高考數(shù)學(xué)二輪總復(fù)習(xí)思想方法訓(xùn)練4轉(zhuǎn)化與化歸思想文

思想方法訓(xùn)練4轉(zhuǎn)化與化歸思想能力突破訓(xùn)練1.已知M={(x,y)|y=x+a},N={(x,y)|x2+y2=2},且M∩N=?,則實(shí)數(shù)a的取值范圍是()A.a>2 B.a<2C.a>2或a<2 D.2<a<22.(2022廣西岑溪中學(xué)模擬)設(shè)△ABC的內(nèi)角A,B,C所對(duì)邊的長(zhǎng)分別為a,b,c,若△ABC的面積為S,且43S=(a+b)2c2,則sinC-π6=A.1 B.12 C.22 D3.設(shè)P為曲線(xiàn)C:y=x2+2x+3上的點(diǎn),且曲線(xiàn)C在點(diǎn)P處的切線(xiàn)傾斜角的取值范圍為0,π4,則點(diǎn)PA.-1,-1C.[0,1] D.14.已知函數(shù)f(x)在區(qū)間[0,2]上單調(diào)遞增,且函數(shù)f(x+2)是偶函數(shù),則下列結(jié)論成立的是()A.f(1)<f52<f7B.f72<f(1)<fC.f72<f52D.f52<f(1)<f5.(2022遼寧沈陽(yáng)三模)已知函數(shù)g(x)=exlnxm的圖象恒在f(x)=(em1)x的圖象的上方,則實(shí)數(shù)m的取值范圍是()A.(∞,1) B.(∞,e1)C.(0,1) D.(0,e1)6.平面上動(dòng)點(diǎn)M到定點(diǎn)F(3,0)的距離比到直線(xiàn)l:x+1=0的距離大2,則動(dòng)點(diǎn)M的軌跡方程為.

7.(2022四川成都七中三模)已知函數(shù)f(x)=sinπx,x∈[0,2],8.設(shè)f(x)是定義在R上的奇函數(shù),且當(dāng)x≥0時(shí),f(x)=x2,若對(duì)任意x∈[a,a+2],不等式f(x+a)≥f(3x+1)恒成立,則實(shí)數(shù)a的取值范圍是.

9.已知△ABC的內(nèi)角A,B,C的對(duì)邊分別為a,b,c,且13acosA5ccosB=5bcosC.(1)求sinA;(2)若a=27,且△ABC的面積為6,求△ABC的周長(zhǎng).10.(2022浙江,20)已知等差數(shù)列{an}的首項(xiàng)a1=1,公差d>1.記{an}的前n項(xiàng)和為Sn(n∈N*).(1)若S42a2a3+6=0,求Sn;(2)若對(duì)于每個(gè)n∈N*,存在實(shí)數(shù)cn,使an+cn,an+1+4cn,an+2+15cn成等比數(shù)列,求d的取值范圍.思維提升訓(xùn)練11.已知拋物線(xiàn)y2=4x的焦點(diǎn)為F,點(diǎn)P(x,y)為拋物線(xiàn)上的動(dòng)點(diǎn),又點(diǎn)A(1,0),則|PF||A.12 B.22 C.3212.函數(shù)f(x)=2sinxcosxsinxcosx(x∈R)的最小值是()A.1 B.14 C.1 D.13.若函數(shù)f(x)=x2ax+2在區(qū)間[0,1]上至少有一個(gè)零點(diǎn),則實(shí)數(shù)a的取值范圍是.

14.(2022廣西河池一中高三月考)若向量a=(x,2),b=(3,y),c=(1,2),且(ac)⊥(b+c),則|ab|的最小值為.

15.(2022西北工業(yè)大學(xué)附屬中學(xué)模擬)已知函數(shù)f(x)=ax+1+x22x+1+(x1)lna(a>0,且a≠1).(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若對(duì)?x1,x2∈[0,2],使|f(x1)f(x2)|≤1a1恒成立,求實(shí)數(shù)a的取值范圍答案：能力突破訓(xùn)練1.C解析:M∩N=?等價(jià)于方程組y=x把y=x+a代入到方程x2+y2=2中,消去y,得關(guān)于x的一元二次方程2x2+2ax+a22=0,①由題易知一元二次方程①無(wú)實(shí)根,即Δ=(2a)24×2×(a22)<0,由此解得a>2或a<2.2.B解析:由題意可知,S=12absinC,a2+b2c2=2abcosC又43S=(a+b)2c2=a2+b2c2+2ab,∴23absinC=2abcosC+2ab.∵ab≠0,∴3sinC=cosC+1,即3sinCcosC=1,∴32sinC12cosC=12,∴sinC3.A解析:設(shè)P(x0,y0),曲線(xiàn)C在點(diǎn)P處的切線(xiàn)的傾斜角為α,則0≤tanα≤1,令y=f(x)=x2+2x+3,則f'(x)=2x+2,于是0≤2x0+2≤1,1≤x0≤12,故選A4.B解析:因?yàn)楹瘮?shù)f(x+2)是偶函數(shù),所以f(x+2)=f(x+2),即函數(shù)f(x)的圖象關(guān)于直線(xiàn)x=2對(duì)稱(chēng).又因?yàn)楹瘮?shù)f(x)在區(qū)間[0,2]上單調(diào)遞增,所以函數(shù)f(x)在區(qū)間[2,4]上單調(diào)遞減.因?yàn)閒(1)=f(3),72>3>52,所以f72<f(3)<f52,即f725.A解析:由題意可得(em1)x<exlnxm,故x·em+m+lnx<ex+x,即em+lnx+m+lnx<ex+x.令φ(x)=ex+x,而φ(x)=ex+x為增函數(shù),于是原不等式可化為φ(m+lnx)<φ(x),所以m+lnx<x,即m<xlnx,令h(x)=xlnx,則h'(x)=11x=x-1x,當(dāng)0<x<1時(shí),h'(x)<0,當(dāng)x>1時(shí),所以函數(shù)h(x)在區(qū)間(0,1)上單調(diào)遞減,在區(qū)間(1,+∞)上單調(diào)遞增,所以h(x)min=h(1)=1,所以m<1.6.y2=12x解析:因?yàn)閯?dòng)點(diǎn)M到定點(diǎn)F(3,0)的距離比到直線(xiàn)l:x+1=0的距離大2,所以動(dòng)點(diǎn)M到定點(diǎn)F(3,0)的距離與到直線(xiàn)x+3=0的距離相等,所以點(diǎn)M的軌跡是以點(diǎn)F(3,0)為焦點(diǎn),直線(xiàn)x=3為準(zhǔn)線(xiàn)的拋物線(xiàn),故動(dòng)點(diǎn)M的軌跡方程是y2=12x.7.3解析:函數(shù)y=f(x)ln(x1)的零點(diǎn)個(gè)數(shù)即函數(shù)f(x)的圖象與函數(shù)y=ln(x1)的圖象的交點(diǎn)個(gè)數(shù),作出函數(shù)f(x)與y=ln(x1)的圖象,如圖.由圖可知,函數(shù)f(x)的圖象與函數(shù)y=ln(x1)的圖象有3個(gè)交點(diǎn),故函數(shù)y=f(x)ln(x1)的零點(diǎn)個(gè)數(shù)為3.8.(∞,5]解析:當(dāng)x≥0時(shí),f(x)=x2,此時(shí)函數(shù)f(x)單調(diào)遞增.∵f(x)是定義在R上的奇函數(shù),∴函數(shù)f(x)在R上單調(diào)遞增.若對(duì)任意x∈[a,a+2],不等式f(x+a)≥f(3x+1)恒成立,則x+a≥3x+1恒成立,即a≥2x+1恒成立.∵x∈[a,a+2],∴(2x+1)max=2(a+2)+1=2a+5,即a≥2a+5,解得a≤5,∴實(shí)數(shù)a的取值范圍是(∞,5].9.解:(1)因?yàn)?3acosA5ccosB=5bcosC,所以13sinAcosA=5sinBcosC+5sinCcosB,即13sinAcosA=5sin(B+C),所以13sinAcosA=5sinA,因?yàn)?<A<π,所以sinA>0,所以cosA=513所以sinA=1-(2)因?yàn)椤鰽BC的面積S△ABC=12bcsinA=613bc=6,所以由余弦定理得a2=b2+c22bccosA=b2+c210=(b+c)22bc10=(b+c)236,即28=(b+c)236,解得b+c=8,所以△ABC的周長(zhǎng)為8+27.10.解:(1)∵S42a2a3+6=0,∴4a1+4×32d2(a1+d)(a1+2d)+6又a1=1,∴4d2+12d=0,解得d=3或d=0(舍去),∴Sn=na1+n(n-(2)∵對(duì)每個(gè)n∈N*,存在實(shí)數(shù)cn使得an+cn,an+1+4cn,an+2+15cn成等比數(shù)列,∴(an+1+4cn)2=(an+cn)(an+2+15cn),即cn2+(8an+1an+215an)cn+an+12an而8an+1an+215an=8(a1+nd)[a1+(n+1)d]15[a1+(n1)d]=8+(148n)d,an+12anan+2=(an+d)2an(an+2d)∴cn2+[8+(148n)d]cn+d2∴Δ=[8+(148n)d]24d2≥0,即[(2n)d+1][(32n)d+2]≥0,當(dāng)n=1時(shí),顯然成立;當(dāng)n=2時(shí),有d+2≥0,∴d≤2,∴1<d≤2;當(dāng)n≥3時(shí),原式=[(n2)d1][(2n3)d2]>0恒成立.∴1<d≤2,即d的取值范圍是(1,2].思維提升訓(xùn)練11.B解析:顯然點(diǎn)A為準(zhǔn)線(xiàn)與x軸的交點(diǎn),如圖,過(guò)點(diǎn)P作PB垂直準(zhǔn)線(xiàn)于點(diǎn)B,則|PB|=|PF|.∴|PF||PA|設(shè)過(guò)點(diǎn)A的直線(xiàn)AC與拋物線(xiàn)切于點(diǎn)C,則0<∠BAC≤∠PAB≤π2∴sin∠BAC≤sin∠PAB.設(shè)點(diǎn)C的坐標(biāo)為(x0,y0),則y02=4x又y0x0∴x0=1,y0=2,∴C(1,2),|AC|=22.∴sin∠BAC=22∴|PF||PA|12.D解析:令t=sinx+cosx=2sinx+π4,則t∈[2,2],∴2sin∴f(x)=2sinxcosxsinxcosx可轉(zhuǎn)化為g(t)=t2t1=t-∴當(dāng)t=12時(shí),g(t)的最小值為54,即f(x)的最小值為13.[3,+∞)解析:由題意,知關(guān)于x的方程x2ax+2=0在[0,1]上有實(shí)數(shù)解.又易知x=0不是方程x2ax+2=0的解,所以根據(jù)0<x≤1可將方程x2ax+2=0變形為a=x2+2x從而問(wèn)題轉(zhuǎn)化為求函數(shù)g(x)=x+2x(0<x≤1)的值域易知函數(shù)g(x)在區(qū)間(0,1]上單調(diào)遞減,所以g(x)∈[3,+∞).故所求實(shí)數(shù)a的取值范圍是a≥3.14.2解析:由題設(shè),得ac=(x+1,4),b+c=(4,y2),又(ac)⊥(b+c),∴(ac)·(b+c)=4(x+1)+4(y2)=0,則xy+3=0,又ab=(x+3,2y),則|ab|=(x∴要求|ab|的最小值,即求定點(diǎn)(3,2)到直線(xiàn)xy+3=0的距離,∴|ab|min=|-315.解:(1)f(x)的定義域?yàn)镽,f'(x)=ax+1lna+2x2+lna=2(x1)+(1ax+1)lna(a>0,且a≠1),顯然可見(jiàn),f'(1)=0.①當(dāng)x>1時(shí),2(x1)>0,x+1<0.若0<a<1,則lna<0,ax+1>1,得1ax+1<0,于是f'(x)>0;若a>1,則lna>0,0<ax+1<1,得1ax+1>0,于是f'(x)>0.∴當(dāng)x>1時(shí),f'(x)>0,即f(x)在區(qū)間(1,+∞)上單調(diào)遞增.②當(dāng)x<1時(shí),2(x1)<0,x+1>0,若0<a<1,則lna<0,0<ax+1<1,得1ax+1>0,于是f'(x)<0;若a>1,則lna>0,ax+1>1,得1ax+1<0,于是f'(x)<0.∴當(dāng)x<1時(shí),f'(x)<0,即f(x)在區(qū)間(∞,1)上單調(diào)遞減.綜上所述,f(x)的單調(diào)遞增區(qū)間為(1,+∞),單調(diào)遞減區(qū)間為(∞,1).(2)對(duì)?x1,x2∈[0,2],使|f(x1)f(x2)|≤1a即對(duì)?x∈[0,2],f(x)maxf(x)min≤1a1成立由(1)知f(x)在區(qū)間[0,1]上單調(diào)遞減,在區(qū)間[1,2]上單調(diào)遞增,∴f(x)min=f(1)=1,f(x)max為f(0)和f(2)中的較大者.又f(0)=a+1lna,f(2)=a1+1+lna,f(0)f(2)=a1a2lna設(shè)φ(a)=a1a2lna,則φ'(a)=1+1a2-2a=∴φ(a)在區(qū)間(0,+∞)上單調(diào)遞增.注意到φ(1)=0,∴當(dāng)0<a<1時(shí),φ(a)<0,f(0)<f(2);當(dāng)a>1時(shí),φ(a)>0,f(0)>f(2).①當(dāng)0<a<1時(shí),f(x)maxf(x)min=f