《2022年高考數(shù)學(xué)大二輪復(fù)習(xí) 專題二 函數(shù)與導(dǎo)數(shù) 2.3(二)導(dǎo)數(shù)的綜合應(yīng)用練習(xí)》由會員分享,可在線閱讀,更多相關(guān)《2022年高考數(shù)學(xué)大二輪復(fù)習(xí) 專題二 函數(shù)與導(dǎo)數(shù) 2.3(二)導(dǎo)數(shù)的綜合應(yīng)用練習(xí)(6頁珍藏版)》請在裝配圖網(wǎng)上搜索。
1、2022年高考數(shù)學(xué)大二輪復(fù)習(xí) 專題二 函數(shù)與導(dǎo)數(shù) 2.3(二)導(dǎo)數(shù)的綜合應(yīng)用練習(xí)1(2018昆明市高三摸底調(diào)研測試)若函數(shù)f(x)2xx21,對于任意的xZ且x(,a),都有f(x)0恒成立,則實數(shù)a的取值范圍為()A(,1 B(,0C(,4 D(,5解析:對任意的xZ且x(,a),都有f(x)0恒成立,可轉(zhuǎn)化為對任意的xZ且x(,a),2xx21恒成立令g(x)2x,h(x)x21,當(dāng)x0時,g(x)h(x),當(dāng)x0或1時,g(x)h(x),當(dāng)x2或3或4時,g(x)h(x)綜上,實數(shù)a的取值范圍為(,5,故選D.答案:D2已知函數(shù)yf(x)是R上的可導(dǎo)函數(shù),當(dāng)x0時,有f(x)0,則函數(shù)F
2、(x)xf(x)的零點(diǎn)個數(shù)是()A0 B1C2 D3解析:由F(x)xf(x)0,得xf(x),設(shè)g(x)xf(x),則g(x)f(x)xf(x),因為x0時,有f(x)0,所以x0時,0,即當(dāng)x0時,g(x)f(x)xf(x)0,此時函數(shù)g(x)單調(diào)遞增,此時g(x)g(0)0,當(dāng)x0時, g(x)f(x)xf(x)g(0)0,作出函數(shù)g(x)和函數(shù)y的圖象,(直線只代表單調(diào)性和取值范圍),由圖象可知函數(shù)F(x)xf(x)的零點(diǎn)個數(shù)為1個答案:B3定義1:若函數(shù)f(x)在區(qū)間D上可導(dǎo),即f(x)存在,且導(dǎo)函數(shù)f(x)在區(qū)間D上也可導(dǎo),則稱函數(shù)f(x)在區(qū)間D上存在二階導(dǎo)數(shù),記作f(x),即f
3、(x)f(x).定義2:若函數(shù)f(x)在區(qū)間D上的二階導(dǎo)數(shù)恒為正,即f(x)0恒成立,則稱函數(shù)f(x)在區(qū)間D上為凹函數(shù)已知函數(shù)f(x)x3x21在區(qū)間D上為凹函數(shù),則x的取值范圍是_解析:f(x)x3x21,f(x)3x23x,f(x)6x3.令f(x)0,即6x30,解得x.x的取值范圍是.答案:4已知函數(shù)f(x),g(x)(x1)2a2,若當(dāng)x0時,存在x1,x2R,使得f(x2)g(x1)成立,則實數(shù)a的取值范圍是_解析:由題意得存在x1,x2R,使得f(x2)g(x1)成立,等價于f(x)ming(x)max.因為g(x)(x1)2a2,x0,所以當(dāng)x1時,g(x)maxa2.因為f
4、(x),x0,所以f(x).所以f(x)在(0,1)上單調(diào)遞減,在(1,)上單調(diào)遞增,所以f(x)minf(1)e.又g(x)maxa2,所以a2ea或a.故實數(shù)a的取值范圍是(,)答案:(,)5(2018武漢市武昌區(qū)調(diào)研考試)已知函數(shù)f(x)ln x,aR.(1)討論函數(shù)f(x)的單調(diào)性;(2)當(dāng)a0時,證明f(x).解析:(1)f(x)(x0)當(dāng)a0時,f(x)0,f(x)在(0,)上單調(diào)遞增當(dāng)a0時,若xa,則f(x)0,函數(shù)f(x)在(a,)上單調(diào)遞增;若0xa,則f(x)0時,f(x)minf(a)ln a1.要證f(x),只需證ln a1,即證ln a10.令函數(shù)g(a)ln a1
5、,則g(a)(a0),當(dāng)0a1時,g(a)1時,g(a)0,所以g(a)在(0,1)上單調(diào)遞減,在(1,)上單調(diào)遞增所以g(a)ming(1)0.所以ln a10恒成立,所以f(x).6(2018南昌市第一次模擬測試卷)已知函數(shù)f(x)exaln xe(aR),其中e為自然對數(shù)的底數(shù)(1)若f(x)在x1處取得極小值,求a的值及函數(shù)f(x)的單調(diào)區(qū)間;(2)若當(dāng)x1,)時,f(x)0恒成立,求a的取值范圍解析:(1)易知f(x)的定義域為(0,)由f(x)exaln xe(aR),得f(x)ex.由題意可知f(1)0,所以ae,所以f(x)ex.令g(x)xexe,則g(x)ex(1x)當(dāng)x0
6、時,g(x)0,所以g(x)在(0,)上單調(diào)遞增,且g(1)0.所以當(dāng)x(0,1)時,g(x)0,所以當(dāng)x(0,1)時,f(x)0.故函數(shù)f(x)的減區(qū)間為(0,1),增區(qū)間為(1,)(2)由f(x)exaln xe,得f(x)ex.當(dāng)a0時,f(x)ex0,所以f(x)在1,)上單調(diào)遞增,f(x)minf(1)0.(符合題意)當(dāng)a0時,f(x)ex,當(dāng)x1,)時,exe.()當(dāng)a(0,e時,因為x1,),所以e,f(x)ex0,所以f(x)在1,)上單調(diào)遞增,f(x)minf(1)0.(符合題意)()當(dāng)a(e,)時,存在x01,),滿足f(x0)e0,所以f(x)在1,x0)上單調(diào)遞減,在(
7、x0,)上單調(diào)遞增,故f(x0)0恒成立,f(x)的單調(diào)遞增區(qū)間為(,),無單調(diào)遞減區(qū)間;當(dāng)a0時,令f(x)0,得x0,得xln a,f(x)的單調(diào)遞減區(qū)間為(,ln a),單調(diào)遞增區(qū)間為(ln a,)(2)令g(x)0,得f(x)0或x,先考慮f(x)在區(qū)間0,1上的零點(diǎn)個數(shù),當(dāng)a1時,f(x)在(0,)上單調(diào)遞增且f(0)0,f(x)在0,1上有一個零點(diǎn);當(dāng)ae時,f(x)在(,1)上單調(diào)遞減,f(x)在0,1上有一個零點(diǎn);當(dāng)1ae時,f(x)在(0,ln a)上單調(diào)遞減,在(ln a,1)上單調(diào)遞增,而f(1)ea1,當(dāng)ea10,即1ae1時,f(x)在0,1上有兩個零點(diǎn),當(dāng)ea10,
8、即e1ae1或a2(1)時,g(x)在0,1上有兩個零點(diǎn);當(dāng)10,a1)(1)當(dāng)ae(e是自然對數(shù)的底數(shù))時,求函數(shù)f(x)的單調(diào)區(qū)間;(2)若存在x1,x21,1,使得|f(x1)f(x2)|e1,求實數(shù)a的取值范圍解析:(1)f(x)axln a2xln a2x(ax1)ln a.當(dāng)ae時,f(x)2xex1,在R上是增函數(shù),又f(0)0,f(x)0的解集為(0,),f(x)1時,ln a0,y(ax1)ln a在R上是增函數(shù),當(dāng)0a1時,ln a1或0a0),g(a)120,g(a)a2ln a在(0,)上是增函數(shù)而g(1)0,故當(dāng)a1時,g(a)0,即f(1)f(1);當(dāng)0a1時,g(a)0,即f(1)1時,f(x)maxf(x)minf(1)f(0)e1,即aln ae1,函數(shù)yaln a在(1,)上是增函數(shù),解得ae;當(dāng)0a1時,f(x)maxf(x)minf(1)f(0)e1,即ln ae1,函數(shù)yln a在(0,1)上是減函數(shù),解得0a.綜上可知,實數(shù)a的取值范圍為e,)