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

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)=的最小正周期是()A.B.C.D.22.已知函數(shù)f(x)=2sin(x+)對(duì)任意x都有f=f,則f等于()A.2或0B.-2或2C.0D.-2或03.已知函數(shù)f(x)=sin(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=對(duì)稱D.函數(shù)f(x)在區(qū)間上是增加的4.當(dāng)x=時(shí),函數(shù)f(x)=sin(x+)取得最小值,則函數(shù)y=f()A.是奇函數(shù),且圖像關(guān)于點(diǎn)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=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)=2sin(>0)的圖像與函數(shù)g(x)=cos(2x+)的圖像的對(duì)稱中心完全相同,則為()A.B.-C.D.-6.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)“=”是“函數(shù)y=cos 2x與函數(shù)y=sin(2x+)在區(qū)間上的單調(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=tan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)間上遞增,在區(qū)間上遞減,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)為的交點(diǎn),則的值是. 綜合提升組11.(2018天津,理6)將函數(shù)y=sin的圖像向右平移個(gè)單位長(zhǎng)度,所得圖像對(duì)應(yīng)的函數(shù)()A.在區(qū)間上遞增B.在區(qū)間上遞減C.在區(qū)間上遞增D.在區(qū)間上遞減12.已知函數(shù)f(x)=sin(x+),A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.,kZB.,kZC.,kZD.,kZ13.函數(shù)f(x)=sin的遞減區(qū)間為. 14.設(shè)函數(shù)f(x)=Asin(x+)A>0,>0,|<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=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)=sin+1的圖像在區(qū)間上恰有一條對(duì)稱軸和一個(gè)對(duì)稱中心,則實(shí)數(shù)的取值范圍為()A.B.C.D.16.(2018江西南昌三模,9)將函數(shù)f(x)=sin的圖像上所有點(diǎn)的橫坐標(biāo)壓縮為原來(lái)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)=,故f(x)的最小正周期為.2.B由f=f知,函數(shù)圖像關(guān)于x=對(duì)稱,f是函數(shù)f(x)的最大值或最小值.故選B.3.Cf(x)=sin=-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)在上是增加的,D正確.故選C.4.C由題意,得sin =-1,=2k-(kZ).f(x)=sin=sin.y=f=sin(-x)=-sin x.y=f是奇函數(shù),且圖像關(guān)于直線x=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)=2sin,由2x+=k,kZ,即x=-,kZ,f(x)的對(duì)稱中心為,kZ,g(x)的對(duì)稱中心為,kZ,g=cos=cos=±cos=0,kZ,即-=k+,kZ,則=k+,kZ,當(dāng)k=-1時(shí),=-+=-,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)間上遞減.當(dāng)=時(shí),函數(shù)y=sin,x,可得2x+.函數(shù)y=sin在區(qū)間上遞減.當(dāng)=+2時(shí),函數(shù)y=sin(2x+)=sin在區(qū)間上遞減,“=”是函數(shù)“y=cos 2x與函數(shù)y=sin(2x+)在區(qū)間上的單調(diào)性相同”的充分不必要條件.故選A.8.(kZ)2由k-<+<k+,kZ,得2k-<x<2k+,kZ.最小正周期T=2.9.f(x)=sin x(>0)過原點(diǎn),當(dāng)0x,即0x時(shí),y=sin x是增加的;當(dāng)x,即x時(shí),y=sin x是減少的.由題意知=,=.10.由題意cos=sin,即sin=,+=k+(-1)k·(kZ),因?yàn)?<,所以=.11.A將函數(shù)y=sin的圖像向右平移個(gè)單位長(zhǎng)度,所得圖像對(duì)應(yīng)的函數(shù)解析式為y=sin+=sin 2x.當(dāng)-+2k2x+2k,kZ,即-+kx+k,kZ時(shí),y=sin 2x遞增.當(dāng)+2k2x+2k,kZ,即+kx+k,kZ時(shí),y=sin 2x遞減.結(jié)合選項(xiàng),可知y=sin 2x在區(qū)間上遞增.故選A.12.D由題意,得(2)2+=42,即12+=16,求得=.再根據(jù)·+=k,kZ,且-<<,可得=-,則f(x)=sin.令2k-x-2k+,kZ,求得4k-x4k+,kZ,故f(x)的遞增區(qū)間為,4k+,kZ,故選D.13.(kZ)由已知函數(shù)為y=-sin,欲求函數(shù)的遞減區(qū)間,只需求y=sin的遞增區(qū)間.由2k-2x-2k+,kZ,得k-xk+,kZ.故所給函數(shù)的遞減區(qū)間為k-,k+(kZ).14.,kZ由題意,得A=3,T=,=2,f(x)=3sin(2x+).又f=3或f=-3,2×+=k+,kZ,=+k,kZ.|<,=,f(x)=3sin.令-+2k2x+2k,kZ,化簡(jiǎn),得-+kx+k,kZ,函數(shù)f(x)的遞增區(qū)間為,kZ.15.C由題意,知x,2x+,+,函數(shù)f(x)的圖象在區(qū)間0,上恰有一條對(duì)稱軸和一個(gè)對(duì)稱中心,+,+,+,即+<,即<.故選C.16.C由題意知g(x)=sin,x1,x2-2,2,2x1+,2x2+-4+,4+.g(x1)+g(x2)=2,g(x1)=g(x2)=1,要使x1-x2的值最大,則2x1+=2+,2x2+=-4+,-=2(x1-x2)=-=6,x1-x2=3.9