2025版高考數(shù)學(xué)一輪總復(fù)習(xí)10年高考真題分類(lèi)題組2.5函數(shù)的圖象

.5函數(shù)的圖象考點(diǎn)一函數(shù)圖象的識(shí)辨1.(2017課標(biāo)Ⅰ文,8,5分)函數(shù)y=sin2x1-cos答案C本題考查函數(shù)圖象的識(shí)辨.易知y=sin2x1-cosx為奇函數(shù),圖象關(guān)于原點(diǎn)對(duì)稱(chēng),故解除B選項(xiàng);sin2≈sin120°=3則f(1)=sin21-cos1=3,故解除f(π)=sin2π1-cosπ=0,故解除D選項(xiàng)方法總結(jié)已知函數(shù)解析式推斷函數(shù)圖象的方法:(1)依據(jù)函數(shù)的定義域推斷圖象的左右位置,依據(jù)函數(shù)的值域推斷圖象的上下位置;(2)依據(jù)函數(shù)的單調(diào)性推斷圖象的改變趨勢(shì);(3)依據(jù)函數(shù)的奇偶性推斷圖象的對(duì)稱(chēng)性;(4)依據(jù)函數(shù)的周期性推斷圖象的循環(huán)往復(fù).2.(2017課標(biāo)Ⅲ文,7,5分)函數(shù)y=1+x+sinxx2的部分圖象大致為答案D當(dāng)x∈(0,1)時(shí),sinx>0,∴y=1+x+sinxx2>1+x>1,解除令f(x)=x+sinxx2,則∴f(x)=x+sinxx∴y=1+x+sinxx2的圖象關(guān)于點(diǎn)(0,1)對(duì)稱(chēng),故選D.解后反思函數(shù)圖象問(wèn)題,一般從定義域、特別點(diǎn)的函數(shù)值、單調(diào)性、奇偶性等方面入手進(jìn)行分析.選擇題通常采納解除法.3.(2016浙江,3,5分)函數(shù)y=sinx2的圖象是()答案D解除法.由y=sinx2為偶函數(shù)推斷函數(shù)圖象的對(duì)稱(chēng)性,解除A,C;當(dāng)x=π2時(shí),y=sinπ22=sinπ24≠4.(2015安徽文,10,5分)函數(shù)f(x)=ax3+bx2+cx+d的圖象如圖所示,則下列結(jié)論成立的是()A.a>0,b<0,c>0,d>0B.a>0,b<0,c<0,d>0C.a<0,b<0,c>0,d>0D.a>0,b>0,c>0,d<0答案A由f(x)的圖象易知d>0,且f'(x)=3ax2+2bx+c的圖象是開(kāi)口向上的拋物線,與x軸正半軸有兩個(gè)不同的交點(diǎn),則a>0,-評(píng)析本題考查導(dǎo)數(shù)的應(yīng)用及運(yùn)用圖象解題的實(shí)力.5.(2015浙江,5,5分)函數(shù)f(x)=x-1xcosx(-π≤x≤π且x≠0)的圖象可能為答案D因?yàn)閒(-x)=-x+1xcos(-x)=-x-1xcosx=-f(x),所以函數(shù)f(x)為奇函數(shù),解除A、B.當(dāng)0<x<1時(shí),x-16.(2013課標(biāo)Ⅰ文,9,5分)函數(shù)f(x)=(1-cosx)sinx在[-π,π]的圖象大致為()答案C因?yàn)閒(-x)=[1-cos(-x)]sin(-x)=-(1-cosx)·sinx=-f(x),所以函數(shù)f(x)為奇函數(shù),圖象關(guān)于原點(diǎn)對(duì)稱(chēng),解除選項(xiàng)B;當(dāng)x∈(0,π)時(shí),1-cosx>0,sinx>0,所以f(x)>0,解除選項(xiàng)A;又函數(shù)f(x)的導(dǎo)函數(shù)f'(x)=sinx·sinx+(1-cosx)·cosx,所以f'(0)=0,解除D.故選C.評(píng)析本題考查對(duì)函數(shù)圖象的識(shí)辨實(shí)力,考查綜合運(yùn)用所學(xué)學(xué)問(wèn)的意識(shí),體現(xiàn)了數(shù)形結(jié)合的思想方法;難點(diǎn)是推斷選項(xiàng)C中f'(0)=0.7.(2012課標(biāo)理,10,5分)已知函數(shù)f(x)=1ln(x+1)-x,答案B令g(x)=ln(x+1)-x,則g'(x)=1x+1-1=∴當(dāng)-1<x<0時(shí),g'(x)>0,當(dāng)x>0時(shí),g'(x)<0,∴g(x)max=g(0)=0.∴f(x)<0,解除A、C,又由定義域可解除D,故選B.評(píng)析本題考查了函數(shù)的圖象,考查了利用導(dǎo)數(shù)推斷函數(shù)單調(diào)性,求值域,考查了數(shù)形結(jié)合的數(shù)學(xué)思想.8.(2013北京理,5,5分)函數(shù)f(x)的圖象向右平移1個(gè)單位長(zhǎng)度,所得圖象與曲線y=ex關(guān)于y軸對(duì)稱(chēng),則f(x)=()A.ex+1B.ex-1C.e-x+1D.e-x-1答案D與曲線y=ex關(guān)于y軸對(duì)稱(chēng)的圖象對(duì)應(yīng)的函數(shù)為y=e-x,將函數(shù)y=e-x的圖象向左平移1個(gè)單位長(zhǎng)度即得y=f(x)的圖象,∴f(x)=e-(x+1)=e-x-1,故選D.9.(2012湖北文,6,5分)已知定義在區(qū)間[0,2]上的函數(shù)y=f(x)的圖象如圖所示,則y=-f(2-x)的圖象為()答案B當(dāng)x=1時(shí),y=-f(1)=-1,解除A、C.當(dāng)x=2時(shí),y=-f(0)=0,故選B.評(píng)析結(jié)合選項(xiàng),利用特別值求解.考點(diǎn)二函數(shù)圖象的應(yīng)用1.(2017天津文,8,5分)已知函數(shù)f(x)=|x|+2,x<1,x+2x,x≥1.設(shè)a∈R,若關(guān)于A.[-2,2]B.[-23,2]C.[-2,23]D.[-23,23]答案A令g(x)=x2當(dāng)a≤0時(shí),如圖1所示,若f(x)≥g(x)恒成立,則g(0)≤2,得a≥-2,∴-2≤a≤0;圖1當(dāng)a>0,x≥1時(shí),如圖2所示,f(x)=x+2x則f'(x)=1-2x2,由f'(x)=12,得即點(diǎn)B(2,3),則g(2)=22+a≤得a≤2,∴0<a≤2.圖2綜上可知,-2≤a≤2.思路分析作出函數(shù)y=f(x)的圖象,借助于圖象的直觀性求出f(x)≥x2+a在R上恒成立時(shí)a方法總結(jié)解決含肯定值不等式恒成立的問(wèn)題,往往將不等式問(wèn)題轉(zhuǎn)化為兩函數(shù)圖象的上、下位置關(guān)系問(wèn)題,從而利用數(shù)形結(jié)合得出滿意條件的不等式,進(jìn)而求出參數(shù)a的值.2.(2016課標(biāo)Ⅱ,12,5分)已知函數(shù)f(x)(x∈R)滿意f(-x)=2-f(x),若函數(shù)y=x+1x與y=f(x)圖象的交點(diǎn)為(x1,y1),(x2,y2),…,(xm,ym),則∑i=1m(xiA.0B.mC.2mD.4m答案B由f(-x)=2-f(x)可知f(x)的圖象關(guān)于點(diǎn)(0,1)對(duì)稱(chēng),又易知y=x+1x=1+1x的圖象關(guān)于點(diǎn)(0,1)對(duì)稱(chēng),所以?xún)珊瘮?shù)圖象的交點(diǎn)成對(duì)出現(xiàn),且每一對(duì)交點(diǎn)都關(guān)于點(diǎn)(0,1)對(duì)稱(chēng),則x1+xm=x2+xm-1=…=0,y1+ym=y2+ym-1∴∑i=1m(xi+yi)=0×m2+2×m3.(2016山東,15,5分)已知函數(shù)f(x)=|x|,x≤m,x2-2mx+4m,x>m,答案(3,+∞)解析f(x)的圖象如圖所示,若存在實(shí)數(shù)b,使得關(guān)于x的方程f(x)=b有三個(gè)不同的根,只需4m-m2<m,解之得m>3或m<0,又m>0,所以m>3.方法總結(jié)分段函數(shù)問(wèn)題、函數(shù)零點(diǎn)個(gè)數(shù)問(wèn)題或方程根的個(gè)數(shù)問(wèn)題通常采納數(shù)形結(jié)合的思想方法來(lái)解決.4.(2015安徽文,14,5分)在平