與名師對(duì)話高三數(shù)學(xué)文一輪復(fù)習(xí)課時(shí)跟蹤訓(xùn)練:第二章 函數(shù)的概念與基本初等函數(shù) 課時(shí)跟蹤訓(xùn)練8 Word版含解析
課時(shí)跟蹤訓(xùn)練(八)基礎(chǔ)鞏固一、選擇題1函數(shù)yx的圖象是()解析函數(shù)圖象過(guò)(1,1)點(diǎn),排除A、D;又當(dāng)x(0,1)時(shí),y>x,故選B.答案B2函數(shù)yx2ax6在上是增函數(shù),則a的取值范圍為()A(,5 B(,5C5,) D5,)解析對(duì)稱軸x,解得a5.答案C3(2018鄭州外國(guó)語(yǔ)學(xué)校期中)已知1,1,2,3,則使函數(shù)yx的值域?yàn)镽,且為奇函數(shù)的所有的值為()A1,3B1,1C1,3D1,1,3解析因?yàn)楹瘮?shù)yx為奇函數(shù),故的可能值為1,1,3.又yx1的值域?yàn)閥|y0,函數(shù)yx,yx3的值域都為R.所以符合要求的的值為1,3.答案A4(2017山東菏澤模擬)已知a,b,cR,函數(shù)f(x)ax2bxc.若f(0)f(4)>f(1),則()Aa>0,4ab0 Ba<0,4ab0Ca>0,2ab0 Da<0,2ab0解析由f(0)f(4)得f(x)ax2bxc圖象的對(duì)稱軸為x2,4ab0,又f(0)>f(1),f(x)先減后增,于是a>0.故選A.答案A5若函數(shù)f(x)x2axa在區(qū)間0,2上的最大值為1,則實(shí)數(shù)a等于()A1B1C2D2解析函數(shù)f(x)x2axa的圖象為開(kāi)口向上的拋物線,函數(shù)的最大值在區(qū)間的端點(diǎn)取得,f(0)a,f(2)43a,或解得a1.答案B6(2017湖南長(zhǎng)沙一模)已知函數(shù)f(x)x,則()Ax0R,使得f(x0)<0Bx(0,),f(x)0Cx1,x20,)(x1x2),使得<0Dx10,),x20,),使得f(x1)>f(x2)解析由f(x)x的定義域?yàn)?,),且在0,)上,f(x)0恒成立,故A錯(cuò)誤,B正確;易知f(x)是0,)上的增函數(shù),x1,x20,)(x1x2),>0,故C錯(cuò)誤;在D中,當(dāng)x10時(shí),不存在x20,)使得f(x1)>f(x2),故D錯(cuò)誤故選B.答案B二、填空題7二次函數(shù)的圖象過(guò)點(diǎn)(0,1),對(duì)稱軸為x2,最小值為1,則它的解析式為_(kāi)解析依題意可設(shè)f(x)a(x2)21,又其圖象過(guò)點(diǎn)(0,1),4a11,a.f(x)(x2)21.答案f(x)(x2)218(2017安徽安慶模擬)已知P2,Q3,R3,則P,Q,R的大小關(guān)系是_解析P23,根據(jù)函數(shù)yx3是R上的增函數(shù),且>>,得3>3>3,即P>R>Q.答案P>R>Q9若f(x)x22ax與g(x)在區(qū)間1,2上都是減函數(shù),則a的取值范圍是_解析由f(x)x22ax在1,2上是減函數(shù)可得1,2a,),a1.y在(1,)上為減函數(shù),由g(x)在1,2上是減函數(shù)可得a>0,故0<a1.答案(0,1三、解答題10已知冪函數(shù)f(x)x(m2m)1(mN*)的圖象經(jīng)過(guò)點(diǎn)(2,),試確定m的值,并求滿足條件f(2a)>f(a1)的實(shí)數(shù)a的取值范圍解冪函數(shù)f(x)的圖象經(jīng)過(guò)點(diǎn)(2,),2(m2m)1,即22(m2m)1.m2m2.解得m1或m2.又mN*,m1.f(x)x,則函數(shù)的定義域?yàn)?,),并且在定義域上為增函數(shù)由f(2a)>f(a1)得解得1a<.a的取值范圍為.能力提升11若冪函數(shù)y(m23m3)xm2m2的圖象不過(guò)原點(diǎn),則m的取值是()A1m2 Bm1或m2Cm2 Dm1解析由冪函數(shù)性質(zhì)可知m23m31,m2或m1.又冪函數(shù)圖象不過(guò)原點(diǎn),m2m20,即1m2.m2或m1.答案B12(2016全國(guó)卷)已知函數(shù)f(x)(xR)滿足f(x)f(2x),若函數(shù)y|x22x3|與yf(x)圖象的交點(diǎn)為(x1,y1),(x2,y2),(xm,ym),則xi()A0BmC2mD4m解析由f(x)f(2x)知f(x)的圖象關(guān)于直線x1對(duì)稱,又函數(shù)y|x22x3|(x1)24|的圖象也關(guān)于直線x1對(duì)稱,所以這兩個(gè)函數(shù)的圖象的交點(diǎn)也關(guān)于直線x1對(duì)稱不妨設(shè)x1<x2<<xm,則1,即x1xm2,同理有x2xm12,x3xm22,又xixmxm1x1,所以2xi(x1xm)(x2xm1)(xmx1)2m,所以xim.取特殊函數(shù)f(x)0(xR),它與y|x22x3|的圖象有兩個(gè)交點(diǎn)(1,0),(3,0),此時(shí)m2,x11,x23,故xi2m,只有B選項(xiàng)符合答案B13當(dāng)x(1,2)時(shí),不等式x2mx4<0恒成立,則m的取值范圍是_解析解法一:設(shè)f(x)x2mx4,當(dāng)x(1,2)時(shí),f(x)<0恒成立m5.解法二:不等式x2mx4<0對(duì)x(1,2)恒成立,mx<x24對(duì)x(1,2)恒成立,即m<對(duì)x(1,2)恒成立,令yx,則函數(shù)yx在(1,2)上是減函數(shù),4<y<5,5<<4,m5.答案(,514(2018河北“五個(gè)一名校聯(lián)盟”質(zhì)量監(jiān)測(cè))設(shè)f(x)與g(x)是定義在同一區(qū)間a,b上的兩個(gè)函數(shù),若函數(shù)yf(x)g(x)在xa,b上有兩個(gè)不同的零點(diǎn),則稱f(x)和g(x)在a,b上是“關(guān)聯(lián)函數(shù)”,區(qū)間a,b稱為“關(guān)聯(lián)區(qū)間”若f(x)x23x4與g(x)2xm在0,3上是“關(guān)聯(lián)函數(shù)”,則m的取值范圍為_(kāi)解析由題意知,yf(x)g(x)x25x4m在0,3上有兩個(gè)不同的零點(diǎn)在同一直角坐標(biāo)系下作出函數(shù)ym與yx25x4(x0,3)的圖象如圖所示,結(jié)合圖象可知,當(dāng)x2,3時(shí),yx25x4,故當(dāng)m時(shí),函數(shù)ym與yx25x4(x0,3)的圖象有兩個(gè)交點(diǎn)答案15(2017蘭州調(diào)研)已知函數(shù)f(x)x22ax3,x4,6(1)當(dāng)a2時(shí),求f(x)的最值;(2)求實(shí)數(shù)a的取值范圍,使yf(x)在區(qū)間4,6上是單調(diào)函數(shù);(3)當(dāng)a1時(shí),求f(|x|)的單調(diào)區(qū)間解(1)當(dāng)a2時(shí),f(x)x24x3(x2)21,由于x4,6,f(x)的最小值是f(2)1,又f(4)35,f(6)15,故f(x)的最大值是35.(2)由于函數(shù)f(x)的圖象開(kāi)口向上,對(duì)稱軸是xa,所以要使f(x)在4,6上是單調(diào)函數(shù),應(yīng)有a4或a6,即a6或a4.故a的取值范圍是(,64,)(3)當(dāng)a1時(shí),f(|x|)x22|x|3其圖象如圖所示,又x4,6,f(|x|)在區(qū)間4,1)和0,1)上為減函數(shù),在區(qū)間1,0)和1,6上為增函數(shù)16已知函數(shù)f(x)x24xa3,aR.(1)若函數(shù)f(x)在(,)上至少有一個(gè)零點(diǎn),求a的取值范圍;(2)若函數(shù)f(x)在a,a1上的最大值為3,求a的值解(1)依題意,函數(shù)yf(x)在R上至少有一個(gè)零點(diǎn),即方程f(x)x24xa30至少有一個(gè)實(shí)數(shù)根,所以164(a3)0,解得a1.(2)函數(shù)yf(x)x24xa3的圖象的對(duì)稱軸方程是x2.當(dāng)a2,即a時(shí),ymaxf(a)a23a33.解得a0或a3.又因?yàn)閍,所以a0.當(dāng)a>2,即a>時(shí),ymaxf(a1)a2a3,解得a.又因?yàn)閍>,所以a.綜上,a0或a.延伸拓展(2018西安模擬)對(duì)二次函數(shù)f(x)ax2bxc(a為非零整數(shù)),四位同學(xué)分別給出下列結(jié)論,其中有且僅有一個(gè)結(jié)論是錯(cuò)誤的,則錯(cuò)誤的結(jié)論是()A1是f(x)的零點(diǎn)B1是f(x)的極值點(diǎn)C3是f(x)的極值D點(diǎn)(2,8)在曲線yf(x)上解析A項(xiàng)中,1是f(x)的零點(diǎn),則有abc0;B項(xiàng)中,1是f(x)的極值點(diǎn),則有b2a;C項(xiàng)中,3是f(x)的極值;則有3;D項(xiàng)中,點(diǎn)(2,8)在曲線yf(x)上,則有4a2bc8.聯(lián)立解得a,b,c;聯(lián)立解得a5,b10,c8,由a為非零整數(shù)可判斷A項(xiàng)錯(cuò)誤,故選A.答案A