新高考?jí)狠S題中函數(shù)的新定義問(wèn)題 學(xué)生版

=·1·(2)任意x>0，恒有sinhx-kx>0成立，求實(shí)數(shù)·1· (2024·山東濱州·二模)定義：函數(shù)f(x)滿足對(duì)于任意不同的x1,x2∈[a,b]，都有|f(x1(-f(x2(|<k|x1-x2|，則稱f(x)為[a,b[上的“k類函數(shù)”.·2·，|f(x1(-f(x2(|<1·2· +的函數(shù)稱為n次置換.滿足對(duì)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(=+·3··3· 【題型訓(xùn)練-刷模擬】(2024(2024·浙江紹興·三模)若函數(shù)f(x(在區(qū)間I上有定義，且?x∈I，f(x(∈I，則稱I是f(x(的一個(gè)“封(1)已知函數(shù)f(x(=x+sinx，區(qū)間I=[0,r[(r>0(且f(x(的一個(gè)“封閉區(qū)間”，求r的取值集合；(2)已知函數(shù)g(x(=ln(x+1(+設(shè)集合P={x|g(x(=x{.(i)求集合P中元素的個(gè)數(shù)；(ii)用b-a表示區(qū)間[a,b[(a<b(的長(zhǎng)度，設(shè)m為集合P中的最大元素．證明：存在唯一長(zhǎng)度為m的閉區(qū) 且x1≠x2f(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(t)”.(3)若函數(shù)f(x)=+2lnx-3與y=g(x)“具有性質(zhì)H(1)”，且函數(shù)y=g(x)在區(qū)間(0,+∞)上存在兩個(gè)·4··4· (1)用初等函數(shù)構(gòu)造區(qū)間[0,1(到區(qū)間[0,+∞(的一個(gè)完美對(duì)應(yīng)f(x(；(3)若f(x(=x3-kx2+1，k∈R，且f(x(是某區(qū)間A到區(qū)間[-3,2[的一個(gè)完美對(duì)應(yīng)，求k的取值范圍.·5··5· (x)=sinx-x2與φ2(x(=ex-x是否具有性質(zhì)φ1-φ2?x0>0？并說(shuō)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.k((k∈N*)，并探究φ(pq(與φ(p(和φ(q(的關(guān)系；·6·(3)設(shè)數(shù)列{an{的通項(xiàng)公式為φ(3m(求該數(shù)列的前·6· (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)稱為和弦成立，則稱函數(shù)y=f(x(為增函數(shù)”.(3)設(shè)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·7·=g(x(是增函數(shù)”.·7· ·8··8· -β|≤1，則稱f(x(和g(x(互為“零點(diǎn)相鄰函數(shù)”.設(shè)f(x(=ln(a+x((a∈R(，g(x(=(2)令h(x(=g/(x(-f(x((g/(x(為g(x(的導(dǎo)函數(shù))，分析h(x(與g(x(是否互·9··9· 面積(f2(x(-f1(x((dx，其中(dx=(1)如圖，連續(xù)函數(shù)y=f(x(在區(qū)間[-3,-2[與[2,3[的圖形分別為直徑為1的上、下半圓周，在區(qū)間[-2,0[與[0,2[的圖形分別為直徑為2的下、上半圓周，設(shè)F(x(=dt．求F(2(-F(3(的值；·10·(2)在曲線f(x(=x2(x≥0)上某一個(gè)點(diǎn)處作切線，便之與曲線和x軸所圍成的面積為求切線方程； ·10· -f(x1)+m[?[kx2-f(x2)+m[≥0，則稱(k,m)為f(x)f(x)+m≤0. 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·11·(3)若x1+x2+x3=1,x+x+x=2,x+x+x=3，求x+x+x·11· 么稱d(A,B)=|x1-x2|+|y1-y2|為A，B兩點(diǎn)間的曼哈頓距離.d(M,N1(，d(M,N2(，求d(M,N1(和d(M,N2(的最小值；(2)已知點(diǎn)N是直線x+