).綜上,a的取值范圍是1e21.解(1)∵f'(x)=3x2-1,∴f'(-1)=2.當(dāng)x1=-1時(shí),f(-1)=0,故y=f(x)在點(diǎn)(-1,0)處的切線方程為y=2x+2.又y=2x+2與y=g(x)相切,將直線y=2x+2代入g(x)=x2+a,得x2-2x+a-2=0.由Δ=4-4(a-2)=0,得a=3.(2)∵f'(x)=3x2-1,∴f'(x1)=3x12-1,則曲線y=f(x)在點(diǎn)x1,fx1處的切線為y-x由g(x)=x2+a,得g'(x)=2x.設(shè)曲線y=g(x)在點(diǎn)(x2,g(x2))處的切線為y-(x22+a)=2x2(x-