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1、課時(shí)跟蹤訓(xùn)練(十)基礎(chǔ)鞏固一、選擇題1(20xx湖北省仙桃中學(xué)月考)計(jì)算2log63log64的結(jié)果是()Alog62B2Clog63D3解析2log63log64log69log64log6362.故選B.答案B2(20xx臨川二中月考)若函數(shù)f(x)logax(0a1)在區(qū)間a,2a上的最大值是最小值的3倍,則a的值為()A.BCD解析0a0且a1)滿足ff,則f0的解為()A0x1Bx1Dx0解析因?yàn)楹瘮?shù)f(x)logax(a0且a1)在(0,)上為單調(diào)函數(shù),而f,所以f(x)logax(a0且a1)在(0,)上單調(diào)遞減,從而f0011,所以01.故選C.答案C4(20xx江西南昌調(diào)研)
2、函數(shù)y2log4(1x)的圖象大致是()解析函數(shù)y2log4(1x)的定義域?yàn)?,1),排除A、B;又函數(shù)y2log4(1x)在定義域內(nèi)單調(diào)遞減,排除D,選C.答案C5(20xx河南鄭州質(zhì)量預(yù)測(cè))已知函數(shù)f(x)則ff(1)f的值是()A5B3C1D解析由題意可知f(1)log210,ff(1)f(0)3012,f313log321213,所以ff(1)f5.答案A6若lgxlgy2lg(2x3y),則log的值為()A0B2C0或2D或1解析依題意,可得lg(xy)lg(2x3y)2,即xy4x212xy9y2,整理得:421390,解得1或.x0,y0,2x3y0,log2.選B.答案B二
3、、填空題7(20xx杭州調(diào)研)計(jì)算:log2_;_.解析log2log2log221;答案38設(shè)f(x)lg是奇函數(shù),則使f(x)0的x的取值范圍是_解析由f(x)是奇函數(shù)可得a1,f(x)lg,定義域?yàn)?1,1)由f(x)0,可得01,1x0,且a1),若f(x)1在區(qū)間1,2上恒成立,則實(shí)數(shù)a的取值范圍是_解析當(dāng)a1時(shí),f(x)loga(8ax)在1,2上是減函數(shù),由f(x)1在區(qū)間1,2上恒成立,則f(x)minloga(82a)1,解之得1a.當(dāng)0a1在區(qū)間1,2上恒成立,則f(x)minloga(8a)1,且82a0.a4,且a4,故不存在綜上可知,實(shí)數(shù)a的取值范圍是.答案三、解答題
4、10(20xx日照模擬)已知函數(shù)f(x)log(a為常數(shù))(1)若常數(shù)a0,當(dāng)0a2時(shí),解得x;當(dāng)a0時(shí),解得x1.故當(dāng)0a2時(shí),f(x)的定義域?yàn)閤;當(dāng)a0時(shí),f(x)的定義域?yàn)閤.(2)令u,因?yàn)閒(x)logu為減函數(shù),故要使f(x)在(2,4)上是減函數(shù),只需u(x)a在(2,4)上單調(diào)遞增且為正故由得1af(),則a的取值范圍是()A(,)B(0,)C(,)D(1,)解析f(x)是定義在R上的偶函數(shù),且在區(qū)間(,0上單調(diào)遞增,f(x)在區(qū)間0,)上單調(diào)遞減又f()f(),f(2 log3a)f()2 log3a 0,f(x)在區(qū)間0,)上單調(diào)遞減,02log3a log3a0a,故選
5、B.答案B12(20xx福建省福州市高三質(zhì)量檢測(cè))已知aln8,bln5,clnln,則()AabcBacbCcabDcba解析因?yàn)閍ln8,bln5,clnln,所以aln,bln,clnln.又對(duì)數(shù)函數(shù)ylnx在(0,)上為單調(diào)遞增函數(shù),由,得lnlnln,所以ac0時(shí),由|f(a)|2可得|1log2a|2,所以1log2a2或1log2a2,解得01,b1,則alnb的最大值為_(kāi)解析由題意知b,則alnbaa(2lna),令ta(2lna)(t0),則lntlna(2lna)(lna)22lna(lna1)211,當(dāng)lna1時(shí),“”成立,此時(shí)lnt1,所以te,即alnb的最大值為e.
6、 答案e15(20xx山西運(yùn)城期中)已知函數(shù)f(x)(log2x2).(1)當(dāng)x1,4時(shí),求該函數(shù)的值域;(2)若f(x)mlog2x對(duì)x4,16恒成立,求m的取值范圍解(1)令tlog2x,t0,2,f(t)(t2)(t2)(t1),f(0)f(t)f,f(t)1,故該函數(shù)的值域?yàn)?(2)同(1)令tlog2x,x4,16,t2,4,(t2)(t1)mt,t32m恒成立令g(t)t,其在(,)上單調(diào)遞增,g(t)g(4),32m,m.16(20xx瀘州二診)已知函數(shù)f(x)lg(a0)為奇函數(shù),函數(shù)g(x)1x(bR)(1)求函數(shù)f(x)的定義域;(2)當(dāng)x時(shí),關(guān)于x的不等式f(x)lgg(
7、x)有解,求b的取值范圍解(1)由f(x)lg(a0)為奇函數(shù),得f(x)f(x)0,即lglglg0,所以1,解得a1(a1舍去),故f(x)lg,所以f(x)的定義域是(1,1)(2)不等式f(x)lgg(x)有解,等價(jià)于1x有解,即bx2x在上有解,故只需b(x2x)min,函數(shù)yx2x2在區(qū)間上單調(diào)遞增,所以ymin2,所以b的取值范圍是.延伸拓展(20xx東北三省四市一模)已知點(diǎn)(n,an),(nN*)在yex的圖象上,若滿足Tnlna1lna2lnank時(shí)n的最小值為5,則k的取值范圍是()Ak15Bk10C10k15D10k15解析由題意得,anen,Tnlna1lna2lnan12n,T4kT5,10k15,故選C.答案C