新版高考數(shù)學(xué)復(fù)習(xí) 第三章 第三節(jié)

1 1課時(shí)提升作業(yè)(十九)一、選擇題1.(20xx·福州模擬)已知函數(shù)f(x)=3cos(2x-4)在0,2上的最大值為M,最小值為m,則M+m等于( )(A)0(B)3+322(C)3-322(D)322.(20xx·岳陽(yáng)模擬)函數(shù)y=-12cos2x+12的遞增區(qū)間是( )(A)(k,k+2)(kZ)(B)(k+2,k+)(kZ)(C)(2k,2k+)(kZ)(D)(2k+,2k+2)(kZ)3.已知函數(shù)f(x)=sin(2x-6),若存在a(0,),使得f(x+a)=f(x-a)恒成立,則a的值是( )(A)6(B)3(C)4(D)24.(20xx·咸陽(yáng)模擬)已知函數(shù)y=Asin(x+)在同一周期內(nèi),當(dāng)x=3時(shí)有最大值2,當(dāng)x=0時(shí)有最小值-2,那么函數(shù)的解析式為( )(A)y=2sin32x(B)y=2sin(3x+2)(C)y=2sin(3x-2)(D)y=12sin3x5.(20xx·景德鎮(zhèn)模擬)下列命題正確的是( )(A)函數(shù)y=sin(2x+3)在區(qū)間(-3,6)內(nèi)單調(diào)遞增(B)函數(shù)y=cos4x-sin4x的最小正周期為2(C)函數(shù)y=cos(x+3)的圖像是關(guān)于點(diǎn)(6,0)成中心對(duì)稱(chēng)的圖形(D)函數(shù)y=tan(x+3)的圖像是關(guān)于直線(xiàn)x=6成軸對(duì)稱(chēng)的圖形6.(20xx·銅川模擬)已知函數(shù)f(x)=f(-x),且當(dāng)x(-2,2)時(shí),f(x)=x+sinx,設(shè)a=f(1),b=f(2),c=f(3),則()(A)a<b<c(B)b<c<a(C)c<b<a(D)c<a<b7.函數(shù)y=2sin(2x+3)的圖像關(guān)于點(diǎn)P(x0,0)對(duì)稱(chēng),若x0-2,0,則x0等于()(A)-2(B)-6(C)-4(D)-38.函數(shù)y=lg(sinx)+cosx-22的定義域?yàn)?)(A)(2k,2k+2(kZ)(B)(2k,2k+34(kZ)(C)(2k,2k+4(kZ)(D)2k,2k+4(kZ)9.(20xx·撫州模擬)設(shè)f(x)=xsinx,x-2,2,若f(x1)>f(x2),則()(A)x1+x2>0(B)x12>x22(C)x1>x2(D)x1<x210.(20xx·西安模擬)已知函數(shù)y=sin(4-2x),則其圖像的下列結(jié)論中,正確的是()(A)關(guān)于點(diǎn)(-8,1)中心對(duì)稱(chēng)(B)關(guān)于直線(xiàn)x=8軸對(duì)稱(chēng)(C)向左平移8后得到奇函數(shù)(D)向左平移8后得到偶函數(shù)二、填空題11.(20xx·宿州模擬)若函數(shù)y=a-bsin(4x-3)(b>0)的最大值是5,最小值是1,則a2-b2=.12.(能力挑戰(zhàn)題)已知直線(xiàn)y=b(b<0)與曲線(xiàn)f(x)=sin(2x+2)在y軸右側(cè)依次的三個(gè)交點(diǎn)的橫坐標(biāo)成等比數(shù)列,則b的值是.13.給出如下五個(gè)結(jié)論:存在(0,2),使sin+cos=13;存在區(qū)間(a,b),使y=cosx為減少的而sinx<0;y=tanx在其定義域內(nèi)為增加的;y=cos2x+sin(2-x)既有最大值和最小值,又是偶函數(shù);y=sin|2x+6|的最小正周期為.其中正確結(jié)論的序號(hào)是 .14.對(duì)于函數(shù)f(x)=sinx,sinxcosx,cosx,sinx>cosx,給出下列四個(gè)命題:該函數(shù)是以為最小正周期的周期函數(shù);當(dāng)且僅當(dāng)x=+k(kZ)時(shí),該函數(shù)取得最小值-1;該函數(shù)的圖像關(guān)于x=54+2k(kZ)對(duì)稱(chēng);當(dāng)且僅當(dāng)2k<x<2+2k(kZ)時(shí),0<f(x)22.其中正確命題的序號(hào)是.(請(qǐng)將所有正確命題的序號(hào)都填上)三、解答題15.(能力挑戰(zhàn)題)已知a>0,函數(shù)f(x)=-2asin(2x+6)+2a+b,當(dāng)x0,2時(shí),-5f(x)1.(1)求常數(shù)a,b的值.(2)設(shè)g(x)=f(x+2)且lgg(x)>0,求g(x)的單調(diào)區(qū)間.答案解析1.【解析】選C.由x0,2得2x-4-4,34,故M=f(8)=3cos 0=3,m=f(2)=3cos34=-322,故M+m=3-322.2.【解析】選A.由2k<2x<2k+,kZ得,k<x<k+2,kZ.所以函數(shù)y=-12cos2x+12的遞增區(qū)間是(k,k+2)(kZ).3.【解析】選D.因?yàn)楹瘮?shù)滿(mǎn)足f(x+a)=f(x-a),所以函數(shù)是周期函數(shù),且周期為2a,又a(0,),所以2a=22,所以a=2.【方法技巧】對(duì)周期函數(shù)的理解(1)周期函數(shù)定義中的等式:f(x+T)=f(x)是定義域內(nèi)的恒等式,即對(duì)定義域內(nèi)的每個(gè)x值都成立,若只是存在個(gè)別x滿(mǎn)足等式的常數(shù)T不是周期.(2)每個(gè)周期函數(shù)的定義域是一個(gè)無(wú)限集,其周期有無(wú)窮多個(gè),對(duì)于周期函數(shù)y=f(x),T是周期,則kT(kZ,k0)也是周期,但并非所有周期函數(shù)都有最小正周期.【變式備選】已知函數(shù)f(x)=Asin(x+)(A>0,>0)滿(mǎn)足條件f(x+12)+f(x)=0,則的值為( )(A)2(B) (C)2 (D)4【解析】選A.由f(x+12)+f(x)=0得f(x+12)=-f(x),所以f(x+1)=f(x),故函數(shù)的周期是1,又由2=1得=2.4.【解析】選C.由條件知A=2,T2=3,所以T=23,因此=2T=3,所以f(x)=2sin(3x+).把x=0,y=-2代入上式得-2=2sin,得sin=-1,所以=2k-2(kZ),因此f(x)=2sin(3x+2k-2)(kZ)=2sin(3x-2).5.【解析】選C.對(duì)于A,當(dāng)x(-3,6)時(shí),2x+3(-3,23),故函數(shù)y=sin(2x+3)不單調(diào),故A錯(cuò)誤;對(duì)于B,y=cos4x-sin4x=(cos2x-sin2x)(cos2x+sin2x)=cos2x-sin2x=cos2x,最小正周期為,故錯(cuò)誤;對(duì)于C,當(dāng)x=6時(shí),cos(6+3)=0,所以(6,0)是對(duì)稱(chēng)中心,故C正確;對(duì)于D,正切函數(shù)的圖像不是軸對(duì)稱(chēng)圖形,故錯(cuò)誤.6.【思路點(diǎn)撥】利用函數(shù)y=f(x)的單調(diào)性比較.【解析】選D.由條件知f(x)=x+sinx在(-2,2)上是增加的,又b=f(2)=f(-2),c=f(3)=f(-3),而1,-2,-3(-2,2),且-3<1<-2,所以f(-3)<f(1)<f(-2),即c<a<b.7.【解析】選B.由題意可知2x0+3=k,kZ,故x0=k2-6,kZ,故k=0時(shí),x0=-6-2,0,故選B.8.【解析】選C.由sinx>0,cosx-220,得2k<x<2k+,kZ,2k-4x2k+4,kZ,所以,2k<x2k+4,kZ,故函數(shù)的定義域?yàn)?2k,2k+4(kZ).9.【思路點(diǎn)撥】根據(jù)f(x)=xsinx的奇偶性和在0,2上的單調(diào)性求解.【解析】選B.由f(-x)=-xsin(-x)=xsinx=f(x)知,函數(shù)y=f(x)為偶函數(shù).又f'(x)=sinx+xcosx,當(dāng)x(0,2)時(shí),f'(x)>0,故f(x)在0,2上是增加的.因?yàn)閒(x1)>f(x2),故f(|x1|)>f(|x2|),所以|x1|>|x2|,因此x12>x22.10.【解析】選C.對(duì)于A,由題意知函數(shù)圖像的對(duì)稱(chēng)中心應(yīng)在x軸上,故A不正確.對(duì)于B,由4-2x=k+2(kZ),得x=-k2-8(kZ),故B不正確.對(duì)于C,將函數(shù)向左平移8后得到f(x)=sin4-2(x+8)=sin(-2x)=-sin2x,為奇函數(shù),故C正確.從而D不正確.11.【解析】-1sin(4x-3)1,b>0,-b-bsin(4x-3)b,a-ba-bsin(4x-3)a+b,由題意知a-b=1,a+b=5,解得a=3,b=2.a2-b2=5.答案:512.【思路點(diǎn)撥】化簡(jiǎn)函數(shù)式之后數(shù)形結(jié)合可解.【解析】設(shè)三個(gè)交點(diǎn)的橫坐標(biāo)依次為x1,x2,x3,由圖及題意有:f(x)=sin(2x+2)=cos2x.且x1+x2=,x2+x3=2,x22=x1x3,解得x2=23,所以b=f(23)=-12.答案:-1213.【解析】中(0,2)時(shí),如圖,由三角函數(shù)線(xiàn)知OM+MP>1,得sin+cos>1,故錯(cuò).由y=cosx的減區(qū)間為(2k,2k+)(kZ),故sinx>0,因而錯(cuò).正切函數(shù)的單調(diào)區(qū)間是(k-2,k+2),kZ.故y=tanx在定義域內(nèi)不單調(diào),故錯(cuò).y=cos2x+sin(2-x)=cos2x+cosx=2cos2x+cosx-1=2(cosx+14)2-98.ymax=2,ymin=-98.故函數(shù)既有最大值和最小值,又是偶函數(shù),故正確.結(jié)合圖像可知y=sin|2x+6|不是周期函數(shù),故錯(cuò).答案:14.【解析】畫(huà)出函數(shù)f(x)的圖像.由圖像可得函數(shù)的最小正周期為2,故錯(cuò)誤;當(dāng)x=+2k(kZ)或x=32+2k(kZ)時(shí),函數(shù)取得最小值-1,故不正確;結(jié)合圖像可得正確.答案:15.【解析】(1)x0,2,2x+66,76.sin(2x+6)-12,1,-2asin(2x+6)-2a,a.f(x)b,3a+b.又-5f(x)1,b=-5,3a+b=1,因此a=2,b=-5.(2)由(1)得a=2,b=-5,f(x)=-4sin(2x+6)-1,g(x)=f(x+2)=-4sin(2x+76)-1=4sin(2x+6)-1,又由lgg(x)>0得g(x)>1,4sin(2x+6)-1>1,sin(2x+6)>12,2k+6<2x+6<2k+56,kZ,其中當(dāng)2k+6<2x+62k+2,kZ時(shí),g(x)是增加的,即k<xk+6,kZ.g(x)的遞增區(qū)間為(k,k+6,kZ.又當(dāng)2k+2<2x+6<2k+56,kZ時(shí),g(x)是減少的,即k+6<x<k+3,kZ.g(x)的遞減區(qū)間為(k+6,k+3),kZ.