廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練19 三角函數(shù)的圖象與性質(zhì) 文.docx

考點(diǎn)規(guī)范練19三角函數(shù)的圖象與性質(zhì)一、基礎(chǔ)鞏固1.函數(shù)y=|2sin x|的最小正周期為()A.B.2C.2D.4答案A解析由圖象(圖象略)知T=.2.已知直線y=m(0<m<2)與函數(shù)f(x)=2sin(x+)(>0)的圖象相鄰的三個(gè)交點(diǎn)依次為A(1,m),B(5,m),C(7,m),則=()A.3B.4C.2D.6答案A解析由題意,得函數(shù)f(x)的相鄰的兩條對(duì)稱軸分別為x=1+52=3,x=5+72=6,故函數(shù)的周期為2(6-3)=2,得=3,故選A.3.若函數(shù)f(x)=3cosx-4(1<<14)的圖象關(guān)于x=12對(duì)稱,則等于()A.2B.3C.6D.9答案B解析f(x)=3cosx-4(1<<14)的圖象關(guān)于x=12對(duì)稱,12-4=k,kZ,即=12k+3.1<<14,由此求得=3,故選B.4.已知函數(shù)f(x)=sinx+4(>0)的最小正周期為,則函數(shù)f(x)的圖象()A.關(guān)于直線x=4對(duì)稱B.關(guān)于直線x=8對(duì)稱C.關(guān)于點(diǎn)4,0對(duì)稱D.關(guān)于點(diǎn)8,0對(duì)稱答案B解析函數(shù)f(x)的最小正周期為,2=.=2.f(x)=sin2x+4.函數(shù)f(x)圖象的對(duì)稱軸為2x+4=k+2,kZ,即x=8+k2,kZ.故函數(shù)f(x)的圖象關(guān)于直線x=8對(duì)稱,故選B.5.y=cos(x+1)圖象上相鄰的最高點(diǎn)和最低點(diǎn)之間的距離是()A.2+4B.C.2D.2+1答案A解析因?yàn)閥=cos(x+1)的周期是2,最大值為1,最小值為-1,所以y=cos(x+1)圖象上相鄰的最高點(diǎn)和最低點(diǎn)之間的距離是2+4,故選A.6.已知曲線f(x)=sin 2x+3cos 2x關(guān)于點(diǎn)(x0,0)成中心對(duì)稱,若x00,2,則x0=()A.12B.6C.3D.512答案C解析由題意可知f(x)=2sin2x+3,其對(duì)稱中心為(x0,0),故2x0+3=k(kZ),即x0=-6+k2(kZ).又x00,2,故k=1,x0=3,故選C.7.已知函數(shù)y=sin x的定義域?yàn)閍,b,值域?yàn)?1,12,則b-a的值不可能是()A.3B.23C.D.43答案A解析畫(huà)出函數(shù)y=sinx的草圖分析,知b-a的取值范圍為23,43.8.(2018廣東深圳模擬)已知函數(shù)f(x)=sin(2x+)0<<2的圖象的一個(gè)對(duì)稱中心為38,0,則函數(shù)f(x)的單調(diào)遞減區(qū)間是()A.2k-38,2k+8(kZ)B.2k+8,2k+58(kZ)C.k-38,k+8(kZ)D.k+8,k+58(kZ)答案D解析由題意知,sin238+=0,又0<<2,所以=4.所以f(x)=sin2x+4.由2+2k2x+432+2k(kZ),得f(x)的單調(diào)遞增區(qū)間是k+8,k+58(kZ).9.(2018陜西高三質(zhì)檢)已知函數(shù)f(x)=sin xsin(x+3)是奇函數(shù),其中0,2,則f(x)的最大值為()A.12B.22C.1D.2答案A解析函數(shù)f(x)=sinxsin(x+3)是奇函數(shù),y=sinx是奇函數(shù),y=sin(x+3)是偶函數(shù),3=k+2,kZ,=6,f(x)=sinxsinx+2=12sin2x,則f(x)的最大值為12.10.已知函數(shù)f(x)=sin(x+)>0,|<2的最小正周期為4,且f3=1,則f(x)圖象的對(duì)稱中心是.答案2k-23,0(kZ)解析由題意得2=4,解得=12,故f(x)=sin12x+,由f3=1可得123+=2k+2,kZ,由|<2可得=3,故f(x)=sin12x+3,由12x+3=k可得x=2k-23,kZ.f(x)的對(duì)稱中心為2k-23,0,kZ.11.已知函數(shù)y=cos x與y=sin(2x+)(0<),它們的圖象有一個(gè)橫坐標(biāo)為3的交點(diǎn),則的值是.答案6解析由題意cos3=sin23+,即sin23+=12,23+=2k+6(kZ)或23+=2k+56(kZ).因?yàn)?<,所以=6.12.已知>0,在函數(shù)y=2sin x與y=2cos x的圖象的交點(diǎn)中,距離最短的兩個(gè)交點(diǎn)的距離為23,則=.答案2解析如圖所示,在同一直角坐標(biāo)系中,作出函數(shù)y=2sinx與y=2cosx的圖象.A,B為符合條件的兩個(gè)交點(diǎn).則A4,2,B-34,-2.由|AB|=23,得2+(22)2=23,解得=2,即=2.二、能力提升13.(2018安徽合肥二模)已知函數(shù)f(x)=2sin(x+)(>0,0<<)相鄰兩條對(duì)稱軸間的距離為32,且f2=0,則下列說(shuō)法正確的是()A.=2B.函數(shù)y=f(x-)為偶函數(shù)C.函數(shù)f(x)在-,-2上單調(diào)遞增D.函數(shù)y=f(x)的圖象關(guān)于點(diǎn)34,0對(duì)稱答案C解析由題意可得,函數(shù)f(x)的周期為T(mén)=232=3,則=2T=23,A說(shuō)法錯(cuò)誤;當(dāng)x=2時(shí),x+=232+=k,=k-3(kZ),0<<,故取k=1可得=23,函數(shù)的解析式為f(x)=2sin23x+23,y=f(x-)=2sin23(x-)+23=2sin23x,函數(shù)為奇函數(shù),B說(shuō)法錯(cuò)誤;當(dāng)x-,-2時(shí),23x+23-3,3,故函數(shù)f(x)在-,-2上單調(diào)遞增,C說(shuō)法正確;f34=2sin2334+23=2sin760,則函數(shù)y=f(x)的圖象不關(guān)于點(diǎn)34,0對(duì)稱,D說(shuō)法錯(cuò)誤.14.已知函數(shù)f(x)=3sin(x+)>0,-2<<2,A13,0為f(x)圖象的對(duì)稱中心,B,C是該圖象上相鄰的最高點(diǎn)和最低點(diǎn),若BC=4,則f(x)的單調(diào)遞增區(qū)間是()A.2k-23,2k+43,kZB.2k-23,2k+43,kZC.4k-23,4k+43,kZD.4k-23,4k+43,kZ答案C解析由題意,得(23)2+T22=42,即12+22=16,求得=2.再根據(jù)213+=k,kZ,且-2<<2,可得=-6,f(x)=3sin2x-6.令2k-22x-62k+2,kZ,求得4k-23x4k+43,故f(x)的單調(diào)遞增區(qū)間為4k-23,4k+43,kZ,故選C.15.已知函數(shù)y=sin x+cos x,y=22sin xcos x,則下列結(jié)論正確的是()A.兩個(gè)函數(shù)的圖象均關(guān)于點(diǎn)-4,0成中心對(duì)稱B.兩個(gè)函數(shù)的圖象均關(guān)于直線x=-4對(duì)稱C.兩個(gè)函數(shù)在區(qū)間-4,4內(nèi)都是單調(diào)遞增函數(shù)D.可以將函數(shù)的圖象向左平移4個(gè)單位長(zhǎng)度得到函數(shù)的圖象答案C解析函數(shù)y=sinx+cosx=2sinx+4,y=22sinxcosx=2sin2x,由于的圖象不關(guān)于點(diǎn)-4,0成中心對(duì)稱,故A不正確.由于函數(shù)的圖象不可能關(guān)于直線x=-4成軸對(duì)稱,故B不正確.由于這兩個(gè)函數(shù)在區(qū)間-4,4內(nèi)都是單調(diào)遞增函數(shù),故C正確.由于將函數(shù)的圖象向左平移4個(gè)單位長(zhǎng)度得到函數(shù)y=2sin2x+4,而y=2sin2x+42sinx+4,故D不正確,故選C.16.已知函數(shù)f(x)=3sinx-6(>0)和g(x)=3cos(2x+)的圖象的對(duì)稱中心完全相同,若x0,2,則f(x)的取值范圍是.答案-32,3解析由兩個(gè)三角函數(shù)的圖象的對(duì)稱中心完全相同,可知它們的周期相同,則=2,即f(x)=3sin2x-6.當(dāng)x0,2時(shí),-62x-656,解得-12sin2x-61,故f(x)-32,3.三、高考預(yù)測(cè)17.已知函數(shù)f(x)=sin2x+6,其中x-6,a.當(dāng)a=3時(shí),f(x)的值域是;若f(x)的值域是-12,1,則a的取值范圍是.答案-12,16,2解析若-6x3,則-62x+656,此時(shí)-12sin2x+61,即f(x)的值域是-12,1.若-6xa,則-62x+62a+6.因?yàn)楫?dāng)2x+6=-6或2x+6=76時(shí),sin2x+6=-12,所以要使f(x)的值域是-12,1,則22a+676,即32a,所以6a2,即a的取值范圍是6,2.