大一輪答案完整

高考數(shù)學(xué)大一輪復(fù)習(xí)講義答案冊(cè)第一章預(yù)備知識(shí)1.1集合與常用邏輯用語考向1集合【典例分析】1.【解答】解:∵集合A={BA故選:C.2.【解答】解:∵集合A={∴∴A?B故選:C.3.【解答】a=-4.【解答】解:對(duì)于①,根據(jù)集合的子集關(guān)系得到{a,b對(duì)于②,兩個(gè)集合的元素完全相同,所以{a,b對(duì)于③,{0}含有運(yùn)算0,而Φ沒有任何元素;故{0對(duì)于④,根據(jù)集合與元素的關(guān)系,0∈{0}對(duì)于5,Φ與{0}都是集合而∈是元素與集合的關(guān)系對(duì)于⊙,空集是任何集合的子集,所以Φ?{0}故選:B.5.【解答】解:因?yàn)镸中:x=N中:x=所以N?故選:C.6.【解答】解:根據(jù)題意,滿足題意題意條件的集合M中必須有1,2,3這三個(gè)元素,且至少含有4、則M的個(gè)數(shù)應(yīng)為集合{4,5集合{4,5,6,7,8}故選:D.7.【解答】解:依題意,a-2=0當(dāng)a-2=0時(shí),此時(shí)A={0,-2當(dāng)2a-2=0時(shí)此時(shí)A={0,-1故選:B.8.【解答】解:根據(jù)題意得,?B=?時(shí),∴m②B≠?時(shí),m+1≤2綜上:m≤故選:D.9.【解答】解:∵A∴A∪B={x∣x=3k+1∴C故選:A.10.【解答】解:當(dāng)a>1時(shí),若A?B=R,∴1當(dāng)a=1時(shí),易得A=R,當(dāng)a<1時(shí),若A?B=R,則a∴a綜上,a的取值范圍是(-∞,2故選:B.11.【解答】解:由A?B=A,由x-3≤m,故有4≤m+3-2≥-m+3即m的最小值為5.故答案為:5.12.【解答】解:設(shè)只喜歡足球的百分比為x,只喜歡游泳的百分比為y,兩個(gè)項(xiàng)目都喜歡的百分比為z,由題意,可得x+z=60,x∴該中學(xué)既喜歡足球又喜歡游泳的學(xué)生數(shù)占該校學(xué)生總數(shù)的比例是46%故選:C.13.【解答】解:由已知中的Venn圖可得,陰影部分的元素屬于M,屬于P,但不屬于S,所以陰影部分表示的集合為M∩故選:B.14.【解答】解:全集U={x∣x為不大于20∵A∴A【精選練習(xí)】1.【解答】解:由已知得,若2a+1=3,解得a=1,此時(shí)A={1,3},B={0,1,3},若2a+1=3a-2,解得a=綜上所述:a=故選:C.2.【解答】解:集合A=由1x-1≥0,得x>1,故故選:D.3.【解答】解:由已知A∩?RB=A又因?yàn)?R所以a≥即實(shí)數(shù)a的取值范圍是{a故選:C.4.【解答】解:依題意,B?A,若a4=16,解得a=-2a=2舍去,a4故選:B.5.【解答】解:因?yàn)楹螦=1,3,所以a+2=3解得a=1或a=-1當(dāng)a=±1時(shí)a2=1,集合A不滿足元素的互異性當(dāng)a=2時(shí)A={故選:C.6.【解答】解:由B=A則A?故選:C.7.【解答】解:因?yàn)榧螧={x∣x?A},B={?,{太平洋},{大西洋},{太平洋,大西洋}},8.【解答】解:因?yàn)槿疷={1,所以M?故選:C.9.【解答】解:因?yàn)榧螦={1,2,3,4},A:結(jié)合選項(xiàng)可知,A顯然不符合題意;B:x∣x2>1={C:x∣logD:x∣2故選:D.10.【解答】解:根據(jù)描述法表示集合,A=∴A故選:B.11.【解答】解:A={-則A?B={0,1故A∩B的真子集個(gè)數(shù)為故選:B.12.【解答】解:A??UB,說明A的所有元素都不屬于B,等價(jià)于B的所有元素都不屬于A,如圖,故選:B.13.【解答】解:A={x∈R∣-5<故A?故選:C.14.【解答】解:集合A=x?1xB所以?R故選:C.15.【解答】解:∵集合A={x∥x-B∴A故選:A.16.【解答】解:集合M={N∴M故選:B.17.【解答】解:因?yàn)榧蟂={x∣x<-所以a<-1a+8>則實(shí)數(shù)a的取值范圍為-3故選:A.18.【解答】解:當(dāng)M?N?A:M?B:?UC:M?D:N?故選:D.19.【解答】解:由于A={x,故求A∩B的元素個(gè)數(shù)即為求y=解方程y=2xy=x3,可得x=0故A∩B的元素個(gè)數(shù)是3故選:C.20.【解答】解:集合A=B所以?R所以?R故選:D.考向2常用邏輯用語【典例分析】1.【解答】解:∵sin①當(dāng)sinx=1時(shí),則cos②當(dāng)cosx=0時(shí),則sin∴sinx=1是cos故選:A.2.【解答】解:∵x∵x∵0<x<50∴0<x<5是即x2-5x<0故選:B.3.【解答】解:由直線l1:mx+4y-可得:m2解得m=±2,經(jīng)過驗(yàn)證m=2時(shí),∴m∴“m=-2”是“直線l1:mx+4故選:C.4.【解答】解:若?x∈1,3,使得x2可得a<x+3x因?yàn)楹瘮?shù)fx=x+3x在1,3上單調(diào)遞減,在故當(dāng)x∈1,3時(shí),fxmax所以p的一個(gè)必要不充分條件是a<故選:A.5.【解答】解:命題:p:?x∈R,x+1>1,x=-1時(shí),不成立,所以命題:p是假命題;則?p是真命題.命題所以?p和q都是真命題故選:B.6.【解答】解:“?x∈R,?n∈N*,使得n≥x故選:D.【點(diǎn)評(píng)】本題考查命題的否定,解本題的關(guān)鍵是掌握住特稱命題的否定是全稱命題,書寫答案是注意量詞的變化.7.【解答】解:命題的否定是:?x故選:C.【精選練習(xí)】1.【解答】解:2a當(dāng)a<0或b<0時(shí),反之由log2a>log2b即:a>∴?alog2a>log故選:A.2.【解答】解:當(dāng)a>0且b>0時(shí)則ab+ba≥2ab?ba=2所以充分性成立;當(dāng)a<0且b<0時(shí)則ab+ba≥2ab?ba=2所以必要性不成立;所以“a>0且b>0”是“故選:A.3.【解答】解:a=1.1,b=0.1,滿足a+b≤a>0,b>0,a+b≤1時(shí),得出所以“a+b≤2”是“故選:B.4.【解答】解:若a1=-1,q=1,則Sn=n∵S則Sn∴若Sn是遞增數(shù)列∴S則a1∴滿足必要性,故甲是乙的必要條件但不是充分條件,故選:B.5.【解答】解:A,當(dāng)l//β且α⊥β時(shí),則l⊥α或l//B,當(dāng)l⊥β且α⊥β時(shí),則l//αC,當(dāng)l?β且α⊥β時(shí),則l⊥α或l//α或l?α或l與D,當(dāng)l⊥β且α//β時(shí),則故選:D.6.【解答】解:設(shè)函數(shù)fx=cosx+b則“b=0”?“fx“fx為偶函數(shù)”?“b∴函數(shù)fx=cosx+b則“b=0”是“fx故選:C.7.【解答】解:由全稱命題的否定形式可知:命題“?x>0,x2>2x”1.2不等式考向1二次不等式【典例分析】1.【解答】解:不等式x2-12x+20=x-2x不等式x2-5x+6=不等式式9x2-6x+1由等式-2x2+2x-3>0故選:ABD.2.【解答】解:(1)∵不等式1+ax2+∴-3和1是對(duì)應(yīng)方程1+a當(dāng)x=1時(shí),1+a+a故不等式3x2+2-ax+4a>0等價(jià)于3x2+故不等式的解集為x?（2）ax2-2a+當(dāng)a=0時(shí),-3x+6當(dāng)a<0時(shí),x-3ax-2>0當(dāng)3a=2,即a=32時(shí),x-2當(dāng)3a>2,即0<a<當(dāng)0<3a<2,即a>32時(shí),綜上所述,當(dāng)a=0時(shí),解集為當(dāng)a<0時(shí),解集為x∣x>當(dāng)a=32時(shí),解集為當(dāng)0<a<32時(shí)當(dāng)a>32時(shí),解集為3.【解答】解:∵y=fx=-x（1）當(dāng)0≤a2≤1時(shí),即0≤由14a2-a+2=2得a=-2或（2）當(dāng)a2<0,即a<0時(shí),fx在0,1上單調(diào)遞減得-a4+12=（3）當(dāng)a2>1,即a>2時(shí),fx在0由f1=2得:-1+a綜上所述,a=-6或4.【解答】解:因?yàn)閒x=x2-4x當(dāng)t+1≤2,即t≤1時(shí),fx在當(dāng)t≥2時(shí),fx在t,t+當(dāng)t<2<t+1,即1<t<2時(shí),f故gt5.【解答】設(shè)函數(shù)fx（1）當(dāng)x∈R時(shí),fx≥0恒成立,則m（2）當(dāng)x∈0,+∞時(shí),fx≥0恒成立,則（3）當(dāng)x∈-1,1時(shí),fx≥0恒成立,6.【解答】解:(1)原式可化為ax2+1-ax+a≥0有實(shí)數(shù)解①當(dāng)a=0時(shí),x≥0有實(shí)數(shù)解,②當(dāng)a>0時(shí),取x=0,則ax③當(dāng)a<0時(shí),二次函數(shù)y=ax2+1-當(dāng)且僅當(dāng)△=1-a2-4綜上所述:a≥-1,所以實(shí)數(shù)a的取值范圍為（2）不等式fx≥-2對(duì)于實(shí)數(shù)a∈-1,顯然x2-x+1>0,函數(shù)g(a)=從而得g-1≥0,即-x2所以實(shí)數(shù)x的取值范圍是{17.【解答】解:當(dāng)a=0時(shí),x=12.當(dāng)a≠0時(shí),①若方程有一正一負(fù)根,則x1②若方程有兩個(gè)正根,則λ≥0x綜上得:實(shí)數(shù)a的取值范圍是[-8.【解答】解:令fx=x2+m求得0<故選:D.【精選練習(xí)】1.【解答】解:對(duì)于A:若a=-1,-3x2-2x+1>0,即即不等式的解集為-1,13,故對(duì)于B:若不等式的解集為-2,43,∴a<0,且-2與∴-2+43=-23,-2×對(duì)于C:若不等式的解集為x1,x2,∴a<0,且x1與∴x1+x2=-23x1?x2對(duì)于D:由不等式恒成立,即不等式的解集為R,當(dāng)a=0,當(dāng)a≠0時(shí),a>0Δ=綜上所述:a∈[0,3)故故選:ABC.2.【解答】解:通過畫二次函數(shù)圖象觀察圖象,欲使得閉區(qū)間0,m上有最大值3,區(qū)間0,m且在2的左側(cè)(否則最大值會(huì)超過3)∴知m∈答案:13.【解答】解:∵二次函數(shù)fx=x對(duì)于任意x∈m,m+1,∴即-22<m<2故選:A.4.【解答】解:由f=x令gm由題意知在-1,1上,g∴g-1=x-2×-1故當(dāng)x∈-∞,對(duì)任意的m∈-1,1,函數(shù)5.【解答】解:∵2∴m由于函數(shù)y=2x-1x在區(qū)間故選:D.考向2幾種常用不等式的解法【典例分析】1.【解答】解:不等式2-3xx-解得23故選:D.2.【解答】解:2x解得,-7故答案為:-73.【解答】解:原不等式等價(jià)于x+1x-3≤0?1-2xx≤故答案為:x?4.【解答】解:因?yàn)閤+5x-12≥2,所以2x2-5x-3≤0且x-所以不等式的解集為-1故選:D.5.【解答】解:關(guān)于x的不等式ax-1x+1>由不等式的解集是-∞,-則-1和12是方程ax-1所以12a-1=0故選:D.6.【解答】解:(1)不等式x3∵方程x2-3x+3=0∴x2-∴x∴x∴不等式x3-2x2（2）∵x∴當(dāng)x<-2時(shí),∴x∴x<-2當(dāng)-2≤x≤-1時(shí),同理可得當(dāng)-1<x<0時(shí)∴-1<x當(dāng)x≥0時(shí),xx綜上所述,不等式xx-12x+13x（3）∵x∴x-2x2+16由穿根法知,原不等式的解集為x?或者:x-62x-5解1得x∈?解2得52≤x<3∴原不等式的解集為:x?7.【解答】解:(1)由2x若x≤0,則不等式2x則x>0,原不等式即為解得13故不等式的解集為13（2）當(dāng)x≤-12時(shí),原不等式即為解得-7當(dāng)-12<x<2解得-1當(dāng)x≥2時(shí),原不等式即為解得2≤綜上,不等式的解集為-7【精選練習(xí)】1.【解答】解:由于x2+2x+3=x+12故選:B.2.【解答】解:(1)原不等式等價(jià)于x+∴x≠5x+4x-2>0,解得如圖所示:∴原不等式解集為{x∣x<-4或x（2）將4x2--∴原不等式等價(jià)于不等式組4x則2x2x-5>∴原不等式的解集為x?-12<（3）由x2-4x+1∴原不等式等價(jià)于2x-1x-1如圖所示:∴原不等式解集為x∣x<133.【解答】解:不等式1≤2x-1<5,可得1≤2x-1<5或-5<2不等式2x+5>7的解集為{x∣x<-6不等式x<aa>0解集為{x∣-a不等式x>aa>0解集為{x∣x<-a或故選:A.4.【解答】解:方法一:不等式等價(jià)轉(zhuǎn)化為x+1≥x-3,兩邊平方得x故不等式的解集為[1故答案為[1方法二:不等式等價(jià)轉(zhuǎn)化為x+1≥x-3,根據(jù)絕對(duì)值的幾何意義可得數(shù)軸上點(diǎn)x到點(diǎn)-1對(duì)應(yīng)點(diǎn)的距離大于或等于它到點(diǎn)3對(duì)應(yīng)點(diǎn)的距離,到這兩點(diǎn)-1和3距離相等時(shí),x=1,考向3基本不等式【典例分析】1.【解答】解:當(dāng)x>0時(shí),當(dāng)且僅當(dāng)3x+3=4x+1故答案為:432.【解答】解:∵a>1,b>所以a-即a+b≥3+22,當(dāng)且僅當(dāng)a-故選:C.3.【解答】解:對(duì)于A,所以函數(shù)的最小值為3,故選項(xiàng)A錯(cuò)誤;對(duì)于B,因?yàn)?<sinx≤1當(dāng)且僅當(dāng)sinx=4sinx,即因?yàn)閟inx≤1所以y=sinx+4sinx>對(duì)于C,因?yàn)?x>0,所以當(dāng)且僅當(dāng)2x=2,即x所以函數(shù)的最小值為4,故選項(xiàng)C正確;對(duì)于D,因?yàn)楫?dāng)x=1e時(shí),所以函數(shù)的最小值不是4,故選項(xiàng)D錯(cuò)誤.故選:C.4.【解答】解:因?yàn)閤>所以x2-x≤x+2-x22=1,當(dāng)且僅當(dāng)x因?yàn)閤+1x≥2x?1x=2,所以2-x-令t=4+x2原式可化為y=t2+1t=t+1t,在x+4x+1=x+1+4x+1-1≥2故選:C.5.【解答】解:對(duì)于命題①ab≤1:由2=a+b對(duì)于命題2a+b≤2:令a=1,對(duì)于命題③a2+b2對(duì)于命題④a3+b3≥3:令對(duì)于命題51a+1b≥2:16.【解答】解:由題意,a+1+b+故答案為:327.【解答】解:因?yàn)锳B為直徑,故AD⊥根據(jù)圖形,在直角三角形ADB中,利用射影定理得,CD所以CD即CD=又OD由OD≥得a+同理,在直角三角形OCD中,由射影定理得,CD所以DE=由于CD≥所以ab≥故答案為:①③.8.【解答】解:方法一:由x2+y2-xy令x-y2=cosθ3∴x+y=3sinθ+cosθ=2sinθ∵x故C對(duì),D錯(cuò),方法二:對(duì)于A,B,由x2+y2-xy=1∴x+y2≤4,∴-2≤x+y≤對(duì)于C,D,由x2+y2∴x2+y2≤2∵-xy∴x2+y2≥23故選:BC.9.【解答】解:a2+b2+ab因?yàn)閍,b>0,從而a+b2-a+b因?yàn)閍≠b,所以即a+b2-a+b綜上可知,1<a+b<43ab從而ab<12,故C錯(cuò)誤,D故選:ABD.10.【解答】解:因?yàn)橹本€ax+by-2=所以a+則4a當(dāng)且僅當(dāng)a=4b,即b故選:B.11.【解答】解:令t=sin2x,=當(dāng)且僅當(dāng)t1-t故答案為:9.12.【解答】解:a>0,b>0則a-所以1a-1+3b-2故選:C.13.【解答】解:∵b∴a當(dāng)且僅當(dāng)b=a-ba2=8故答案為:1614.【解答】解:因?yàn)?當(dāng)且僅當(dāng)1a=1b,且1ab=ab,即a=b故選:C.15.【解答】解:∵a則a4當(dāng)且僅當(dāng)a4即a=142b=1故選:B.16.【解答】解:由x+2yx當(dāng)且僅當(dāng)2xy=6yx時(shí),所以x+1y+1故選:C.17.【解答】解:∵a+2b=2,則a2由基本不等式可得1+當(dāng)且僅當(dāng)4ba=9a4b所以,1+4a+3故答案為:25218.【解答】解:∵a>0,b>0,∴4=a+b∴a+b2-2故選:D.19.【解答】解:設(shè)m=x+y,n∴2當(dāng)且僅當(dāng)m2=13n∴2x2+y故答案為:2320.【解答】解:由題意得x2所以x2當(dāng)且僅當(dāng)15y2=4y25故選:B.21.【解答】解:∵a,b為正數(shù),且∴a∴9∵==≥當(dāng)且僅當(dāng)9ba=∴故選:B.22.【解答】解:由柯西不等式得x2∵是當(dāng)且僅當(dāng)x1=y2=z2時(shí)∴x2+【點(diǎn)評(píng)】本題主要考查不等式的證明,利用柯西不等式是解決本題的關(guān)鍵.,23.【解答】解:x≥當(dāng)且僅當(dāng)1即xy=±22,取得最小值故答案為:9.24.【解答】解:根據(jù)題意,若2x則4x又由2x+y=2則4x當(dāng)且僅當(dāng)y+1=2x即4x2y+1故答案為:4525.【解答】解:令fx則f'令f'x=0,又因?yàn)閟in2x+cos2x所以當(dāng)x∈0,arcsin255時(shí)當(dāng)x∈arcsin255,π2所以x=arcsin255是函數(shù)fx所以fx的最小值為f故答案為:55【精選練習(xí)】1.【解答】解:由題意可得:OC=a+b2,CD∴a+b2≤a2+因此該圖形可以完成的無字證明為B.故選:B.2.【解答】解:①已知a>0,b>0,且a+b=1,所以a+b2②利用分析法:要證2a-b>12,只需證明a-b>-1即可,即a>b-1,由于a>0,b>③log2a+log2b=log④由于a>0,b>0利用分析法:要證a+b≤2成立,只需對(duì)關(guān)系式進(jìn)行平方,整理得a+b+2ab≤2,即2ab≤1,故ab≤12故選:ABD.3.【解答】解:因?yàn)?<所以x2故x2+y2-故選:C.4.【解答】解:因?yàn)閍2所以a+b2=3ab+4≤3整理可得a+b2≤16故選:B.5.【解答】解:因?yàn)?<x<1,則1x當(dāng)且僅當(dāng)1-xx=2x1-x,即x故選:D.6.【解答】解:2==≥當(dāng)且僅當(dāng)a-5如取a=2,故選:B.7.【解答】解:由題意a>則5a當(dāng)且僅當(dāng)9ab=ba且a+b=即5a+b故答案為:12.8.【解答】解:因?yàn)閷?shí)數(shù)x,y滿足x>y>0所以x>1-x>0則2x==當(dāng)且僅當(dāng)22x-故選:B.9.【解答】解:∵實(shí)數(shù)x,y滿足2x2即4x+4y+8≥16,求得x+y故x+y故答案為:2.10.【解答】解:已知a>1,b>0,則1a當(dāng)且僅當(dāng)a-12b則1a-1+12故選:A.第二章函數(shù)2.1函數(shù)三要素考向1函數(shù)的定義域【典例分析】1.【解答】解:要使原式有意義,需要log0.5即0<4x-3<1所以原函數(shù)的定義域?yàn)?4故選:A.2.【解答】解:由題知:log2x-1≠0,且x-1>0又因?yàn)閤-2-1≥0,解得:所以x≥故答案為:{x【點(diǎn)評(píng)】函數(shù)的定義域是高考的必考題,經(jīng)常以選擇或填空的形式出現(xiàn),應(yīng)給以重視.3.【解答】解:由題意得:2cosx-1>0,即cos故選:A.4.【解答】解:(1)已知y=fx的定義域?yàn)閯t①由0≤x2≤2得0≤x≤2或-2②由0≤2x-1≤2得③由0≤x-2≤2得2（2）已知函數(shù)fx2-1則0≤x≤1,即fx的定義域?yàn)?（3）已知函數(shù)f2x+1則0<x<1,即fx的定義域?yàn)?由1<2x-1<3,得1<x<（4）已知函數(shù)fx+1的定義域?yàn)閯t-2≤x≤3由-1≤1x+2解得x≥12或即f1x+2的定義域是x?（5）已知函數(shù)fx的定義域?yàn)?則0≤由0≤x+m≤∵m∴當(dāng)1-m=m時(shí),即此時(shí)x=若0<m<12若m>12∴當(dāng)0<m<12時(shí)當(dāng)m=12時(shí),函數(shù)的定義域?yàn)楫?dāng)m>12時(shí),函數(shù)定義域?yàn)榭占?此時(shí)不成立綜上:故當(dāng)0<m<12時(shí)當(dāng)m=12時(shí),函數(shù)的定義域?yàn)椋?）設(shè)μ1=ax,μ2則Fx=fμ1∴①當(dāng)a≥1時(shí),-a2≤-1②當(dāng)0<a<1時(shí)∴當(dāng)a≥1時(shí),Fx的定義域?yàn)楫?dāng)0<a<1時(shí),Fx5.【解答】解:由題意知,2+x2-x故:-2<x2解得-4<x<-故選:B.6.【解答】解:∵函數(shù)fx的定義域?yàn)?,3,則對(duì)于函數(shù)應(yīng)有1≤2x-1≤33x-3≥1,求得43≤x≤7.【解答】解:函數(shù)fx=xmx即mx2+當(dāng)m=0時(shí),1>當(dāng)m≠0時(shí),需m>0Δ=綜上可得,0≤故選:A.8.【解答】解:∵fx=lga2∴a2-當(dāng)a2-1=0時(shí),得當(dāng)a2-1≠0解得a>53或當(dāng)a=-1時(shí),函數(shù)y=lga2-1x實(shí)數(shù)a的取值范圍是a>53或【精選練習(xí)】1.【解答】解:要使函數(shù)有意義,則log2即log2x>1解得x>2或即函數(shù)的定義域?yàn)?,故選:C.2.【解答】解:要使函數(shù)有意義,則4-即-4x-2x-3x-3>0等價(jià)為①x②x<3x-2x-3<0即2<x<3∵-4∴解得3<x≤4即函數(shù)的定義域?yàn)?,故選:C.3.【解答】解:∵原函數(shù)的定義域?yàn)?1∴-1<2x+1∴則函數(shù)f2x+1故選:B.4.【解答】解:因?yàn)閒x的定義域?yàn)?,2,所以對(duì)gx,0≤2x≤2且x5.【解答】解:fx的定義域?yàn)镽,即ax2當(dāng)a=0時(shí),ax2當(dāng)a≠0時(shí),要使fx的定義域?yàn)镽,則需要滿足解得0<綜上,a∈[故答案為:[0【點(diǎn)評(píng)】本題考查對(duì)數(shù)函數(shù)的性質(zhì)和應(yīng)用,是中檔題.考向2函數(shù)的解析式【典例分析】1.【解答】解:由題意f1ff…f故f故答案為:x2.【解答】解:函數(shù)fln設(shè)t=ln則x=那么:ft即函數(shù)fx的表達(dá)式為f故選:D.3.【解答】解:fsin則fcos故選:C.4.【解答】解:令1-x1+x=∴fx=2x15.【解答】解:設(shè)y=ax2+bx+ca≠0∵f即2ax+a+b=2x…(8分)∴26.【解答】解:由已知可得:f1x+2fx=1x,聯(lián)立fx故答案為:fx7.【解答】解:∵fx為定義在R∴f-x=fx又∵gx為定義在R∴f-x+g-8.【解答】解:(I)由fx令x=1,y=0又f1=0,則(II)由fx令y=0,得由f0=-2,則9.【解答】解:令x=12,y=又f12≠0,故1+f令x=12,y=-即f0+f12f-12=-1又f12≠0,故f-12=令y=-12,則有即fx-12=-2x有fx+1-12即函數(shù)fx+1令x=1,有f12=-2×1=-2,故B故選:ABD.10.【解答】解:∵函數(shù)fx圖象關(guān)于直線x=1對(duì)稱∵當(dāng)x≤1,fx=∴fx=f2-x=e2∴曲線y=fx在點(diǎn)P2,f2處的切線方程為故答案為:2x11.【解答】解:設(shè)x,y是函數(shù)y=則它關(guān)于y=-x的對(duì)稱點(diǎn)為-y,-x,即-y,-x在即y=-log2-x+a,∴f-2+f12.【解答】解:因?yàn)閒x∵x∈(0,1]∴x∈(1,2]∴x∈(2,3]當(dāng)x∈(2,3]時(shí),由4x-2x若對(duì)任意x∈(-∞,m],都有fx≥-89,則13.【解答】解:∵fx是定義在0,+∞上的單調(diào)函數(shù),且∴fx-1x是常數(shù),設(shè)fx∴fc=1c+f∴f∴f故選:D.【精選練習(xí)】1.【解答】解:f1f又2018=∴f故選:D.2.【解答】解:任意的x屬于R都有有ff而函數(shù)是單調(diào)的,所以對(duì)任何的x,fx-3即fxf而fc所以3c解得:c=而fx故選:B.3.【解答】解:(1)因?yàn)楹瘮?shù)y=fx設(shè)fx因?yàn)閒x所以ax+故a2則a2=42ab-3a=-10所以fx=-2x+（2）解:設(shè)二次函數(shù)的解析式為fx由f0=1,可得故fx因?yàn)閒x則ax+12+b所以2a=2a+b所以fx（3）設(shè)t=2x+1則ft所以fx（4）因?yàn)閒x所以fx（5）由fx則f1由①②可得,fx（6）令x=y=0,可得f0=令y=0,可得fx=f0（7）令t=1-x1+所以ft故fx4.【解答】解:設(shè)Px,y為y=fx的反函數(shù)圖象上的任意一點(diǎn),則P關(guān)于y=x的對(duì)稱點(diǎn)又∵函數(shù)y=fx的圖象與函數(shù)y=gx的圖象關(guān)于直線x+y=0對(duì)稱,∴P'y,x∴必有-y=g-x,即y=-g-x∴y5.【解答】解:當(dāng)-1≤x≤0時(shí),0≤故答案為:-1考向3函數(shù)的值域1.【解答】y=-=-故函數(shù)的值域?yàn)閥?2【解答】.y=令x=sin故y=(其中sinθ因?yàn)?π2≤a≤π所以當(dāng)a+θ=-π2+arcsin當(dāng)a+θ=π2故函數(shù)的值域?yàn)?23.【解答】解:∵∴====∴當(dāng)log即x=22時(shí),函數(shù)fx故答案為:-4.【解答】解析:令5+4cosx=t當(dāng)0≤x≤π時(shí)所以f當(dāng)且僅當(dāng)t=3時(shí)取等號(hào).同理可得當(dāng)π<x≤2綜上可知fx的值域?yàn)?12,125.【解答】3x2+2y令x=故z=-1∵1∴0故函數(shù)z=x2+y6.【解答】解:∵函數(shù)fx=1-x2x2+ax+∴f∴f令t=x2+4x-5=x故當(dāng)t=-4時(shí),fx取得最大值16.7.【解答】解:解法一:畫出y=2x,y=x+2,y=10-當(dāng)2≤x≤4時(shí),fx=x+2,當(dāng)x>4時(shí),fx解法二:由x+2-10-x0<x≤22<x≤4時(shí)由2x+x-x>x1時(shí)2x>10-x,∴fxmax=f(4)=6.8.【解答】作函數(shù)fx=2當(dāng)x=-12時(shí),故函數(shù)的值域?yàn)?99.【解答】解:解析式表示過Acosx,sinx,B2,1的直線的斜率所以設(shè)切線得斜率為k,則直線方程為y-1=kx-2,即kx-y-2k+1=0,1-2k10.【解答】解:由原函數(shù)得,2y∴sinx∴1+y∴sinx∴3∴3-2y2≤∴原函數(shù)的值域?yàn)?2-233,2+233.11.【解答】fx=2x+3x-2,x∈3,8,可化解為:fx=2x-2+12.【解答】013.【解答】解:y=∵-1∴-1∴1∴1∴4∴1∴函數(shù)y=2+cosx214.【解答】-∞15【解答】解:要求最大值,顯然x-1∴令x-1=t∵t當(dāng)且僅當(dāng)t=2時(shí)等號(hào)成立.故選:16.【解答】解:令x-則x=y=由于t+2t的導(dǎo)數(shù)為在t≤-2上導(dǎo)數(shù)大于0,則有t+即有t+即有-1則函數(shù)的值域?yàn)?117.【解答】(18.【解答】y19.【解答】220.【解答】0【精選練習(xí)】1.【解答】解:(1)函數(shù)y=x+1的定義域?yàn)閇0,+∞),又函數(shù)為增函數(shù),（2）令1-x=tt≥0,則∴函數(shù)y=2x+4（3）由y=∵833x-4≠0,∴y(4)由y=x∴y≠1且綜上,函數(shù)y=x2+4x-5x(5)由y=x2+4x當(dāng)y=1時(shí),方程化為5x=7當(dāng)y≠1時(shí),由△=y+42-4y-1綜上,函數(shù)y=x2+4（6）由絕對(duì)值的幾何意義可知,y=x+1+x-2為數(shù)軸上的動(dòng)點(diǎn)x到兩定點(diǎn)-1、2的距離和,∴函數(shù)y=2.【解答】解(1)y由二次函數(shù)的性質(zhì)可知,當(dāng)x=16時(shí),故函數(shù)的值域?yàn)?3(2)y∵∴故函數(shù)的值域0(3)y故函數(shù)的值域-∞（4）令1-x=t則ty=x+41-x=1-t2+4t=-t-2∴函數(shù)的值域?yàn)?-∞,(5)令x=cosα,則當(dāng)sinα≥0取α∈0,當(dāng)sinα<0取α∈π,∴-（6）y∴故函數(shù)的值域[(7)∵∴①當(dāng)y=2時(shí),x②當(dāng)y≠2時(shí),Δ=y解可得1≤y≤5綜上可得,1故函數(shù)的值域?yàn)閧(8)∵∴∴故函數(shù)的值域?yàn)閇（9）∵y=1-sinx2-cosx=sinx-∴故函數(shù)的值域?yàn)?(10)∵∴∴y≠-15∴函數(shù)的值域?yàn)閥(11)∵令1-x=t,則x∴根據(jù)二次函數(shù)的性質(zhì)可知,當(dāng)t=1時(shí),函數(shù)的值域?yàn)?-∞,(12)y①當(dāng)x=0時(shí),②當(dāng)x∵∴③當(dāng)x<0時(shí),∵∴綜上可得,函數(shù)的值域?yàn)?-1(13)∵y=4-令f則0∴∴2≤fx(14)∵y=x-1-2x的定義域?yàn)椤喈?dāng)x=12時(shí),故函數(shù)的值域-∞(15)∵∴當(dāng)y=2時(shí),故y≠∴△=即y∴故函數(shù)的值域(2.2基本初等函數(shù)考向1指數(shù)、對(duì)數(shù)運(yùn)算【典例分析】1.【解答】解:由2a可得8b則4a故選:C.2.【解答】由題1log8a-1log?log2a=-1或log2所以log2a=6=故答案為:64.3.【解答】解:由題意得a=則0<1a<1b對(duì)A,根據(jù)對(duì)數(shù)函數(shù)y=lgx在0,+∞上單調(diào)遞增,則lga>lgb,故對(duì)B,因?yàn)?a+1b=log153+log155=1,即對(duì)C,因?yàn)閍>b>0,根據(jù)指數(shù)函數(shù)y=12x在R上單調(diào)遞減,則1對(duì)D,因?yàn)閍>b>0,1a成立,而顯然a≠b,則a+b>4,故選:ABD.4.【解答】解:由已知可得K1+e-0.23t兩邊取對(duì)數(shù)有-0.23解得t*故選:C.5.【解答】解:因?yàn)楫?dāng)n越來越大時(shí),1+1nn所以當(dāng)n越來越大時(shí),1+12n2n趨向于常數(shù)e,1+12nn=1+1當(dāng)x越來越大時(shí),1+1xx趨向于常數(shù)e,1+1x3x當(dāng)x越來越大時(shí),1+13x3x趨向于常數(shù)e,故選:BD.6.【解答】解:由題意得,60≤5020可得p1≥pp2≤10p3=100p1≤10故選:ACD.7.【解答】解:fx=3x是指數(shù)函數(shù)滿足fx+fx=log2x是對(duì)數(shù)函數(shù)滿足ffx=tanx滿足fx+y故選:B.8.【解答】解:y=lnx定義域{x∣y=x3為奇函數(shù)y=2x定義域?yàn)镽且為偶函數(shù),當(dāng)x>0時(shí),又fx1x2fx=x定義域?yàn)镽又fx1x2故選:D.【精選練習(xí)】1.【解答】解:∵2∴1故選:C.2.【解答】解:因?yàn)閍s+t所以2lgxy=2故選:D.3.【解答】解:設(shè)3x=5y=15∴對(duì)于選項(xiàng)A:∵x+y=log3t+對(duì)于選項(xiàng)B:∵∴xz+yz=zx+y對(duì)于選項(xiàng)C:x3=log3t3∵t>1,∴函數(shù)y=xlogt∴3即x3>y5>z15對(duì)于選項(xiàng)D:∵l∴l(xiāng)g215>4∴xy>4z2,故選項(xiàng)故選:BCD.4.【解答】解:由題意得,a=所以abc=則log4故選:B.5.【解答】解:設(shè)太陽的星等是m1=-26.7,天狼星的星等是由題意可得:-1.45∴l(xiāng)gE1E2=50.5故選:A.考向2指數(shù)、對(duì)數(shù)圖像性質(zhì)【典例分析】1.【解答】解:對(duì)于選項(xiàng)A:當(dāng)x=1時(shí),y=1a=1對(duì)于選項(xiàng)B:當(dāng)x=1時(shí),y=a0=1對(duì)于選項(xiàng)C:當(dāng)x=1時(shí),y=loga1+1對(duì)于選項(xiàng)D:當(dāng)x=1時(shí),y=a+1,圖象不經(jīng)過定點(diǎn)1,1故選:ABC.2.【解答】解:函數(shù)fx令x-2=0得,x=2∴函數(shù)fx的圖像恒過定點(diǎn)2,2,即又∵點(diǎn)P在直線mx+ny-∴2∴1m+1n=2m+n1m+1立,∴1m+故選:D.3.【解答】解:∵x=-1時(shí),∴函數(shù)y=logax+即A-∵點(diǎn)A在直線mx+ny+∴-m-3n+1∵mn∴m∴=當(dāng)且僅當(dāng)m=n=14時(shí)取等號(hào)4.【解答】解:冪函數(shù)a=2,b=在第一象限內(nèi),x=1的右側(cè)部分的圖象,圖象由下至上,冪指數(shù)增大,所以a>b>c5.【解答】解:由函數(shù)y=當(dāng)a>1時(shí),可得y=1ax是遞減函數(shù),函數(shù)y=logax+12當(dāng)1>a>0時(shí),可得y=1ax函數(shù)y=logax+12∴滿足要求的圖象為:D故選:D.6.【解答】解:因?yàn)閒x所以函數(shù)為偶函數(shù),當(dāng)a=12時(shí),fx=12x函數(shù)在0,+∞上單調(diào)遞減,gx=x12=當(dāng)a=3時(shí),fx=3x函數(shù)在0,+∞上單調(diào)遞增,gx=x3,函數(shù)定義域?yàn)楫?dāng)a=2時(shí),fx=2x函數(shù)在0,+∞上單調(diào)遞增,gx=x2,函數(shù)定義域?yàn)?∞,+∞且在-∞,0對(duì)于C,由題可知關(guān)于y軸對(duì)稱的函數(shù)為fx=ax,且在0,+∞上單調(diào)遞減,故a=1數(shù)定義域?yàn)閇0,+∞)且單調(diào)遞增,故C故選:C.7.【解答】解:∵fx∴m解得m=3或由當(dāng)x∈0,+∞則m2解得2<∴m故選:C.8.【解答】解:依題意,設(shè)fx=xα,則有18α所以α=12,于是由于函數(shù)fx=x12在定義域[0,+∞)內(nèi)單調(diào)遞增,所以當(dāng)x從而有x1fx1<x又因?yàn)閒x1x1,fx2x圖象,容易得出直線OP的斜率大于直線OQ的斜率,故fx1x1答案239.【解答】解:∵a>0且∴內(nèi)層函數(shù)t=1-要使函數(shù)y=loga1-ax則0<a<11-∴a的取值范圍是0故選:D.10.【解答】解:對(duì)于A:當(dāng)a=0時(shí),fx=lgx2+1,由x2+1>對(duì)于B:fx的定義域?yàn)镽,則x2+ax+解得-2<a<2,故對(duì)于C:fx的值域?yàn)镽,則t=x2+解得a∈-∞,-2]?[2,+∞對(duì)于D:因?yàn)閺?fù)合函數(shù)fx=lgx2+ax+1是由y=lgt,t單調(diào)遞增,又fx=lgx2+ax+所以t=x2+ax+1在2,+∞上單調(diào)遞增,則有又x2+ax+1>0在2,+∞上恒成立,則有綜上,a≥-52,故D故選:ABC.11.【解答】解:對(duì)于A,若a<0,b<0,c<0而函數(shù)fx=2x-1在區(qū)間-∞,0上是減函數(shù),故fa>f?b>fc,與題設(shè)矛盾,所以A不正確;對(duì)于B,若a<0,b≥對(duì)于C,取a=0,c=3,同樣fc=f3=對(duì)于D,因?yàn)閍<c,且f(a)>fia、c位于函數(shù)的減區(qū)間-∞,0,此時(shí)a<b<(ii)a、c不在函數(shù)的減區(qū)間-∞,0,所以fa=1-2a>2c-1=fc,化簡(jiǎn)整理,故選:D.12.【解答】解:對(duì)于A:fm+n=am對(duì)于B:fmn=amn,而fmf對(duì)于C:fm+n2由m+n2≥mn,而當(dāng)0<a<1時(shí),fm+n2故選:AD.13.【解答】解法一:由題意可知:fx的定義域?yàn)?令x+a=0解得x=-a;令若-a≤-b,當(dāng)x∈-b,此時(shí)fx<0若-b<-a<1-b,當(dāng)x此時(shí)fx<0若-a=1-b,當(dāng)x∈-b,1-b當(dāng)x∈[1-b,+∞)時(shí),可知x+可知若-a=1-若-a>1-b,當(dāng)x∈1此時(shí)fx<0綜上所述:-a=1-b則a2+b2=a2+a+所以a2+b2解法二:由題意可知:fx的定義域?yàn)?令x+a=0解得x=-a;令則當(dāng)x∈-b,1-b時(shí),lnx+b<x∈1-b,+∞時(shí),lnx+b>0故1-b+a=0當(dāng)且僅當(dāng)a=-12,b所以a2+b2故選:C.14.【解答】解:因?yàn)閒a=f?b,所以lga=lgb,所以a=b(舍去),或b=1a,所以a+2b=a+2a又0<a<b所以fa>f1=1+21=3,即15.【解答】解:作出函數(shù)fx的圖象,如圖所示設(shè)fx由圖可知,當(dāng)0<t≤1時(shí),直線y=t交點(diǎn)的橫坐標(biāo)分別為x1,x2,x當(dāng)x>1時(shí),令fx=log2x-1=由圖可知,x1由fx3=fx4,可得-則有x3所以4x令gx易知gx在(2,3]上為減函數(shù)故4≤且4∈故選:BC.16.【解答】解:設(shè)A,B,C,則-log∴x∴b又m>0,∴m+82m+1∴ba≥27217.【解答】解:設(shè)Px,y為y則P關(guān)于y=x的對(duì)稱點(diǎn)P'y,x又∵函數(shù)y=fx的圖象與函數(shù)y=gx∴P'y,x關(guān)于直線x+y=0的對(duì)稱點(diǎn)∴必有-y=g-x∴y=fx故選:D.18.【解答】解:∵y∴3x+1=e∴x∴所求反函數(shù)為y=故選:D.19.【解答】解:解法1:由題意2x2②所以2xx1=log2令2x1=7-∴5-2t=2于是2即x1解法2:將2分別變?yōu)?x令f1因?yàn)?x與log2x關(guān)于y所以f1x,f2x從而它們與f3x的交點(diǎn)也關(guān)于y=易求出f3x與y=x-所以x1解法3:x1滿足2x+2x2滿足2x+2log2則3<x1+x2<4,故選:C.【精選練習(xí)】1.【解答】解:把R0=3.28,T=6代入當(dāng)t=0時(shí),I0=1兩邊取對(duì)數(shù)得0.38t=ln2,解得故選:B.2.【解答】解:(1)fx令log2x=t,因此當(dāng)t=12時(shí),y=t當(dāng)t=2時(shí),y=即當(dāng)x=2時(shí),函數(shù)fx取得最小值-94;當(dāng)x=4時(shí)（2）fx<m即log22x令log2x=t,則令gt則g0<0g2<解得m>所以實(shí)數(shù)m的取值范圍為0,+∞3.【解答】解:由x2-4x-5>0,令t=x2-4x-∴要使函數(shù)fx=lgx2-4x則需內(nèi)層函數(shù)t=x2-4x-5在a,+∞上單調(diào)遞增且恒大于0,∴a的取值范圍是[5,+∞).故選:4.【解答】解:因?yàn)閘oga23>1當(dāng)a>1時(shí),1<a<2當(dāng)0<a<1時(shí),對(duì)數(shù)函數(shù)是減函數(shù),故選:C.5.【解答】解:∵函數(shù)fx=log令t=2x+b-t=2x∴a∵當(dāng)x=0時(shí),∴0又∵f∴b∴0故選:A.6.【解答】解:畫出函數(shù)fx的圖象,如圖所示如圖,0<t<2時(shí),方程存在4個(gè)不同根,故A正確,當(dāng)t=2∴l(xiāng)og2x=t時(shí),log2x1=log由正弦函數(shù)對(duì)稱性知x3+x4=12,故B∵fx3=-x3-62+36在4,5上單調(diào)遞增,∵fx1=x1+1x1+12在14,1故選:AC.考向3比較大小問題【典例分析】1.【解答】解:因?yàn)閥=2x是定義域R上的單調(diào)增函數(shù),所以20.7>2因?yàn)閥=13x是定義域R上的單調(diào)減函數(shù),所以130.7<130=因?yàn)閥=log2x是定義域0,+∞上的單調(diào)增函數(shù),所以log2所以a>故選:C.2.【解答】解:∵log∵log∵0∴a故選:D.3.【解答】解:∵abc∴a故選:A.4.【解答】解:a=而log32>log525.【解答】解:∵y=x25在x>0時(shí)是增函數(shù)∴a>c又∵y=25x6.【解答】解:∵abc綜上可得:b<故選:A.7.【解答】解:a=而log23>log2ec=5-12=所以c<a,綜上故選:C.8..【解答】解:∵a>b>0,0<c<1,∴l(xiāng)ogca<logcb,故B正確;∴當(dāng)a>b>1時(shí),0>logac>logbc,故9.【解答】解:由已知可得1a作出y=2x,y=log2x,y=log3x故選:B.10.【解答】解:分別作出四個(gè)函數(shù)y=y=2x,y由圖象知:∴a故選:A.11.【解答】解:法一、∵=log0.32×0.2=∴ab<0,可得a+b<0,結(jié)合0<a+b法二、∵a∴a∵lg103>lg52,12.【解答】解法一:由34∵log5534>log53,∵5∵134<85,∴4解法二:∵a∴<∵5∴a<b<c.13.【解答】解:對(duì)于A,由x>0lnx≠0,可得函數(shù)f'x=lnx-1lnx2所以fx在x=e處的切線方程為y-e=0,即y=對(duì)于B,由f'x=0,當(dāng)x∈0,l,x∈1,e時(shí),f'x<0,fx為減函數(shù),對(duì)于C,fx在1,+∞上的值域包含gx在由x∈1,e時(shí),fx為減函數(shù),當(dāng)x∈e,+∞時(shí),fx為增函數(shù),所以域?yàn)閇e,+∞),由gx在R上的值域?yàn)閇a,+∞),所以a≥e對(duì)于D,由π>3>e,所以fπ>f(3),所以πl(wèi)nπ>3故ππ>3π>π3>33,故14.【解答】解:構(gòu)造函數(shù)fx導(dǎo)數(shù)f'當(dāng)0<x<e時(shí),x>e時(shí),f'可得f(e)取得最大值1e由2<5<e,即有l(wèi)n22<ln55,即由e<π<e,即有l(wèi)nee<lnππ,即ln設(shè)gx=2x-x在2<x<4時(shí)即有211<11,故由fx=lnxx的最大值1e,看設(shè)x即eln22<22則3eln2<4故選:B.15.【解答】解法一:x、y、z為正數(shù),令2x∴3∴l(xiāng)g3解法二:x、y、z為正數(shù),令2x∴2x3y=23×lg3lg2=lg9解法三:令2x=3y=5z=k>1,則2fx在e,+∞上單調(diào)遞減,3lnkln3<2lnk16.【解答】解:∵9a=10m-11=∴f∴fx=xm-x-1在1,+∞單調(diào)遞增,∴f10>f(8),17.【解答】解:設(shè)fx=xlnx,則f'x=1+lnx,當(dāng)x∈1e,+∞時(shí),則f'x>0,fx單調(diào)遞增,當(dāng)x∈0,∴f12>f1e?118.【解答】解:∵a令fx=2ln1+x-1+∴x∴g't=4tt∴gt>g(1)=2ln4-1再令1+4x=t∴φ't=2tt2+1-1∴hx<0,∴c19.【解答】解:構(gòu)造函數(shù)fx則f'當(dāng)f'x=0時(shí)0<x<1時(shí),x>1時(shí),f'∴fx在x=1∴l(xiāng)nx∴l(xiāng)n0.9∵-ln∴0.1設(shè)gx則g'令hx當(dāng)0<x<2-1時(shí),h'x當(dāng)2-1<x<1時(shí),h'x∵h(yuǎn)0=0,∴當(dāng)0<當(dāng)0<x<2-1時(shí)∴g∴c故選:C.20.【解答】解:設(shè)fx=cosx+12設(shè)gx故gx在0,1單調(diào)遞增,即即f'x>0,故fx在所以f14>f0=0,可得利用三角函數(shù)線可得x∈0,π2∴tan14>14,即sin1綜上:c>故選:A.【精選練習(xí)】1.【解答】解:令a=則logblogloglog故最大的是logb故選:A.2.【解答】解:∵1∴0∴a故選:C.3.【解答】解:∵log∴a<c<b.4.【解答】解:如圖,作函數(shù)y=12x與設(shè)12①當(dāng)M>1時(shí),結(jié)合圖象可得,a<b<0,②當(dāng)M=1時(shí),結(jié)合圖象可得,③當(dāng)0<M<1時(shí),結(jié)合圖象可得,a>b>0故選:CD.5.【解答】解:∵=2∴a∵a∴a∵b∴b故c<故選:B.6.【解答】解:畫出函數(shù)y=1ex,由圖象可知:x2故選:D.7.【解答】解:方法一:由2x-2y<3令fx=2x-3-x,則fx在R上單調(diào)遞增,且fx<fy,所以故lny方法二:取x=-1,y=0,滿足2x-2y<3-x-32.3函數(shù)四大性質(zhì)考向1函數(shù)單調(diào)性【典例分析】1.【解答】解:對(duì)于A,y=lnx2+1的定義域?yàn)镽,在-∞,0上,從而得出y=lnx2+1在從而在定義域R上該函數(shù)不是增函數(shù),即該選項(xiàng)錯(cuò)誤,故A錯(cuò)誤;對(duì)于B,y=lgx0.5該函數(shù)的定義域?yàn)閇0,+∞),在[0,+∞)上t=x0.5+∴該函數(shù)在定義域上是增函數(shù),故B正確;對(duì)于C,y'=ex-e-x,∴y=ex+e-x-2在定義域?qū)τ贒,由AC錯(cuò)誤,得D錯(cuò)誤.故選:B.2.【解答】解:對(duì)于A,∵在0,+∞內(nèi),函數(shù)y=4x2+1為增函數(shù),∴y=4x2+1+2∴y=14x2+1+2x在對(duì)于B,∵函數(shù)y=5x+2在∴函數(shù)fx=2-5x+2在0,+∞對(duì)于C,由對(duì)勾函數(shù)的性質(zhì)可知,函數(shù)y=x+3x在0,3上單調(diào)遞減∴函數(shù)fx=2x+3x在0,3上單調(diào)遞減,在3對(duì)于D,函數(shù)fx=1x+1∵函數(shù)y=1x+1在-1,+∞上單調(diào)遞減,y∴函數(shù)fx=1x+1-lnx+1在故選:B.3.【解答】解:∵函數(shù)fx=ax-1x+1∴-a+1x+1,在-∞,-1上是減函數(shù),∴a+1x+1,故答案為:-∞,-4.【解答】解:設(shè)t=xx-a=x2-∵y=2t是∴要使fx在區(qū)間0,1則t=x2-ax在區(qū)間即a2≥1,即故實(shí)數(shù)a的取值范圍是[2故選:D.5.【解答】解:根據(jù)題意,令t=x2-ax-1,則因?yàn)閥=log13t在1,+∞上單調(diào)遞減,所以t=則滿足a2≤112-a×1-1分析選項(xiàng):a≤0的一個(gè)充分不必要條件是故選:A.6.【解答】解:已知fx為增函數(shù)且m當(dāng)m>0,由復(fù)合函數(shù)的單調(diào)性可知fmx和mf此時(shí)不符合題意.當(dāng)m<0時(shí),有因?yàn)閥=2x2在x∈[1,+∞)上的最小值為2,所以1+1m2<2,即m故答案為:m<-7.【解答】因?yàn)閒x在R上單調(diào)遞增,且x≥0時(shí),f則需滿足--2a2×-即a的范圍是-1故選:B.8.【解答】解:∵函數(shù)fx=cosπx3a,當(dāng)x=a時(shí),cosπx3a=cosπ3=129.【解答】解:當(dāng)fx=2-x時(shí),函數(shù)exfx=e2x在R故選:A.10.【解答】解:由fx=1x+2+lg1-x1故函數(shù)fx的定義域?yàn)?1,1,且函數(shù)由于f0=12,則由不等式fxx求得x>12,或再結(jié)合函數(shù)fx的定義域?yàn)?1,1,可得x故答案為:-111.【解答】解:由函數(shù)fx①當(dāng)x>0時(shí),fx=ex-e-x+2,則f所以fx②當(dāng)x≤0時(shí),fx=x2+1為減函數(shù),所以fx≥f又因?yàn)?0.01>30=1,所以a因?yàn)?2log32=log所以f0<f?b<因?yàn)閘og30.5>log313=-1,所以-1<所以fa故選:A.12.【解答】解:由題意可得f1對(duì)任意x1<x2,則fx1-fx令gx=fx+x,則可得gx在R由flog22x-1故2x解可得,0<故選:D.13.【解答】解:依題意,?x1,x2∈0,+∞,當(dāng)A選項(xiàng),取x1=1,x2則x2fx下面利用構(gòu)造函數(shù)法來進(jìn)行證明:依題意,?x1,x2∈0,+∞,當(dāng)構(gòu)造函數(shù)gx=fxxx>0,任取此時(shí)x2所以gx所以gx同理可證得:任取0<x2<x1綜上所述,gx=fxxx>對(duì)于B選項(xiàng),gx因?yàn)閥=x2+1和y=x所以gx在0,+∞上遞增,符合題意,B對(duì)于C選項(xiàng),gx=fxx=x2,在0,+∞對(duì)于D選項(xiàng),gx=fxx=x-1x3所以gx在0,+∞上遞增,符合題意,D故選:BCD.14.【解答】解:若x≤0,則則fx+fx-12>1等價(jià)為x+1+x-12當(dāng)x>0時(shí),fx=2x>1,x-12>-12,當(dāng)0≥x-12>-12,即12≥x>0綜上x>-14,故答案為:15.【解答】解:對(duì)于A,在fx+f令x=y=0得解得f0=1,故A對(duì)于B,令x<0,則x+y<y即fx即fx為增函數(shù),故B錯(cuò)誤對(duì)于C,令x=-y得,fx+f-x=對(duì)于D,由基本不等式得2x+2-x≥22由B選項(xiàng)分析可知fx為增函數(shù)所以f2所以f2即f2x+f2-x≥故選:ACD.16.【解答】解:方法一:由2x-2y<3令fx=2x-3-x,則fx在所以x<y,即y-x>0故lny方法二:取x=-1,y=0此時(shí)lny-x+1=ln故選:A.17.【解答】B18.【解答】解:令fx則f'當(dāng)x趨近于0時(shí),xex-1<0,當(dāng)x因此在0,1上必然存在因此函數(shù)fx在0,1上先遞減后遞增,故A令gxg當(dāng)0<x<1時(shí)∴gx在0,∵0∴即x2∴選項(xiàng)C正確而D不正確.故選:C.19.【解答】解:因?yàn)閤1,x2在不等式中地位等價(jià),不妨設(shè)x1<x2,則因?yàn)閤1,x2>0,在不等式x1lnx令fx=lnxx,則fxf'x=1-ln故選:A.【精選練習(xí)】1.【解答】解:∵fx=-x22+1-mx+3,開口向下,對(duì)稱軸為:x=1-m,又2.【解答】解:∵f∴根據(jù)復(fù)合函數(shù)的單調(diào)性我們易得在區(qū)間(-∞,1]在區(qū)間1,2故選:D.3.【解答】解:若函數(shù)fx=2x2-8則2解得1故答案為:14.【解答】解:∵對(duì)任意的x1,x2∈Rx1∴函數(shù)fx=a-x,x∴解得,14故選:C.5.【解答】解:解法一:∵函數(shù)fx∴不等式即lnx令t=x2-4>令ht=lnt+2t,顯然函數(shù)ht在0∴由不等式①可得t<1,即x2-4<由x2-4>0x2<5故答案為:-5解法二:由于函數(shù)fx再根據(jù)函數(shù)fx=lnx+2x在定義域0,+∞上式增函數(shù),∴由求得-5<x<-2故答案為:-56.【解答】解:對(duì)任意x1,x2∈R,x1≠x2則fx令gx=fx-x,則函數(shù)gxg1則不等不等式f3∴3解得x<∴不等式f3x-2<故答案為:-∞,7.【解答】解:可令t=由fx=log13t在0,+∞遞減,可得t=4x2-4ax+3a可得a2≥1且4-4a+3a≥0,且3a≥0,解得2≤8.【解答】解:由題意當(dāng)x2>x1時(shí),有fx1-fx2x1x2>ex2x1-ex1x則gx在0,+∞上單調(diào)遞減,由于f1=e,而flna>2e-alna,即有flna+alna>f1+1×e1,即考向2函數(shù)奇偶性【典例分析】1.【解答】解:∵fx=xexeax-1的定義域?yàn)椤鄁∴-∴x∴ax-x故選:D.2.【解答】解:由2x-12x+1>0由fx是偶函數(shù)∴f得-x即-x即-x則x-∴x-a=x+得a=故選:B.3.【解答】解:根據(jù)題意,設(shè)fx若fx為偶函數(shù),則f變形可得a-2x=0在R上恒成立故答案為:2.4.【解答】解:∵函數(shù)fx=log∴fx+f∴l(xiāng)ogax+x2∴a=±22又a對(duì)數(shù)式的底數(shù),a5.【解答】解:fx若a=0,則函數(shù)fx的定義域?yàn)閧x∣x≠1}由函數(shù)解析式有意義可得,x≠1且∴x≠1且x≠1+1a,∵函數(shù)∴1+1a=-1,解得a=-12,∴由f0=0得,ln12+b6.【解答】解:因?yàn)閒x所以fln2=ln因?yàn)閒x所以f-x=lnex+e-x=fx當(dāng)x>0時(shí),y'=ex-e-x>0,即y=ex+e-x由C的判斷及偶函數(shù)關(guān)于y軸對(duì)稱可知,當(dāng)x=0時(shí),函數(shù)取得最小值f0故選:ACD.7.【解答】解:因?yàn)楹瘮?shù)fx,gx的定義域都為R,且fx是奇函數(shù),所以f-令hx=fxgx,則h-x=f-xg令Hx=fxgx,則H-x=f-xg若gx-fx=x3+若函數(shù)fx在-∞,+∞上單調(diào)遞減且f1=-1不等式-1≤fx-所以-1≤x-2≤1故選:ACD.8.【解答】解:函數(shù)g滿足g-x=ln1+x2+函數(shù)fx=ln1+x2-x+1,則f-a=-ln19.【解答】解:∵lglog∴l(xiāng)glog210與lglg2互為相反數(shù)則設(shè)lg令fx=gx+4,即gx=∴fm=gm+10.【解答】解:函數(shù)可化為fx令gx=2x+sinxx2+1,則∴函數(shù)fx=x+1即M+m=211.【解答】解:∵函數(shù)y=fx∴必有b=0,y=f∴3再根據(jù)sinx+φ=2y-2a9+再根據(jù)最大值與最小值之和為6,可得2a=6,即a=3故選:C.12.【解答】解:∵定義在R的奇函數(shù)fx在-∞,0單調(diào)遞減,且f∴fx在0,+∞上單調(diào)遞減,且故f-當(dāng)x=0時(shí),不等式xfx當(dāng)x=1時(shí),不等式xfx當(dāng)x-1=2或x-1=-2時(shí),即x=3或當(dāng)x>0時(shí),不等式xfx-1此時(shí)x>00<x-當(dāng)x<0時(shí),不等式xfx-1即x<0-2≤x綜上-1≤x≤0即實(shí)數(shù)x的取值范圍是-1故選:D.13.【解答】解:由2x+1≠02又f-∴fx由fx∵2可得內(nèi)層函數(shù)t=2x在-∞,-12上單調(diào)遞減,在-則12,+∞又對(duì)數(shù)式y(tǒng)=lnt由復(fù)合函數(shù)的單調(diào)性可得,fx在-∞,-1故選:D.14.【解答】解:∵函數(shù)fx=ln1且在x≥0時(shí),導(dǎo)數(shù)為f'即有函數(shù)fx在[0,+∞)∴fx>f2即x>平方得3x解得:13所求x的取值范圍是13故選:B.15.【解答】解:fx的定義域?yàn)镽∵f∴fx∵函數(shù)y=x3,y=lgx∴fx在[0,+∞)上為增函數(shù),所以fx在由ftsin2θ+f∴f∴tsin2θ>-4當(dāng)θ∈0,π2令gx當(dāng)t=0時(shí),gx=-當(dāng)t≠0時(shí),對(duì)稱軸為當(dāng)0<12t<1,即t>12時(shí),則有g(shù)12t=4t-當(dāng)12t<0,即t<0時(shí),有g(shù)1=t-1+4t>0當(dāng)12t≥1,即0<t≤12時(shí),有g(shù)1=綜上,t∈故選:D.16.【解答】解:根據(jù)題意,由函數(shù)fx為奇函數(shù),有f0=-1-m可得fx經(jīng)檢驗(yàn)f-x=-1故m=依次分析選項(xiàng):對(duì)于A,由于m=1,則A對(duì)于B,由f'可得函數(shù)fx單調(diào)遞增,故B選項(xiàng)正確對(duì)于C,由函數(shù)fx單調(diào)遞增且為奇函數(shù),故只有f若ffx=0,必有fx=0,可得x=對(duì)于D,由fx+1+fa有x+1≤2-a當(dāng)a<01+4a≤由上知當(dāng)a≤-14時(shí),不等式恒成立,故D故選:BCD.17.【解答】解:對(duì)于A,aln1設(shè)px則p'x=2a1從而0<2a<12,0對(duì)于B,fx=0?aln則它的定義域?yàn)?1,1且h-所以hx是奇函數(shù)由題意hx=0有三個(gè)根x1,x2,x3,則對(duì)于C,由fx+kf-x=所以aln所以aln1即aln1+x1-x+當(dāng)k>0時(shí),令1-kex=0,則x=lnk,只需保證lnk對(duì)于D,由B可知,x1=-x3,又f'e所以f=aln=ex3f'-x3故選:BCD.18.【解答】解:由fxy取x=y=0,可得f0=0取x=y=1,可得f1=2f1,即取x=y=-1,得f1=取y=-1,得f-x=fx,可得fx由上可知,f-1=f不妨取fx=0,滿足常數(shù)函數(shù)fx=0無極值,故D故選:ABC.19.【解答】解:A.令fx=sin2πx+x但fx在R上不是增函數(shù),故A錯(cuò)誤B.令x1=x,x2∴fx+f-x∴函數(shù)fx是奇函數(shù),故B正確C.∵?x1,x2∴f即fx=fx+2π,∴fD.任取-1<x1<x2<1,∵∴sin∴sinx∴fx2-sinx∴函數(shù)fx+sinx在區(qū)間-函數(shù)fx-sinx在區(qū)間-1,1上單調(diào)遞減故選:BCD.20.【解答】解:∵對(duì)任意x1,f∴令x1=x2=∴令x1=x,x2∴f∴fx+故選:C.21.【解答】解:因?yàn)楹瘮?shù)fx的定義域?yàn)镽,f1=1,若對(duì)任意x所以ex令x=y=0可得f0=2f令x=1,?y=-所以f-1=-令x+y=0,所以exfx為奇函數(shù),C令x=y=12所以f1令x=y=1,則f2所以fx在0,+∞上不具有單調(diào)性,D故選:ABC.22.【解答】解:令x=y=1,所以f1令x=y=-1,所以f-令y=-1,則有所以f-所以y=fx為奇函數(shù),故A因?yàn)閒x是R上的減函數(shù),且f所以fx不是R上的減函數(shù),故B錯(cuò)誤當(dāng)x≠0時(shí),令則f1所以xf1x+1xfx=令x=則有f2k+所以f2所以f2所以f2所以k=110f2k=2+22+23【精選練習(xí)】1.【解答】解:∵fx=xlnx+∴-ln-∴l(xiāng)na+x22.【解答】解:若fx=lne3x+1即lne3x+1+即2a=-3,解得a=-323.【解答】解:因?yàn)閒x=x所以f-x=fx,即-x3g-x=x3gx對(duì)于A,定義域?yàn)?1,1對(duì)于B,定義域?yàn)镽,g-x對(duì)于C,定義域?yàn)镽,g-x對(duì)于D,定義域?yàn)镽,g-x=lnx2+1故選:BD.4.【解答】解:fx=sinx是奇函數(shù),但其在區(qū)間-1,1上單調(diào)遞增,∵fx=-x+1,∴f-x=--∵a>1時(shí),y=ax在-1,1上單調(diào)遞增,y=a-x-1,1上單調(diào)遞減,∴fx5.【解答】解:∵fx是奇函數(shù),gx∴f-x=-fx,g-xf-x?g-x=fxf-x?g-x?-fxf-x?g-x=fx?gx為偶函數(shù)6.【解答】解:根據(jù)題意,函數(shù)fx若fx為奇函數(shù),則f-x=-fx,即e-x+ae導(dǎo)數(shù)f若fx是R上的增函數(shù),則fx的導(dǎo)數(shù)f'x=ex變形可得:a≤e2x恒成立,分析可得a≤0,即故答案為:-17.【解答】解:∵fx是定義在R上的奇函數(shù),gx是定義在由fx得f-①②聯(lián)立解得f由已知g∴∴故選:B.8.【解答】解:∵偶函數(shù)fx在[0,+∞)單調(diào)遞減,∴不等式fx-1>0等價(jià)為fx-1解得-1<x<39.【解答】解:奇函數(shù)fx在R上是增函數(shù),當(dāng)x>0,fx∴gx=xfx∴gx在0,+∞單調(diào)遞增,且g∴a則2<由gx在0,+∞單調(diào)遞增,則∴b故選:C.10.【解答】解:由題意可知:fx的定義域?yàn)閧x∣x<-1或由fx=lgx-1+故fx為偶函數(shù)當(dāng)x>1時(shí),由于函數(shù)t=2023x,?y=lgx-1均為1,+∞因此fx為1,+∞所以不等式f3x<fx解得13即原不等式的解集為13故選:C.11.【解答】解:函數(shù)fx=xf可得fx在R上遞增又f-可得fx為奇函數(shù)則fa即有f由f-f即有2a解得-1故答案為:-112.【解答】解;因?yàn)閒x所以fx是奇函數(shù),又fx=ex-1故不等式fex+fax<0有解等價(jià)于fex<-fax令gx=ex+ax,則g'x=ex+aa>0時(shí),g'x>0,gx是增函數(shù),當(dāng)x當(dāng)a<0時(shí),當(dāng)x>ln-a時(shí),g'x>0,gx∴gxmin=gln-a=-a當(dāng)-x>1,x<-1時(shí),p'x>0,px并且當(dāng)-1<x<0∴∴當(dāng)x<-e時(shí)px<0,即當(dāng)a<-e所以a的取值范圍是-∞,-e∪0,+∞.13.【解答】解:設(shè)gx=fx-2=x5+x,易得gx由fx2-3+即gx因?yàn)間x為奇函數(shù)所以gx則x2-3<-2x故選:B.考向3函數(shù)周期性與對(duì)稱性【典例分析】1.【解答】解:由fx+4=f∴fx為周期為6的周期函數(shù),f919=f153×6+1=f1,由fx是定義在R上的偶函數(shù),則f1=2.【解答】解:因?yàn)閒x所以f6=-f4=f2=-f0,又fx是定義在R上的奇函數(shù),所以f3.【解答】解:∵fx?f∴f∴f99=f2×4.【解答】解:已知函數(shù)fx滿足f令x=1,y=0,則有1=1+f01-令x=y=1,則f2=1+1令y=-x,則f0=fx+f-x1-ff所以fx+4=-1fx+故選:ABD.5.【解答】解:根據(jù)題意,函數(shù)fx滿足f則有fx+1=fx+2+fx則有fx+6=-fx+3=fx,fx是周期為6的周期函數(shù),則有f2024在fx=fx+1+fx-1中,令x=1可得故答案為:2.6.【解答】解:令y=1,則fx+1+∴f∴fx+3=-fx令x=1,y=0得f又fx∴fff6∴k故選:A.7.【解答】解:(1)若y=fx+a是偶函數(shù),則fx（2）若y=fx+a是奇函數(shù),則fx（3）fa+x=fb-x?函數(shù)(4)fa+x=fb-x=2c8.【解答】解:(1)函數(shù)y=2x與y=2-x（2）函數(shù)y=2x-1與y=2（3）函數(shù)y=fx-1與y=f（4）函數(shù)y=fx-a與y=(5)函數(shù)y=fa+x與y=9.【解答】D10.【解答】解:因?yàn)楹瘮?shù)y=fx-2所以y=fx的圖象的圖象關(guān)于-因?yàn)閥=fx和y=gx所以y=gx的圖象關(guān)于2故選:D.11.【解答】fx=lnx2-設(shè)Pm,n為yPm,n關(guān)于1,a因?yàn)镻m,n在y=fx而f2=-所以Q2-m,2a-由P的任意性可得y=fx圖象為中心對(duì)稱圖形,且對(duì)稱中心為12.【解答】解:因?yàn)閒x=3ex1所以fx因?yàn)閒x因?yàn)閥=ex+1在0,+∞上單調(diào)遞增,所以y=3ex+1在0單調(diào)遞增,又因?yàn)椴坏仁絝4-ax+fx2≥3在0,+∞上恒成立,所以4-ax≥-x2在0,+∞上恒成立,即a≤由雙勾函數(shù)的性質(zhì)可知y=所以a≤4,即a的取值范圍為故答案為:3;(-∞,13.【解答】解:函數(shù)fxx∈R即為fx+f-x=2,可得fx關(guān)于點(diǎn)0,1對(duì)稱,函數(shù)y=x+1x,即y=1+1x的圖象關(guān)于點(diǎn)0,1對(duì)稱,即有…則有i==m故選:B.14.【解答】解:由fx+y=fx+fy-4取y=-x,得f0=fx+f-x-4,即故hx為奇函數(shù),∵gx=2x則gx=φx+hx+4,∵φ-x=-故函數(shù)gx在-2023,故選:B.15.【解答】解:因?yàn)閒x所以函數(shù)fx有唯一零點(diǎn)等價(jià)于方程1-x等價(jià)于函數(shù)y=1-x-12①當(dāng)a=0時(shí),fx=x2②當(dāng)a<0時(shí),由于y=1-x-12在且y=aex-1+1ex-所以函數(shù)y=1-x-12的圖象的最高點(diǎn)為由于2a<0<1,此時(shí)函數(shù)y=1-x③當(dāng)a>0時(shí),由于y=1-x-