新高考压轴题中函数的新定义问题 解析版

=·1·(2)任意x>0，恒有sinhx-kx>0成立，求实数·1· *(coshx(/=/==sinhx，(coshx(/=/==sinhx，2(coshx(2-1=2×2-1=-1==cosh(2x(；(coshx(2-(sinhx(2=22=-=1；①当k≤1时，由coshx=≥ex⋅e-x=1，所以F/(x(=coshx-k>0，即F(x)为增函数，此时F(x)>F(0)=0，对任意x>0，sinhx>kx恒成立，满足题意；x(=sinhx>0，可知G(x)是增函数，0类比双曲余弦函数的二倍角公式cosh(n=cosh(2n-1m(，·2·n=cosh(2n-1m(·2· 2024=cosh(22023m(=，设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(，则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，则|f(x1(-f(x2(|=+1+1(|=|x1-x2|<2|x1-x2|，·3·<x·3· 则|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[内单调递增，x(≤3[1,e[内恒成立；即-3≤f/(x(≤3在[1,e[内恒成立，又因为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，符合题意；<x2≤2，①若0<x2-x1≤,则|f(x1(-f(x2(|<2|x1-x2|≤1；②若<x2-x1≤1，则|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·+的函数称为n次置换.满足对任意i∈f1(f(⋯f((，记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当f(i(=当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【题型训练-刷模拟】·6··6· (2)已知函数g(x(=ln(x+1(+x3，设集合P={x(2)(i)对h(x(=ln(x+1(+x3-x(x>-1(求导得出函数h(x(的单调性，利用零点存在定理即可求得集合P中元素的个数为2个；r[，=1+cosx≥0恒成立，所以f(x(在[0,r[上单调递增，可得f(x(的值域为[0,r+sinr[，即可得r+sinr≤r，即sinr≤0(r>0(，(2)(i)记函数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(则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，假设长度为m的闭区间D=[a,a+x+x0[+x0[，·7·当-1<a<0时，由(i)得h(x(在(-1,0·7· ∴h(a+x0(=g(a+x0(-(a+x0(>h(x0(=0，，而g(x(显然在(-1,+∞(单调递增，∴g(0(≤g(x(≤g(x0(，由(i)可得g(0(=h(0(+0=0，g(x0(=h(x0(+x0=D，满足要求. 2=D，若x1f(x1(-f(x2(|≤t|g(x1(-g(x2(|(t>0)成立，则称函数y=f(x)与y=g(x)“具有性质H，求证x+x>2.得2<x1+x2<4，即|x1+x2|<4，则|x1+x2|⋅|x1-x2|<4|x1-x2f(x1(-f(x2(|≤2|g(x1(-g(x2(|，·8··8· ,1[与g(x(=“具有性质H(t)”，即|2+x-2-x|≤t-，x2，则0<x1+x2<2，0<x1x2<1，即0<x1x2(x1+x2)<2，又函数f(x(=+2lnx-3与y=g(x)“具有性质H(1)”，则|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，记h(x(=+lnx-3，即h(t1(=h(t2(，因为h/(x(=-+=，要证x+x>2，即证t1+t2>2，不妨设0<t1<1<t2，即证t2>2-t1>1，只需证h(t2(>h(2-t1(，即证h(t1(>h(2-t1(，设H(x(=h(x(-h(2-x(，即H(x(=+lnx---ln(2-x(，因为H/(x(=-+--+-=-≤0，所以函数y=H(x(在(0,+∞(是减函数，且H(1)=0，又0<t1<1，则H(t1(>H(1(=0，即h(t1(-h(2-t1(>0，则h(t1(>h(2-t1(得证，故x+x>2.·9··9· 故可取f(x)=tanx(.(2)假设有f(x)是集合Z到Q的一个完美对应，(3)f(x(=x3-kx2+1，f/(x(=3x2-2kx=0，得x=0或，f/又f(0)=1<2，故只有极小值f≤-3才满足题意，即3-k2+1≤-3，k≥3，f/·10·又f(0)=1>-3，故只有极大值f≥2才满足题意，·10· 即3-k2+1≥2，即k≤-.后对k进行合理分类讨论即可.*对于f2(x)=lnx,f(x)=(x>0)是严格减函数，则f2(x)在不同点处的切线斜率不同，=-1==tan2x≥0恒成立，且仅当x=0时g/(x)=0，令g1由y=g1(x)的图象是连续曲线，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· 则-x1tan2x1+tanx1-x1=x1tan2x1-tanx1+x1，即x1(1+tan2x1)=tanx1，则有x1=sinx1cosx1，即2x1=sin2x1，令φ(x)=x-sinx,0<x<π,φ/(x)=1-cosx>0，又h(x)在点(t,sint)处的切线方程为y-sint=cost(x-t)，即y=xcost+sint-tcost，h(x)在点(s,sins)处的切线方程为y=xcoss+sins-scoss，-t(t，coss=cost且tans=-tant，从而存在n∈N*，使得s=2nπ-t，代入tans-s=tant-t，可得tant-t+nπ=0，则xn=t，即t是数列{xn{中的项；反之，若t是数列{xn{中的项，则存在n∈N*，使得xn=t，即tant-t+nπ=0，令s=2nπ-t，则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,+∞)，使得点(s,sins)与(t,sint)是函数y=sinx的图象的一是数列{xn{中的项”.【点睛】结论点睛：函数y=f(x)是区间D上的可导函数，则曲线y=f(x)在点(x0,f(x0))(x0∈D)方程为：y-f(x0)=f/(x0)(x-x0).(x)=sinx-x2与φ2(x(=ex-x是否具有性质φ1-φ2⎳x0>0？并说明理由.(2)已知函数f(x(=aex-ln(x+1(与g(x(=ln(x+a(-ex+1具有性质f-g⎳x1>x2.(2)(i)a∈(0,1(∪(1(x)=sinx-x2与φ2(x(=ex-x具有性质φ1-φ2⎳x0>0，理由如下：φ(x)=cosx-2x，令h(x(=φ(x(=cosx-2x，·12·则h/(x(=-sinx-2<0，故φ(x·12· 则φ1(x(在(-∞,x0(上单调递增，在(x0,+∞(上单调递减，(x)在(-∞,0(上单调递减，在(0,+∞(上单调递增，(x)=sinx-x2与φ2(x(=ex-x具有性质φ1-φ2⎳x0>0；(2)(i)f(x(=aex-，又x+1>0，故x>-1，当a≤0时，f(x(=aex-<0，此时f(x(没有极值点，故舍去，当a>0时，令m(x(=f(x(=aex-故f(x(在(-1,+∞(上单调递增，g(x(=-ex，x+a>0，故x>-a，故g(x(在(-a,+∞(上单调递减，f(0(=ae0-=a-1<0，又x→+∞时，f(x(→+故此时存在x1∈(0,+∞(，使f(x(在(-1,x1(上单调递减，在(x1,+∞(上单调递增，则f(x(有唯一极值点x1∈(0,+∞(，故此时存在x2∈(0,+∞(，使g(x(在(-a,x2(上单调递增，在(x2,+∞(上单调递减，则g(x(有唯一极值点x2∈(0,+∞(，x1=x2=此时需满足x1>x2>0，则ex1>ex2，当a∈(1,+∞(时，f(0(=ae0-=a-1>0，又x→-1时，f(x(→-∞,·13·故此时存在x1∈(-1,0(，使f(x(在(-1,x1(上单调递减，在(x1,+∞(上单调递增，则f(x(有唯一极值点x1∈(-1,0·13· 当a=1时，有f(0(=ae0-=a-1=0，g(0(=-e0=-1=0，>x2>0，则ex2=>e0=1，故0<x2+a<1，g(x(在(-a,x2(上单调递增，在(x2,+∞(上单调递减，则g(x1(<g(x2(=ln(x2+a(-ex2+1=ln(x2+a(-+1，则μ(t(=+>0，故μ(t(在(0,1(则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，则ex2=<e0=1，即x2+a>1，则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)把满足任意x,y∈R总有f(x+y(+f(x-y(=2f(x(f(y(的函数称为和弦·15··15· f(2(=187令x=n,y=1,n∈N+，则f(n+1(+f(n-1(=2f(n(f(1(=f(n(，2f(n+1(-f(n(=2[2f(n(-f(n-1([，则f(y(+f(-y(=2f(0(f(y(=2f(y(，即f(y(=f(-y(，g(x(是偶函数，x2|=即Cn+1+Cn-1>2Cn，Cn+1>2Cn-Cn-1=Cn+(Cn-Cn-1(>Cn，又g(x(是偶函数，所以有g(x2(>g(x1(.=,|x2|=将问题转化为判断Cn=g(的增减性.成立，则称函数y=f(x(为增函数”.·16··16· (3)设g(x(=ex-ln(x+1(-1，若曲线y=g(x(在x=x0处的切线方程为y=0，求x0的值，并证明函数y=g(x(是增函数”.故函数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则-a≤0，即a≥.x0(=ex0-可得方程的一个解x0=0，x(=ex+>0，故g/(x(在(0,+∞(上是严格增函数，所以x0=0是唯一解，设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,+∞(上是严格增函数以及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,+∞(上是严格增函数，因为s>0，则w(s(>w(0(=g(t(-g(0(-g(t(=0，故g(s+t(>g(s(+g(t(，得证，所以函数y=g·17··17· 如新环序号an-m+1对应的是原环中的a0，⋯,新环序号an-1对应的是原环中的am-2.这就需要用模取余，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)时推测成立，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，此时2k+t+1=2k+1，即f(2k+1,2(=2(2k+1mod2k+1(.故当n+1=2k+t+1时，推测成立.推测成立.-β|≤1，则称f(x(和g(x(互为“零点相邻函数”.设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)，则h/(x(=2-=，令h/(x(=0·19·当-a<x<-a时，h/(x(<0,h(x(单调递减，·19· 所以h(x)min=h-a(=2-2a-ln=2+ln2-2a，min>0,h(x(无零点，min<0，又因为h(1(>0，min<0，又因为h(-1(<0,h(1(>0，(3)当x>0,a=1时，f-<0⇔ln(1+-<0，设t=，则t>0，则ln(1+-<0⇔ln(1+t(-<0⇔·1+tln(1+t(-t<0，21+t设F(t(=1+tln(1+t(-t(t>0)，则F/(t(=ln(1+t(+2-21+21+t令p(t(=ln(1+t(+2-2、1+t,t>0，则p/(t(=-1=<0所以p(t(在(0,+∞(上单调递减，·20· 面积(f2(x(-f1(x((dx，其中(dx=-F(3(的值； (2)y=2x-1∴F(2)-F(3)=⋅-π=.·21··21· 则切线方程为y-x=2x0(x-x0(，所以切线与轴的交点为B,0(，联立⇒an=令f(x(=-dx2+，F/(x(=f(x(⇒F(x(=-+C(C为常数), -f(x1)+m[⋅[kx2-f(x2)+m[≥0，则称(k,m)为f(x)f(x)+m≤0.(2)x=-2-2、2为函数f(x(的一个极大值点,x=-2、2为f(x(的一个极小值点.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为函数f(x(的一个极大值点,x=-2、2为f(x(的一个极小值点.∈R,使得kx0-f(x0(+m>0,则对任意x0'∈R,都有[kx0-f(x0(+m[⋅[kx0'-f(x0'(+m[≥0.∴对任意x∈R,kx-f(x(+m≥0恒成立.令F(x(=(x2+ax+、2a(ex-kx-m，则F(x(≤0在R上恒成由二次函数性质可知,必存在t0>0使得当x>t0时,x2+ax+、2a>0恒成立,且此时ex>1,∴当x>t0时有F(x(=(x2+ax+、2a(ex-kx-m>x2+ax+、2a-kx-m，由二次函数性质可知,必存在x1>t0使得当x>x1时，F(x(>x2+ax+、2a-kx-m>0.这与F(x(≤0在R上恒成立矛盾.∴对任意x∈R,都有kx-f(x(+m≤00∈R,使得kx0-f(x0(+m>0,根据和谐数组的定义转化得存在x0∈R,使得kx0-f(x0(+m>0,设F(x(=(x2+ax+、2a(ex-kx-m，通过二次 0=1,e1=x1+x2+x3,e2=x1x2+x2x3+x1x3,e3=x1x2x3.已知三次函数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.【详解】(1)证明：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，