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1、課時(shí)跟蹤檢測(cè)(十四)高考基礎(chǔ)題型得分練12017湖南岳陽(yáng)一模下列函數(shù)中,既是奇函數(shù)又存在極值的是()Ayx3 Byln(x)Cyxex Dyx答案:D解析:由題意知,B,C選項(xiàng)中的函數(shù)不是奇函數(shù),A選項(xiàng)中,函數(shù)yx3單調(diào)遞增(無(wú)極值),而D選項(xiàng)中的函數(shù)既為奇函數(shù)又存在極值2已知函數(shù)f(x)的導(dǎo)函數(shù)f(x)ax2bxc的圖象如圖所示,則f(x)的圖象可能是()A BC D答案:D解析:當(dāng)x0時(shí),由導(dǎo)函數(shù)f(x)ax2bxc0時(shí),由導(dǎo)函數(shù)f(x)ax2bxc的圖象可知,導(dǎo)函數(shù)在區(qū)間(0,x1)上的值是大于0的,則在此區(qū)間內(nèi)函數(shù)f(x)單調(diào)遞增3函數(shù)yx2ln x的單調(diào)遞減區(qū)間為()A(0,1)B(
2、0,) C(1,)D(0,2)答案:A解析:對(duì)于函數(shù)yx2ln x,易得其定義域?yàn)閤|x0,yx,令0,所以x210,解得0x1,即函數(shù)yx2ln x的單調(diào)遞減區(qū)間為(0,1)42017江西南昌模擬已知函數(shù)f(x)(2xx2)ex,則()Af()是f(x)的極大值也是最大值Bf()是f(x)的極大值但不是最大值Cf()是f(x)的極小值也是最小值Df(x)沒(méi)有最大值也沒(méi)有最小值答案:A解析:由題意,得f(x)(22x)ex(2xx2)ex(2x2)ex,當(dāng)x0,函數(shù)f(x)單調(diào)遞增;當(dāng)x時(shí),f(x)0,在x處取得極小值f()2(1)e0.又當(dāng)x0時(shí),f(x)(2xx2)ex0;當(dāng)x(1,e時(shí),
3、f(x)0,所以f(x)的單調(diào)遞增區(qū)間是(0,1),單調(diào)遞減區(qū)間是(1,e,所以當(dāng)x1時(shí),f(x)取得最大值ln 111.6已知函數(shù)f(x)x在(,1)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是()A1,) B(,0)(0,1C(0,1 D(,0)1,)答案:D解析:函數(shù)f(x)x的導(dǎo)數(shù)為f(x)1,由于f(x)在(,1)上單調(diào)遞增,則f(x)0在(,1)上恒成立,即x2在(,1)上恒成立由于當(dāng)x1,則有1,解得a1或a0,得函數(shù)的增區(qū)間是(,2),(2,),由y0,得函數(shù)的減區(qū)間是(2,2),由于函數(shù)在(k1,k1)上不是單調(diào)函數(shù),所以k12k1或k12k1,解得3k1或1k1f(x),f(0)6,f
4、(x)是f(x)的導(dǎo)函數(shù),則不等式exf(x)ex5(其中e為自然對(duì)數(shù)的底數(shù))的解集為()A(0,)B(,0)(3,)C(,0)(1,)D(3,)答案:A解析:設(shè)g(x)exf(x)ex,(xR),則g(x)exf(x)exf(x)exexf(x)f(x)1,f(x)1f(x),f(x)f(x)10,g(x)0,yf(x)在定義域上單調(diào)遞增,exf(x)ex5,g(x)5,又g(0)e0f(0)e0615,g(x)g(0),x0,不等式的解集為(0,)故選A.4.若函數(shù)f(x)x3x22ax在上存在單調(diào)遞增區(qū)間,則a的取值范圍是_答案:解析:對(duì)f(x)求導(dǎo),得f(x)x2x2a22a.當(dāng)x時(shí),
5、f(x)的最大值為f2a.令2a0,解得a.所以a的取值范圍是.52017湖北武漢調(diào)研已知函數(shù)f(x)ax2bxln x(a0,bR)(1)設(shè)a1,b1,求f(x)的單調(diào)區(qū)間;(2)若對(duì)任意的x0,f(x)f(1),試比較ln a與2b的大小解:(1)由f(x)ax2bxln x,x(0,),得f(x).a1,b1,f(x)(x0)令f(x)0,得x1.當(dāng)0x1時(shí),f(x)0,f(x)單調(diào)遞減;當(dāng)x1時(shí),f(x)0,f(x)單調(diào)遞增f(x)的單調(diào)遞減區(qū)間是(0,1),f(x)的單調(diào)遞增區(qū)間是(1,)(2)由題意可知,f(x)在x1處取得最小值,即x1是f(x)的極值點(diǎn),f(1)0,2ab1,即
6、b12a.令g(x)24xln x(x0),則g(x).令g(x)0,得x.當(dāng)0x時(shí),g(x)0,g(x)單調(diào)遞增;當(dāng)x時(shí),g(x)0,g(x)單調(diào)遞減g(x)g1ln 1ln 40,g(a)0,即24aln a2bln a0,故ln a2b.6已知函數(shù)f(x)(1)求f(x)在區(qū)間(,1)上的極小值和極大值點(diǎn);(2)求f(x)在1,e(e為自然對(duì)數(shù)的底數(shù))上的最大值解:(1)當(dāng)x1時(shí),f(x)3x22xx(3x2),令f(x)0,解得x0或x.當(dāng)x變化時(shí),f(x),f(x)的變化情況如下表.x(,0)0f(x)00f(x)極小值極大值故當(dāng)x0時(shí),函數(shù)f(x)取得極小值為f(0)0,函數(shù)f(x)的極大值點(diǎn)為x.(2)當(dāng)1x1時(shí),由(1)知,函數(shù)f(x)在1,0和上單調(diào)遞減,在上單調(diào)遞增因?yàn)閒(1)2,f,f(0)0,所以f(x)在1,1)上的最大值為2.當(dāng)1xe時(shí),f(x)aln x,當(dāng)a0時(shí),f(x)0;當(dāng)a0時(shí),f(x)在1,e上單調(diào)遞增,則f(x)在1,e上的最大值為f(e)a.綜上所述,當(dāng)a2時(shí),f(x)在1,e上的最大值為a;當(dāng)a2時(shí),f(x)在1,e上的最大值為2.