適用于新高考新教材2024版高考數(shù)學(xué)二輪復(fù)習(xí)考點(diǎn)突破練15函數(shù)的圖象與性質(zhì)

考點(diǎn)突破練15函數(shù)的圖象與性質(zhì)一、必備知識(shí)夯實(shí)練1.(2023四川成都七中模擬)已知函數(shù)f(x)=f(x+1),x≤0,x2-A.-6 B.0 C.4 D.62.(2023山東濰坊一模)存在函數(shù)f(x)滿足:對(duì)任意x∈R都有()A.f(|x|)=x3 B.f(sinx)=x2C.f(x2+2x)=|x| D.f(|x|)=x2+13.(2023廣東廣州一模)函數(shù)f(x)=x-sinxx3在區(qū)間[-π,π]上的圖象大致為4.(2023新高考Ⅱ,4)若f(x)=(x+a)ln2x-12x+1為偶函數(shù)A.-1 B.0 C.12 D.5.(2023河南適應(yīng)性考試)已知函數(shù)f(x)=3x+1-1,x≥1,-log3(x+5A.-16 B.16 C.26 D.276.(2023安徽師大附中模擬)已知函數(shù)f(x)的定義域?yàn)?1,+∞),則函數(shù)F(x)=f(2x-3)+3-x的定義域?yàn)?A.(2,3] B.(-2,3]C.[-2,3] D.(0,3]7.(2023青海西寧一模)已知函數(shù)f(x)=logax,x>1,ax-2,x≤1,對(duì)任意x1A.(0,1) B.(1,2] C.(0,1] D.(1,2)8.(2023安徽皖南八校三模)函數(shù)f(x)=-x+4,x9.(2023山東棗莊模擬)已知函數(shù)y=f(x)是定義在[-4,4]上的減函數(shù),且f(a+1)>f(2a),則a的取值范圍是.

二、關(guān)鍵能力提升練10.(2023山東濱州二模)函數(shù)f(x)=3+cosxax2-bxA.a>0,b=0,c<0 B.a<0,b=0,c<0C.a<0,b<0,c=0 D.a>0,b=0,c>011.(2021新高考Ⅱ,8)已知函數(shù)f(x)的定義域?yàn)镽,f(x+2)為偶函數(shù),f(2x+1)為奇函數(shù),則()A.f-12=0 B.f(-1)C.f(2)=0 D.f(4)=012.(2023福建泉州模擬)如圖是下列四個(gè)函數(shù)中某個(gè)函數(shù)的大致圖象,則該函數(shù)是()A.y=|x| B.C.y=ln1+x1-x D13.(多選題)(2022新高考Ⅰ,12)已知函數(shù)f(x)及其導(dǎo)函數(shù)f'(x)的定義域均為R,記g(x)=f'(x).若f32-2x,g(2+x)均為偶函數(shù),A.f(0)=0 B.g-12C.f(-1)=f(4) D.g(-1)=g(2)14.(2023山東淄博模擬)函數(shù)y=11-x的圖象與函數(shù)y=2sinπx(-2≤x≤4)的圖象所有交點(diǎn)的橫坐標(biāo)之和等于三、核心素養(yǎng)創(chuàng)新練15.(2023四川成都樹德中學(xué)模擬)已知函數(shù)f(x)=lnxx,x>0,xex16.(多選題)(2023山東煙臺(tái)二模)定義在R上的函數(shù)f(x)滿足f(x)+f(4+x)=0,f(2+2x)是偶函數(shù),f(1)=1,則()A.f(x)是奇函數(shù)B.f(2023)=-1C.f(x)的圖象關(guān)于直線x=1對(duì)稱D.∑k=1100kf(2k-

考點(diǎn)突破練15函數(shù)的圖象與性質(zhì)1.A解析因?yàn)?4<0,所以f(-4)=f(-4+5)=f(1)=1-3-4=-6,所以f(f(-4))=f(-6)=f(-6+7)=f(1)=-6.2.D解析對(duì)于A,當(dāng)x=1時(shí),f(|1|)=f(1)=1;當(dāng)x=-1時(shí),f(|-1|)=f(1)=-1,不符合函數(shù)定義,A錯(cuò)誤;對(duì)于B,令x=0,則f(sinx)=f(0)=0,令x=π,則f(sinπ)=f(0)=π2,不符合函數(shù)定義,B錯(cuò)誤;對(duì)于C,令x=0,則f(0)=0,令x=-2,則f(0)=f((-2)2+2×(-2))=2,不符合函數(shù)定義,C錯(cuò)誤;對(duì)于D,f(|x|)=x2+1=|x|2+1,x∈R,則|x|≥0,則存在x≥0時(shí),f(x)=x2+1,符合函數(shù)定義,即存在函數(shù)f(x)=x2+1(x≥0),滿足對(duì)任意x∈R都有f(|x|)=x2+1,D正確.3.B解析函數(shù)f(x)=x-sinxx3的定義域?yàn)?-∞,0)∪(0,而f(-x)=-x-sin(-x)(-x)3=-x-sinxx3≠f(x),即函數(shù)f(x)既不是奇函數(shù)也不是偶函數(shù),故排除選項(xiàng)C,D;當(dāng)x=π時(shí),f(x)=f(π)=π>0,故排除選項(xiàng)A.4.B解析(方法一)易知函數(shù)f(x)的定義域?yàn)?∞,-12∪12,+∞,∵函數(shù)f(x)為偶函數(shù),不妨令x=1,則有f(-1)=f(1),∴(-1+a)ln3=(1+a)ln13,∴-1+a=-1-a,∴a=0此時(shí)f(x)=xln2x-12x+1,f(-x)=-xln-2x-1-2x+1=-x(方法二)設(shè)g(x)=ln2x-12x+1,函數(shù)gg(-x)=ln-2x-1-2x+1=ln∴函數(shù)g(x)是奇函數(shù).而f(x)=(x+a)g(x)為偶函數(shù),有f(-x)=(-x+a)g(-x)=-(-x+a)g(x)=(x-a)g(x)=f(x),故x-a=x+a,則a=0.故選B.5.C解析當(dāng)m≥1時(shí),f(m)=-2,即3m+1-1=-2,則3m+1=-1,無解;當(dāng)m<1時(shí),f(m)=-2,即-log3(m+5)-2=-2,解得m=-4,所以f(m+6)=f(2)=32+1-1=26.6.A解析由題可知,2x-3>1,3-x≥0,解得27.B解析因?yàn)閷?duì)任意x1≠x2,都有f(x1)-所以函數(shù)f(x)在定義域內(nèi)為增函數(shù).因?yàn)閒(x)=loga解得1<a≤2,故選B.8.[2,+∞)解析當(dāng)x≤2時(shí),f(x)=-x+4≥2;當(dāng)x>2時(shí),f(x)=1+log2x>2.故函數(shù)f(x)的值域?yàn)閇2,+∞).9.(1,2]解析函數(shù)y=f(x)是定義在[-4,4]上的減函數(shù),且f(a+1)>f(2a),則-4≤a+1<2a≤4,解得1<a≤2.10.A解析由圖可得函數(shù)圖象關(guān)于y軸對(duì)稱,即函數(shù)f(x)為偶函數(shù),所以f(-x)=3+cosxax2+bx+c=f(x由圖象可知f(0)=4c<0,則c<0,故D錯(cuò)誤因?yàn)槎x域不連續(xù),所以ax2-bx+c=0有兩個(gè)實(shí)數(shù)根,則Δ=b2-4ac=-4ac>0,即a,c異號(hào),故a>0,故B錯(cuò)誤,A正確.11.B解析因?yàn)楹瘮?shù)f(x+2)為偶函數(shù),則f(2+x)=f(2-x),可得f(x+3)=f(1-x).因?yàn)楹瘮?shù)f(2x+1)為奇函數(shù),則f(1-2x)=-f(2x+1),所以f(1-x)=-f(x+1),所以f(x+3)=-f(x+1)=f(x-1),即f(x)=f(x+4),故函數(shù)f(x)是以4為周期的周期函數(shù).因?yàn)楹瘮?shù)F(x)=f(2x+1)為奇函數(shù),則F(0)=f(1)=0,故f(-1)=-f(1)=0.故選B.12.D解析觀察圖象知,圖象對(duì)應(yīng)函數(shù)的定義域?yàn)镽,值域?yàn)?-a,a)(a為正常數(shù)),函數(shù)在R上單調(diào)遞增,其圖象過原點(diǎn).對(duì)于A,函數(shù)y=|x|的定義域?yàn)镽,值域?yàn)閇0,+∞),不符合題意,A對(duì)于B,函數(shù)y=2xx2+1的定義域?yàn)镽,當(dāng)x∈(0,+∞)時(shí),y=2xx2+1≤2x2x=1,當(dāng)且僅當(dāng)x=1時(shí),等號(hào)成立,因此函數(shù)對(duì)于C,函數(shù)y=ln1+x1-x有意義,1+x1-x>0,解得-1<x<1,即函數(shù)y=ln1+對(duì)于D,函數(shù)y=2x-2-x2x+2-x的定義域?yàn)镽,y=22x-122x+1=1-222x+1,當(dāng)x=0時(shí),y=0.因?yàn)楹瘮?shù)22x+1在R上單調(diào)遞增,則函數(shù)222x+1在R上單調(diào)遞減,因此y=2x-2-x2x+2-x在所以函數(shù)y=2x-2-13.BC解析∵f(32-2x)是偶函數(shù),∴f(32+2x)=f(32-∴函數(shù)f(x)的圖象關(guān)于直線x=32對(duì)稱∴f(-1)=f(4).故C正確;∵g(2+x)為偶函數(shù),∴g(2-x)=g(2+x),∴g(x)的圖象關(guān)于直線x=2對(duì)稱.∵g(x)=f'(x),g(x)的圖象關(guān)于直線x=2對(duì)稱,∴f(x)的圖象關(guān)于點(diǎn)(2,t)(t∈R)對(duì)稱.∵f(x)的圖象關(guān)于直線x=32對(duì)稱∴g(x)的圖象關(guān)于點(diǎn)32,∴f(x)與g(x)均是周期為2的函數(shù).∴f(0)=f(2)=t(不恒等于0),故A錯(cuò)誤;g-12=g32=0,∴構(gòu)造函數(shù)f(x)=sin(πx)符合題目要求,g(x)=πcos(πx),而g(-1)=πcos(-π)=-π,g(2)=πcos2π=π,故D錯(cuò)誤.故選BC.14.8解析由y=f(x)=2sinπx,則f(2-x)=2sinπ(2-x)=-2sinπx=-f(x),即y=2sinπx的圖象關(guān)于點(diǎn)(1,0)對(duì)稱.由y=g(x)=11-x在區(qū)間(-∞,1)上單調(diào)遞增且值域?yàn)?0,+∞),在區(qū)間(1,+∞)上單調(diào)遞增且值域?yàn)?-∞,0),即y=11-x的圖象關(guān)于點(diǎn)(1,0)對(duì)稱.又f12=2sinπ2=2=g12=所以y=g(x),y=f(x),x∈[-2,4]的圖象如圖所示,所以在x=1的兩側(cè)各有4個(gè)交點(diǎn),且4對(duì)交點(diǎn)分別關(guān)于點(diǎn)(1,0)對(duì)稱,故任意兩個(gè)對(duì)稱的交點(diǎn)橫坐標(biāo)之和為2,則所有交點(diǎn)的橫坐標(biāo)之和為8.15.B解析當(dāng)1-x>0,即x<1時(shí),y=f(1-x)=ln(1-x令y'>0,得x<1-e;令y'<0,得1-e<x<1,所以函數(shù)y=f(1-x)在區(qū)間(-∞,1-e)上單調(diào)遞增,在區(qū)間(1-e,1)上單調(diào)遞減,由此得A和C和D不正確;當(dāng)1-x≤0,即x≥1時(shí),y=f(1-x)=(1-x)e1-x,y'=(1-x)'e1-x+(1-x)(e1-x)'=-e1-x-(1-x)e1-x=-e1-x(2-x).令y'>0,得x>2;令y'<0,得1≤x<2,所以函數(shù)y=f(1-x)在區(qū)間(2,+∞)上單調(diào)遞增,在區(qū)間[1,2)上單調(diào)遞減,故B正確.故選B.16.ABD解析對(duì)于選項(xiàng)A,∵f(2+2x)是偶函數(shù),∴f(2-2x)=f(2+2x),∴函數(shù)f(x)關(guān)于直線x=2對(duì)稱,∴f(-x)=f(4+x).∵f(x)+f(4+x)=0,∴f(-x)=-f(x),∴f(x)是奇函數(shù),故A正確.對(duì)于選項(xiàng)B,∵f(4+x)=-f(x),∴f(8+x)=-f(4+x),∴f(8+x)=f(x),∴f(x)的周期為8,∴f(2023)=f(253×8-1)=f(-1)=-f(1)=-1,故B正確.對(duì)于選項(xiàng)C,若f(x)的圖象關(guān)于直線x=1對(duì)稱,則f(3)=f(-1),但是f(-1)=-f(1)=-1,f(3)=f(1)=1,即f(3)≠f(-1),這與假設(shè)條件矛盾,故C錯(cuò)誤;對(duì)于選項(xiàng)D,將x=12代入f(2-2x)=f(2+2x),得f(3)=f(1)=1將x=1代入f(