2025版高考數(shù)學(xué)一輪總復(fù)習(xí)10年高考真題分類題組2.1函數(shù)及其性質(zhì)

.1函數(shù)及其性質(zhì)考點(diǎn)一函數(shù)的概念及表示1.(2015湖北文,7,5分)設(shè)x∈R,定義符號(hào)函數(shù)sgnx=1,x>0A.|x|=x|sgnx|B.|x|=xsgn|x|C.|x|=|x|sgnxD.|x|=xsgnx答案D由已知可知xsgnx=x,x>0,0,2.(2014江西理,3,5分)已知函數(shù)f(x)=5|x|,g(x)=ax2-x(a∈R).若f[g(1)]=1,則a=()A.1B.2C.3D.-1答案A由已知條件可知:f[g(1)]=f(a-1)=5|a-1|=1,∴|a-1|=0,得a=1.故選A.評(píng)析本題主要考查函數(shù)的解析式,正確理解函數(shù)的定義是解題關(guān)鍵.3.(2015重慶文,3,5分)函數(shù)f(x)=log2(x2+2x-3)的定義域是()A.[-3,1]B.(-3,1)C.(-∞,-3]∪[1,+∞)D.(-∞,-3)∪(1,+∞)答案D由x2+2x-3>0,解得x<-3或x>1,故選D.4.(2015湖北文,6,5分)函數(shù)f(x)=4-|x|+lgx2A.(2,3)B.(2,4]C.(2,3)∪(3,4]D.(-1,3)∪(3,6]答案C要使函數(shù)f(x)有意義,需滿意4即|x|≤4,(x-35.(2014山東理,3,5分)函數(shù)f(x)=1(log2A.0,1C.0,12∪(2,+∞)D.答案C要使函數(shù)f(x)有意義,需使(log2x)2-1>0,即(log2x)2>1,∴l(xiāng)og2x>1或log2x<-1.解之得x>2或0<x<12故f(x)的定義域?yàn)?,16.(2016課標(biāo)Ⅱ文,10,5分)下列函數(shù)中,其定義域和值域分別與函數(shù)y=10lgx的定義域和值域相同的是()A.y=xB.y=lgxC.y=2xD.y=1答案D函數(shù)y=10lgx的定義域、值域均為(0,+∞),而y=x,y=2x的定義域均為R,解除A,C;y=lgx的值域?yàn)镽,解除B,故選D.易錯(cuò)警示利用對(duì)數(shù)恒等式將函數(shù)y=10lgx變?yōu)閥=x,將其值域認(rèn)為是R是失分的主要緣由.評(píng)析本題考查函數(shù)的定義域和值域,嫻熟駕馭基本初等函數(shù)的圖象和性質(zhì)是解題的關(guān)鍵.7.(2016江蘇,5,5分)函數(shù)y=3-2x答案[-3,1]解析若函數(shù)有意義,則3-2x-x2≥0,即x2+2x-3≤0,解得-3≤x≤1.考點(diǎn)二分段函數(shù)1.(2024天津理,8,5分)已知a∈R.設(shè)函數(shù)f(x)=x2-2ax+2a,x≤1,x-alnx,x>1.A.[0,1]B.[0,2]C.[0,e]D.[1,e]答案C本題主要考查分段函數(shù)及不等式恒成立問題,考查學(xué)生推理論證實(shí)力及運(yùn)算求解實(shí)力,將恒成立問題轉(zhuǎn)化為求最值問題,考查了學(xué)生化歸與轉(zhuǎn)化思想及分類探討思想.(1)當(dāng)x≤1時(shí),f(x)=x2-2ax+2a=(x-a)2+2a-a2,①若a>1,則f(x)在(-∞,1]上是減函數(shù),所以f(x)≥f(1)=1>0恒成立;②若a≤1,則f(x)≥f(a)=2a-a2,要使f(x)≥0在(-∞,1]上恒成立,只需2a-a2≥0,得0≤a≤2,∴0≤a≤1,綜合①②可知,a≥0時(shí),f(x)≥0在(-∞,1]上恒成立.(2)當(dāng)x>1時(shí),lnx>0,f(x)=x-alnx≥0恒成立,即a≤xlnx令g(x)=xlnx,g'(x)=lnx-1(lnx)2,令g'(x)=0,得x=e,當(dāng)x∈(1,e)時(shí),g'(x)<0,g(x)為減函數(shù),當(dāng)x綜合(1)(2)可知,a的取值范圍是0≤a≤e,故選C.解后反思求不等式恒成立時(shí)的參數(shù)取值范圍的方法:一是分別參數(shù)法,不等式f(x)≥a在R上恒成立?f(x)min≥a,f(x)≤a在R上恒成立?f(x)max≤a;二是探討分析法,依據(jù)參數(shù)取值狀況進(jìn)行分類探討,從而確定參數(shù)的取值范圍.2.(2024天津文,8,5分)已知函數(shù)f(x)=2x,0≤x≤1,1x,x>1.若關(guān)于x的方程f(x)=-14x+a(aA.54,9C.54,94∪{1}答案D本題以分段函數(shù)和方程的解的個(gè)數(shù)為背景,考查函數(shù)圖象的畫法及應(yīng)用.畫出函數(shù)y=f(x)的圖象,如圖.方程f(x)=-14x+a的解的個(gè)數(shù),即為函數(shù)y=f(x)的圖象與直線l:y=-14x+a當(dāng)直線l經(jīng)過點(diǎn)A時(shí),有2=-14×1+a,a=9當(dāng)直線l經(jīng)過點(diǎn)B時(shí),有1=-14×1+a,a=5由圖可知,a∈54,94時(shí),函數(shù)y=f(x)另外,當(dāng)直線l與曲線y=1x,x>1相切時(shí),恰有兩個(gè)公共點(diǎn),此時(shí)聯(lián)立y=1x,y=-14x+a由Δ=a2-4×14×1=0,得a=1(舍去負(fù)根綜上,a∈54,94一題多解令g(x)=f(x)+14x=2x+x4(0≤x≤1),1x+x4(x>1),當(dāng)0≤x≤1時(shí),g(x)=2x+x4為增函數(shù),其值域?yàn)?,94;當(dāng)x>1時(shí),g(x)=1x+x4,對(duì)g(x)求導(dǎo)得g'(x)=-1x2+14,令g'(x)=0,得x=2,當(dāng)方程f(x)=-14x+a恰有兩個(gè)互異的實(shí)數(shù)解,即函數(shù)y=g(x)的圖象與直線y=a有兩個(gè)不同的交點(diǎn),由圖可知54≤a≤94或a=1滿意條件易錯(cuò)警示本題入手時(shí),簡(jiǎn)單分段探討方程2x=-14x+a(0≤x≤1)與1x=-14x+a(x>1)的解,陷入相對(duì)困難的運(yùn)算過程.利用數(shù)形結(jié)合時(shí),3.(2015課標(biāo)Ⅰ文,10,5分)已知函數(shù)f(x)=2x-1-2,A.-74B.-54C.-34答案A當(dāng)a≤1時(shí),f(a)=2a-1-2=-3,即2a-1=-1,不成立,舍去;當(dāng)a>1時(shí),f(a)=-log2(a+1)=-3,即log2(a+1)=3,得a+1=23=8,∴a=7,此時(shí)f(6-a)=f(-1)=2-2-2=-74.故選評(píng)析本題主要考查分段函數(shù),指數(shù)與對(duì)數(shù)的運(yùn)算,考查分類探討的思想,屬中等難度題.4.(2015陜西文,4,5分)設(shè)f(x)=1-x,x≥0A.-1B.14C.12答案C∵f(-2)=2-2=14,∴f(f(-2))=f14=1-14=15.(2015山東文,10,5分)設(shè)函數(shù)f(x)=3x-b,x<1,2A.1B.78C.34答案Df56=3×56-b=當(dāng)52-b≥1,即b≤32時(shí),f52即252-b=4=22,得到52當(dāng)52-b<1,即b>32時(shí),f52-b即152-4b=4,得到b=78<32綜上,b=12,故選6.(2014江西文,4,5分)已知函數(shù)f(x)=a·2x,x≥0,2-x,A.14B.12C.1答案A由f[f(-1)]=f(2)=4a=1,得a=14,故選7.(2014課標(biāo)Ⅰ文,15,5分)設(shè)函數(shù)f(x)=ex-1,x<1,x1答案(-∞,8]解析f(x)≤2?x<1,ex-1≤2或x≥1,x13≤2?x<1考點(diǎn)三函數(shù)的單調(diào)性與最值1.(2016北京文,4,5分)下列函數(shù)中,在區(qū)間(-1,1)上為減函數(shù)的是()A.y=11-xB.y=cosxC.y=ln(x+1)答案D選項(xiàng)A中,y=11-x=1-(x-1)的圖象是將y=-1x的圖象向右平移1個(gè)單位得到的,故y=11-x在(-1,1)上為增函數(shù),不符合題意;選項(xiàng)B中,y=cosx在(-1,0)上為增函數(shù),在(0,1)上為減函數(shù),不符合題意;選項(xiàng)C中,y=ln(x+1)的圖象是將y=lnx的圖象向左平移1個(gè)單位得到的,評(píng)析本題考查了基本函數(shù)的圖象和性質(zhì)以及圖象的變換,屬中檔題.2.(2015課標(biāo)Ⅱ文,12,5分)設(shè)函數(shù)f(x)=ln(1+|x|)-11+x2,則使得f(x)>f(2x-1)成立的x的取值范圍是A.13,1B.C.-13,13答案A當(dāng)x>0時(shí),f(x)=ln(1+x)-11+x2,∴f'(x)=11+x+2x(1+x2)∴|x|>|2x-1|,即3x2-4x+1<0,解得13<x<1,故選3.(2016浙江,7,5分)已知函數(shù)f(x)滿意:f(x)≥|x|且f(x)≥2x,x∈R.()A.若f(a)≤|b|,則a≤bB.若f(a)≤2b,則a≤bC.若f(a)≥|b|,則a≥bD.若f(a)≥2b,則a≥b答案B依題意得f(a)≥2a,若f(a)≤2b,則2a≤f(a)≤2b,∴2a≤2b,又y=2x是R上的增函數(shù),∴a≤b.故選B.4.(2024課標(biāo)Ⅲ文,12,5分)已知函數(shù)f(x)=sinx+1sinx,則(A.f(x)的最小值為2B.f(x)的圖象關(guān)于y軸對(duì)稱C.f(x)的圖象關(guān)于直線x=π對(duì)稱D.f(x)的圖象關(guān)于直線x=π2答案D對(duì)于A,令sinx=t,t∈[-1,0)∪(0,1],則g(t)=t+1t,當(dāng)t∈(0,1]時(shí),g(t)=t+1t≥2,當(dāng)且僅當(dāng)t=1時(shí),取“=”,故g(t)∈[2,+∞),又∵g(t)=-g(-t),∴g(t)∴g(t)的值域?yàn)?-∞,-2]∪[2,+∞),故A錯(cuò)誤;對(duì)于B,由f(x)≠f(-x),知f(x)不是偶函數(shù),故B錯(cuò)誤;對(duì)于C,f(2π-x)=sin(2π-x)+1sin(2π-x)=-sinx-1對(duì)于D,f(π-x)=sin(π-x)+1sin(π-x)=sinx+1sinx=f(x),故f(x)的圖象關(guān)于直線x=π5.(2016北京文,10,5分)函數(shù)f(x)=xx-1(x≥2)答案2解析解法一:∵f'(x)=-1(x-1)2∴f(x)在[2,+∞)上單調(diào)遞減,∴f(x)在[2,+∞)上的最大值為f(2)=2.解法二:∵f(x)=xx-1=x∴f(x)的圖象是將y=1x的圖象向右平移1個(gè)單位,再向上平移1個(gè)單位得到的.∵y=1x在[2,+∞)∴f(x)在[2,+∞)上單調(diào)遞減,故f(x)在[2,+∞)上的最大值為f(2)=2.解法三:由題意可得f(x)=1+1x∵x≥2,∴x-1≥1,∴0<1x-∴1<1+1x-1≤2,即1<故f(x)在[2,+∞)上的最大值為2.評(píng)析本題考查函數(shù)的最值,有多種解法,屬中檔題.6.(2015浙江文,12,6分)已知函數(shù)f(x)=x2,x≤1,x+6答案-12;26解析f(-2)=(-2)2=4,f(f(-2))=f(4)=4+64-6=-1當(dāng)x≤1時(shí),f(x)=x2≥0,當(dāng)x>1時(shí),f(x)=x+6x-6≥26當(dāng)且僅當(dāng)x=6時(shí),等號(hào)成立,又26-6<0,所以f(x)min=26-6.考點(diǎn)四函數(shù)的奇偶性1.(2015北京文,3,5分)下列函數(shù)中為偶函數(shù)的是()A.y=x2sinxB.y=x2cosxC.y=|lnx|D.y=2-x答案BA中函數(shù)為奇函數(shù),B中函數(shù)為偶函數(shù),C與D中函數(shù)均為非奇非偶函數(shù),故選B.2.(2014課標(biāo)Ⅰ,理3,文5,5分)設(shè)函數(shù)f(x),g(x)的定義域都為R,且f(x)是奇函數(shù),g(x)是偶函數(shù),則下列結(jié)論中正確的是()A.f(x)g(x)是偶函數(shù)B.|f(x)|g(x)是奇函數(shù)C.f(x)|g(x)|是奇函數(shù)D.|f(x)g(x)|是奇函數(shù)答案C由題意可知f(-x)=-f(x),g(-x)=g(x),對(duì)于選項(xiàng)A,f(-x)·g(-x)=-f(x)·g(x),所以f(x)g(x)是奇函數(shù),故A項(xiàng)錯(cuò)誤;對(duì)于選項(xiàng)B,|f(-x)|g(-x)=|-f(x)|g(x)=|f(x)|g(x),所以|f(x)|g(x)是偶函數(shù),故B項(xiàng)錯(cuò)誤;對(duì)于選項(xiàng)C,f(-x)|g(-x)|=-f(x)|g(x)|,所以f(x)|g(x)|是奇函數(shù),故C項(xiàng)正確;對(duì)于選項(xiàng)D,|f(-x)g(-x)|=|-f(x)g(x)|=|f(x)g(x)|,所以|f(x)g(x)|是偶函數(shù),故D項(xiàng)錯(cuò)誤,選C.評(píng)析本題考查函數(shù)奇偶性的定義及其應(yīng)用,考查學(xué)生的學(xué)問應(yīng)用實(shí)力及邏輯推理論證實(shí)力,精確理解函數(shù)奇偶性的定義是解決本題的關(guān)鍵.3.(2011課標(biāo),理2,文3,5分)下列函數(shù)中,既是偶函數(shù)又在(0,+∞)單調(diào)遞增的函數(shù)是()A.y=x3B.y=|x|+1C.y=-x2+1D.y=2-|x|答案By=x3是奇函數(shù),y=-x2+1和y=2-|x|在(0,+∞)上都是減函數(shù),故選B.評(píng)析本題考查函數(shù)的奇偶性和單調(diào)性的判定,屬簡(jiǎn)單題.4.(2024課標(biāo)Ⅲ文,16,5分)已知函數(shù)f(x)=ln(1+x2-x)+1,f(a)=4,則f(-a)=答案-2解析本題考查函數(shù)的奇偶性.易知f(x)的定義域?yàn)镽,令g(x)=ln(1+x則g(x)+g(-x)=0,∴g(x)為奇函數(shù),∴f(a)+f(-a)=2,又f(a)=4,∴f(-a)=-2.解題關(guān)鍵視察出函數(shù)g(x)=ln(1+x2-x)5.(2017課標(biāo)Ⅱ文,14,5分)已知函數(shù)f(x)是定義在R上的奇函數(shù),當(dāng)x∈(-∞,0)時(shí),f(x)=2x3+x2,則f(2)=.

答案12解析本題主要考查運(yùn)用函數(shù)的奇偶性求函數(shù)值.由題意可知f(2)=-f(-2),∵x∈(-∞,0)時(shí),f(x)=2x3+x2,∴f(2)=-f(-2)=-[2×(-8)+4]=-(-12)=12.6.(2016天津,13,5分)已知f(x)是定義在R上的偶函數(shù),且在區(qū)間(-∞,0)上單調(diào)遞增.若實(shí)數(shù)a滿意f(2|a-1|)>f(-2),則a的取值范圍是.

答案1解析由題意知函數(shù)f(x)在(0,+∞)上單調(diào)遞減.因?yàn)閒(2|a-1|)>f(-2),f(-2)=f(2),所以f(2|a-1|)>f(2),所以2|a-1|<21解之得12<a<37.(2014課標(biāo)Ⅱ文,15,5分)偶函數(shù)y=f(x)的圖象關(guān)于直線x=2對(duì)稱,f(3)=3,則f(-1)=.

答案3解析∵函數(shù)y=f(x)的圖象關(guān)于直線x=2對(duì)稱,∴f(2+x)=f(2-x)對(duì)隨意x恒成立,令x=1,得f(1)=f(3)=3,∴f(-1)=f(1)=3.8.(2012課標(biāo)文,16,5分)設(shè)函數(shù)f(x)=(x+1)2+sinxx2答案2解析f(x)=x2+1+2x+sinxx2+1=1+2x+sinxx2+1,令g(x)=2x考點(diǎn)五函數(shù)的周期性1.(2016山東,9,5分)已知函數(shù)f(x)的定義域?yàn)镽.當(dāng)x<0時(shí),f(x)=x3-1;當(dāng)-1≤x≤1時(shí),f(-x)=-f(x);當(dāng)x>12時(shí),fx+12=fx-A.-2B.-1C.0D.2答案D當(dāng)x>12時(shí),由fx+12=fx-12可得f(x)=f(x+1),所以f(6)=f(1),而f(1)=-f(-1),f(-1)=(-1)2.(2016