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1、課時(shí)跟蹤檢測(cè)(十九) 三角函數(shù)的圖象與性質(zhì)一抓基礎(chǔ),多練小題做到眼疾手快1(2019南通調(diào)研)已知函數(shù)ycos ax(a0)的最小正周期為2,則實(shí)數(shù)a_.解析:函數(shù)ycos ax(a0)的最小正周期為2,a1.答案:12(2018南京名校聯(lián)考)函數(shù)ytan x,x的值域是_解析:函數(shù)ytan x在區(qū)間上單調(diào)遞增,所以值域是0,1答案:0,13(2018南京調(diào)研)如圖,已知A,B分別是函數(shù)f(x)sin x(0)在y軸右側(cè)圖象上的第一個(gè)最高點(diǎn)和第一個(gè)最低點(diǎn),且AOB,則該函數(shù)的最小正周期是_解析:連結(jié)AB,設(shè)AB與x軸的交點(diǎn)為C,則由AOB,得COCACB.又OACA,所以AOC是高為的正三角形
2、,從而OC2,所以該函數(shù)的最小正周期是4.答案:44(2018蘇北四市調(diào)研)函數(shù)y3sin xcos x的單調(diào)遞增區(qū)間是_解析:化簡(jiǎn)可得y2sin,由2kx2k(kZ),得2kx2k(kZ),又x,所以函數(shù)的單調(diào)遞增區(qū)間是.答案:5已知函數(shù)f(x)sin,其中x.若f(x)的值域是,則的取值范圍是_解析:若x,則2x2.因?yàn)楫?dāng)2x或2x時(shí),sin,所以要使f(x)的值域是,則2,即2,所以,即的取值范圍是.答案:6下列正確命題的序號(hào)為_(kāi)ytan x為增函數(shù);ytan(x)(0)的最小正周期為;在x,上ytan x是奇函數(shù);在上ytan x的最大值是1,最小值為1.解析:函數(shù)ytan x在定義域
3、內(nèi)不具有單調(diào)性,故錯(cuò)誤;函數(shù)ytan(x)(0)的最小正周期為,故正確;當(dāng)x,時(shí),ytan x無(wú)意義,故錯(cuò)誤;由正切函數(shù)的圖象可知正確答案:二保高考,全練題型做到高考達(dá)標(biāo)1(2018如東中學(xué)檢測(cè))函數(shù)ysin2xsin x1的值域?yàn)開(kāi)解析:由ysin2xsin x1,令tsin x,t1,1,則有yt2t12,畫出函數(shù)圖象如圖所示,從圖象可以看出,當(dāng)t 及t1時(shí),函數(shù)取最值,代入yt2t1,可得y.答案:2設(shè)偶函數(shù)f(x)Asin(x)(A0,0,0)的部分圖象如圖所示,KLM為等腰直角三角形,KML90,KL1,則f_.解析:由題意知,點(diǎn)M到x軸的距離是,根據(jù)題意可設(shè)f(x)cos x,又由
4、題圖知1,所以,所以f(x)cos x,故fcos.答案:3函數(shù)f(x)2sin(x)(0)對(duì)任意x都有ff,則f_.解析:因?yàn)楹瘮?shù)f(x)2sin(x)對(duì)任意x都有ff,所以該函數(shù)圖象關(guān)于直線x對(duì)稱,因?yàn)樵趯?duì)稱軸處對(duì)應(yīng)的函數(shù)值為最大值或最小值,故f2.答案:2或24(2018通州期末)已知f(x)sin(x)(0,0)是R上的偶函數(shù),其圖象關(guān)于M對(duì)稱,在區(qū)間上是單調(diào)函數(shù),則_,_.解析:由f(x)是R上的偶函數(shù),得k,kZ.0,.f(x)sincos x.函數(shù)f(x)的圖象關(guān)于M對(duì)稱,k,kZ,即k,kZ.又f(x)在區(qū)間上是單調(diào)函數(shù),即T,02.故2或.答案:2或5(2019海安模擬)函數(shù)
5、f(x)sin的圖象在區(qū)間上的對(duì)稱軸方程為_(kāi)解析:對(duì)于函數(shù)f(x)sin的圖象,令2xk,kZ,得x,kZ,令k0,可得函數(shù)f(x)在區(qū)間上的對(duì)稱軸方程為x.答案:x6(2018鎮(zhèn)江一中測(cè)試)已知角的終邊經(jīng)過(guò)點(diǎn)P(4,3),函數(shù)f(x)sin(x)(0)圖象的相鄰兩條對(duì)稱軸之間的距離等于,則f_.解析:由于角的終邊經(jīng)過(guò)點(diǎn)P(4,3),所以cos .再根據(jù)函數(shù)f(x)sin(x)(0)圖象的相鄰兩條對(duì)稱軸之間的距離等于,可得2,所以2,所以f(x)sin(2x),所以fsincos .答案:7(2019阜寧中學(xué)檢測(cè))若直線x(|k|1)與函數(shù)ytan的圖象不相交,則k_.解析:直線x(|k|1)
6、與函數(shù)ytan的圖象不相交,等價(jià)于當(dāng)x時(shí),函數(shù)ytan無(wú)意義,即2m,mZ,km,mZ.當(dāng)m0時(shí),k,滿足條件當(dāng)m1時(shí),k,滿足條件當(dāng)m1時(shí),k,不滿足條件故滿足條件的k或.答案:或8(2019常州調(diào)研)如圖,在平面直角坐標(biāo)系xOy中,函數(shù)f(x)sin(x)(0,0)的圖象與x軸的交點(diǎn)A,B,C滿足OAOC2OB,則_.解析:設(shè)函數(shù)f(x)sin(x)(0,0)的圖象與x軸的交點(diǎn)坐標(biāo)分別為A(x1,0),B(x3,0),C(x2,0),則得x33x1,將x33x1代入,得x25x1,所以Tx2x38x1,所以,故f(x)sin.由圖象可知f(x1)0,所以sin0,令k,kZ,得k,kZ.又
7、0,所以.答案:9(2019宿遷中學(xué)調(diào)研)已知函數(shù)f(x)sin 3xcos 3x,xR. (1)求函數(shù)f(x)的單調(diào)遞增區(qū)間;(2)求函數(shù)f(x)在區(qū)間上的最值,并求出取得最值時(shí)x的值解:(1)f(x)sin 3xcos 3x22sin.由2k3x2k(kZ),得x(kZ),故函數(shù)f(x)的單調(diào)遞增區(qū)間為(kZ)(2)x,3x.當(dāng)3x或,即x或時(shí),f(x)min;當(dāng)3x,即x時(shí),f(x)max2.10(2018清江中學(xué)測(cè)試)已知a0,函數(shù)f(x)2asin2ab,當(dāng)x時(shí),5f(x)1.(1)求常數(shù)a,b的值;(2)設(shè)g(x)f且lg g(x)0,求g(x)的單調(diào)區(qū)間解:(1)因?yàn)閤,所以2x
8、.所以sin,又因?yàn)閍0,所以2asin2a,a,所以f(x)b,3ab又因?yàn)?f(x)1,所以b5,3ab1,因此a2,b5.(2)由(1)知a2,b5,所以f(x)4sin1,g(x)f4sin14sin1,又由lg g(x)0,得g(x)1,所以4sin11,所以sin,所以2k2x2k,kZ.當(dāng)2k2x2k,kZ,即kxk,kZ時(shí),g(x)單調(diào)遞增,所以g(x)的單調(diào)遞增區(qū)間為,kZ.當(dāng)2k2x2k,kZ,即kxk,kZ時(shí),g(x)單調(diào)遞減所以g(x)的單調(diào)遞減區(qū)間為,kZ.綜上,g(x)的單調(diào)遞增區(qū)間為,kZ;單調(diào)遞減區(qū)間為,kZ.三上臺(tái)階,自主選做志在沖刺名校1函數(shù)ytan(sin
9、 x)的值域?yàn)開(kāi)解析:因?yàn)?sin x1,所以sin x.又因?yàn)閥tan x在上單調(diào)遞增,所以tan(1)ytan 1,故函數(shù)的值域是tan 1,tan 1答案:tan 1,tan 1 2(2018揚(yáng)州期末)已知函數(shù)f(x)sin(0x),且f()f()(),則_.解析:因?yàn)?x,所以2x,所以由f(x)得2x或,解得x或,由于f()f()(),所以.答案:3(2019揚(yáng)州調(diào)研)已知函數(shù)f(x)1cos 2x2sin2.(1)求f(x)的最小正周期和單調(diào)遞減區(qū)間;(2)若方程f(x)m0在區(qū)間上有兩個(gè)不同的實(shí)數(shù)解,求實(shí)數(shù)m的取值范圍解:(1)f(x)1cos 2x2sin2cos 2xcoscos 2xsin 2x 2sin,T.由2k2x2k,kZ,得kxk,kZ.f(x)的單調(diào)遞減區(qū)間為(kZ)(2)由題意知,函數(shù)yf(x)在區(qū)間上的圖象與直線ym有兩個(gè)不同的交點(diǎn)由(1)知,函數(shù)f(x)在上單調(diào)遞減,在上單調(diào)遞增,f(x)minf2,又f1,f(),當(dāng)2m1時(shí),函數(shù)yf(x)在區(qū)間上的圖象與直線ym有兩個(gè)不同的交點(diǎn),即方程f(x)m0在區(qū)間上有兩個(gè)不同的實(shí)數(shù)解實(shí)數(shù)m的取值范圍為(2,18