新高考壓軸題中函數(shù)的新定義問題 解析版

=·1·(2)任意x>0，恒有sinhx-kx>0成立，求實數(shù)·1· *(coshx(/=/==sinhx，(coshx(/=/==sinhx，2(coshx(2-1=2×2-1=-1==cosh(2x(；(coshx(2-(sinhx(2=22=-=1；①當k≤1時，由coshx=≥ex?e-x=1，所以F/(x(=coshx-k>0，即F(x)為增函數(shù)，此時F(x)>F(0)=0，對任意x>0，sinhx>kx恒成立，滿足題意；x(=sinhx>0，可知G(x)是增函數(shù)，0類比雙曲余弦函數(shù)的二倍角公式cosh(n=cosh(2n-1m(，·2·n=cosh(2n-1m(·2· 2024=cosh(22023m(=，設t=22023mt=4或，即t=±ln4，所以m=±,即a1=coshm==2+2.n+1=2a-1，所以an+1=2(coshxn(2-1=cosh(2xn(，則an+1=cosh(xn+1(=cosh(2xn(，1所以a=a1=cosh(x1(=(ex1+e-x1(=4+4=2+2，f(x1(-f(x2(|<f(x1(-f(x2(|<1.1≤x1<x2≤3，則|f(x1(-f(x2(|=+1+1(|=|x1-x2|<2|x1-x2|，·3·<x·3· 則|f(x1(-f(x2(|<3|x1-x2|=3(x2-x1(恒成立，可得3x1-3x2<f(x1(-f(x2(<3x2-3x1，即f(x1(+3x1<f(x2(+3x2，f(x2(-3x2<f(x1(-3x1均恒成立，x(=f/(x(+3，由f(x1(+3x1<f(x2(+3x2可知g(x(在[1,e[內(nèi)單調(diào)遞增，x(≤3[1,e[內(nèi)恒成立；即-3≤f/(x(≤3在[1,e[內(nèi)恒成立，又因為f/(x)=axex-x-1-lnx，即-3≤axex-x-1-lnx≤3，整理得≤a≤,可得≤a≤,令t=x+lnx，t(≥G(e+1(=；e+1e+11=x2，可得|f(x1(-f(x2(|=0<1，符合題意；<x2≤2，①若0<x2-x1≤,則|f(x1(-f(x2(|<2|x1-x2|≤1；②若<x2-x1≤1，則|f(x1(-f(x2(|=|f(x1(-f(1(+f(1(-f(x2(|=|f(x1(-f(1(+f(2(-f(x2(|≤|f(x1(-f(1(|+|f(2(-f(x2(|≤2(x1-1(+2(2-x2(=2-2(x2-x1(<1；·4·f(x1(-f(x2(|<1·4·+的函數(shù)稱為n次置換.滿足對任意i∈f1(f(?f((，記f(i(=f1(i(,f(f(i((=f2(i(,f(f2(i((=f3(i(,?,f(fk-1(i((=fk(i(,i∈A,k∈N+.f(i(=i(；+i(=i(=，,f3(i(=；i(=44當f(i(=當f(i(=444(或f(i(=.2134(或f(i(=2431(或f(i(=23322334(或f(i(=2431(或f(i(=213444(或f(i(=(或f(i(=·5·； +f1(i(k+=?(，即f(i(=ffffffffffffffffffffffffffffffffffffffffffffffffffff【題型訓練-刷模擬】·6··6· (2)已知函數(shù)g(x(=ln(x+1(+x3，設集合P={x(2)(i)對h(x(=ln(x+1(+x3-x(x>-1(求導得出函數(shù)h(x(的單調(diào)性，利用零點存在定理即可求得集合P中元素的個數(shù)為2個；r[，=1+cosx≥0恒成立，所以f(x(在[0,r[上單調(diào)遞增，可得f(x(的值域為[0,r+sinr[，即可得r+sinr≤r，即sinr≤0(r>0(，(2)(i)記函數(shù)h(x(=g(x(-x=ln(x+1(+x3-x(x>-1(，4+9x2(x+1(-4(x+1(9x2(x+1(-4xx(3x+4((3x-1(則h′(x(=4+9x2(x+1(-4(x+1(9x2(x+1(-4xx(3x+4((3x-1(>0得-1<x<0或x>；由h′(x(<0得0<x<；(ii)由(i)得m=x0，假設長度為m的閉區(qū)間D=[a,a+x+x0[+x0[，·7·當-1<a<0時，由(i)得h(x(在(-1,0·7· ∴h(a+x0(=g(a+x0(-(a+x0(>h(x0(=0，，而g(x(顯然在(-1,+∞(單調(diào)遞增，∴g(0(≤g(x(≤g(x0(，由(i)可得g(0(=h(0(+0=0，g(x0(=h(x0(+x0=D，滿足要求. 2=D，若x1f(x1(-f(x2(|≤t|g(x1(-g(x2(|(t>0)成立，則稱函數(shù)y=f(x)與y=g(x)“具有性質(zhì)H，求證x+x>2.得2<x1+x2<4，即|x1+x2|<4，則|x1+x2|?|x1-x2|<4|x1-x2f(x1(-f(x2(|≤2|g(x1(-g(x2(|，·8··8· ,1[與g(x(=“具有性質(zhì)H(t)”，即|2+x-2-x|≤t-，x2，則0<x1+x2<2，0<x1x2<1，即0<x1x2(x1+x2)<2，又函數(shù)f(x(=+2lnx-3與y=g(x)“具有性質(zhì)H(1)”，則|f(x1(-f(x2(|≤|g(x1(-g(x2(|=0，即f(x1(=f(x2(，即+2lnx1-3=+2lnx2-3，令x=t1,x=t2，即+lnt1-3=+lnt2-3，記h(x(=+lnx-3，即h(t1(=h(t2(，因為h/(x(=-+=，要證x+x>2，即證t1+t2>2，不妨設0<t1<1<t2，即證t2>2-t1>1，只需證h(t2(>h(2-t1(，即證h(t1(>h(2-t1(，設H(x(=h(x(-h(2-x(，即H(x(=+lnx---ln(2-x(，因為H/(x(=-+--+-=-≤0，所以函數(shù)y=H(x(在(0,+∞(是減函數(shù)，且H(1)=0，又0<t1<1，則H(t1(>H(1(=0，即h(t1(-h(2-t1(>0，則h(t1(>h(2-t1(得證，故x+x>2.·9··9· 故可取f(x)=tanx(.(2)假設有f(x)是集合Z到Q的一個完美對應，(3)f(x(=x3-kx2+1，f/(x(=3x2-2kx=0，得x=0或，f/又f(0)=1<2，故只有極小值f≤-3才滿足題意，即3-k2+1≤-3，k≥3，f/·10·又f(0)=1>-3，故只有極大值f≥2才滿足題意，·10· 即3-k2+1≥2，即k≤-.后對k進行合理分類討論即可.*對于f2(x)=lnx,f(x)=(x>0)是嚴格減函數(shù)，則f2(x)在不同點處的切線斜率不同，=-1==tan2x≥0恒成立，且僅當x=0時g/(x)=0，令g1由y=g1(x)的圖象是連續(xù)曲線，且g1(-g1=-1<0，·11·:y-tanx1+x1-a=tan2x1?(x-x1)，l2:y-tanx2+x2-a=tan2x2?(x-x2)，有相同截距，即-x1tan2x1+tanx1-x1+a=-x2tan2x2+tanx2-x2+a，而x2=-x1·11· 則-x1tan2x1+tanx1-x1=x1tan2x1-tanx1+x1，即x1(1+tan2x1)=tanx1，則有x1=sinx1cosx1，即2x1=sin2x1，令φ(x)=x-sinx,0<x<π,φ/(x)=1-cosx>0，又h(x)在點(t,sint)處的切線方程為y-sint=cost(x-t)，即y=xcost+sint-tcost，h(x)在點(s,sins)處的切線方程為y=xcoss+sins-scoss，-t(t，coss=cost且tans=-tant，從而存在n∈N*，使得s=2nπ-t，代入tans-s=tant-t，可得tant-t+nπ=0，則xn=t，即t是數(shù)列{xn{中的項；反之，若t是數(shù)列{xn{中的項，則存在n∈N*，使得xn=t，即tant-t+nπ=0，令s=2nπ-t，則s∈(2π,+∞)且coss=cost,tans-s-(tant-t)=2(t-tant-nπ)=0，即tans-s=tant-t，可得sins-scoss=sint-tcost，所以存在s∈(2π,+∞)，使得點(s,sins)與(t,sint)是函數(shù)y=sinx的圖象的一是數(shù)列{xn{中的項”.【點睛】結(jié)論點睛：函數(shù)y=f(x)是區(qū)間D上的可導函數(shù)，則曲線y=f(x)在點(x0,f(x0))(x0∈D)方程為：y-f(x0)=f/(x0)(x-x0).(x)=sinx-x2與φ2(x(=ex-x是否具有性質(zhì)φ1-φ2?x0>0？并說明理由.(2)已知函數(shù)f(x(=aex-ln(x+1(與g(x(=ln(x+a(-ex+1具有性質(zhì)f-g?x1>x2.(2)(i)a∈(0,1(∪(1(x)=sinx-x2與φ2(x(=ex-x具有性質(zhì)φ1-φ2?x0>0，理由如下：φ(x)=cosx-2x，令h(x(=φ(x(=cosx-2x，·12·則h/(x(=-sinx-2<0，故φ(x·12· 則φ1(x(在(-∞,x0(上單調(diào)遞增，在(x0,+∞(上單調(diào)遞減，(x)在(-∞,0(上單調(diào)遞減，在(0,+∞(上單調(diào)遞增，(x)=sinx-x2與φ2(x(=ex-x具有性質(zhì)φ1-φ2?x0>0；(2)(i)f(x(=aex-，又x+1>0，故x>-1，當a≤0時，f(x(=aex-<0，此時f(x(沒有極值點，故舍去，當a>0時，令m(x(=f(x(=aex-故f(x(在(-1,+∞(上單調(diào)遞增，g(x(=-ex，x+a>0，故x>-a，故g(x(在(-a,+∞(上單調(diào)遞減，f(0(=ae0-=a-1<0，又x→+∞時，f(x(→+故此時存在x1∈(0,+∞(，使f(x(在(-1,x1(上單調(diào)遞減，在(x1,+∞(上單調(diào)遞增，則f(x(有唯一極值點x1∈(0,+∞(，故此時存在x2∈(0,+∞(，使g(x(在(-a,x2(上單調(diào)遞增，在(x2,+∞(上單調(diào)遞減，則g(x(有唯一極值點x2∈(0,+∞(，x1=x2=此時需滿足x1>x2>0，則ex1>ex2，當a∈(1,+∞(時，f(0(=ae0-=a-1>0，又x→-1時，f(x(→-∞,·13·故此時存在x1∈(-1,0(，使f(x(在(-1,x1(上單調(diào)遞減，在(x1,+∞(上單調(diào)遞增，則f(x(有唯一極值點x1∈(-1,0·13· 當a=1時，有f(0(=ae0-=a-1=0，g(0(=-e0=-1=0，>x2>0，則ex2=>e0=1，故0<x2+a<1，g(x(在(-a,x2(上單調(diào)遞增，在(x2,+∞(上單調(diào)遞減，則g(x1(<g(x2(=ln(x2+a(-ex2+1=ln(x2+a(-+1，則μ(t(=+>0，故μ(t(在(0,1(則g(x2(=lnt-+1<μ(1(=ln1-+1=0，g(x1(+x2<0，g(x1(+x2<g(x2(+x2=ln(x2+a(-ex2+1+x2=ln-ex2+1+x2=1-ex2<0，>|x2|；>x1>x2，則ex2=<e0=1，即x2+a>1，則g(x1(>g(0(=ln(0+a(-e0+1=lna>0，g(x1(+x2>0，g(x1(+x2=ln(x1+a(-ex+1+x2>ln(x2+a(-ex+1+x2=ln-ex1+1+x2=-x2-ex1+1+x2=1-ex1>1-e0=0，·14··14· k*k(3)Tm=+-?3m.=6，=6，k(=pk-pk-1=(p-1(pk-1.于是φ(pq(=pq-1-(p-1(-(q-1(=pq-p-q+1=(p-1((q-1(，m=(5m-3(×3m-1，0+72+?+(5m-3(?3m-1，3T1+72+123+?+(5m-3(?3m，2+?+5?3m-1-(5m-3(?3m，所以-2Tm=-(5m-3(?3m+2，故Tm=+-?3m.(2024·安徽合肥·三模)把滿足任意x,y∈R總有f(x+y(+f(x-y(=2f(x(f(y(的函數(shù)稱為和弦·15··15· f(2(=187令x=n,y=1,n∈N+，則f(n+1(+f(n-1(=2f(n(f(1(=f(n(，2f(n+1(-f(n(=2[2f(n(-f(n-1([，則f(y(+f(-y(=2f(0(f(y(=2f(y(，即f(y(=f(-y(，g(x(是偶函數(shù)，x2|=即Cn+1+Cn-1>2Cn，Cn+1>2Cn-Cn-1=Cn+(Cn-Cn-1(>Cn，又g(x(是偶函數(shù)，所以有g(shù)(x2(>g(x1(.=,|x2|=將問題轉(zhuǎn)化為判斷Cn=g(的增減性.成立，則稱函數(shù)y=f(x(為增函數(shù)”.·16··16· (3)設g(x(=ex-ln(x+1(-1，若曲線y=g(x(在x=x0處的切線方程為y=0，求x0的值，并證明函數(shù)y=g(x(是增函數(shù)”.故函數(shù)y=cos故任意的s,t∈(0,+∞(，有3s+t-1-(s+t(-a>3s-1-s-a+3t-1-t-a恒成立，s+t-1-3s-1-3t-1>-a恒成立，所以(3s-1((3t-1(>-a恒成立，t則-a≤0，即a≥.x0(=ex0-可得方程的一個解x0=0，x(=ex+>0，故g/(x(在(0,+∞(上是嚴格增函數(shù)，所以x0=0是唯一解，設w(s(=g(s+t(-g(s(-g(t(，其中s>0,t>0，ws(=g/(s+t(-g/(s(，由y=g/(x(在(0,+∞(上是嚴格增函數(shù)以及s+t>s>0，得g/(s+t(>g/(s(，即w/(s(=g/(s+t(-g/(s(>0，所以w(s(=g(s+t(-g(s(-g(t(在(0,+∞(上是嚴格增函數(shù)，因為s>0，則w(s(>w(0(=g(t(-g(0(-g(t(=0，故g(s+t(>g(s(+g(t(，得證，所以函數(shù)y=g·17··17· 如新環(huán)序號an-m+1對應的是原環(huán)中的a0，?,新環(huán)序號an-1對應的是原環(huán)中的am-2.這就需要用模取余，即f(n+1,m(=(f(n,m(+m(mod(n+1(.·18·f(2,2(=(f(1,2(+2(f(4,2(=(f(3,2(+2(mod4f(6,2(=(f(5,2(+2(mod6=4f(7,2(=(f(6,2(+2(mf(8,2(=(f(7,2(+2(mo·18· ，且0≤t<2k)時推測成立，即f(2k+t,2(=2[(2k+t(mod2k[=2t.k+t+1,2(=(f(2k+t,2(+2(mod(2k+t+1(=(2t+2(mod(2k+t+1(.k-1k+t+1,2(=2t+2=2[(2k+t+1(mod2k[；k-1k+t+1,2(=0，此時2k+t+1=2k+1，即f(2k+1,2(=2(2k+1mod2k+1(.故當n+1=2k+t+1時，推測成立.推測成立.-β|≤1，則稱f(x(和g(x(互為“零點相鄰函數(shù)”.設f(x(=ln(a+x((a∈R(，g(x(=令g(x(=x(x+1(=0，得x1=-1,x2=0，-a-(-1(|≤1，解得1≤a≤3，-a-0|≤1，解得0≤a≤2，(2)h(x(=2x+1-ln(x+a((x>-a)，則h/(x(=2-=，令h/(x(=0·19·當-a<x<-a時，h/(x(<0,h(x(單調(diào)遞減，·19· 所以h(x)min=h-a(=2-2a-ln=2+ln2-2a，min>0,h(x(無零點，min<0，又因為h(1(>0，min<0，又因為h(-1(<0,h(1(>0，(3)當x>0,a=1時，f-<0?ln(1+-<0，設t=，則t>0，則ln(1+-<0?ln(1+t(-<0?·1+tln(1+t(-t<0，21+t設F(t(=1+tln(1+t(-t(t>0)，則F/(t(=ln(1+t(+2-21+21+t令p(t(=ln(1+t(+2-2、1+t,t>0，則p/(t(=-1=<0所以p(t(在(0,+∞(上單調(diào)遞減，·20· 面積(f2(x(-f1(x((dx，其中(dx=-F(3(的值； (2)y=2x-1∴F(2)-F(3)=?-π=.·21··21· 則切線方程為y-x=2x0(x-x0(，所以切線與軸的交點為B,0(，聯(lián)立?an=令f(x(=-dx2+，F(xiàn)/(x(=f(x(?F(x(=-+C(C為常數(shù)), -f(x1)+m[?[kx2-f(x2)+m[≥0，則稱(k,m)為f(x)f(x)+m≤0.(2)x=-2-2、2為函數(shù)f(x(的一個極大值點,x=-2、2為f(x(的一個極小值點.R,代入k=m=0,得:[kx1-f(x1(+m[?[kx2-f(x2(+m[=f(x1(?f(x2(≥0.·22·x(=(x2+(2+42(x+8+42(ex=(x+22((x+2+22(ex于是可列表如下:·22·x(-∞,-2-2√2(-2-2/2((-2-2/2-2/2(-2/2(-2√2,+∞(f/(x(+0-0+f(x(↗↘↗ ∴x=-2-2、2為函數(shù)f(x(的一個極大值點,x=-2、2為f(x(的一個極小值點.∈R,使得kx0-f(x0(+m>0,則對任意x0'∈R,都有[kx0-f(x0(+m[?[kx0'-f(x0'(+m[≥0.∴對任意x∈R,kx-f(x(+m≥0恒成立.令F(x(=(x2+ax+、2a(ex-kx-m，則F(x(≤0在R上恒成由二次函數(shù)性質(zhì)可知,必存在t0>0使得當x>t0時,x2+ax+、2a>0恒成立,且此時ex>1,∴當x>t0時有F(x(=(x2+ax+、2a(ex-kx-m>x2+ax+、2a-kx-m，由二次函數(shù)性質(zhì)可知,必存在x1>t0使得當x>x1時，F(xiàn)(x(>x2+ax+、2a-kx-m>0.這與F(x(≤0在R上恒成立矛盾.∴對任意x∈R,都有kx-f(x(+m≤00∈R,使得kx0-f(x0(+m>0,根據(jù)和諧數(shù)組的定義轉(zhuǎn)化得存在x0∈R,使得kx0-f(x0(+m>0,設F(x(=(x2+ax+、2a(ex-kx-m，通過二次 0=1,e1=x1+x2+x3,e2=x1x2+x2x3+x1x3,e3=x1x2x3.已知三次函數(shù)f(x(=a0x3+a1x2-e1P2+e2P1-e3P0=0；k(3)若x1+x2+x3=1,x+x+x=2,x+x+x=3，求x+x+x.(3)x+x+x=6.【詳解】(1)證明：f(x(=(x-x1((x-x2((x-x3(·23·=x3-(x1+x2+x3(x2+(x1x2+x2x3+x1x3(x-x1x2x3，0=1=e0,a1=-(x1+x2+x3(=-e1,a2=e2,a3=-e3，·23· (i)由①+②+③得P3-e1P2+e2P1-3e3=P3-e1P2+e2P1-e3P0=0.④+⑤+⑥得Pm-e1Pm-1+e2Pm-2-e3Pm-3=0，1=P1=1，