2025高考數(shù)學(xué)考二輪專題突破練2函數(shù)的圖象與性質(zhì)-專項(xiàng)訓(xùn)練【含答案】

2025高考數(shù)學(xué)考二輪專題突破練2函數(shù)的圖象與性質(zhì)專項(xiàng)訓(xùn)練一、單項(xiàng)選擇題1.下列函數(shù)是偶函數(shù)且值域?yàn)閇0,+∞)的是()A.f(x)=x2-1 B.f(x)=xC.f(x)=log2x D.f(x)=|x|2.(2024·新高考Ⅰ,6)已知函數(shù)f(x)=-x2-2ax-aA.(-∞,0] B.[-1,0]C.[-1,1] D.[0,+∞)3.(2022·新高考Ⅱ,8)若函數(shù)f(x)的定義域?yàn)镽,且f(x+y)+f(x-y)=f(x)f(y),f(1)=1,則∑k=122f(k)=A.-3 B.-2 C.0 D.14.已知函數(shù)f(x)=2exex-e-x的圖象與函數(shù)g(x)=-x3+12x+1的圖象交點(diǎn)分別為:P1(x1,y1),P2(x2,y2),…,Pk(xk,yk)(k∈N*),則(x1+x2+…+xk)+(y1+y2+A.-2 B.0 C.2 D.4二、多項(xiàng)選擇題5.已知函數(shù)f(x)的定義域?yàn)?1,+∞),值域?yàn)镽,則()A.函數(shù)f(x2+1)的定義域?yàn)镽B.函數(shù)f(x2+1)-1的值域?yàn)镽C.函數(shù)fex+1D.函數(shù)f(f(x))的定義域和值域都是R6.已知函數(shù)f(x)=(x+1)A.函數(shù)f(x)的圖象的對(duì)稱中心是點(diǎn)(0,1)B.函數(shù)f(x)在R上是增函數(shù)C.函數(shù)f(x)是奇函數(shù)D.方程f(2x-1)+f(2x)=2的解為x=17.(2024·九省聯(lián)考)已知函數(shù)f(x)的定義域?yàn)镽,且f(12)≠0,若f(x+y)+f(x)f(y)=4xy,則(A.f(-12)=B.f(12)=-C.函數(shù)f(x-12D.函數(shù)f(x+12三、填空題8.已知函數(shù)f(x)=sinx,x≥0,f(-9.寫出一個(gè)圖象關(guān)于直線x=2對(duì)稱,且在區(qū)間[0,2]上單調(diào)遞增的偶函數(shù)f(x)=.

10.已知函數(shù)f(x)=ln(4x2+1+2x)-12x+1,若f(log2a)=2,則f(log11.已知函數(shù)f(x)=|xx-1|,x<1,x2-4x+4,x≥1,若關(guān)于x的方程f(x)=m有4個(gè)不同的解x1,x2,x3,x4

專題突破練2函數(shù)的圖象與性質(zhì)答案一、單項(xiàng)選擇題1.D解析對(duì)于A,f(x)=x2-1為偶函數(shù),但值域?yàn)閇-1,+∞),故A不符合題意;對(duì)于B,f(x)=x12的定義域不關(guān)于原點(diǎn)對(duì)稱,為非奇非偶函數(shù),故B不符合題意;對(duì)于C,f(x)=log2x的定義域不關(guān)于原點(diǎn)對(duì)稱,為非奇非偶函數(shù),故C不符合題意;對(duì)于D,f(x)=|x|為偶函數(shù),且值域?yàn)閇0,+∞),故D2.B解析因?yàn)楹瘮?shù)f(x)在R上單調(diào)遞增,且當(dāng)x<0時(shí),f(x)=-x2-2ax-a,所以f(x)=-x2-2ax-a在區(qū)間(-∞,0)上單調(diào)遞增,又圖象的對(duì)稱軸為直線x=-a,所以-a≥0,即a≤0;當(dāng)x≥0時(shí),f(x)=ex+ln(x+1),所以函數(shù)f(x)在區(qū)間[0,+∞)上單調(diào)遞增.若函數(shù)f(x)在R上單調(diào)遞增,則-a≤f(0)=1,即a≥-1.綜上,實(shí)數(shù)a的取值范圍是[-1,0].故選B.3.A解析令y=1,得f(x+1)+f(x-1)=f(x)·f(1)=f(x),即f(x+1)=f(x)-f(x-1).從而f(x+2)=f(x+1)-f(x),f(x+3)=f(x+2)-f(x+1).消去f(x+2)和f(x+1),得到f(x+3)=-f(x),從而f(x+6)=f(x),故f(x)的周期為6.令x=1,y=0,得f(1)+f(1)=f(1)·f(0),得f(0)=2,f(2)=f(1)-f(0)=1-2=-1,f(3)=f(2)-f(1)=-1-1=-2,f(4)=f(3)-f(2)=-2-(-1)=-1,f(5)=f(4)-f(3)=-1-(-2)=1,f(6)=f(5)-f(4)=1-(-1)=2,∑k=122f(k)=3[f(1)+f(2)+…+f(6)]+f(19)+f(20)+f(21)+f(22)=f(1)+f(2)+f(3)+f(4)=1+(-1)+(-2)+(-1)即∑k=122f(k)=-3,故選4.D解析由于f(x)=2exex-e-x=ex+e-xex-e-因?yàn)閥=-x3+12x是奇函數(shù),所以函數(shù)g(x)=-x3+12x+1的圖象關(guān)于點(diǎn)(0,1)對(duì)稱.因?yàn)閒'(x)=-4e2x(e2x-1)2<0,所以f(因?yàn)間'(x)=-3(x2-4),所以函數(shù)g(x)在區(qū)間(-∞,-2),(2,+∞)內(nèi)單調(diào)遞減,在區(qū)間(-2,2)內(nèi)單調(diào)遞增.畫出函數(shù)f(x)和g(x)的大致圖象(圖略),可知,f(x)與g(x)的圖象有4個(gè)交點(diǎn),不妨設(shè)x1<x2<x3<x4,則點(diǎn)P1與P4,點(diǎn)P2與P3分別關(guān)于點(diǎn)(0,1)對(duì)稱,所以x1+x4=0,x3+x2=0,y1+y4=2,y3+y2=2,故所求和為4.二、多項(xiàng)選擇題5.BC解析對(duì)于選項(xiàng)A,令x2+1>1可得x≠0,所以f(x2+1)的定義域?yàn)閧x|x≠0},故選項(xiàng)A不正確;對(duì)于選項(xiàng)B,因?yàn)閒(x)值域?yàn)镽,x2+1≥1,所以f(x2+1)的值域?yàn)镽,可得f(x2+1)-1的值域?yàn)镽,故選項(xiàng)B正確;對(duì)于選項(xiàng)C,因?yàn)閑x+1ex=1+1ex>1對(duì)x∈R恒成立,所以fex+1ex的定義域?yàn)镽,因?yàn)閑x+1對(duì)于選項(xiàng)D,若函數(shù)f(f(x))的值域是R,則f(x)>1,此時(shí)無法判斷其定義域是否為R,故選項(xiàng)D不正確.6.ABD解析由于f(x)=(x+1)2+x3x2+1=x2+2x+1+x3x2+1=1+2x+x3x2+1,對(duì)于選項(xiàng)A,設(shè)g(x)=2x+x3x2+1,則f(x)=1+g(x),g(-x)=-2x-x3x2+1=-g(對(duì)于選項(xiàng)B,由f(x)=1+2x+x3x2+1,則f'(x)=x2+x4+2(x2對(duì)于選項(xiàng)C,f(1)=52,f(-1)=-12,則f(1)≠-f(-1),所以函數(shù)f(x)不是奇函數(shù),故C對(duì)于選項(xiàng)D,因?yàn)閒(x)的圖象關(guān)于點(diǎn)(0,1)成中心對(duì)稱,且f(x)在R上是增函數(shù),所以由方程f(2x-1)+f(2x)=2,得2x-1+2x=0,解得x=14,所以D正確,故選ABD7.ABD解析令x=12,y=0,則有f(12)+f(12)×f(0)=f(12)[1+f(0)]=0,又f(12)≠0,所以1+f(0)=0,所以f(0)=-1.令x=12,y=-12,則有f(12?12)+f(12)f(-12)=4×12×(-12),即f(0)+f(12)f(-12)=-1,所以f(12)f(-12)=0,又令y=-12,則有f(x-12)+f(x)f(-12)=4x×(-12),即f(x-12)=-2x,所以函數(shù)f(x-12)是奇函數(shù),f(12)=f(1-12)=-2×1所以f(x+1-12)=-2(x+1)=-2x-2,所以f(x+12)=-2x-2,所以函數(shù)f(x+12)是減函數(shù).故D正確.三、填空題8.12解析因?yàn)?π6<0,所以f-π6=f--π9.-cosπ2x(答案不唯一)解析如f(x)=-cosπ2x,顯然f(-x)=f(x),即f(x)為偶函數(shù),由π2x=kπ,k∈Z,得x=2k,k當(dāng)k=1時(shí),f(x)=-cosπ2x的圖象關(guān)于直線x=2對(duì)稱由x∈[0,2],得π2x∈[0,π],則由余弦函數(shù)的性質(zhì)可知,函數(shù)f(x)=-cosπ2x在區(qū)間[0,2]10.-3解析根據(jù)題意,函數(shù)f(x)=ln(4x2+1+2x)-12x+1,則f(-x)=ln(4x2+1-2x)-12-x+1=-ln(4x2+1+2x)-2x2x+1,于是f(x)+f(-x)=-1,所以f(log12a11.2(3,103)解析當(dāng)x<1時(shí),函數(shù)f(x)=1+1x-1在區(qū)間(-∞,0]上單調(diào)遞減,函數(shù)值的取值集合為[0,1),在區(qū)間[0,1)上單調(diào)遞增,函數(shù)值的取值集合為當(dāng)x≥1時(shí),f(x)=x2-4x+4在區(qū)間[1,2]上單調(diào)遞減,函數(shù)值的取值集合為[0,1],在區(qū)間[2,+∞)上單調(diào)遞增,函數(shù)值的取值集合為[0,+∞).在平面直角坐標(biāo)系中畫出函數(shù)f(