《2020屆高考數(shù)學(xué)一輪總復(fù)習(xí) 課時(shí)跟蹤練(十四)導(dǎo)數(shù)與函數(shù)的單調(diào)性(基礎(chǔ)課) 理(含解析)新人教A版》由會員分享,可在線閱讀,更多相關(guān)《2020屆高考數(shù)學(xué)一輪總復(fù)習(xí) 課時(shí)跟蹤練(十四)導(dǎo)數(shù)與函數(shù)的單調(diào)性(基礎(chǔ)課) 理(含解析)新人教A版(6頁珍藏版)》請?jiān)谘b配圖網(wǎng)上搜索。
1、課時(shí)跟蹤練(十四)A組基礎(chǔ)鞏固1函數(shù)f(x)cos xx在(0,)上的單調(diào)性是()A先增后減 B先減后增C單調(diào)遞增 D單調(diào)遞減解析:易知f(x)sin x1,x(0,),所以f(x)0;在(0,)上,f(x)0,選項(xiàng)D滿足答案:D3(2019龍泉二中月考)若函數(shù)f(x)x312x在區(qū)間(k1,k1)上不是單調(diào)函數(shù),則實(shí)數(shù)k的取值范圍是()Ak3或1k1或k3B不存在這樣的實(shí)數(shù)kC2k2D3k1或1k3解析:因?yàn)閒(x)x312x,所以f(x)3x212,令f(x)0,解得x2或x2,若函數(shù)f(x)x312x在(k1,k1)上不是單調(diào)函數(shù),則方程f(x)0在(k1,k1)內(nèi)有解所以k12k1或k
2、12k1,解得3k1或1k3.答案:D4若f(x),eaf(b) Bf(a)f(b)Cf(a)1解析:f(x),當(dāng)xe時(shí),f(x)f(b)答案:A5(2019保定一中模擬)函數(shù)f(x)的定義域?yàn)镽,f(1)2,對任意xR,f(x)2,則f(x)2x4的解集為()A(1,1) B(1,)C(,1) D(,)解析:由f(x)2x4,得f(x)2x40,設(shè)F(x)f(x)2x4,則F(x)f(x)2,因?yàn)閒(x)2,所以F(x)0在R上恒成立,所以F(x)在R上單調(diào)遞增又F(1)f(1)2(1)42240,故不等式f(x)2x40等價(jià)于F(x)F(1),所以x1.答案:B6(2017山東卷)若函數(shù)e
3、xf(x)(e2.718 28是自然對數(shù)的底數(shù))在f(x)的定義域上單調(diào)遞增,則稱函數(shù)f(x)具有M性質(zhì)下列函數(shù)中具有M性質(zhì)的是()Af(x)2x Bf(x)x2Cf(x)3x Df(x)cos x解析:若f(x)具有性質(zhì)M,則exf(x)exf(x)f(x)0在f(x)的定義域上恒成立,即f(x)f(x)0在f(x)的定義域上恒成立對于選項(xiàng)A,f(x)f(x)2x2xln 22x(1ln 2)0,符合題意經(jīng)驗(yàn)證,選項(xiàng)B,C,D均不符合題意故選A.答案:A7已知函數(shù)f(x)是函數(shù)f(x)的導(dǎo)函數(shù),f(1),對任意實(shí)數(shù)都有f(x)f(x)0,設(shè)F(x),則不等式F(x)0,知F(x)0.所以F(
4、x)在定義域R上單調(diào)遞減由F(x)1.所以不等式F(x)0),所以當(dāng)x(0,1)時(shí),f(x)0,f(x)為減函數(shù),因此f(x)的單調(diào)遞減區(qū)間是(0,1)答案:(0,1)9若函數(shù)f(x)ax33x2x恰好有三個(gè)單調(diào)區(qū)間,則實(shí)數(shù)a的取值范圍是_解析:由題意知f(x)3ax26x1,由函數(shù)f(x)恰好有三個(gè)單調(diào)區(qū)間,得f(x)有兩個(gè)不相等的零點(diǎn),需滿足a0,且3612a0,解得a3,所以實(shí)數(shù)a的取值范圍是(3,0)(0,)答案:(3,0)(0,)10(2019昆明調(diào)研)已知函數(shù)f(x)(xR)滿足f(1)1,f(x)的導(dǎo)數(shù)f(x),則不等式f(x2)的解集為_解析:設(shè)F(x)f(x)x,所以F(x)
5、f(x),因?yàn)閒(x),所以F(x)f(x)0,即函數(shù)F(x)在R上單調(diào)遞減因?yàn)閒(x2),所以f(x2)f(1),所以F(x2)1,解得x1,即不等式的解集為x|x1答案:x|x111討論函數(shù)f(x)(a1)ln xax21的單調(diào)性解:f(x)的定義域?yàn)?0,)f(x)2ax.(1)當(dāng)a1時(shí),f(x)0,故f(x)在(0,)上單調(diào)遞增;(2)當(dāng)a0時(shí),f(x)0,故f(x)在(0,)上單調(diào)遞減;(3)當(dāng)0a1時(shí),令f(x)0,解得x .故當(dāng)x時(shí),f(x)0.所以f(x)在上單調(diào)遞減,在上單調(diào)遞增12(2019徐州調(diào)研)設(shè)函數(shù)f(x)ax2aln x,g(x),其中aR,e2.718為自然對數(shù)
6、的底數(shù)(1)討論f(x)的單調(diào)性;(2)證明:當(dāng)x1時(shí),g(x)0.(1)解:由題意得f(x)2ax(x0)當(dāng)a0時(shí),f(x)0時(shí),由f(x)0得x,當(dāng)x時(shí),f(x)0,f(x)單調(diào)遞增(2)證明:令s(x)ex1x,則s(x)ex11.當(dāng)x1時(shí),s(x)0,所以s(x)s(1),即ex1x,從而g(x)0.故當(dāng)x1時(shí),g(x)0.B組素養(yǎng)提升13(2016全國卷改編)若函數(shù)f(x)xsin 2xasin x在(,)單調(diào)遞增,則a的取值范圍是_解析:f(x)1cos 2xacos x1(2cos2x1)acos xcos2xacos x,f(x)在R上單調(diào)遞增,則f(x)0在R上恒成立令cos
7、 xt,t1,1,則t2at0在1,1上恒成立,即4t23at50在1,1上恒成立,令g(t)4t23at5,則解得a.答案:14(2019天津?yàn)I海新區(qū)八校聯(lián)考)設(shè)函數(shù)f(x)x2ex.(1)求在點(diǎn)(1,f(1)處的切線方程;(2)求函數(shù)f(x)的單調(diào)區(qū)間;(3)當(dāng)x2,2時(shí),使得不等式f(x)2a1能成立的實(shí)數(shù)a的取值范圍解:(1)因?yàn)閒(x)2xexx2ex,所以kf(1)3e,且f(1)e.故切線方程為3exy2e0.(2)令f(x)0,即xex(x2)0,解得x0或x2.令f(x)0,得2x0.所以f(x)的單調(diào)增區(qū)間是(,2)和(0,),單調(diào)減區(qū)間是(2,0)(3)由(2)知f(x)在區(qū)間(2,0)上單調(diào)遞減,在區(qū)間(0,2)上單調(diào)遞增所以f(x)minf(0)0.當(dāng)x2,2時(shí),不等式f(x)2a1能成立,須2a1f(x)min,即2a10,故a,所以實(shí)數(shù)a的取值范圍為.6