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1、新編高考數(shù)學(xué)復(fù)習(xí)資料第三節(jié)導(dǎo)數(shù)的應(yīng)用(二)來(lái)源:全盤(pán)鞏固1已知f(x)x2sin,f(x)為f(x)的導(dǎo)函數(shù),則f(x)的圖象是()解析:選Af(x)x2sinx2cos x,f(x)xsin x.易知該函數(shù)為奇函數(shù),所以排除B、D.當(dāng)x時(shí),fsin0.來(lái)源:3已知函數(shù)f(x)x3x2x,則f(a2)與f(1)的大小關(guān)系為()Af(a2)f(1) Bf(a2)f(1)Cf(a2)f(1) Df(a2)與f(1)的大小關(guān)系不確定解析:選A由題意可得f(x)x22x,令f(x)(3x7)(x1)0,得x1或x.當(dāng)x0,f(x)為增函數(shù);當(dāng)1x時(shí),f(x)0,f(x)為減函數(shù)所以f(1)是函數(shù)f(x
2、)在(,0上的最大值,又因?yàn)閍20,所以f(a2)f(1)4(2014青島模擬)若函數(shù)yaex3x(xR,aR),有大于零的極值點(diǎn),則實(shí)數(shù)a的取值范圍是()A(3,0) B(,3)來(lái)源:C. D.解析:選A由題可得yaex3,若函數(shù)在xR上有大于零的極值點(diǎn),即yaex30有正根,顯然有a0,得參數(shù)a的范圍為a3.綜上知,3a0.5f(x)是定義在(0,)上的非負(fù)可導(dǎo)函數(shù),且滿足xf(x)f(x)0,對(duì)任意正數(shù)a,b,若a0),則F(x).因?yàn)閤0,xf(x)f(x)0,所以F(x)0,故函數(shù)F(x)在(0,)上為減函數(shù)又0a0時(shí)有0,則不等式xf(x)0的解集為()Ax|1x1或1x0 Dx|
3、1x0時(shí)有0,即0,在(0,)上單調(diào)遞增f(x)為R上的偶函數(shù),xf(x)為R上的奇函數(shù)xf(x)0,x20,0.在(0,)上單調(diào)遞增,且0,當(dāng)x0時(shí),若xf(x)0,則x1.又xf(x)為R上的奇函數(shù),當(dāng)x0,則1x1或1x0,得x2,由f(x)0,得1x2,所以函數(shù)f(x)在(,1),(2,)上單調(diào)遞增,在(1,2)上單調(diào)遞減,從而可知f(x)的極大值和極小值分別為f(1),f(2),若欲使函數(shù)f(x)恰好有兩個(gè)不同的零點(diǎn),則需使f(1)0或f(2)0,解得a5或a4.答案:5或48已知函數(shù)f(x)x24x3ln x在t,t1上不單調(diào),則t的取值范圍是_解析:由題意知f(x)x4,由f(x
4、)0,得函數(shù)f(x)的兩個(gè)極值點(diǎn)為1,3,則只要這兩個(gè)極值點(diǎn)有一個(gè)在區(qū)間(t,t1)內(nèi),函數(shù)f(x)在區(qū)間t,t1上就不單調(diào),由t1t1或t3t1,得0t1或2t1,則函數(shù)f(x)在0,1上單調(diào)遞減,故只要f(0)f(1)1,即只要a2,即1|a|;若|a|1,此時(shí)f(x)minf(|a|)|a|3a2|a|a2|a|,由于f(0)0,f(1)a2,故當(dāng)|a|時(shí),f(x)maxf(1),此時(shí)只要a2a2|a|1即可,即a2,由于|a|,故|a|110,故此式成立;當(dāng)0,即aa,又f(0)a.所以要使方程f(x)k在0,)上有兩個(gè)不相等的實(shí)數(shù)根,k的取值范圍是.11(2014杭州模擬)天目山某景
5、區(qū)為提高經(jīng)濟(jì)效益,現(xiàn)對(duì)某一景點(diǎn)進(jìn)行改造升級(jí),從而擴(kuò)大內(nèi)需,提高旅游增加值經(jīng)過(guò)市場(chǎng)調(diào)查,旅游增加值y萬(wàn)元與投入x(x10)萬(wàn)元之間滿足:yf(x)ax2xbln,a,b為常數(shù)當(dāng)x10萬(wàn)元時(shí),y19.2萬(wàn)元;當(dāng)x20萬(wàn)元時(shí),y35.7萬(wàn)元(參考數(shù)據(jù):ln 20.7,ln 31.1,ln 51.6)(1)求f(x)的解析式;(2)求該景點(diǎn)改造升級(jí)后旅游利潤(rùn)T(x)的最大值(利潤(rùn)旅游增加值投入)解:(1)由條件解得a,b1,則f(x)xln(x10)(2)由T(x)f(x)xxln(x10),得T(x).令T(x)0,得x1(舍)或x50.當(dāng)x(10,50)時(shí),T(x)0,因此T(x)在(10,50)
6、上是增函數(shù);當(dāng)x(50,)時(shí),T(x)0,因此T(x)在(50,)上是減函數(shù)則x50為T(mén)(x)的極大值點(diǎn),也是最大值點(diǎn)即該景點(diǎn)改造升級(jí)后旅游利潤(rùn)T(x)的最大值為T(mén)(50)24.4萬(wàn)元12已知函數(shù)f(x)axln x,g(x)ex.(1)當(dāng)a0時(shí),求f(x)的單調(diào)區(qū)間;(2)若不等式g(x)2.解:(1)f(x)的定義域是(0,),f(x)a(x0),當(dāng)a0時(shí),f(x)0,f(x)在(0,)上單調(diào)遞增;當(dāng)a0,f(x)單調(diào)遞增;當(dāng)x時(shí),f(0)0,f(x)單調(diào)遞減,綜上所述:當(dāng)a0時(shí),f(x)在(0,)上單調(diào)遞增;當(dāng)a0時(shí),f(x)在上單調(diào)遞增,在上單調(diào)遞減(2)由題意:ex有解,即exxm有
7、解,因此只需m1,且當(dāng)x(0,)時(shí),ex1,所以1ex0,即h(x)0.故h(x)在(0,)上單調(diào)遞減,所以h(x)0,所以m(x)在(0,)上單調(diào)遞增故m(x)m(0)1,又設(shè)n(x)ln xx,x(0,),則n(x)1,當(dāng)x(0,1)時(shí),n(x)0,n(x)單調(diào)遞增;當(dāng)x(1,)時(shí),n(x)1(1)2.沖擊名校設(shè)函數(shù)f(x)ln xax,g(x)exax ,其中a為實(shí)數(shù)(1)若f(x)在(1,)上是單調(diào)減函數(shù),且g(x)在(1,)上有最小值,求a的取值范圍;(2)若g(x)在(1,)上是單調(diào)增函數(shù),試求f(x)的零點(diǎn)個(gè)數(shù),并證明你的結(jié)論解:(1)令f(x)a0,進(jìn)而解得xa1,即f(x)在
8、(a1,)上是單調(diào)減函數(shù)同理,f(x)在(0,a1)上是單調(diào)增函數(shù)由于f(x)在(1,)上是單調(diào)減函數(shù),故(1,)(a1,),從而a11,即a1.令g(x)exa0,得xln a.當(dāng)0xln a時(shí),g(x)ln a時(shí),g(x)0,所以xln a是g(x)的極小值點(diǎn)又g(x)在(1,)上有最小值,所以ln a1,即ae.綜上,a的取值范圍為(e,)(2)當(dāng)a0時(shí),g(x)必為單調(diào)增函數(shù);當(dāng)a0時(shí),令g(x)exa0,解得aln a,因?yàn)間(x)在(1,)上是單調(diào)增函數(shù),類似(1)有l(wèi)n a1,即00,得f(x)存在唯一的零點(diǎn)()當(dāng)a0時(shí),由于f(ea)aaeaa(1ea)0,且函數(shù)f(x)在ea
9、,1上的圖象不間斷,所以f(x)在(ea,1)上存在零點(diǎn)另外,當(dāng)x0時(shí),f(x)a0,故f(x)在(0,)上是單調(diào)增函數(shù),所以f(x)只有一個(gè)零點(diǎn)()當(dāng)0ae1時(shí),令f(x)a0,解得xa1.當(dāng)0x0,當(dāng)xa1時(shí),f(x)0,即0ae1時(shí),f(x)有兩個(gè)零點(diǎn)實(shí)際上,對(duì)于0ae1,由于f(e1)1ae10,且函數(shù)f(x)在e1,a1上的圖象不間斷,所以f(x)在(e1,a1)上存在零點(diǎn)另外,當(dāng)x(0,a1)時(shí),f(x)a0,故f(x)在(0,a1)上是單調(diào)增函數(shù),所以f(x)在(0,a1)上只有一個(gè)零點(diǎn)下面考慮f(x)在(a1,)上的情況先證f(ea1)a(a2ea1)e時(shí),exx2.設(shè)h(x)
10、exx2,則h(x)ex2x,再設(shè)l(x)h(x)ex2x,則l(x)ex2.當(dāng)x1時(shí),l(x)ex2e20,所以l(x)h(x)在(1,)上是單調(diào)增函數(shù)故當(dāng)x2時(shí),h(x)ex2xh(2)e240,從而h(x)在(2,)上是單調(diào)增函數(shù),進(jìn)而當(dāng)xe時(shí),h(x)exx2h(e)eee20,即當(dāng)xe時(shí),exx2.當(dāng)0ae時(shí),f(ea1)a1aea1a(a2ea1)0,且函數(shù)f(x)在a1,ea1上的圖象不間斷,所以f(x)在(a1,ea1)上存在零點(diǎn)又當(dāng)xa1時(shí),f(x)a0,故f(x)在(a1,)上是單調(diào)減函數(shù),所以f(x)在(a1,)上只有一個(gè)零點(diǎn)綜合()()(),當(dāng)a0或ae1時(shí),f(x)的零點(diǎn)個(gè)數(shù)為1,當(dāng)0ae1時(shí),f(x)的零點(diǎn)個(gè)數(shù)為2.