數(shù)零點(diǎn)和不等式相結(jié)合進(jìn)行考2.高考對(duì)函數(shù)的考查重點(diǎn)關(guān)注以基本初等函數(shù)組成的復(fù)合函數(shù)以及抽象函數(shù)為載體，對(duì)函數(shù)內(nèi)容和性質(zhì)進(jìn)行考查，考查函數(shù)的定分段函數(shù)、三次函數(shù)的圖像A.(-∞,0]B.[-1,0]C.[-1,1]D.[0,+∞)【詳解】因?yàn)閒(x(在R上單調(diào)遞增，且x≥0時(shí)，f(x(=ex+ln(x+1(單調(diào)遞增，即a的范圍是[-1,0].【題2】(2024新高考Ⅰ卷·8)已知函數(shù)為f(x)的定義域?yàn)镽，f(x)>f(x-1)+f(x-2)，且當(dāng)x<3時(shí)f(x)=x，A.f(10)>100B.又因?yàn)閒(x)>f(x-1)+f(x-2)，則f(3)>f(2)+f(1)=3,f(4)>f(3)+f(2)>5，f(5)>f(4)+f(3)>8,f(6)>f(5)+f(4)>13,f(7)>f(6)+f(5)>21，f(8)>f(7)+f(6)>34,f(9)>f(8)+f(7)>55,f(10)>f(9)+f(8)>89，f(11)>f(10)+f(9)>144,f(12)>f(11)+f(10)>233,f(13)>f(12)+f(11)>377f(14)>f(13)+f(12)>610,f(15)>f(14)+f(13)>987，且無證據(jù)表明ACD一定正確.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題的關(guān)鍵是利用f(1)=1,f(2)=2，再利用題目所給的函數(shù)性質(zhì)f(x)>f(x-【題3】(2024新高考Ⅱ卷·8)設(shè)函數(shù)f(x)=(x+a)令x+a=0解得x=-a；令ln若-a≤-b，當(dāng)x∈(-b,1-b(時(shí)，可知x+a>0,ln(x+b(<0，此時(shí)f(x)<0，不合題意；若-b<-a<1-b，當(dāng)x∈(-a,1-b(時(shí)，可知x+a>0,ln(x+b(<0，此時(shí)f(x)<0，不合題意；若-a=1-b，當(dāng)x∈(-b,1-b(時(shí)，可知x+a<0,ln(x+b(<0，此時(shí)f(x)>0；-b,+∞(時(shí)，可知x+a≥0,ln(x+b(≥0，此時(shí)f(x)≥0；此時(shí)f(x)<0，不合題意；解法二：由題意可知：f(x)的定義域?yàn)?-b,+∞(，令x+a=0解得x=-a；令ln(x+b)=0解得x=1-b；A.x=3是f(x)的極小值點(diǎn)B.當(dāng)0<x<1時(shí)，f(x)<f(x2(C.當(dāng)1<x<2時(shí)，-4<f(2x-1)<0D.當(dāng)-1<x<0時(shí)，f(2-x)>f(x)f(x(在(1,3(上的值域即可判斷C；直接作差可判斷D.【詳解】對(duì)A，因?yàn)楹瘮?shù)f(x(的定義域?yàn)镽，而f(x(=2(x-1((x-4(+(x-1(2=3(x-1((x-3(，f(x(<0，當(dāng)x∈(-∞,1(或x∈(3,+∞(時(shí)，f(x(>0函數(shù)f(x(在(-∞,1(上單調(diào)遞增，在(1,3(上單調(diào)遞減，在(3,+∞(上單調(diào)遞增，故x=3是函數(shù)而由上可知，函數(shù)f(x(在(0,1(上單調(diào)遞增，所以f(x(>f(x2(，錯(cuò)誤；所以f(1(>f(2x-1(>f(3(，即-4<f(2x-1(<0，正確；對(duì)D，當(dāng)-1<x<0時(shí)，f(2-x)-f(x)=(1-x(2(-2-x(-(x-1(2(x-4(=(x-1(2(2-2x(>0，所以f(2-x)>f(x)，正確；D.存在a，使得點(diǎn)(1,f(1((為曲線y=f(x)的對(duì)稱中心f(x)在(-1,0),(0,a),(a,2a)上各有一個(gè)零點(diǎn)；B選項(xiàng)，根據(jù)極值和導(dǎo)函數(shù)符號(hào)的關(guān)系進(jìn)行分析；C故x∈(-∞,0(∪(a,+∞(時(shí)f/(x)>0，故f(x)在(-∞,0(,(a,+∞(上單調(diào)遞增，x)<0，f(x)單調(diào)遞減，由f(0)=1>0，f(a)=1-a3<0，則f(0)f(a)<0，根據(jù)零點(diǎn)存在定理f(x)在(0,a)上有一個(gè)零點(diǎn)，又f(-1)=-1-3a<0，f(2a)=4a3+1>0，則f(-1)f(0)<0,f(a)f(2a)<0，x∈(0,+∞)時(shí)f/(x)>0，f(x)單調(diào)遞此時(shí)f(x)在x=0處取到極小值，B選即存在這樣的a,b使得f(x)=f(2b-x)，即2x3-3ax2+1=2(2b-x)3-3a(2b-x)2+1，根據(jù)二項(xiàng)式定理，等式右邊(2b-x)3展開式含有x3的項(xiàng)為2C(2b)0(-x)3=-2x3，D選項(xiàng)，則f(x)+f(2-x)=6-6a，事實(shí)上，f(x)+f(2-x)=2x3-3ax2+1+2(2-x)3-3a(2-x)2+1=(12-6a)x2+(12a-24)x+18-(12-6a=0L18-12a=6-6af(x)=2x3-3ax2+1，f/(x)=6x2-6ax，f”(x)=12x-6a，f(2a-x)=2b；(3)任何三次函數(shù)f(x)=ax3+bx2+cx+d都有對(duì)稱中心，對(duì)稱中心是三次函數(shù)的f(-是三次函數(shù)的對(duì)稱中心A.-3B.-2C.0D.1因?yàn)閒(x+y(+f(x-y(=f(x(f(y(，令x=1,y=0可得，2f(1(=f(1(f(0(，所以f(0(=2，令x=0可得，f(y(+f(-y(=2f(y(，即f(y(=f(-y(，所以函數(shù)f(x(為偶函數(shù)，令y=1得，f(x+1(+f(x-1(=f(x(f(1(=f(x(，即有f(x+2(+f(x(=f(x+1(，從而可知f(x+2(=-f(x-1(，f(x-1(=-f(x-4(，故f(x+2(=f(x-4(，即f(x(=f(x+6(，所以函數(shù)f(x(的一個(gè)周期為6．因?yàn)閒(2(=f(1(-f(0(=1-2=-1，f(3(=f(2(-f(1(=-1-1=-2，f(4(=f(-2(=f(2(=-1，f(5(=f(-1(=f(1(=1，f(6(=f(0(=2，所以一個(gè)周期內(nèi)的f(1(+f(2(+?+f(6(=0．由于22除以6余4，所以f(k(=f(1(+f(2(+f(3(+f(4(=1-1-2-1=-3．故選：A.由f(x+y(+f(x-y(=f(x(f(y(cos(x+y(+cos(x-y(=2cosxcosy，可設(shè)f(x(=acosωx，則由方法一中f(0(=2,f(1(=1知a=所以f(x(=2cos則f(x+y(+f(x-y(=2cosx+y(+2cosx-y(=4cosxcosy=f(x(f(y(，所以f(x(=2cos的周期T==6，f(0(=2,f(1(=1，且f(2(=-1,f(3(=-2,f(4(=-1,f(5(=1,f(6(=2，所以f(1)+f(2)+f(3)+f(4)+f(5)+f(6)=0，所以f(k(=f(1(+f(2(+f(3(+f(4(=1-1-2-1=-3．故選：A.【題8】(2023新高考Ⅱ卷·4)若f(x(=(x+a(ln為偶函數(shù)，則a=().故此時(shí)f(x(為偶函數(shù).【題9】(2022新高考Ⅰ卷·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，若f(-2x(，=0B.g(-(=0C.f逐項(xiàng)判斷即可得解.對(duì)于f(x)，因?yàn)閒-2x(為偶函數(shù)，所以f-2x(=f+2x(即f-x(=f+x(①,所以f(3-x(=f(x(，所以f(x)關(guān)于x=對(duì)稱，則f(-1)=f(4)，故C正確；f 若函數(shù)f(x)滿足題設(shè)條件，則函數(shù)f(x)+C(C為常數(shù))也滿足題設(shè)條件，所以無法確定f(x)的函數(shù)因?yàn)閒-2x(，g(2+x)均為偶函數(shù)，所以f-2x(=f+2x(即f-x(=f+x(，g(2+x)=g(2-x)，所以f(3-x(=f(x(，g(4-x)=g(x)，則f(-1)=f(4)，故C正確；函數(shù)f(x)，g(x)的圖象分別關(guān)于直線x=,x=2對(duì)稱，又g(x)=f/(x)，且函數(shù)f(x)可導(dǎo)，所以g=0,g(3-x(=-g(x(，所以g(4-x)=g(x)=-g(3-x(，所以g(x+2)=-g(x+1)=g(x(，若函數(shù)f(x)滿足題設(shè)條件，則函數(shù)f(x)+C(C為常數(shù))也滿足題設(shè)條件，所以無法確定f(x)的函數(shù)p0>0(是聽覺下限閾值，p是實(shí)際聲壓．下表為不同聲源的聲壓級(jí)：與聲源的距離/m/dBA.p1≥p2B.p2>10p3C.p3=100p0D.p1≤100p2p=40， 當(dāng)且僅當(dāng)Lp=50p3=202>0≤100p2【題11】(2023新高考Ⅰ卷·11)已知函數(shù)f(x(的定義域?yàn)镽，f(xy(=y2f(x(+x2f(y(，則().B.f(1(=0D.B.f(1(=0D.x=0為f(x(的極小值點(diǎn)C.f(x(是偶函數(shù)除選項(xiàng)D.可.因?yàn)閒(xy)=y2f(x)+x2f(y)，對(duì)于C，令x=y=-1，f(1)=f(-1)+f(-1)=2f(-1)，則f(-1)=0，令y=-1,f(-x)=f(x)+x2f(-1)=f(x)，因?yàn)閒(xy)=y2f(x)+x2f(y)，對(duì)于C，令x=y=-1，f(1)=f(-1)+f(-1)=2f(-1)，則f(-1)=0，令y=-1,f(-x)=f(x)+x2f(-1)=f(x)，當(dāng)x>0肘，f(x)=x2lnx，則f/(x(=2xlnx+x2?=x(2lnx+1)，令f/(x(<0，得0<x<e-2；令f/(x(>0，得x>e-2；--上單調(diào)遞減，②任意兩個(gè)自變量x1，x2且x1<③都有f(x1)<f(x2)或f(x1)>f(x2)；如果對(duì)于函數(shù)f(x)的定義域內(nèi)任意一個(gè)x，都有f(-x)=f(x)，那么函數(shù)f(x)就叫做偶函數(shù)關(guān)于y軸對(duì)稱如果對(duì)于函數(shù)f(x)的定義域內(nèi)任意一個(gè)x，都有f(-x)=-f(x)，那么函數(shù)f(x)就叫做奇函數(shù)判斷f(-x)與f(x)的關(guān)系時(shí)，也可以使用如下結(jié)論：如果f(-x)-f(x)=0或=1(f(x)≠0)，則函數(shù)f(x)為偶函數(shù)；如果f(-x)+f(x)=0或=-1(f(x)≠0)，則函數(shù)f(x)為奇函數(shù).定義域內(nèi)(即定義域關(guān)于原點(diǎn)對(duì)稱).(1)若函數(shù)y=f(x+a)為偶函數(shù)，則函數(shù)(3)若f(x)=f(2a-x)，則函數(shù)f(x)關(guān)于x=a對(duì)稱.(4)若f(x)+f(2a-x)=2b，則函數(shù)f(x)關(guān)于點(diǎn)(a，b)對(duì)稱.如果在周期函數(shù)f(x)的所有周期中存在一個(gè)最小的正數(shù)，那么稱這個(gè)最小整數(shù)叫做f(x)的最小正周期.y=xy=x2y=x31y=x2y=x-1RRR{x|x≥0}{x|x≠0}R{y|y≥0}R{y|y≥0}{y|y≠0}奇偶奇奇增在(-∞,0)調(diào)遞增增在(-∞,0)?nan表示指數(shù)個(gè)底數(shù)相乘.y=ax0<a<1a>1⑤x<0時(shí)，ax>1；x>0時(shí)，0<ax<1>1②alog=N(其中a>0且a≠1，N>0)；④loga(MN)=logaM+logaN；⑤loga=logaM-logaN；⑦alogb=b和logaab=b；⑧l(xiāng)ogab=；a>10<a<1當(dāng)0<x<1時(shí)，y<0，當(dāng)x≥1時(shí)，y≥0當(dāng)0<x<1時(shí)，y>0，當(dāng)x≥1時(shí)，y≤0對(duì)于函數(shù)y=f(x(，我們把使f(x(=0的實(shí)數(shù)x叫做函數(shù)y=f(x(的零點(diǎn).方程f(x(=0有實(shí)數(shù)根?函數(shù)y=f(x(的圖像與x軸有公共點(diǎn)?函數(shù)y=f(x(有零點(diǎn).如果函數(shù)y=f(x(在區(qū)間[a,b[上的圖像是連續(xù)不斷的一條曲線，并且有f(a(?f(b(<0，那么函數(shù)y=f(x(f(c(=0,c也就是方程f(x(=0的根.對(duì)于區(qū)間[a,b[上連續(xù)不斷且f(a(?f(b(<0的函數(shù)f(x(，通過不斷地把函數(shù)f(x(的零點(diǎn)f(x(=0的近似解就是求函數(shù)f(x(零點(diǎn)的近似值.5、用二分法求函數(shù)f(x(零點(diǎn)近似值的步驟f(a(?f(b(<0，給定精度ε.(3)計(jì)算f(x1(.若f(x1(=0,則x1就是函數(shù)f(x(的零點(diǎn)；若f(a(?f(x1(<0，則令b=x1(此時(shí)2是f(x)定義域內(nèi)一個(gè)區(qū)間上的任意兩個(gè)量，且x1<x2；④得出結(jié)論.③若f(x)>0且f(x)為增函數(shù)，則函數(shù)f(x)為增函數(shù)為減函數(shù)；④若f(x)>0且f(x)為減函數(shù)，則函數(shù)f(x)為減函數(shù)為增函數(shù).函數(shù)f(x)是奇函數(shù)?函數(shù)f(x)的圖象關(guān)于原點(diǎn)中心對(duì)稱.偶函數(shù)y=f(x)必滿足f(x)=f(|x|).個(gè)區(qū)間上單調(diào)性相同.或函數(shù)②函數(shù)f(x)=±(ax-a-x).③函數(shù)f=loga=loga(1+(或函數(shù)f=loga=loga(1-④函數(shù)f(x)=loga(、x2+1+x)或函數(shù)f(x)=loga(、x2+1-x).偶函數(shù)：①函數(shù)f(x)=±(ax+a-x).②函數(shù)f=loga③函數(shù)f(|x|)類型的一切函數(shù).(2)若函數(shù)y=f(x)的圖象有兩個(gè)對(duì)稱中心(a,c),(b,c)(a<b)，則函數(shù)y=f(x)是周期函數(shù)，且T=2(b-a)；(b-a).(1)若函數(shù)y=f(x)關(guān)于直線x=a對(duì)稱，則f(a+x)=f(a-x).(2)若函數(shù)y=f(x)關(guān)于點(diǎn)(a，b)對(duì)稱，則f(a+x)+f(a-x)=2b.(3)函數(shù)y=f(a+x)與y=f(a-x)關(guān)于y軸對(duì)稱，函數(shù)y=f(a+x)與y=-f(a-x)關(guān)于原點(diǎn)對(duì)稱.【分析】由已知f(x(為偶函數(shù)，可得f(x(=f(-x(，列方程求解即可.由fsinx，因?yàn)閒(x(為偶函數(shù)，所以f(-x(=f(x(，【題2】(2024·湖南邵陽·三模)“0<a<1”是“函數(shù)f(x(=ax-a(a>0且a≠1)在R上單調(diào)遞減”的()可知f(x(在R上單調(diào)遞增；可知f(x(在R上單調(diào)遞減；可得M=lg5000-lg0.002=lg-lg=4-lg2-(lg2-3(=7-2lg2≈6.4.【題4】(2024·河北·二模)已知函數(shù)y=f(x-1(為奇函數(shù)，則函數(shù)y=f(x(+1的圖象()A.關(guān)于點(diǎn)(1,1(對(duì)稱B.關(guān)于點(diǎn)(1,-1(對(duì)稱C.關(guān)于點(diǎn)(-1,1(對(duì)稱D.關(guān)于點(diǎn)(-1,-1(對(duì)稱【詳解】函數(shù)y=f(x-1(為奇函數(shù)，圖象關(guān)于(0,0(對(duì)稱，將函數(shù)y=f(x-1(向左平移一個(gè)單位可得函數(shù)y=f(x(，則函數(shù)y=f(x(關(guān)于(-1,0(對(duì)稱，所以函數(shù)y=f(x(+1的圖象關(guān)于(-1,1(對(duì)稱.x2-2ax,x≥1【題s】(2024陜西渭南二模)已知函數(shù)f(x)x2-2ax,x≥1ra≤1ra≤1【題6】(2024·湖北·二模)已知函數(shù)f(x(=log5(ax-2(在[1,+∞(上單調(diào)遞增，則a的取值范圍是()A.(1,+∞(B.[ln2,+∞(C.(2,+∞(D.[2,+∞(【詳解】若f(x(=log5(ax-2(在[1,+∞(上單調(diào)遞增，x-2≥a-2>0，所以f(x(在[1,+∞(上有定義，再由a>1知ax-2在R上單調(diào)遞增，所以f(x(在[1,+∞(上單調(diào)遞增.A.函數(shù)f(x(單調(diào)遞增B.函數(shù)f(x(值域?yàn)?0,2(C.函數(shù)f(x(的圖象關(guān)于(0,1(對(duì)稱D.函數(shù)f(x(的圖象關(guān)于(1,1(對(duì)稱CD.又內(nèi)層函數(shù)t=2x-1+1在R上單調(diào)遞增，外層函數(shù)y=2-在(1,+∞(上單調(diào)遞增，所以函數(shù)f(x(的值域?yàn)?0,2(，故B正確；所以函數(shù)f(x(關(guān)于點(diǎn)(1,1(對(duì)稱，故C錯(cuò)誤，D正確.【題8】(2023·遼寧葫蘆島·二模)已知函數(shù)f(x)=x3-x+1，則(A.f(x)有一個(gè)極值點(diǎn)B.f(x)有兩個(gè)零點(diǎn)C.點(diǎn)(0，1)是曲線y=f(x)的對(duì)稱中心D.直線y=2x是曲線y=f(x)的切線數(shù)的幾何意義判斷D.,(x(=3x2-1，令f,(x(>0得x>或x<-，令f,(x)<0得-<x<，所以f(x)在(-∞,-,+∞(上單調(diào)遞增，(-,上單調(diào)遞減，所以x=±是極所以，函數(shù)f(x(在(-∞,-上有一個(gè)零點(diǎn)，當(dāng)時(shí)，f(x(≥f((>0，即函數(shù)f(x(在(,+∞(上無零點(diǎn)，令h(x)=x3-x，該函數(shù)的定義域?yàn)镽，h(-x(=(-x(3-(-x(=-x3+x=-h(x(，將h(x)的圖象向上移動(dòng)一個(gè)單位得到f(x)的圖象，令f,(x(=3x2-1=2，可得x=±1，又f(1)=f(-1(=1，①x1>0②x3<43成等差數(shù)列析，根據(jù)g(x(=x(x-2+|可判斷①,根據(jù)函數(shù)的極大值可判斷②,根據(jù)三次函數(shù)的對(duì)稱性可判斷③,舉例可判斷④.(則g(x(的極小值為g 設(shè)g(x=x3-7x2+14x，則g,(x(=3x2-14x+14=3(x-(則g(x(的極小值為g /73/73對(duì)①,因?yàn)間(x(=x(x2-7x+14(=x(x-2+，(a-1(x-1x≤1(a-1(x-1x≤1<2；②若a>2，③當(dāng)a=2時(shí)，f,x>1,，綜上所述≤a≤2.【題11】(2024·河南·三模)設(shè)函數(shù)f(x(的定義域?yàn)镽,y=f(x-1(+1為奇函數(shù)，y=f(x-2(為偶函數(shù)，若f(2024(=1，則f(-2(=A.1B.-1【詳解】因?yàn)閥=f(x-1(+1為奇函數(shù)，所以f(-x-1(+1=-1-f(x-1(，所以f(x(的圖象關(guān)于點(diǎn)(-1,-1(中心對(duì)稱，則f(-1(=-1.因?yàn)閥=f(x-2(為偶函數(shù)，所以f(-x-2(=f(x-2(，所以f(x(的圖象關(guān)于直線x=-2軸對(duì)稱.由f(-x-1(+1=-1-f(x-1(，得f(-x-2(=-2-f(x(，所以f(x-2(=-2-f(x(，則f(x-4(=-2-f(x-2(=-2-f(-x-2(=f(x(，f(2024(=f(0(=-2-f(-2(=1，則f(-2(=-3.(1)f(x+a(=f(b-x(?f(x(關(guān)于軸對(duì)稱，(2)f(x+a(+f(b-x(=2c?f(x(關(guān)于中心對(duì)稱，(3)f(x+a(=f(x+b(?f(x(的一個(gè)周期為T=|a-b|，(4)f(x+a(=-f(x+b(?f(x(的一個(gè)周期為T=2|a-b|.可以類比三角函數(shù)的性質(zhì)記憶以上結(jié)論.【題12】(2024·四川·三模)已知定義在R上的函數(shù)f(x(在區(qū)間[-1,0[上單調(diào)遞增，且滿足f(4-x(=f(x(，f(2-x(=-f(x(，則()A.f(k(=0B.f(0.9(+f(1.2(>0C.f(2.5(>f(log280(D.f(sin1(<f(ln即可得結(jié)論.【詳解】對(duì)于A，因?yàn)閒(4-x(=f(x(，則函數(shù)f(x(關(guān)于直線x=2對(duì)稱，由f(2-x(=-f(x(，則函數(shù)f(x(關(guān)于點(diǎn)(1,0(對(duì)稱，所以f(4-x(=-f(2-x(，所以得f(2-x(=-f(-x(，則f(4-x(=f(-x(，故函數(shù)f(x(的周期為4，且f(-x(=f(x(，故函數(shù)f(x(為偶函數(shù)，因?yàn)楹瘮?shù)f(x(在區(qū)間[-1,0[上單調(diào)遞增，則函數(shù)f(x(的大致圖象如下圖：令x=1，由f(2-1(=-f(1(，所以f(1(=0，且f(4-1(=f(1(=0，令x=0，由f(4-0(=f(0(=f(4(，由f(2-x(=-f(x(得f(2(=-f(0(，所以f(1(+f(2(+f(3(+f(4(=0，根據(jù)對(duì)稱性，f(x(在[1,2[單調(diào)遞減，而f(1(=0，所以f(2(<0，因?yàn)楹瘮?shù)f(x(的周期為4，所以f(k(=[f(1(+f(2(+f(3(+f(4([×2+f(9(+f(10(=0+f(1(+f(2(=f(2(≠0，故A不對(duì)于B，由于f(1(=0，f(0.9(+f(1.1(=0，f(x(在[1,2[單調(diào)遞減，所以f(1.1(>f(1.2(，所以f(0.9(+f(1.2(<0，故B不正確；對(duì)于C，又f(log280(=f(log216+log25(=f(4+log25(=f(log25(=log22=log2>log25根據(jù)圖象f(x(在[2,2.5[上單調(diào)遞增，所以f(2.5(>f(log280(，故C不正確；對(duì)于C，f(ln(=f(-ln2(=f(ln2(，且0<ln2<0.7，因?yàn)?，因?yàn)閒(x(在[0,1[上單調(diào)遞減，所以f(sin1(<f(ln(，故D正確.g(2x-1(+1為奇函數(shù)，則函數(shù)y=f(x(圖象的對(duì)稱中心是【詳解】因?yàn)閥=g(2x-1(+1為奇函數(shù)，所以g(-2x-1(+1=-g(2x-1(-1，即g(-2x-1(+g(2x-1(=-2，由于函數(shù)y=f(x(與y=g(x(的圖象關(guān)于直線x=1對(duì)稱，且(-1,-1(關(guān)于x=1的對(duì)稱點(diǎn)為(3,-1(，故y=f(x(的對(duì)稱中心為(3,-1(.【題14】(2024·全國·模擬預(yù)測)已知函數(shù)f(x)=x3+ax2+bx+c下列結(jié)論中正確的是()=0C.若x0是f(x)的極小值點(diǎn)，則f(xD.函數(shù)y=f(x)的圖象是中心對(duì)稱圖形f(x)=2f(-成立即可判斷D.【詳解】A：因?yàn)閒(x)=x3+ax2+bx+c，所以f(x)=3x2+2ax+b，A錯(cuò)誤；所以若x0為f(x(的極小值點(diǎn)時(shí)，f(x(在(-∞,x0(上先遞增再遞減，故C錯(cuò)誤；D：f(--x(+f(x)=(--x(3+a(--x(2+b(--x(+c+x3+ax2+bx+c=a3-+2c，而f(-=(-3+a(-2+b(-+c=a3-+c，則f(--x(+f(x)=2f(-，確.-t-t左右同時(shí)取對(duì)數(shù)得log2=-，故得t=-12.43log2，故A錯(cuò)誤，而當(dāng)t=62.15時(shí)，N=N0?2-=2-5?N0=N0，由題意得0.4N0=N0?2-=-12.43log2=-12.43log2，=-12.43(log22-log25)=-12.43(1-log25)=-12.43(1-=-12.43(1-，A.f(0(=0B.f(-1(=e2C.exf(x(為奇函數(shù)D對(duì)C：令y=-x，則有f(x-x(=-+，即f(0(=exf(x(+e-xf(-x(，xf(x(=-e-xf(-x(，又函數(shù)f(x(的定義域?yàn)镽，則函數(shù)exf(x(的定義域?yàn)镽，故f(x(在(0,+∞(上不具有單調(diào)性，故D錯(cuò)誤.xA.y=|f(x)|的圖象也關(guān)于直線x=1對(duì)稱B.y=f(x)的圖象關(guān)于(1,2)中心對(duì)稱所以，|f(1-x)-2|=|f(1+x)-2|，所以f(1-x)-2=f(1+x)-2或f(1-x)-2=-f(1+x)+2，當(dāng)f(1-x)-2=f(1+x)-2時(shí)，f(1-x)=f(1+x)，y=f(x)的圖象關(guān)于直線x=1對(duì)稱，此時(shí)，a(1+x(3+b(1+x(2+c(1+x(+d=a(1-x(3+b(1-x(2+c(1-x(+d，∴a[(1+x(3-(1-x(3[+b[(1+x(2-(1-x(2[+c[(1+x(-(1-x([=0，[(1+x(2+(1+x((1-x(+(1-x(2[+b[(1+x(+(1-x([+c=0,3=-，∴此等式不成立，即f(1-x)-2=f(1+x)-2不成立,∴f(1-x)-2=-f(1+x)+2，即f(1-x)+f(1+x)=4，所以y=f(x)的圖象關(guān)于(1,2)中心對(duì)(0,f(0)(與(2,f(2)(關(guān)于(1,2)對(duì)稱，但此時(shí)，|f(-1)|=|-a+b-c+d|=|-4a-c+2|=|-6a+2|，|f(3)|=|6a+2|所以由|f(-1(|=|f(3(|可得a=0，但這與已知矛盾，所以y=|f(x)|的圖象不可能關(guān)于直線x=1對(duì)稱【題18】(2024·浙江紹興·二模)已知定義在R上的函數(shù)f(x(在區(qū)間[-1,0[上單調(diào)遞增，且滿足f(4-x(=f(x(，f(2-x(=-f(x(，則()A.f(k(=0B.f(0.9(+f(1.2(<0C.f(2.5(>f(log280(D.f(sin1(<f(ln項(xiàng)判斷即可得結(jié)論.【詳解】對(duì)于函數(shù)f(x(有，f(4-x(=f(x(，則函數(shù)f(x(關(guān)于直線x=2對(duì)稱，由f(2-x(=-f(x(，則函數(shù)f(x(關(guān)于點(diǎn)(1,0(對(duì)稱，所以f(4-x(=-f(2-x(，所以得f(2-x(=-f(-x(，則f(4-x(=f(-x(，故函數(shù)f(x(的周期為4，且f(-x(=f(x(，故函數(shù)f(x(為偶函數(shù)，因?yàn)楹瘮?shù)f(x(在區(qū)間[-1,0[上單調(diào)遞增，則函數(shù)f(x(的大致圖象如下圖：由對(duì)稱性可得f(1(+f(2(+f(3(+f(4(=0，所以f(k(=[f(1(+f(2(+f(3(+f(4([×2+f(9(+f(10(=0+f(1(+f(2(=f(2(≠0，故A不由于f(0.9(+f(1.1(=0，f(1.1(>f(1.2(，所以f(0.9(+f(1.2(<0，故B正確；又f(log280(=f(log216+log25(=f(4+log25(=f(log25(，=log22=log2>log25>2，所以f(2.5(>f(log280(，故C正確；f(ln(=f(-ln2(=f(ln2(，且0<ln2<0.7，所以f(sin1(<f(ln(，故D正確.=f(x)為奇函數(shù)．有同學(xué)發(fā)現(xiàn)可以將其推廣為：函數(shù)y=f(x)的圖象關(guān)于點(diǎn)P(a,b)成中心對(duì)稱圖形的充要條件是函數(shù)y=f(x+a)-b為奇函數(shù)．已知函數(shù)則下列結(jié)論正確的有()C.函數(shù)f(x)的導(dǎo)函數(shù)f(x)的圖象關(guān)于直線x=1對(duì)稱,則=4048x>0-h(x)，對(duì)于C，由選項(xiàng)B知，f(-x+1)-1=-[f(x+1)-1]，即f(1-x)+f(1+x)=2，兩邊求導(dǎo)得-f(1-x)+f(1+x)=0，即f(1-x)=f(1+x)，①存在常數(shù)a，b使得f(x)+f(2a-x)=2b?f(a+x)+f(a-x)=2b，則函數(shù)y=f(x)圖象關(guān)于②存在常數(shù)a使得f(x)=f(2a-x)?f(a+x)=f(a-x)，則函數(shù)y=f(x)圖象關(guān)于直線x=a對(duì)稱.【題20】(2024·湖北荊州·三模)已知函數(shù)f(x(的定義域?yàn)镽，且f(x+y)+f(x-y)=f(x)f(y)，f(1(=1，則C.f(x(是周期函數(shù)D.f(x(的解析式可能為f(x(=2cos【詳解】由f(x+y)+f(x-y)=f(x)f(y)，且函數(shù)f(x(的定義域?yàn)镽，對(duì)于選項(xiàng)C：令x=0，則f(y)+f(-y)=f(0)f(y)=2f(y(，則f(y)=f(-y)，即f(x)=f(-x)，可知f(x)為偶函數(shù)，令y=1，則f(x+1)+f(x-1)=f(x)f(1(=f(x(，可知f(x+2)+f(x)=f(x+1)，f(x+3)+f(x+1)=f(x+2)，可得f(x+3)+f(x)=0，則f(x+6)+f(x+3)=0，則f(1(=f(-1(=f(5(=1，不滿足f(1(+f(5(=0，且f(x+y)+f(x-y)=2cos+y(+2cosy(=(2cosx((2cosy(=f(x(f(y(，質(zhì)解決問題.【題21】(2024·江蘇宿遷·三模)已知定義在R上不為常數(shù)的函數(shù)f(x)滿足f(2x)+f(x+y)f(x-y)=0，則A.f(0)=-1B.f(3)=[f(1)]3C.f(x)f(-x)=2D.f(x)+f(-x)≤-2【詳解】對(duì)于A，令x=y，則f(2x令x=0,y=1，則f(0(+f(1(?f(-1(=0，得f(1(=--，令x=1