2020版高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練19 三角函數(shù)的圖像與性質(zhì) 理 北師大版.doc

課時(shí)規(guī)范練19三角函數(shù)的圖像與性質(zhì)基礎(chǔ)鞏固組1.函數(shù)f(x)=sinx2cosx2的最小正周期是()A.4B.2C.D.22.已知函數(shù)f(x)=2sin(x+)對(duì)任意x都有f6+x=f6-x,則f6等于()A.2或0B.-2或2C.0D.-2或03.已知函數(shù)f(x)=sin2x+32(xR),下面結(jié)論錯(cuò)誤的是()A.函數(shù)f(x)的最小正周期為B.函數(shù)f(x)是偶函數(shù)C.函數(shù)f(x)的圖像關(guān)于直線x=4對(duì)稱D.函數(shù)f(x)在區(qū)間0,2上是增加的4.當(dāng)x=4時(shí),函數(shù)f(x)=sin(x+)取得最小值,則函數(shù)y=f34-x()A.是奇函數(shù),且圖像關(guān)于點(diǎn)2,0對(duì)稱B.是偶函數(shù),且圖像關(guān)于點(diǎn)(,0)對(duì)稱C.是奇函數(shù),且圖像關(guān)于直線x=2對(duì)稱D.是偶函數(shù),且圖像關(guān)于直線x=對(duì)稱5.(2018河南六市聯(lián)考一,5)已知函數(shù)f(x)=2sinx+6(>0)的圖像與函數(shù)g(x)=cos(2x+)|<2的圖像的對(duì)稱中心完全相同,則為()A.6B.-6C.3D.-36.函數(shù)y=xcos x-sin x的部分圖像大致為()7.(2018四川雙流中學(xué)考前模擬)“=34”是“函數(shù)y=cos 2x與函數(shù)y=sin(2x+)在區(qū)間0,4上的單調(diào)性相同”的()A.充分不必要條件B.必要不充分條件C.充要條件D.既不充分也不必要條件8.函數(shù)y=tanx2+3的遞增區(qū)間是,最小正周期是.9.若函數(shù)f(x)=sin x(>0)在區(qū)間0,3上遞增,在區(qū)間3,2上遞減,則=.10.已知函數(shù)y=cos x與y=sin(2x+)(0<),它們的圖像有一個(gè)橫坐標(biāo)為3的交點(diǎn),則的值是.綜合提升組11.(2018天津,理6)將函數(shù)y=sin2x+5的圖像向右平移10個(gè)單位長(zhǎng)度,所得圖像對(duì)應(yīng)的函數(shù)()A.在區(qū)間34,54上遞增B.在區(qū)間34,上遞減C.在區(qū)間54,32上遞增D.在區(qū)間32,2上遞減12.已知函數(shù)f(x)=3sin(x+)>0,-2<<2,A13,0為f(x)圖像的對(duì)稱中心,B,C是該圖像上相鄰的最高點(diǎn)和最低點(diǎn),若BC=4,則f(x)的遞增區(qū)間是()A.2k-23,2k+43,kZB.2k-23,2k+43,kZC.4k-23,4k+43,kZD.4k-23,4k+43,kZ13.函數(shù)f(x)=sin-2x+3的遞減區(qū)間為.14.設(shè)函數(shù)f(x)=Asin(x+)A>0,>0,|<2與直線y=3的交點(diǎn)的橫坐標(biāo)構(gòu)成以為公差的等差數(shù)列,且x=6是f(x)圖像的一條對(duì)稱軸,則函數(shù)f(x)的遞增區(qū)間為.創(chuàng)新應(yīng)用組15.(2018河北衡水中學(xué)考前仿真,6)已知函數(shù)f(x)=2sin2x+4+1的圖像在區(qū)間0,12上恰有一條對(duì)稱軸和一個(gè)對(duì)稱中心,則實(shí)數(shù)的取值范圍為()A.38,58B.38,58C.34,54D.34,5416.(2018江西南昌三模,9)將函數(shù)f(x)=sinx+6的圖像上所有點(diǎn)的橫坐標(biāo)壓縮為原來的,縱坐標(biāo)保持不變,得到g(x)的圖像,若g(x1)+g(x2)=2,且x1,x2-2,2,則x1-x2的最大值為()A.B.2C.3D.4參考答案課時(shí)規(guī)范練19三角函數(shù)的圖像與性質(zhì)1.C由已知得f(x)=|sinx|2,故f(x)的最小正周期為.2.B由f6+x=f6-x知,函數(shù)圖像關(guān)于x=對(duì)稱,f6是函數(shù)f(x)的最大值或最小值.故選B.3.Cf(x)=sin2x+32=-cos 2x,故其最小正周期為,A正確;易知函數(shù)f(x)是偶函數(shù),B正確;由函數(shù)f(x)=-cos 2x的圖像可知,函數(shù)f(x)的圖像關(guān)于直線x=不對(duì)稱,C錯(cuò)誤;由函數(shù)f(x)的圖像易知,函數(shù)f(x)在0,2上是增加的,D正確.故選C.4.C由題意,得sin 4+=-1,=2k-34(kZ).f(x)=sinx+2k-34=sinx-34.y=f34-x=sin(-x)=-sin x.y=f34-x是奇函數(shù),且圖像關(guān)于直線x=2對(duì)稱.5.D兩個(gè)函數(shù)圖像的對(duì)稱中心完全相同,則它們的周期相同,=2,即f(x)=2sin2x+6,由2x+6=k,kZ,即x=k2-12,kZ,f(x)的對(duì)稱中心為k2-12,0,kZ,g(x)的對(duì)稱中心為k2-12,0,kZ,gk2-12=cos2k2-12+=cosk-6+=cos-6=0,kZ,即-6=k+2,kZ,則=k+23,kZ,當(dāng)k=-1時(shí),=-+23=-3,故選D.6.C函數(shù)y=f(x)=xcos x-sin x滿足f(-x)=-f(x),即該函數(shù)為奇函數(shù),圖像關(guān)于原點(diǎn)對(duì)稱,故排除B;當(dāng)x=時(shí),y=f()=cos -sin =-<0,故排除A,D.故選C.7.A由題意可得函數(shù)y=cos 2x在區(qū)間0,4上遞減.當(dāng)=34時(shí),函數(shù)y=sin2x+34,x0,4,可得2x+3434,54.函數(shù)y=sin2x+34在區(qū)間0,4上遞減.當(dāng)=34+2時(shí),函數(shù)y=sin(2x+)=sin2x+34在區(qū)間0,4上遞減,“=34”是函數(shù)“y=cos 2x與函數(shù)y=sin(2x+)在區(qū)間0,4上的單調(diào)性相同”的充分不必要條件.故選A.8.2k-53,2k+3(kZ)2由k-<+<k+,kZ,得2k-53<x<2k+,kZ.最小正周期T=12=2.9.f(x)=sin x(>0)過原點(diǎn),當(dāng)0x2,即0x2時(shí),y=sin x是增加的;當(dāng)2x32,即2x32時(shí),y=sin x是減少的.由題意知2=3,=32.10.由題意cos=sin23+,即sin23+=12,23+=k+(-1)k6(kZ),因?yàn)?<,所以=6.11.A將函數(shù)y=sin2x+5的圖像向右平移10個(gè)單位長(zhǎng)度,所得圖像對(duì)應(yīng)的函數(shù)解析式為y=sin2x-10+5=sin 2x.當(dāng)-2+2k2x2+2k,kZ,即-4+kx4+k,kZ時(shí),y=sin 2x遞增.當(dāng)2+2k2x32+2k,kZ,即4+kx34+k,kZ時(shí),y=sin 2x遞減.結(jié)合選項(xiàng),可知y=sin 2x在區(qū)間34,54上遞增.故選A.12.D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,則f(x)=3sin2x-6.令2k-22x-62k+2,kZ,求得4k-23x4k+43,kZ,故f(x)的遞增區(qū)間為4k-23,4k+43,kZ,故選D.13.k-12,k+512(kZ)由已知函數(shù)為y=-sin2x-3,欲求函數(shù)的遞減區(qū)間,只需求y=sin2x-3的遞增區(qū)間.由2k-22x-32k+2,kZ,得k-12xk+512,kZ.故所給函數(shù)的遞減區(qū)間為k-12,k+512(kZ).14.-3+k,6+k,kZ由題意,得A=3,T=,=2,f(x)=3sin(2x+).又f6=3或f6=-3,26+=k+2,kZ,=6+k,kZ.|<2,=6,f(x)=3sin2x+6.令-2+2k2x+62+2k,kZ,化簡(jiǎn),得-3+kx6+k,kZ,函數(shù)f(x)的遞增區(qū)間為-3+k,6+k,kZ.15.C由題意,知x0,12,2x+,+,函數(shù)f(x)的圖象在區(qū)間0,12上恰有一條對(duì)稱軸和一個(gè)對(duì)稱中心,24,+4,4,+4,324,+4,+42,+4,+4<32,即+4<32,即34<54.故選C.16.C由題意知g(x)=sin2x+6,x1,x2-2,2,2x1+6,2x2+6-4+6,4+6.g(x1)+g(x2)=2,g(x1)=g(x2)=1,要使x1-x2的值最大,則2x1+6=2+2,2x2+6=-4+2,2x1+6-2x2+6=2(x1-x2)=2+2-4+2=6,x1-x2=3.