2025高考數(shù)學(xué)·魔鬼特訓(xùn)營(yíng)(-(-x≥0，x≤2，所以A=(-∞,2]．對(duì)于函數(shù)y=、2-x，由于2-x≥0，所以y=2-x≥0，所以B=[0,+∞)，所以A∩B=[0,2[．故選：B.【題4】A【解析】∵X={1,2,3,4}，Y=∴X-Y={1,2}，Y-X={5}，∴(X-Y(∪(Y-X(={1,2,5}，∴|(X-Y(∪(Y-X(|=3，故選：A.2+(a-4(x-2a≤0，即(2x+a((x-2(≤0，當(dāng)-≥2，即a≤-4時(shí)，2≤x≤-，此時(shí)A∩B=?,不合題意；故<2，即a>-4時(shí)，則B=≤x≤2，由于A={x∣-2≤x≤1}，A∩B={x∣-1≤x≤1}，所以-=-1，解得a=2，故選：C.-1，+a2<a1+a3<a1+a4<a2+a4<a3+a4，所以集合Y中至少有以上5個(gè)元素，則顯然x1x2<x1x3<x1x4<x1x5<x2x5<x3x5<x4x5，則集合S中至少有7個(gè)元素，對(duì)于C選項(xiàng)而言，?i≠j，xi≠xj，則與一定成對(duì)出現(xiàn)，-1-1(<0，所以|T|一定是11-|-x+(-可得M?{x∣x≤3}，{x∣x<0或x>1}∩{x∣x≤3}={x∣x<0或1<x≤3}，C.不等式x2+ax-b<0的解集為(x1,x2(，x1x2=-b<0，故C錯(cuò)誤.|x1-x2|=41+x2=-a，x1x2=b-c，則|x1-x2|2=(x1+x2(2-4x1x2=a2 ∈N ∈N{所以Q={1,2,3,6}，由Q={1,2,3,6}，P={1,3,4}，得U=P∪Q2+k(n1,n2∈Z(，22所以則有a-b=7(n1-n2(∈[0[，故正確；④當(dāng)a-b∈[0[時(shí)，則a-b=7n3(n3∈Z(，+k(n4∈Z(2∈C，之積為-8.若m2=m,m2{2=-8.∴A∪B={x∣x≥2}；(2)∵M(jìn)-N={x∣x∈M且x?N}，A={x∣x≥2}，B={x∣3<x<5}，∴A-B={x∣2≤x≤3或x≥5}.2-2ax-3a2<0可化為x2-2x-3<0，得-1<x<3.2-2ax-3a2<0(a>0(，得-a<x<3a.∵q是p的充分不必要條件，a≥{.∴{x∣a≥{.3a≥4,{x∣0≤x≤5}=[0,5[，所以{U∪A∩B=[0,1).x+≥1，解得x≥或x≤-，所以A=(-∞,-∪,+∞(，5+3a(x+5a=(3x+5((x+a(<0，當(dāng)-a<-，即a>，-a<x<-；當(dāng)-a>-，即a<時(shí)，-<x<-a，33又A=(-∞,-∪,+∞(，所以要使A∩B中包含2021個(gè)正整數(shù)，則a<，B=(-,-a(，A∩B=(-,-|∪,-a(，所以a∈[-2022,-2021).,p(2)正整數(shù)集合A={a1,a2,?,an{(1≤a1<a2<?<an,n為正整數(shù)，且n≥2)具有性質(zhì)P，即對(duì)任意的j(1≤i<j≤n，i、j均為正整數(shù))，兩數(shù)aiaj與中至少有一個(gè)屬ai都是an的因數(shù).j(1≤i≤j≤n(，aiaj與兩數(shù)中至少有一個(gè)屬于A44【題1】C【解析】因?yàn)锳={x|x2-4x-5<0{={x∣-1<x<5}，B={-3,-1,1,3,5}所以A∩B={1,3}．故選：C.3x+18<0可化為x2-3x-18>0，即(x-6((x+3(>0，即x>6或x<-3.所以不等式的解集為{x∣x>6或x<-3}．故選：A.2-kx-k的值恒為正值，則Δ<0?k2+4k<0?-4<k<0，∵(-4,0(?(-5,0(，【題4】A【解析】因?yàn)镸=2a(a-2(+7，N=(a-2((a-3(，所以M-N=(2a2-4a+7(-(a2-5a+6(=a2+a+1=(a+2+>0，因?yàn)閥=(x-1(2+5在區(qū)間2,|為增函數(shù)，55 min=-+1=1-≥0ra<0,{|則-1+3=-{|則a<0,即{ba<0,c=-3a,解得-<x<1，∴cx2-bx+a<0的解集為<x<1，D錯(cuò)誤；(t+3(ln(t+2(>(t+2(ln(t+3(，A正確；66所以(t+2(ln(t+1(>(t+1(ln(t+2(，所以(t+1(t+2>(t+2(t+1，B錯(cuò)誤；故g(x(在[2,+∞)上單調(diào)遞減，g(t+1(>g(t+2(，所以>故log(t+1((t+2(>log(t+2((t+3(，D正確；2+【題15】(-,+∞(【解析】因?yàn)椴坏仁?2a-b(x+所以x=是方程(2a-b(x+3a-4b=0的根，且2a-b<0，則不等式(a-4b(x+2a-3b>0，可化為-bx-b>0，-x-<0，解得x>-，即不等式(a-4b(x+2a-3b>0的解集為(-,+∞(．故答案為：(-,+∞(.72-(3m-2(x+2m2-m-3=0得，[x-(m+1([[x-(2m-3([=0，所以x=m+1或x=2m-3，r-5<m+1<4,-5<2m-3<4,得-1<m<3.所以實(shí)數(shù)m的取值范圍為{r-5<m+1<4,(2)設(shè)集合A={m∣-1<m<3}，集合B={m∣1-a<m<1+a}，1-a<1+a,當(dāng)B≠?時(shí)，1-a<1+a,1+a≤3,或{1-a≥-1,1或{1-a≥-1,1+a<3,所以存在實(shí)數(shù)a∈(-∞,2(，滿足條件.所以-1和是方程ax2+bx-2a+5=0的兩個(gè)根，,∴2020年的利潤(rùn)y=1.5x×-8-16x-m=36--m(m≥0(.2<0的解集為(-1,2(，∴方程ax2+bx-2=0的兩根分別為-1和2，r-1+2r-1+2=-b|-1×2=-288解得,1,∴f(x(=x2-x-2，令x+1=x2-x-2，解得x=-1或x=3，作出M(x(的圖象如下圖所示：r2k-1|k+1>0,k>1或k<-1,∴{≤r2k-1|k+1>0,k>1或k<-1,(2)設(shè)發(fā)射點(diǎn)與飛行物之間的水平距離為a千米，要使炮彈能擊中飛行物，則ka-a2=3.2，即則必有正數(shù)解.a2≤∵-==<0，9∴點(diǎn)P(a+c,b+d(是點(diǎn)(c,d由上式對(duì)?m∈{t∣0<t<2019,t∈Z}恒成立.即n-<k<n-，3-x≥0-1<x≤3，3-x≥0【題2】A【解析】由題設(shè)M={x∣x≥1或x≤-1}，N={y∣y≤0}，所以M∩N=(-∞,-1]．故選：A.rx+1,x≤0,-100,x>0,所以f=-100=0，所以f(t(=+1(t≠0(，因此f(x(=+1(x≠0(．故選：B.【題5】C【解析】函數(shù)f(x(為偶函數(shù)，則f(-2(=f(2(，f(-3(=f(3(，則f(2(>f(3(>f(π(，則f(-2(>f(-3(>f(π(，故選：C.f(x+1(為奇函數(shù)，即其圖象關(guān)于(0,0(點(diǎn)對(duì)稱(chēng)，所以f(x(的圖象關(guān)于(1,0(點(diǎn)對(duì)稱(chēng)，f(x+2(為偶函數(shù)，即其圖象關(guān)于y軸對(duì)稱(chēng)，因此f(x(的圖象所以f(1(=0，f(0(=-f(2(，f(3(=f(1(，所以f(1(=2a+b=0，f(0(+f(3(=-f(2(=-(4a+b(=所以x∈[1,2[時(shí)，f(x(=-3?2x+6，由對(duì)稱(chēng)性得f(x+2(=f(2-x(=-f(1-(1-x((=-f(x(，所以f(x+4(=-f(x+2(=f(x(，f(x(是周f(log296(=f(log296-4(=f(4-log296+4(=f(log2=f(log2=-3×+6=-2，【題7】B【解析】由f(x+1(=2f(x(知：f(x(=2f(x-1(，∴f(x(=2kf(x-k((k∈Z(，當(dāng)x∈(0,1]時(shí)，f(x(=x(x-1(=x2-x，則f(x(min=f=-；f(x(=2f(x-1(=2(x-1((x-2(=2x2-6x+4，則f(x(min=f=-；f(x(=4f(x-2(=4(x-2((x-3(=4x2-20x+24，則f(x(min=f=-1<-；作出函數(shù)f(x(的大致圖象如圖所示.∵f(x(≥-對(duì)任意x∈(-∞,m]恒成立，∴f(x(min≥-時(shí)，m≤,由f(1-x(=f(1+x(得f(-x(=f(x+2(，所以f(x(+f(x+2(=2，f(x+1(+f(x+3(=2，即f(x(+f(x+1(+f(x+2(+f(x+3(=4，所以[f(-1(+f(0(+f(1(+f(2([+[f(3(+f(4(+?+f(2021(+f(2022([=4×506=2024，所以f(1(+f(2(+f(3(+?+f(2022(=2024-f(-1(-f(0(=2023．故選：A.-4a≥0?0<a≤4；由②,令x=y=0得f(0(=0，令x=-y得f(-x(=-f(x(，所以函數(shù)為奇函數(shù)；故AB正確，C錯(cuò)誤；因?yàn)閒(x+1(-f(2-x(≤0?f(x+1(≤f(2-x(?x+1≤2-x，解得x≤,故D正確；所以x+y=[x[+a+[y[+b，[x+y[=[x[+[y[+[a+b]，所以[x[+[y[≤[x+y[，即選項(xiàng)B正確；-1<[x[≤x，所以0≤x-[x[<1，所以函數(shù)f(x(=x-[x[的值域?yàn)閇0,1)，t3[=1t5[=3，?,[tn[=n-2同時(shí)成立，nn-2≤t<nn-1，故選BCD.2y又奇函數(shù)f(x+2(是(-3,-1(上的減函數(shù)，所以f(x(是(-1,1(上的奇函數(shù)，且在(-1,1(上單調(diào)遞減.由f(m-1(+f(3-2m(<0，得f(m-1(<-f(3-2m(，所以f(m-1(<f(2m-3(，所以m-1>【題15】-2【解析】因?yàn)閒(x-2(以f(-x-2(=-f(x-2(，f(1-x(=f(1+x(，即f(-x-4(=-f(x(，f(2-x(=f(x(，所以f(-x-4(=-f(2-x(，f[-(-x+2(-4[=-f[2-(-x+2([，即f(x-6(=-f(x(，故f(x+6(=-f(x(，所以f(x+12(=f(x(，即周期為12.令x=0，由f(x+6(=-f(x(得f(6(=-f(0(，又f(0(-f(6(=4，得f(6(=-2，所以f(2034(=f(169×12+6(=f(6(=-2.故答案為：-2.2-18tb+4t2-1=0t(2-4×24×(4t2-x；(2)∵2>1，∴f(x(是增函數(shù)，f(2m-1(-f(m+3(<0，即f(2m-1(<f(m+3(，<x21-x2<0，x1x2>0，f(x1(-f(x2(=(x1-x2-=(x1-x2(+-=(x1-x2(+所以f(x1(<f(x2(，所以f(x(在區(qū)間(0,+∞(上單調(diào)遞增.(2)不等式f(2x(≤m?2x-5在x∈(-∞,0(上恒成立，等價(jià)于-++1≤m在x∈(-∞,0(上恒則有m≥-2t2+5t+1在t∈(1,+∞(上恒成立，則s(t(=-2t2+5t+1=-2(t-2+，【題19】【解析】(1)∵f(x(滿足f(x(=f(x+2(，∴f(1(=f(2-1(=f(-1(=-f(1(，∴f(1(=0，f(-1(=0.由f(x(是奇函數(shù)，∴f(x(=-f(-x(=-=-，x|4x+1,0<x<1,|--1<x<0.綜上，在x|4x+1,0<x<1,|--1<x<0.x+1,f(x(-g(x(=f(x(是奇函數(shù)，g(x(為偶函數(shù)，則f(-x(-g(-x(=，即-f(x(-g(x(=②,由①②得f(x(=，g(x(=.即g(x+5(<g(x-1(，(x+5(2+1(x-1(2+1(x+5(2+1(x-1(2+1，又定義域?yàn)閤∈(-∞,0(∪(0,1(∪(1,+∞x-1(+log2(x-2(>1，(x-1((x-2(>2,:{x(x-1((x-2(>2,x-2>0,:f(x(>1的解集為(3,+∞(.n,故0<a<1，:f(x(=loga(x-(x-a(L|Γ,而y=(x-(x-a(在x∈[m,n[上遞增，y=logax遞減，:f(x(在x∈[m,n[上遞減，故loga(m-(m-a(=logam,loga(n-(n-a(|=logan,:(m-(m-a(=m,n-(n-a(=n,即m，n是(x-(x-a(=x的兩個(gè)不同的實(shí)根，:g(x(=x2+1(x+在(a,+∞(上有m，n兩個(gè)不【題22】【解析】(1)“函數(shù)f(x(=ax-q.a-x(a>0且a≠1(是定義域?yàn)镽的奇函數(shù)，代入原函數(shù)，則有f(-x(=-f(x(，所以f(x(=ax-a-x，“f(1(=，:a-=，2a2-3a-2=0，解得a=2或a“a>0，:a=2，f(x(=2x-2-x，任取實(shí)數(shù)x1<x2，則f(x1(-f(x2(=2x-2-x-(2x-:f(x1(<f(x2(，:f(x(是單調(diào)增函數(shù)；(2)g(x(=log(m-2([a2x+a-2x-mf(x(+1[=log(m-2([22x+2-2x-m(2x-2-x(+1[=log(m-2([(2x-2-x(2-m(2x-2-x(+3[，設(shè)t=2x-2-x，x-2-x(2-m(2x-2-x(+3=t2-mt+3，記h(t(=t2-mt+3，t(min=1，∴h(t(在,|上單調(diào)遞增，-x=1，所以f(0(=20-m=1-m=0，m=1，所以f(-2(=-f(2(=-(22-1(=-3．故選：C.【題3】B【解析】因?yàn)閥=lnx，y=-為(0,+∞(上的單調(diào)遞增函數(shù)，因?yàn)閒(1(=ln1-=-3<0，f(2(=ln2-<0，f(3(=ln3->0ax在定義域(0,+∞(上單調(diào)遞增，【題5】D【解析】因?yàn)楹瘮?shù)f(x(是奇函數(shù)，所以f(2-x(=-f(x-2(，因?yàn)閒(x(=f(2-x(，所以f(x(=-f(x-2(，所以f(x-2(=x-2，所以f(x(=-f(x-2(=2-x．故選：D.∴f(x(的圖象既有對(duì)稱(chēng)中心又有對(duì)稱(chēng)軸，但f(x(不一定具由f(-x(+f(x(=0，則f(x(為奇函數(shù)，故選項(xiàng)A錯(cuò)誤；由f(x-1(=f(1-x(，可得函數(shù)f(x(圖象關(guān)于x=0對(duì)稱(chēng)，故選項(xiàng)B錯(cuò)誤；∴f(2a+x(=f(-x(，f(-x(=-f(2b+x(，∴f(2a+x(=-f(2b+x(，∴f(2a+x-2b(=-f(2b+x-2b(=-f(x(，∴f(x+4a-4b(=-f(x+2a-2b(=f(x(，∴f(x(的一個(gè)周期T=4(a-b(，故選項(xiàng)C正確．故選：C.f(x(=，f(-x(=-f(x(，得f(x(為奇函數(shù)，其圖象關(guān)于原點(diǎn)對(duì)稱(chēng)，當(dāng)a=-1，c=-1，即有f(x(=-(x≠±1(，f(-x(=-當(dāng)b=-1，c=-1，即有f(x(=-(x≠±1(，f(-x(=f(x(，可得f(x(為偶函數(shù)，其圖象關(guān)于y軸f(x(有2個(gè)根，當(dāng)<t<時(shí)，t=f(x(有6個(gè)根，故關(guān)于x的方程m[f(x([2+nf(x(+1=0恰有7個(gè)不同的實(shí)數(shù)根，則t1=，t2=需為mt2+nt+122=>y22-x，log2x=2-x，因?yàn)閒(x(=2x+x-2在R上單調(diào)遞增，+(2-a(2=2(a-1(2+2<a<1(，故選ABD.x=3y=z=t(t>1(，故選AC.所以y=ax+1(a>0,a≠1(恒過(guò)定點(diǎn)(-1,1(.故答案為：(-1,1(.-x，f(x(=-1=，f(x(>0，即當(dāng)x=1時(shí)，f(x(取得極大值為f(1(=-1；當(dāng)x≤0時(shí)，f(x(=-x-2為減函數(shù)，且f(x(≥-2，函數(shù)f(x(的圖象如圖.設(shè)f(x1(=f(x2(=t，由題可知-2≤t≤-1，由f(x1(=t得-x1-2=t，則x1=-t-2，則x1f(x2(=t(-t-2(=-(t+1(2+1，∵-2≤t≤-1，∴當(dāng)t=-1時(shí)，x1f(x2(取得最大值為1．故答案為：1.當(dāng)0<a<1時(shí)，此時(shí)f(x(，g(x(遞減，g(x(在[3,4[上單調(diào)遞減，所以f(x1(min=f(2(=loga(9-a2(，g(x2(min=g(4(=loga(16-4a(，所以loga(9-a2(≥loga(16-4a(?9-a2≤16-當(dāng)1<a<3時(shí)，由復(fù)合函數(shù)單調(diào)性可知f(x(在[1,2[上單調(diào)遞減，g(x(在[3,4[上單調(diào)遞增，所以f(x1(min=f(2(=loga(9-a2(，g(x2(min=g(3(=loga(9-3a(，所以loga(9-a2(≥loga(9-3a(?9-a2≥9-3a，即a當(dāng)x≤-2時(shí)，g(x(=-x-x-2=-2x-2，當(dāng)-2<x<0時(shí)，g(x(=-x+x+2=2，-∞,-2(，g(x(的值域?yàn)閇2,+∞).f(x(-2=-2=x+，-x(=-x=-g(x(，所以g(x(為奇函數(shù)，當(dāng)a≠0時(shí)，定義域?yàn)?-∞,0(∪(0,+∞(，且g(-x(=-x-=-g(x(，所以g(x(為奇函數(shù)，即f(x(==x++2>0在x∈[整理為a>-x2-2x在x∈[1,+∞)上恒成立，令h(x(=-x2-2x=-(x+1(2+1，所以a>-3，故實(shí)數(shù)a的取值范圍為(-3,+∞(.∴可設(shè)y=mx(m>0,x>0(，∴生產(chǎn)A芯片的毛收入y(千萬(wàn)元)與投入的資金x(千萬(wàn)1,,∴生產(chǎn)B芯片的毛收入y(千萬(wàn)元)與投入的資金x(2)設(shè)對(duì)B芯片投入的資金為x千萬(wàn)元，則對(duì)A芯片投入的資金為(40-x(千萬(wàn)元，設(shè)凈利潤(rùn)為W千萬(wàn)則W=x+(40-x(-2(0<x<40(，千萬(wàn)元.-4m(實(shí)數(shù)m∈Z)的圖象關(guān)于y軸對(duì)稱(chēng)，且f(2(>f(3(，所以f(x(在區(qū)間(0,+∞(上為單調(diào)遞減函數(shù)，又由m∈Z，且函數(shù)f(x(=xm-4m(實(shí)數(shù)m∈Z)的圖象關(guān)于y軸對(duì)稱(chēng)，所以m2-4m為偶數(shù)，所以m=2，所以f(x(=x-4.解得-<a<或<a<3，x+x+x=，((2)假設(shè)存在實(shí)數(shù)a，使得f(x(與g(x(在區(qū)間[a+2,a+3[上是“友好”的，則|f(x(-g(x(|=|loga(x2-4ax+3a2(|≤1，即-1≤loga(x2-4ax+3a2(≤1，由復(fù)合函數(shù)的單調(diào)性可得y=loga(x2-4ax+3a2(在區(qū)間[a+2,a+3[上為減函數(shù)，min=loga(9-6a(，解得0<a≤,當(dāng)<a<1時(shí)，f(x(與g(x(在區(qū)間[a+2,a2-4≤0，得A={x∣-2≤x≤2}，,(t(=4t+所以f(x(的單調(diào)增區(qū)間是(2,+∞(．故選：B.,(x(=3x2x3有“巧值點(diǎn)”；對(duì)于B選項(xiàng)，f,(x(=-，由f(x(=f,(x(可得-lnx=0，其中x>0，對(duì)于C選項(xiàng)，f,(x(=-e-x，由f(x(=f,(x(可得e-x=0，這與e-x>0矛盾，所以函數(shù)f(x(=e-x沒(méi)有對(duì)于D選項(xiàng)，f,(x(=-cosx，因?yàn)閒=-=f,，所以函數(shù)f(x(=-sinx有“巧值點(diǎn)”．故f,(x(=3x2-4cx+c2=3(x-(x-c(，,(x(=3(x-(x-2(，,(x(>0，,(x(=3(x-2((x-6(，,(x(>0,(x(<0，所以f,(x(=3ax2+2(3a+1(x+4=(3ax+2((x+2(，故f(x(=-x3-2x2+4x．故選：B.f1(x(=2cos2x-2sin2x，所以f2022(x(=22022(-sin2x-cos2x(．故選B.令f(x(=x-lnx，x>1，則f(x(=1->0，函數(shù)f(x(在(1,+∞(上單調(diào)遞增，有f(x(>f(1(=1t(=<0，g(t(在(1,+∞(上單調(diào)遞減，根據(jù)平均變化率的概念可知若函數(shù)f(x(從x1到x2平均變化率即為割線AB的斜率，即AB的斜率，從而f(x0+△x(≈f(x0(+Δxf(x0(，確.【題10】BC【解析】因?yàn)閒(x(為R上的奇函數(shù)，故f(-x(=-f(x(，所以-f(-x(=-f(x(，即f(-x(=f(x(，所以f(x(為偶函數(shù)，又f(-2(=-f(2(，由圖可得f(2(>2，f(1(=2，故f(-2(<-2，故A錯(cuò)誤.而f(-1(?f(-2(=-f(1(×[-f(2([=f(1(?f(2(>2×2=4，故B正確.f(-1(?f(-2(=f(1(?f(2(，由圖可得f(1(>0，f(2(<0，故f(-1(?f(-2(<0，所以C正確.因?yàn)閒(x(=3x2-6=3(x+、2((x-、2(，所以當(dāng)x∈(-∞,-、2(時(shí)，f(x(>0；f(x(>0，C正確.∴f(x(+、2sinx，又函數(shù)f(x(=-1(-、2cosx在區(qū)間(0,的最小值為-1，又f=a(e--1(-、2cos=-1，可得原條件的一個(gè)必要條件f=0，令g(x(=-e=-e，f=-e-、2>0，∴函數(shù)f(x(在區(qū)間(0,的最小值為f=-1，∵f(x(=-e+、2sinx，又f點(diǎn)x0，f(x0(=0-1(-、2cosx0=、2sinx0-1-、2cosx0=2sin(x0--1，0-∴f(x0(=2sin(x0--1<2-1=1.-e【解析】設(shè)P(x,lnx(，設(shè)g(x(=x2+2?(lnx(2，此時(shí)直線OP的方程為y=-x，設(shè)直線y=-x與曲線y=alnx相切于點(diǎn)(x0,alnx0(，由y=alnx?y/=?=-1?x0=-a，顯然(x0,alnx0(在直線y=-x上，則alnx0=-x0，因此有aln(-a(=a?a=-e，故答案為：-e.得g/(x(=-2x2+3x-1=-(2x-1((x-1(，而g=--<g(1(=-，所以g(x(max=g(1(=-2∈,2f(x1(≥g(x2(成立，則只需f(x(min≥g(x(max即可，由函數(shù)f(x(=+lnx-，可得f/(x(=-=，f/(x(>0，f(x(在,2|遞增，f(x(min=f=3a-ln3-，故3a-ln3-≥-，解得a≥ln3+(舍去)；所以f(x(在(0,a(上單調(diào)遞減，在(a,+∞(上單調(diào)遞增，f(x(<0，f(x(在,2|遞減，f(x(min=f(2(=+ln2-，所以.f(x(在點(diǎn)(1,f(1((處的切線方程為y-9=12(x-1(，即12x-y-3=0.則f(x(=-3x2+6x+9=-3(x+1((x-3(，列表如下：x(-∞,-1(-1(-1,3(3f(x(-0+0-f(x(減增減所以函數(shù)f(x(的極小值為f(-1(=-7，極大值為f(3(=25.f(0(=c=0,f(0(=c=0,所以函數(shù)f(x(的解析式是f(x(=x3-3x2+3；f(1(=1，則有切線l的方程為y-1=-3(x-1(，于是得x2(2ax+b(-(2x-1((ax2+bx+c(=1，化簡(jiǎn)得(a-b(x2+(b-2c(x+c=1,所以函數(shù)f(x(的解析式為f(x(=2x2+2x+1；8x-4a，f(2(=-4-4a，由于曲線y=f(x(在點(diǎn)(2,f(2((處的切線與直線2x-4y+3=0垂直，故-4-4a=-2，解得a=-；(2)函數(shù)f(x(在區(qū)間[0,6[上單調(diào)遞減，則fl(x(=3x2-8x-4a≤0在區(qū)間[0,6[上恒成立，故a≥x2-2x，當(dāng)x∈[0,6[時(shí)，x2-AO2=、52-32=4，所以SO1=r4-r，所以V(r(=πr2(4-r(=π(3r2-r3(，0<r<3.所以Vl(r(=π(6r-3r2(，令Vl當(dāng)r∈(0,2(時(shí)，Vl(r(>0，所以V(r(在(0,2(上單調(diào)遞增；當(dāng)r∈(2,3(時(shí)，Vl(r(<0，所以V(r(在(2,3(上單調(diào)遞減.所以當(dāng)r=2時(shí)，V(r(取得最大x-1-2x，fl(x(=2ex-1-2，k=fl(2(=2e-2，切點(diǎn)(2,2e-4(，切線方程為y=(2e-2((x-2(+2e-4，即y=(2e-2(x-2e；(2)設(shè)g(x(=x-lnx-1(x>1(，則gl(x(=1-，x>1所以2ex-1-a(x-lnx-1(-2x>0?a<--，設(shè)h(x(=lnx-x+1+(x>1(，hl(x(=-1+(x-1(=>0，所以h(x(在(1,+∞(上遞增，h(x(>h(1(=0，即lnx-x+1+>0，x-lnx-1<，設(shè)p(x(=ex-1-x-(x>1(，pl(x(=ex-1-1-(x-1(=ex-1-x，再設(shè)q(x(=ex-1-x，則ql(x(=ex-1-1，所以ex-1-x>，所以f(x(=(2e-x(lnx，fl(x(=-1-lnx(x>0(，設(shè)g(x(=-1-lnx(x>0(，則g(x(=所以g(x(單調(diào)遞減，即f(x(在(0,+∞(上單調(diào)遞減，當(dāng)x∈(e,+∞(時(shí)，f(x(<0，f(x(單調(diào)遞減.<x2，設(shè)F(x(=f(x(-f(2e-x(=(2e-x(lnx-xln(2e-x(，1<x≤e，則F(x(=--2-ln[x(2e-x([，因?yàn)?<x(2e-x(=-(x-e(2+e2≤e2，所以F(x(≥-2-lne2=0，所以F(x(單調(diào)遞增，所以F(x(≤F(e(=0，f(x(<f(2e-x(，所以f(x1(<f(2e-x1(，又f(x1(=f(x2(，所以f(x2(<f(2e-x1(，，f(x(在(e,+∞(上單調(diào)遞減，設(shè)φ(x(=(x>1(，則(x(=設(shè)h(x(=1--lnx(x>1(，則h(x(=-=<0(x>1(，所以(x1-1(ln(2e+1-x1(<(2e-x1(lnx1，即[2e-(2e+1-x1([ln(2e+1-x1(<(2e-x1(lnx1，所以f(2e+1-x1(<f(x1(，又f(x1(=f(x2(，所以f(2e+1-x1(<f(x2(，，f(x(在(e,+∞(上單調(diào)遞減，2x<33，因?yàn)閤∈N*因?yàn)锽={1,2,3,4}，所以A∪B={1,2,3,4,5,6,7}，-3x>0?a>3x>0，log(a-3x(=x-2?a-3x=32-x?a=+3x≥2?3x=6，當(dāng)且僅當(dāng)=3x?x=1時(shí)取等【題4】C【解析】∵拋物線y=-x2-2x+m∴m<-2，∴-m>2，-1<n<m，-f(n(>m-n，故f(m(-m>f(n(-n.設(shè)g(x(=f(x(-x，則g(m(-g(n(>0，所以函數(shù)g(x(在區(qū)間(-1,+∞(上單調(diào)遞增，故g,(x(=f,(x(-=2x+-a-≥0在區(qū)間(-1，+∞)上恒成立，即2x+-≥a在區(qū)間(-1,+∞(上恒成立，-=2(x+1(+-2-≥,當(dāng)且僅當(dāng)x=0時(shí)，等號(hào)成立，所以f,(x(=+2bx，所以f(x(=2lnx-x2+3，f,(x(=-2x=，所以f(x(在(0,1(上單調(diào)遞增，在(1,+∞(上單調(diào)遞減，符合題意，所以f(3(=2ln3-6．故選：C.1(a<0(可得f(x(=3x2+6ax，故極小值點(diǎn)為x0=-2a，由f(x1(=f(x0(可得x+3ax+1=(-2a(3+3a?(-2a(2+1，整理得(x1-a((x1+2a(2=0，?f(x1+x0(=af(-a(=a(2a3+1(，f(x(<0，即函數(shù)f(x(在(-∞,0(上單調(diào)遞減，在(0,+∞(上單調(diào)遞增，所以f(x(min=f(0(=0，0.1-1>0.1，∴h(x(=tanx-x在x∈(0,上單調(diào)遞增，令g(x(=lnx-x+1，則g(x(=-1=，x-1-tanx(x∈(0,0.1](，則t(x(=ex-，∴t(x(在(0,0.1]上單調(diào)遞增，∴t(0(<t(0.1(，∴t(0.1(=e0.1-1-tan0.1>e0-1-tan0=0，0.1-1>tan0.1=c，的圖象可得，當(dāng)x<-4時(shí)，f,(x(<0，當(dāng)x>-4時(shí)，f,(x(≥0(僅在x=0處有f,(x(=0)，故函數(shù)y=f(x(在(-∞,-4(上單調(diào)遞減，在(-4,+∞(上單調(diào)遞增．所以x=-4y=f(x(在區(qū)間(-4,1(上單調(diào)遞增，故選項(xiàng)C正確.由函數(shù)y=f,(x(的圖象可得0=f,(0(<f,(1(，所以y=f(x(在x=1處切線的斜率大于零，故選項(xiàng)D正確.a-2=4【題11】BCD【解析】因?yàn)閒(x+2(是偶函數(shù)，所以f(x+2(=f(-x+2(，又因?yàn)閒(x(是奇函數(shù)，所以f(-x+2(=-f(x-2(，所以f(x+2(=-f(x-2(，所以f(x+4(=-f(x(，所以f(x+8(=-f(x+4(=f(x(，所以f(x(的周期為8，故A錯(cuò)誤；又當(dāng)x∈[0,2[時(shí)，f(x(=x2+2x，所以f(-3(=f(5(=-f(1(=-3，選項(xiàng)B正確；f(2022(=f(6+252×8(=f(6(=-f(2(=-8，選項(xiàng)C正確；f(2023(=f(7+252×8(=f(7(=-f(3(=f(-3(=-3，選項(xiàng)D正確.【題12】ABD【解析】由函數(shù)y=f(x(為偶函數(shù)可得，f(-3(=f(3(，因?yàn)閒(x+6(=f(x(+f(3(，令x=-3可得f(3(=f(-3(+f(3(=2f(3(，所以f(3(=0，B正確；所以f(x+6(=f(x(+f(3(=f(x(，即f(x+6(=f(x(，所以f(x(為周期為6的函數(shù)，又f(-2(=-1，所以f(2024(=f(337×6+2(=f(2(=f(-2(=-1，A正確；函數(shù)在[0,3[上為增函數(shù)，由周期性得到函數(shù)在[-6,-3[上是增函數(shù)，又x=-6為對(duì)稱(chēng)軸，則函數(shù)在[-9,-6[上是減函數(shù)，C錯(cuò)誤；因?yàn)閒(3(=0，所以f(-3(=0，f(9(=f(6+3(=0，f(-9(=f(9(=0，結(jié)合函數(shù)的周期為6及函數(shù)的增減性可得方程f(x(=0在[-9,9[上僅有4個(gè)根，D正確.,,≤0,所以f(-1(=a+1，又因?yàn)閍>-1，所以f(-1(=a+1>0，所以f[f(-1([=f(a+1(=2a+1=4=22，得(x-1((x-a(<0，+bx-4≥0，其對(duì)稱(chēng)軸為t=-．該二次函數(shù)在-,+∞(上是增函數(shù).x(=2--==，x∈(0,+∞(，∴曲線y=f(x(在點(diǎn)(1,f(1((處的切線方程為y-5=-2(x-1(，即2x+y-7=0.(2)令f,(x(<0得0<x<；令f,(x(>0得x>，∴f(x(的單調(diào)減區(qū)間是(0,f(x(的單調(diào)增區(qū)間是,+∞(.-x+x-2=+x-2，所以f,(x(=故當(dāng)a≤0時(shí)，f(x(單調(diào)遞增；(2)f(x(≥0等價(jià)于a≥(2-x(ex.令函數(shù)g(x(=(2-x(ex，則g,(x(=(1-x(ex，當(dāng)x∈(-∞,1(時(shí)，g,(x(>0，g,(x(<0，g(x(單調(diào)遞減.所以f(x(在(e,f(e((處的切線方程為y-e=2(x-e(，即y=2x-e；(2)若2f(x(≥g(x(對(duì)任意的x∈,e恒成立，則2,(x(=0得x=1，,(x(<0，h(x(單調(diào)遞增，,(x(=2ax-lnx-1.因?yàn)楹瘮?shù)f(x(在(0,+∞(上單調(diào)遞增，所以當(dāng)x>0時(shí)，令g(x(=(x>0(，則g,(x(=-，,(x(>0,(x(<0，,即ex-ex<lnx+.,則h,(x(=ex-1只需證ex-,即ex-ex<lnx+.,則h,(x(=ex-1所以lnx+≥0.易知φ(x(在(0,1(上單調(diào)遞增，在(1,+故f(x(的單調(diào)遞減區(qū)間為(-2,0(，單調(diào)遞增區(qū)間為(-∞,-2)，(0,+∞(；設(shè)函數(shù)g(x(=(-2≤x≤0(，-1,0(，且em=-m3，所以當(dāng)f(x(在[-2,0[上有2個(gè)零點(diǎn)時(shí)，≤a<-.f(x(的定義域?yàn)?0,+∞(，f(x(=lnx-2ax，由題意得f(x(=0在(0,+∞(上有兩解，令g(x(=(x>0(，即g(x(的圖象與直線y=2ah(x(=xlnx-2x+(x>0(，h(x(=lnx-1-，令m(x(=lnx-1-，則m(x(=，當(dāng)x>0時(shí)，m(x(>0，所以h(x(在(0,+∞(上單調(diào)遞增，1-，lnx0-1-，，h(x(單調(diào)遞減；,+∞(時(shí)，h(x(>0，h(x(單調(diào)遞增，∴h(x(min=h(x0(.0，∴l(xiāng)nx0-1-，∴h(x0(=x0lnx0-2x0+=-x0+=-x0+<-e+<0.又∵h(yuǎn)(1(=0，h(x(在(0,x0(上單調(diào)遞減，∴h(x(在(0,x0(上有一個(gè)零點(diǎn).∵h(yuǎn)(x(在(x0,+∞(上單調(diào)遞增，且h(e2(=，∴h(x(在(x0,+∞(上有一個(gè)零點(diǎn).∴cosθ+cosθ-sinθ=cosθ-sinθα-1=-，cos(2α-=cos2αcos+sin2αsin=-×+(--(2α+=-cos(2α+=-．故選：C.故選B.f(π+x(的定義域?yàn)?-π,π(，且f(π+x(=[ln(π+x(+ln(π-x([?sin(π+x(=-[ln(π+x)+ln(π-x([?sinx，記g(x(=f(π+x(，則有g(shù)(-x(=-[ln(π-x(+ln(π+x([Isin(-x(=[ln(π-x(+ln(π+x([?sinx=-g(x(，f(π-x(=[ln(π-x(+ln(π+x([?sin(π-x(=[ln(π-x(+ln(π+x([?sinx=-f(π+x(，故f(x(的圖象關(guān)于點(diǎn)(π,0(對(duì)稱(chēng)，選項(xiàng)A錯(cuò)誤；令f(x(=0，則有[lnx+ln(2π-x([?sinx=0，即lnx+ln(2π-x(=0或sinx=0，排除AC；y=sin(2x-的圖象上所有點(diǎn)橫坐標(biāo)擴(kuò)大到原來(lái)的2倍， f(x(單調(diào)遞減，故C正確；故f(x(=3sin(2x+φ(因?yàn)閒(-=3sin2×(-+φ=3，一個(gè)單調(diào)遞增區(qū)間為-,，g(x(在0,|上只有一個(gè)零點(diǎn)，選項(xiàng)BD正確.2x<log2yy設(shè)f(x(=log2x，顯然f(x(在區(qū)間(0,+∞BC.2<xy.x<y，∴xy<02<xy不成立.2<xy不成立.y-x>0，則y-x+1>1，ln(y-x+1(>0，故C正確；x-y<20=1∴sinxcosx=，∴f(t(=t+=t2+t-，t∈[-2,2[，∴當(dāng)t=-1時(shí)，f(t(min=-1-=-1，即f(x(的最小值為一1．故答案為：-1.由-π≤x2<x1≤π,得x2=-，x1=，則x1-2x2=-2×(-=．故答案為：.k-1∴sin(π-α(=sinα=；所以f(x(的單調(diào)遞減區(qū)間為kπ-,kπ+，k∈Z.(1(f(x(=2sin(x+、3cos(x++cos(x-所以函數(shù)f(x(的對(duì)稱(chēng)中心為-,1(，k∈Z.令g(x(=f(x(-m2+2m=0，得f(x(=m2-2m，則0≤m2-2m≤3，解得-1≤m≤0或2≤m≤3，所以m的取值范圍是[-1,0[∪[2,3[.根據(jù)圖象可得，(-,0(是f(x(故函數(shù)的對(duì)稱(chēng)中心為-,0(，k∈Z.到y(tǒng)=sin2(x--=sin(2x-=-cos2x的圖象，即g(x(=-cos2x，令2kπ-π≤2x≤-≤x≤kπ,k∈Z-60°-45°=75°,、即PM==2PB=(、3-1(4同理可得PN=(、3-1(PC，所以PM+PN=(3-1((PC+PB(=(3-1(BC=400(3-1(為定值；MN2=PM2+PN2-2PM?PNcos60°,即MN2=(PM+PN(2-3PM?PN≥(PM+PN(2-3×,所以MN2≥,MN≥,故MN≥200(、3-1(，當(dāng)且僅當(dāng)PM=PN=200(、3-1故當(dāng)P點(diǎn)是MN的中點(diǎn)時(shí)，三條小徑(PM,PN,MN( 、△CPN=4?(3、△CPN=所以S△BPM+S△CPN=?(3-1((PB2+PC2(=[(PB+PC(2-2PB?PC[≥(PB+PC(2-2×=(PB+PC(2=BC2=20000(3-、3(.x-(x+1(，令f(x(=ex-(x+1)，x∈R，g(-1(=1-cosx+ex-(x+1(=ex-cosx-x，∵cosx∈[-1,1[，∴ex-cosx-x≥ex-x-1≥0，即g(-1(≥0.(2)g(1(=cosx-1+ex-(x+1(=cosx+ex-x-2，令h(x(=cosx+ex-x-2，求導(dǎo)h,(x(=ex-sinx-1，hⅡ(x(=ex-cosx，h川(x(=ex+sinx-1,1(，可知函數(shù)h川(x(在(-1,1(上單調(diào)遞增，x(單調(diào)遞,(x(單調(diào)遞增，,(x(取得極小值也是最小值，且h,(0(=0，所以h,(x(≥0，即函數(shù)h(x(在(-1,1(上單調(diào)遞增，-isin=22-22i=cos+isin，所以復(fù)數(shù)cos-isin的輻角主值是．故選：D.2-α[=cos(α+β(cosα+sin(α+β(sinα、、∴BO=BG=BG∴BO=BG=BG∵B=B+A=B+52-1A—=B+52-1(B-B(=BA+BO=BA+BO【題6】D【解析】由正弦定理及題意，得(sinA-sinAcosB(sinB=(sinB-sinCcosC(sinA，∴sinAsinBcosB=sinCsinAcosC.∴P=(2cosθ-x,2sinθ-y(=,-.∴(2cosθ-x(2+(2sinθ-y(2=2+(-2=.-2(2xcosθ+2ysinθ(=，則zn=cosnθ+isinnθ,n∈N*；(1+i(2(i(1+i(2(i-1((i+1(故0<sinA<sinB，從而sin2A<sin2B，即1-cos2A<1-cos2B，2A>cos2B.∠ADC，s可以求出AC，再利用∠ACB、AC解直角△ABC得到AB的值．故選ACD.對(duì)于B，當(dāng)x>0時(shí)，f(1+x(=2f(1-x(，對(duì)f(1+x(=2f(12f,(1-x((-1(=-2f,(1-x(，1+x(+f,(1-x(=-2f,(1-x(+f,(1-x)=-f,(1-x(<0，所以f,(1-x(>0，令1-x=μ,∴μ<1，f,(μ(>0，所以f(x(在(-∞,1]上單調(diào)遞增，所以B錯(cuò)；對(duì)于C，由B知，f(x(在(-∞,1]上單調(diào)遞增，同理可得f(x(在(1,+∞(上單調(diào)遞減，，則f(x1(<f(x2(x2<2；-x2<1，而f(x2(=2f(2-x2(，則f(x2(-f(2-x2(=f(2-x2(<0，所以f(x2(<f(2-x2(?f(x1(<f(2-x2(，而x1，2-x2<1，所以x1<2-x2?x1+x2<2，故C正確；對(duì)于D，由f(x(在(-∞,1]上單調(diào)遞增，(1,+∞(上單調(diào)遞減，知f(x(≤f(1(=0，2∈又x1<2-x2?πx1<π(2-x2(，所以f(x1(>f(x2(?,當(dāng)x≥0時(shí)，f(1+x(=2f(1-x(，-x1(?f(x2(2∈2<1+x2<2?x2<2-x1，所以f(x2(-x1(?f(x2(2-b2=3，故t的取值范圍為(-∞,-|U{6}，故答案為：(-∞,-|U{6}.DA.BC=DB.AC=DC.AB=0，:DA丄BC，DB丄AC，DC丄AB，:B=,，:B=2+所以(a-b(×sinB+(sinA+sinC((c-a(=0，由正弦定理得(a-b(×b=(a+c((a-c(，2-4CD因?yàn)閟inB<sinC，故B<C，“A=A，:C=C+A=-A—+A.(2)設(shè)A=A+λB=A+μC，將B=-A+A，C=-A+A，代入A+λB=A+μC，故A=A++A(=A+A，則Rt△ADF三Rt△BCE，:AF=BE，當(dāng)點(diǎn)N在BC上時(shí)，n∈,MN2=m2+n2-2mncos=m2+n2-4≥2mn-4=4，當(dāng)點(diǎn)N在CD上(不包括端點(diǎn)C)時(shí)，四邊形BCN所以tan∠EBG=所以tan∠EBF=tan(∠FBG-∠EBG(2=AB2+AC2-所以DC=BC=，=1AB2+2A.A+A2.2因?yàn)锳=A，所以A=A+A(=A+A，又A.E=，所以A+A(.(μA-λA(=，6μ-λ=(2)，λ=,λ=-,f,(x(>0，f(x(單調(diào)遞增，，f(x(單調(diào)遞減，“4>π>e>1，:f(4(<f(π(<f(e(．即a<c<b.,(x(>0,(x(<0 不妨設(shè)x2>x12>>x1>0，h,(e-2(=1+lne-2=-1，則h(x(在點(diǎn)(e-2,-2e-2(處的切線方程為y+2e-2=-(x-e-2(，即y=-x-e-2，設(shè)直線y=-x-e-2與直線y=-m的交點(diǎn)橫坐標(biāo)為x3，則x3=m-e-2，3≤x1，設(shè)H(x(=xlnx-(-x-e-2(=xlnx+x+e-2，則H(x(=lnx+2，0<x<e-2時(shí)，H(x(<0，H(x(單調(diào)遞減，e-2<x時(shí)，H(x(>0，H(x(單調(diào)遞增，所以H(x(min=H(e-2(=0，所以H(x(≥0，即xlnx≥-x-e-2，即h(x(的圖象在切線y=-x-e-2同理得h(x(在點(diǎn)(1,0(處的切線方程為y=x-1，設(shè)直線y=x-1與直線y=-m的交點(diǎn)橫坐標(biāo)為x4，則x4=-m+1，同理可得，h(x(的圖象在切線y=x-1的上方(只有點(diǎn)(1,0(是公共點(diǎn))，如圖所示，所以x4>x2，2-x1<x4-x3=-m+1+e-2-m，即x2-x1<e-2+1-2m．即|x1-x2|+2m<e-2+1.,n≥2,n=Sn-Sn-1=(3n-2(-(3n-1-2(=2?3n-1.1,n=1,1,n=1,n-1,n≥2,238n{是公差為17的等差數(shù)列，c2-2accosB，可得1=(a+c(2-2ac-2accos=4-(2+、3(ac，因?yàn)閒(x(=asin(ax(，a>0的極大值和極小值分別為a，-a，n=+2(n-1(得，Sn=nan-2n(n-1(，n=Sn-Sn-1=nan-(n-1(an-1-4(n-1(，所以==-，×(1-+-+?+-n=nan-(n-1(an-1+(n-1(an-1-(n-2(an-2+?+2a2-a1+a1=n2，所以an=n．所以bn=[lgan[.所以f(n(=2|sin-1的周期為T(mén)=2.f(n(=-1.n=a4n-3+a4n-2+a4n-1+a4n，則數(shù)n=3+(n-1(×(-5(=8-5n，故C正確；對(duì)于D，設(shè)數(shù)列{bn{中的第k項(xiàng)是數(shù)列{an{中的第m項(xiàng)，則mAC.n+1=1.2an-1000n+1n-2，3n-1，共有(3-1(?3n-1-1{為等比數(shù)列，因此選項(xiàng)A正確；Sn=2+?+n-1，Sn=23+?+n，Sn=4-因此選項(xiàng)D不正確，點(diǎn)(an-2,an(，(an,an+2((n≥3(是函數(shù)g(x(圖象上的兩點(diǎn)，所以an+2-anan-an-22選項(xiàng)A正確；2n-1>,所以an+Sn=-6+4(n-1(-6n+×4=2n2-4n-10，故答案為：2n2-4n-10.∵CA=2CE，∴BE∴A?B=A+AA-A(=AC-2-×22-×2×2×=-.故答案為：-.335π-3sin3則AP335π-3sin3則AP=2+3,4sin32,設(shè)P(x,y(，則x-(-323(=2+32,|y-23=-S(n-S(n-1(=ω(7n-6(+ω(7n-5(+ω(7n-4(+ω(7n-3(+ω(7n-2(+又7n-6=1×70+(n-1(0+(n-1(?7，7n-1=6×70+(n-1(?7，所以ω(7n-1(=6+n-1=n+5，所以S(n(-S(n-1(=7n+15，n≥2，2所以數(shù)列{an{的通項(xiàng)公式為an=2×2n-1=2n=log22n=nn+1-bn=(n+1(-n=1，即數(shù)列{bn{是等差數(shù)列，1+n2n2+所以數(shù)列1+n2n2+sin∠ACBsin∠ABCsin∠ABCAC由正弦定理得AB=AC?sin∠ACBsin∠ACBsin∠ABCsin∠ABCAC因?yàn)椤螦DB+∠ADC=π,即cos∠ADB+cos∠ADC=0，{an{的公差為d(d≠0(，,,4所以an=2n-1，n∈N*.T=(1-+-+?+-Γ|L=1-=-，qqqn-1=n-5.4Tn4-2anan-1(anan-1+an+1(=0，得(an-an-1(2-anan-1(an-an-1(-2aa-1=0，即(an-an-1+anan-1((an-an-1-2anan-1(=0，由{an{是單調(diào)遞減的正項(xiàng)數(shù)列，得an-an-1-2anan-1<0，則an-an-1+anan-1=0，即n-a-1=1，nn=.11111(n+1(n11111(n+1(n>+?+44=1-=1-+?+--+=1-=1-+?+--+-311n+1所以Sn1n+1=an+2n-1，∴an=an-1+2n-1-1(n≥2(，an-1=an-2+2n-2-1，?1-1，上式累加可得：an-a1=2n-2-(n-1