新高考新風(fēng)向-三角函數(shù)新定義問(wèn)題

特別注意：新定義“伴隨函數(shù)”得出函數(shù)f(x)的表達(dá)式+bcosx一般借助輔助角公式進(jìn)行變形，即f(x)=asinx+A,B(=余弦距離為1-cos(AA.7B.C.4D.11 f(x)，若存在m∈R且m?Z，使得f(m(=f([m[(，則稱函 =cos(sinx(=f(x(.可得：π也為函數(shù)f(x(=cos(sinx(的周期.但是否為該函數(shù)的最小正周期呢？我們可以分區(qū)間研究f(x(=cos(sin22A.1-B.1+C.1-D.1- 注ai=a1+a2+a3+???+an，若f=sinx+cosx，g1≤x<0，則關(guān)于函數(shù)y=f(x)、yA.y=f(x)與y=g(x)都是B.y=f(x)是而y=g(x)不是C.y=f(x)不是而y=g(x)是D.y=f(x)與y=g(x)都不是 <tk使得當(dāng)x取任意實(shí)數(shù)時(shí)，有f(x(+f(x+t1(+f(x+t2(+???+f(x+tk(=0，則稱函數(shù)y=f(x(具有“性<0.3344<?<tk)使得當(dāng)x取任意值時(shí)，有f(x+t1(+f(x+t2(+?+f(x+tk(=0則稱函數(shù)y=f(x(為“k級(jí)周天函數(shù)”.x(=sinx；②f2(x(=x+2；<0.55f(x)，若存在m∈R且m?Z，使得f(m(=f([m[(，則稱函66 >sinh(sin∞(.77f(x)=asinx+bcosx.88特別注意：新定義“伴隨函數(shù)”得出函數(shù)f(x)的表達(dá)式+bcosx一般借助輔助角公式進(jìn)行變形，即f(x)=asinx+b2+b2ab2+b2a2+b2A,B(=余弦距離為1-cos(A+β([2+[13cos(α+β([2=13，+=cosαcos(α+β(+sinαsin(α+β(=cos(α+β-α(=cosβ=，因?yàn)?<β<，所以sinβ=、1-cos2β=.因?yàn)?8sinβ(2+(8cosβ(2=8，=cosαcosβ+sinαsinβ=cos(α-β(=，因?yàn)?<α<β<，則-<α-β<0，所以sin(α-β(=-、1-cos2(α-β(=-.因?yàn)閏osα=cos(α-β+β(=cos(α-β(cosβ-sin(α-β(sinβ=，又因?yàn)閏os2β=2cos2β-1=-，sin2β=2sinβcosβ=，11所以d(M,Q(=-+-=. A(x1A.7B.C.4D.cos(P,Q)==sinαsinβ+cosαcosβ=cos(α-β)，cos(Q,R)==sinαsinβ-cosαcosβ=-cos(α+β)，整理得tanαtanβ=7，故選A f(x)，若存在m∈R且m?Z，使得f(m(=f([m[(，則稱函22【解】(1)對(duì)于f(x)=x2-x，則f(m)=f=0，f([m])=f(0)=0，令m=-，則[m]=-1，又f(x(=x+，則f(-=f(-(，所以f(x)是Ω函數(shù).所以必有T≥1， =cos(sinx(=f(x(.可得：π也為函數(shù)f(x(=cos(sinx(的周期.但是否為該函數(shù)的最小正周期呢？我們可33f(-x(=sin[cos(-x([=sin(cosx(=f(x(，所以f(x(是偶函數(shù).(3)f(x+2π(=sin[cos(x+2π([=sin(cosx(=f(x(，x1x2y1y2.若A(-1,2x1x2y1y2x+yx+yx+yx+yBA.1-B.1+C.【解析】由題x+y=(-1(2+22=5，x+y=((2+((2=1，44注ai=a1+a2+a3+???+an，若f=sinx+cosx，g1≤x<0，則關(guān)于函數(shù)y=f(x)、yA.y=f(x)與y=g(x)都是B.y=f(x)是而y=g(x)不是C.y=f(x)不是而y=g(x)是D.y=f(x)與y=g(x)都不是函數(shù)f=sinx+cosx=sin(x+(，顯然f在[-1,0[上單調(diào)遞增，對(duì)區(qū)間[-1,0[上任意劃分：-1=x0<x1<?<xn-1<xn=0，則|f(xi)-f(xi-1)|=f(xi)-f(xi-1)(i=1,2,?,n)，取M=3，對(duì)區(qū)間[-1,0[上任意劃分：-1=x0<x1<?<xn-1<xn=0，恒成立，對(duì)于函數(shù)x<0，對(duì)區(qū)間[-1,0[上的劃分：-1<-<-<?<-<-<?<-<0， 2(1)=-1；55??,an=cosh(2n-1m(則m=±,因此a=|(eln2+(eln2(-=222 f(x(=sinx，g(x(=lgx， <tk使得當(dāng)x取任意實(shí)數(shù)時(shí)，有f(x(+f(x+t1(+f(x+t2(+???+f(x+tk(=0，則稱函數(shù)y=f(x(具有“性<0.(2)①若a<0，此時(shí)取n=m即可；因?yàn)間(x)+g(x++g(x+=a+bcos2x+ccos5x+dcos8x+a+bcos2(x++66ccos5(x++dcos8(x++a+bcos2(x++ccos5(x++dcos8(x+=a+bcos2x+ccos5x+dcos8x+a+bcos(2x++ccos(5x++dcos(8x++a+=a+bcos2x+ccos5x+dcos8x+a-bcos(2x+cos(5x+cos(8x++a+=3a+bcos2x+ccos5x+dcos8x-bcos2x-sin2x(-ccos5x-sin5x(-dcos8x-sin8x(+b-cos2x-sin2x(+c(-cos5x-sin5x(+d(-cos8x-sin8x(=3a=0∴g(x)=g(x+=g(x+=0，再由g(x)+g(x+π)=a+bcos2x+ccos5x+dcos8x+a+bcos(2x+2π(+ccos(5x+5π(+③若a>0，由②中可知g(m)+g(m++g(m+=3a，所以g(m++g(m+=3a-4a=-a<0，命題成立. 【解】(1)h(x)=2cosx-sinxcosx-sinx(=cosx-sinx770≤x≤ππ<x≤2π∴1≤k<3(3)f(x)=asinx+bcosx=、a2+b2sin(x+φ)此時(shí)tan2x0=tan(π-2φ)=-tan2φ=--令taφ==m則由a2-4ab+3b2<0知3m2-4m+1<0解之得<m<1tan2x0=-=-，因?yàn)閥=m-在m∈,1(上單調(diào)遞增， <?<tk)使得當(dāng)x取任意值時(shí)，有f(x+t1(+f(x+t2(+?+f(x+tk(=0則稱函數(shù)y=f(x(為“k級(jí)周天函數(shù)”.x(=sinx；②f2(x(=x+2；<0.=0，t2=π,則f1(x+t1(+f1(x+t2(=sinx+sin(x+π(=sinx-sinx=0，88f2(x+t1(+f2(x+t2(=(x+t1(+2+(x+t2(+2=2x+4+t2=0，不對(duì)任意x都成立，2=，t3=g(x+t1(+g(x+t2(+g(x+t3(+2(x+2nπ++cos(3n+2(x+4nπ+=cos[(3n+2(x[+cos(3n+2(x++cos(3n+2(x+=cos[(3n+2(x[-cos(3n+2(x++cos(3n+2(x+=cos[(3n+2(x[-cos[(3n+2(x[cos-sin[(3n+2(x[sin+cos[(3n+2(x[cos+sin[(3n+2(x[sin=cos[(3n+2(x[-cos[(3n+2(x[-sin[(3n+2(x[-cos[(3n+2(x[+sin[(3n+2(x[=cos[(3n+2(x[-cos[(3n+2(x[=0<0，則h(m(=4a<0，此時(shí)取n=m，則h(n(<0；所以t(x(+t(x++t(x+=0，所以h(x(+h(x++h(x+=3a=0，所以h(x(=h(x+=h(x+=0，再由h(x(+h(x+π(=0?bcos2x+dcos8x=0恒成立，所以b=d=0，+h(m++h(m+=3a，h(m(=4a，得h(m++h(m+=-a<0，99f(x)，若存在m∈R且m?Z，使得f(m(=f([m[(，則稱函【解】(1)對(duì)于f(x)=x2-x，則f(m)=f=0，f([m])=f(0)=0，令m=-，則[m]=-1，又f(x(=x+，此時(shí)f(-=-+=-，f(-Γ|L(=f(-1)=-1+=-，則f(-=f(-(，所以f(x)是Ω函數(shù). 類似的我們有雙曲正弦函數(shù)sinh(x(①[cosh(x([2-[sinh(x([2=1；③cosh(2x(=[cosh(x([2+[sinh(x([2.>sinh(sinx(.-[sinh(x([2=22=-=1；選③,cosh(2x(==2+2=[cosh(x([2+[sinh(x([2.y=cosh(2x(+sinh(x(=+，令t=sinh(x(=，因?yàn)楹瘮?shù)y=、y=-均為R上的增函數(shù)，故函數(shù)y=sinh(x(也為R上的增函數(shù)，所以y=2t2+t+1=2(t+2+≥,當(dāng)且僅當(dāng)t=-時(shí)取“=”，所以y=cosh(2x(+sinh(x(的最小值為.>sinh(sinx(?>cosx+e-cosx>esinx-e-sinx，-π,0[時(shí)，ecosx+e-cosx>0，sinx≤0≤-sinx，所以esinx≤e-sinx，所以esinx-e-sinx≤0，所以ecosx+e-cosx>esinx-e-sinx成立；<x≤-x<，且正弦函數(shù)y=sinx在(0,上為增函數(shù)，cosx=sin-x(≥sinx，所以ecosx≥esinx，-e-sinx<0<e-cosx，所以ecosx+e-cosx>esinx-e-sinx成立，>sinh(sinx(. f(x)=asinx+bcosx.f(x)=asinx+bcosx， =+bcos=+bcos=+令=tanα,上式化為tan(α+=tan，即=tanα=tan=3；所以g(x)=2(λcosα+μcosβ)sinx+2(λ