專題升級(jí)訓(xùn)練 三角函數(shù)的圖象與性質(zhì)(時(shí)間:60分鐘滿分:100分)一、選擇題(本大題共6小題,每小題6分,共36分)1.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2B.函數(shù)f(x)在區(qū)間上是增函數(shù)來(lái)源:C.函數(shù)f(x)的圖象關(guān)于直線x=0對(duì)稱D.函數(shù)f(x)是奇函數(shù)2.(20xx·山東青島模擬,6)若當(dāng)x=時(shí),函數(shù)f(x)=Asin(x+)(A>0)取得最小值,則函數(shù)y=f是()A.奇函數(shù)且圖象關(guān)于點(diǎn)對(duì)稱B.偶函數(shù)且圖象關(guān)于點(diǎn)(,0)對(duì)稱C.奇函數(shù)且圖象關(guān)于直線x=對(duì)稱D.偶函數(shù)且圖象關(guān)于點(diǎn)對(duì)稱3.已知角的終邊過(guò)點(diǎn)P(x,-3),且cos=,則sin的值為()A.-B.C.-或-1D.-4.要得到函數(shù)y=sin 2x的圖象,只需將函數(shù)y=sin的圖象()A.向右平移個(gè)單位長(zhǎng)度B.向左平移個(gè)單位長(zhǎng)度C.向右平移個(gè)單位長(zhǎng)度D.向左平移個(gè)單位長(zhǎng)度5.函數(shù)f(x)=Asin(x+)的部分圖象如圖所示,則,的值分別為()A.2,0B.2,C.2,-D.2,6.已知函數(shù)f(x)=cos x+x,x,sin x0=,x0,那么下面命題中真命題的序號(hào)是()f(x)的最大值為f(x0)f(x)的最小值為f(x0)f(x)在上是增函數(shù)f(x)在上是增函數(shù)A.B.C.D.二、填空題(本大題共3小題,每小題6分,共18分)7.函數(shù)y=sinx(>0)的圖象向左平移個(gè)單位后如圖所示,則的值是. 8.設(shè)函數(shù)f(x)=2sin,若對(duì)任意xR,都有f(x1)f(x)f(x2)成立,則|x1-x2|的最小值為. 9.已知函數(shù)f(x)=sin(x>0)的圖象與x軸的交點(diǎn)從左到右依次為(x1,0),(x2,0),(x3,0),則數(shù)列xn的前4項(xiàng)和為. 三、解答題(本大題共3小題,共46分.解答應(yīng)寫出必要的文字說(shuō)明、證明過(guò)程或演算步驟)10.(本小題滿分15分)已知函數(shù)y=cos2x+asin x-a2+2a+5有最大值2,試求實(shí)數(shù)a的值.11.(本小題滿分15分)(20xx·北京海淀模擬,15)已知函數(shù)f(x)=2-(sin x-cos x)2.來(lái)源:(1)求f的值和f(x)的最小正周期;(2)求函數(shù)f(x)在區(qū)間上的最大值和最小值.12.(本小題滿分16分)已知定義在區(qū)間上的函數(shù)y=f(x)的圖象關(guān)于直線x=-對(duì)稱,當(dāng)x時(shí),函數(shù)f(x)=Asin(x+)的圖象如圖所示. (1)求函數(shù)y=f(x)在上的表達(dá)式;(2)求方程f(x)=的解.#1.D解析:f(x)=sin=-cos x,A,B,C均正確,故錯(cuò)誤的是D.2.C解析:由已知可知+=2k-,kZ,即=2k-,kZ,又y=f=Asin=-Asin x,y=f是奇函數(shù)且關(guān)于x=對(duì)稱,故選C.3.C解析:角的終邊過(guò)點(diǎn)P(x,-3),cos=,解得x=0或x2=7,sin=-或-1.4.B解析:y=sin=sin 2x-,故要得到函數(shù)y=sin 2x的圖象,只需將函數(shù)y=sin的圖象向左平移個(gè)單位長(zhǎng)度.5.D解析:由圖象得T=,則T=,=2.當(dāng)2x+=+2k,kZ時(shí),函數(shù)取最大值,由2×+=+2k,kZ得=+2k.又|<,=.6.A解析:因?yàn)閟in x0=,x0,所以x0=.函數(shù)的導(dǎo)數(shù)為f'(x)=-sin x,由f'(x)=-sin x>0,解得sin x<.又因?yàn)閤,所以-<x<,此時(shí)函數(shù)單調(diào)遞增,由f'(x)=-sin x<0,解得sin x>,又因?yàn)閤,所以<x<,此時(shí)函數(shù)單調(diào)遞減,所以正確,選A.7.2解析:由題中圖象可知T=,T=,=2.8.2解析:若對(duì)任意xR,都有f(x1)f(x)f(x2)成立,則f(x1)f(x)min且f(x2)f(x)max,當(dāng)且僅當(dāng)f(x1)=f(x)min,f(x2)=f(x)max,|x1-x2|的最小值為f(x)=2sin的半個(gè)周期,即|x1-x2|min=2.9.26解析:令f(x)=sin=0,則x+=k,x=3k-1(kN*),x1+x2+x3+x4=3(1+2+3+4)-4=26.10.解:y=-sin2x+asin x-a2+2a+6,令sin x=t,t-1,1.y=-t2+at-a2+2a+6,對(duì)稱軸為t=,當(dāng)<-1,即a<-2時(shí),-1,1是函數(shù)y的遞減區(qū)間,ymax=-a2+a+5=2,得a2-a-3=0,a=,與a<-2矛盾;當(dāng)>1,即a>2時(shí),-1,1是函數(shù)y的遞增區(qū)間,ymax=-a2+3a+5=2,得a2-3a-3=0,a=,而a>2,即a=;當(dāng)-11,即-2a2時(shí),ymax=-a2+2a+6=2,得3a2-8a-16=0,a=4,或-,而-2a2,即a=-;a=-.11.解:(1)因?yàn)閒(x)=2-(sin x-cos x)2=2-(3sin2x+cos2x-2sin xcos x)=2-(1+2sin2x-sin 2x)=1-2sin2x+sin 2x=cos 2x+sin 2x=2sin,所以f=2sin=2sin,所以f(x)的最小正周期為T=.(2)當(dāng)x時(shí),2x,所以當(dāng)x=-時(shí),函數(shù)取得最小值f=-1,當(dāng)x=時(shí),函數(shù)取得最大值f=2.12.解:(1)當(dāng)x時(shí),A=1,T=2,=1.且f(x)=sin(x+)過(guò)點(diǎn),則+=+2k,kZ,來(lái)源:=+2k,kZ.-<<,=.f(x)=sin.當(dāng)-x<-時(shí),-x-,f=sin,而函數(shù)y=f(x)的圖象關(guān)于直線x=-對(duì)稱,則f(x)=f,來(lái)源:即f(x)=sin=-sin x,-x<-.f(x)=(2)當(dāng)-x時(shí),x+,由f(x)=sin,得x+,x=-.當(dāng)-x<-時(shí),由f(x)=-sin x=,sin x=-,得x=-或-.來(lái)源:x=-或-或-.