2018高考數(shù)學(xué)真題導(dǎo)數(shù)專題

..2017年高考真題導(dǎo)數(shù)專題一．解答題〔共12小題1．已知函數(shù)f〔x=ae2x+〔a﹣2ex﹣x．〔1討論f〔x的單調(diào)性；〔2若f〔x有兩個(gè)零點(diǎn),求a的取值范圍．2．已知函數(shù)f〔x=ax2﹣ax﹣xlnx,且f〔x≥0．〔1求a；〔2證明：f〔x存在唯一的極大值點(diǎn)x0,且e﹣2＜f〔x0＜2﹣2．3．已知函數(shù)f〔x=x﹣1﹣alnx．〔1若f〔x≥0,求a的值；〔2設(shè)m為整數(shù),且對(duì)于任意正整數(shù)n,〔1+〔1+…〔1+＜m,求m的最小值．4．已知函數(shù)f〔x=x3+ax2+bx+1〔a＞0,b∈R有極值,且導(dǎo)函數(shù)f′〔x的極值點(diǎn)是f〔x的零點(diǎn)．〔極值點(diǎn)是指函數(shù)取極值時(shí)對(duì)應(yīng)的自變量的值〔1求b關(guān)于a的函數(shù)關(guān)系式,并寫出定義域；〔2證明：b2＞3a；〔3若f〔x,f′〔x這兩個(gè)函數(shù)的所有極值之和不小于﹣,求a的取值范圍．5．設(shè)函數(shù)f〔x=〔1﹣x2ex．〔1討論f〔x的單調(diào)性；〔2當(dāng)x≥0時(shí),f〔x≤ax+1,求a的取值范圍．6．已知函數(shù)f〔x=〔x﹣e﹣x〔x≥．〔1求f〔x的導(dǎo)函數(shù)；〔2求f〔x在區(qū)間[,+∞上的取值范圍．7．已知函數(shù)f〔x=x2+2cosx,g〔x=ex〔cosx﹣sinx+2x﹣2,其中e≈2.17828…是自然對(duì)數(shù)的底數(shù)．〔Ⅰ求曲線y=f〔x在點(diǎn)〔π,f〔π處的切線方程；〔Ⅱ令h〔x=g〔x﹣af〔x〔a∈R,討論h〔x的單調(diào)性并判斷有無極值,有極值時(shí)求出極值．8．已知函數(shù)f〔x=excosx﹣x．〔1求曲線y=f〔x在點(diǎn)〔0,f〔0處的切線方程；〔2求函數(shù)f〔x在區(qū)間[0,]上的最大值和最小值．9．設(shè)a∈Z,已知定義在R上的函數(shù)f〔x=2x4+3x3﹣3x2﹣6x+a在區(qū)間〔1,2內(nèi)有一個(gè)零點(diǎn)x0,g〔x為f〔x的導(dǎo)函數(shù)．〔Ⅰ求g〔x的單調(diào)區(qū)間；〔Ⅱ設(shè)m∈[1,x0∪〔x0,2],函數(shù)h〔x=g〔x〔m﹣x0﹣f〔m,求證：h〔mh〔x0＜0；〔Ⅲ求證：存在大于0的常數(shù)A,使得對(duì)于任意的正整數(shù)p,q,且∈[1,x0∪〔x0,2],滿足|﹣x0|≥．10．已知函數(shù)f〔x=x3﹣ax2,a∈R,〔1當(dāng)a=2時(shí),求曲線y=f〔x在點(diǎn)〔3,f〔3處的切線方程；〔2設(shè)函數(shù)g〔x=f〔x+〔x﹣acosx﹣sinx,討論g〔x的單調(diào)性并判斷有無極值,有極值時(shí)求出極值．11．設(shè)a,b∈R,|a|≤1．已知函數(shù)f〔x=x3﹣6x2﹣3a〔a﹣4x+b,g〔x=exf〔x．〔Ⅰ求f〔x的單調(diào)區(qū)間；〔Ⅱ已知函數(shù)y=g〔x和y=ex的圖象在公共點(diǎn)〔x0,y0處有相同的切線,〔i求證：f〔x在x=x0處的導(dǎo)數(shù)等于0；〔ii若關(guān)于x的不等式g〔x≤ex在區(qū)間[x0﹣1,x0+1]上恒成立,求b的取值范圍．12．已知函數(shù)f〔x=ex〔ex﹣a﹣a2x．〔1討論f〔x的單調(diào)性；〔2若f〔x≥0,求a的取值范圍．2017年高考真題導(dǎo)數(shù)專題參考答案與試題解析一．解答題〔共12小題1．〔2017?新課標(biāo)Ⅰ已知函數(shù)f〔x=ae2x+〔a﹣2ex﹣x．〔1討論f〔x的單調(diào)性；〔2若f〔x有兩個(gè)零點(diǎn),求a的取值范圍．[解答]解：〔1由f〔x=ae2x+〔a﹣2ex﹣x,求導(dǎo)f′〔x=2ae2x+〔a﹣2ex﹣1,當(dāng)a=0時(shí),f′〔x=﹣2ex﹣1＜0,∴當(dāng)x∈R,f〔x單調(diào)遞減,當(dāng)a＞0時(shí),f′〔x=〔2ex+1〔aex﹣1=2a〔ex+〔ex﹣,令f′〔x=0,解得：x=ln,當(dāng)f′〔x＞0,解得：x＞ln,當(dāng)f′〔x＜0,解得：x＜ln,∴x∈〔﹣∞,ln時(shí),f〔x單調(diào)遞減,x∈〔ln,+∞單調(diào)遞增；當(dāng)a＜0時(shí),f′〔x=2a〔ex+〔ex﹣＜0,恒成立,∴當(dāng)x∈R,f〔x單調(diào)遞減,綜上可知：當(dāng)a≤0時(shí),f〔x在R單調(diào)減函數(shù),當(dāng)a＞0時(shí),f〔x在〔﹣∞,ln是減函數(shù),在〔ln,+∞是增函數(shù)；〔2①若a≤0時(shí),由〔1可知：f〔x最多有一個(gè)零點(diǎn),當(dāng)a＞0時(shí),f〔x=ae2x+〔a﹣2ex﹣x,當(dāng)x→﹣∞時(shí),e2x→0,ex→0,∴當(dāng)x→﹣∞時(shí),f〔x→+∞,當(dāng)x→∞,e2x→+∞,且遠(yuǎn)遠(yuǎn)大于ex和x,∴當(dāng)x→∞,f〔x→+∞,∴函數(shù)有兩個(gè)零點(diǎn),f〔x的最小值小于0即可,由f〔x在〔﹣∞,ln是減函數(shù),在〔ln,+∞是增函數(shù),∴f〔xmin=f〔ln=a×〔+〔a﹣2×﹣ln＜0,∴1﹣﹣ln＜0,即ln+﹣1＞0,設(shè)t=,則g〔t=lnt+t﹣1,〔t＞0,求導(dǎo)g′〔t=+1,由g〔1=0,∴t=＞1,解得：0＜a＜1,∴a的取值范圍〔0,1．方法二：〔1由f〔x=ae2x+〔a﹣2ex﹣x,求導(dǎo)f′〔x=2ae2x+〔a﹣2ex﹣1,當(dāng)a=0時(shí),f′〔x=2ex﹣1＜0,∴當(dāng)x∈R,f〔x單調(diào)遞減,當(dāng)a＞0時(shí),f′〔x=〔2ex+1〔aex﹣1=2a〔ex+〔ex﹣,令f′〔x=0,解得：x=﹣lna,當(dāng)f′〔x＞0,解得：x＞﹣lna,當(dāng)f′〔x＜0,解得：x＜﹣lna,∴x∈〔﹣∞,﹣lna時(shí),f〔x單調(diào)遞減,x∈〔﹣lna,+∞單調(diào)遞增；當(dāng)a＜0時(shí),f′〔x=2a〔ex+〔ex﹣＜0,恒成立,∴當(dāng)x∈R,f〔x單調(diào)遞減,綜上可知：當(dāng)a≤0時(shí),f〔x在R單調(diào)減函數(shù),當(dāng)a＞0時(shí),f〔x在〔﹣∞,﹣lna是減函數(shù),在〔﹣lna,+∞是增函數(shù)；〔2①若a≤0時(shí),由〔1可知：f〔x最多有一個(gè)零點(diǎn),②當(dāng)a＞0時(shí),由〔1可知：當(dāng)x=﹣lna時(shí),f〔x取得最小值,f〔xmin=f〔﹣lna=1﹣﹣ln,當(dāng)a=1,時(shí),f〔﹣lna=0,故f〔x只有一個(gè)零點(diǎn),當(dāng)a∈〔1,+∞時(shí),由1﹣﹣ln＞0,即f〔﹣lna＞0,故f〔x沒有零點(diǎn),當(dāng)a∈〔0,1時(shí),1﹣﹣ln＜0,f〔﹣lna＜0,由f〔﹣2=ae﹣4+〔a﹣2e﹣2+2＞﹣2e﹣2+2＞0,故f〔x在〔﹣∞,﹣lna有一個(gè)零點(diǎn),假設(shè)存在正整數(shù)n0,滿足n0＞ln〔﹣1,則f〔n0=〔a+a﹣2﹣n0＞﹣n0＞﹣n0＞0,由ln〔﹣1＞﹣lna,因此在〔﹣lna,+∞有一個(gè)零點(diǎn)．∴a的取值范圍〔0,1．2．〔2017?新課標(biāo)Ⅱ已知函數(shù)f〔x=ax2﹣ax﹣xlnx,且f〔x≥0．〔1求a；〔2證明：f〔x存在唯一的極大值點(diǎn)x0,且e﹣2＜f〔x0＜2﹣2．[解答]〔1解：因?yàn)閒〔x=ax2﹣ax﹣xlnx=x〔ax﹣a﹣lnx〔x＞0,則f〔x≥0等價(jià)于h〔x=ax﹣a﹣lnx≥0,求導(dǎo)可知h′〔x=a﹣．則當(dāng)a≤0時(shí)h′〔x＜0,即y=h〔x在〔0,+∞上單調(diào)遞減,所以當(dāng)x0＞1時(shí),h〔x0＜h〔1=0,矛盾,故a＞0．因?yàn)楫?dāng)0＜x＜時(shí)h′〔x＜0、當(dāng)x＞時(shí)h′〔x＞0,所以h〔xmin=h〔,又因?yàn)閔〔1=a﹣a﹣ln1=0,所以=1,解得a=1；〔2證明：由〔1可知f〔x=x2﹣x﹣xlnx,f′〔x=2x﹣2﹣lnx,令f′〔x=0,可得2x﹣2﹣lnx=0,記t〔x=2x﹣2﹣lnx,則t′〔x=2﹣,令t′〔x=0,解得：x=,所以t〔x在區(qū)間〔0,上單調(diào)遞減,在〔,+∞上單調(diào)遞增,所以t〔xmin=t〔=ln2﹣1＜0,從而t〔x=0有解,即f′〔x=0存在兩根x0,x2,且不妨設(shè)f′〔x在〔0,x0上為正、在〔x0,x2上為負(fù)、在〔x2,+∞上為正,所以f〔x必存在唯一極大值點(diǎn)x0,且2x0﹣2﹣lnx0=0,所以f〔x0=﹣x0﹣x0lnx0=﹣x0+2x0﹣2=x0﹣,由x0＜可知f〔x0＜〔x0﹣max=﹣+=；由f′〔＜0可知x0＜＜,所以f〔x在〔0,x0上單調(diào)遞增,在〔x0,上單調(diào)遞減,所以f〔x0＞f〔=；綜上所述,f〔x存在唯一的極大值點(diǎn)x0,且e﹣2＜f〔x0＜2﹣2．3．〔2017?新課標(biāo)Ⅲ已知函數(shù)f〔x=x﹣1﹣alnx．〔1若f〔x≥0,求a的值；〔2設(shè)m為整數(shù),且對(duì)于任意正整數(shù)n,〔1+〔1+…〔1+＜m,求m的最小值．[解答]解：〔1因?yàn)楹瘮?shù)f〔x=x﹣1﹣alnx,x＞0,所以f′〔x=1﹣=,且f〔1=0．所以當(dāng)a≤0時(shí)f′〔x＞0恒成立,此時(shí)y=f〔x在〔0,+∞上單調(diào)遞增,這與f〔x≥0矛盾；當(dāng)a＞0時(shí)令f′〔x=0,解得x=a,所以y=f〔x在〔0,a上單調(diào)遞減,在〔a,+∞上單調(diào)遞增,即f〔xmin=f〔a,又因?yàn)閒〔xmin=f〔a≥0,所以a=1；〔2由〔1可知當(dāng)a=1時(shí)f〔x=x﹣1﹣lnx≥0,即lnx≤x﹣1,所以ln〔x+1≤x當(dāng)且僅當(dāng)x=0時(shí)取等號(hào),所以ln〔1+＜,k∈N*．一方面,ln〔1++ln〔1++…+ln〔1+＜++…+=1﹣＜1,即〔1+〔1+…〔1+＜e；另一方面,〔1+〔1+…〔1+＞〔1+〔1+〔1+=＞2；從而當(dāng)n≥3時(shí),〔1+〔1+…〔1+∈〔2,e,因?yàn)閙為整數(shù),且對(duì)于任意正整數(shù)n,〔1+〔1+…〔1+＜m成立,所以m的最小值為3．4．〔2017?XX已知函數(shù)f〔x=x3+ax2+bx+1〔a＞0,b∈R有極值,且導(dǎo)函數(shù)f′〔x的極值點(diǎn)是f〔x的零點(diǎn)．〔極值點(diǎn)是指函數(shù)取極值時(shí)對(duì)應(yīng)的自變量的值〔1求b關(guān)于a的函數(shù)關(guān)系式,并寫出定義域；〔2證明：b2＞3a；〔3若f〔x,f′〔x這兩個(gè)函數(shù)的所有極值之和不小于﹣,求a的取值范圍．[解答]〔1解：因?yàn)閒〔x=x3+ax2+bx+1,所以g〔x=f′〔x=3x2+2ax+b,g′〔x=6x+2a,令g′〔x=0,解得x=﹣．由于當(dāng)x＞﹣時(shí)g′〔x＞0,g〔x=f′〔x單調(diào)遞增；當(dāng)x＜﹣時(shí)g′〔x＜0,g〔x=f′〔x單調(diào)遞減；所以f′〔x的極小值點(diǎn)為x=﹣,由于導(dǎo)函數(shù)f′〔x的極值點(diǎn)是原函數(shù)f〔x的零點(diǎn),所以f〔﹣=0,即﹣+﹣+1=0,所以b=+〔a＞0．因?yàn)閒〔x=x3+ax2+bx+1〔a＞0,b∈R有極值,所以f′〔x=3x2+2ax+b=0的實(shí)根,所以4a2﹣12b≥0,即a2﹣+≥0,解得a≥3,所以b=+〔a≥3．〔2證明：由〔1可知h〔a=b2﹣3a=﹣+=〔4a3﹣27〔a3﹣27,由于a＞3,所以h〔a＞0,即b2＞3a；〔3解：由〔1可知f′〔x的極小值為f′〔﹣=b﹣,設(shè)x1,x2是y=f〔x的兩個(gè)極值點(diǎn),則x1+x2=,x1x2=,所以f〔x1+f〔x2=++a〔++b〔x1+x2+2=〔x1+x2[〔x1+x22﹣3x1x2]+a[〔x1+x22﹣2x1x2]+b〔x1+x2+2=﹣+2,又因?yàn)閒〔x,f′〔x這兩個(gè)函數(shù)的所有極值之和不小于﹣,所以b﹣+﹣+2=﹣≥﹣,因?yàn)閍＞3,所以2a3﹣63a﹣54≤0,所以2a〔a2﹣36+9〔a﹣6≤0,所以〔a﹣6〔2a2+12a+9≤0,由于a＞3時(shí)2a2+12a+9＞0,所以a﹣6≤0,解得a≤6,所以a的取值范圍是〔3,6]．5．〔2017?新課標(biāo)Ⅱ設(shè)函數(shù)f〔x=〔1﹣x2ex．〔1討論f〔x的單調(diào)性；〔2當(dāng)x≥0時(shí),f〔x≤ax+1,求a的取值范圍．[解答]解：〔1因?yàn)閒〔x=〔1﹣x2ex,x∈R,所以f′〔x=〔1﹣2x﹣x2ex,令f′〔x=0可知x=﹣1±,當(dāng)x＜﹣1﹣或x＞﹣1+時(shí)f′〔x＜0,當(dāng)﹣1﹣＜x＜﹣1+時(shí)f′〔x＞0,所以f〔x在〔﹣∞,﹣1﹣,〔﹣1+,+∞上單調(diào)遞減,在〔﹣1﹣,﹣1+上單調(diào)遞增；〔2由題可知f〔x=〔1﹣x〔1+xex．下面對(duì)a的范圍進(jìn)行討論：①當(dāng)a≥1時(shí),設(shè)函數(shù)h〔x=〔1﹣xex,則h′〔x=﹣xex＜0〔x＞0,因此h〔x在[0,+∞上單調(diào)遞減,又因?yàn)閔〔0=1,所以h〔x≤1,所以f〔x=〔1﹣xh〔x≤x+1≤ax+1；②當(dāng)0＜a＜1時(shí),設(shè)函數(shù)g〔x=ex﹣x﹣1,則g′〔x=ex﹣1＞0〔x＞0,所以g〔x在[0,+∞上單調(diào)遞增,又g〔0=1﹣0﹣1=0,所以ex≥x+1．因?yàn)楫?dāng)0＜x＜1時(shí)f〔x＞〔1﹣x〔1+x2,所以〔1﹣x〔1+x2﹣ax﹣1=x〔1﹣a﹣x﹣x2,取x0=∈〔0,1,則〔1﹣x0〔1+x02﹣ax0﹣1=0,所以f〔x0＞ax0+1,矛盾；③當(dāng)a≤0時(shí),取x0=∈〔0,1,則f〔x0＞〔1﹣x0〔1+x02=1≥ax0+1,矛盾；綜上所述,a的取值范圍是[1,+∞．6．〔2017?XX已知函數(shù)f〔x=〔x﹣e﹣x〔x≥．〔1求f〔x的導(dǎo)函數(shù)；〔2求f〔x在區(qū)間[,+∞上的取值范圍．[解答]解：〔1函數(shù)f〔x=〔x﹣e﹣x〔x≥,導(dǎo)數(shù)f′〔x=〔1﹣??2e﹣x﹣〔x﹣e﹣x=〔1﹣x+e﹣x=〔1﹣x〔1﹣e﹣x；〔2由f〔x的導(dǎo)數(shù)f′〔x=〔1﹣x〔1﹣e﹣x,可得f′〔x=0時(shí),x=1或,當(dāng)＜x＜1時(shí),f′〔x＜0,f〔x遞減；當(dāng)1＜x＜時(shí),f′〔x＞0,f〔x遞增；當(dāng)x＞時(shí),f′〔x＜0,f〔x遞減,且x≥?x2≥2x﹣1?〔x﹣12≥0,則f〔x≥0．由f〔=e,f〔1=0,f〔=e,即有f〔x的最大值為e,最小值為f〔1=0．則f〔x在區(qū)間[,+∞上的取值范圍是[0,e]．7．〔2017?XX已知函數(shù)f〔x=x2+2cosx,g〔x=ex〔cosx﹣sinx+2x﹣2,其中e≈2.17828…是自然對(duì)數(shù)的底數(shù)．〔Ⅰ求曲線y=f〔x在點(diǎn)〔π,f〔π處的切線方程；〔Ⅱ令h〔x=g〔x﹣af〔x〔a∈R,討論h〔x的單調(diào)性并判斷有無極值,有極值時(shí)求出極值．[解答]解：〔If〔π=π2﹣2．f′〔x=2x﹣2sinx,∴f′〔π=2π．∴曲線y=f〔x在點(diǎn)〔π,f〔π處的切線方程為：y﹣〔π2﹣2=2π〔x﹣π．化為：2πx﹣y﹣π2﹣2=0．〔IIh〔x=g〔x﹣af〔x=ex〔cosx﹣sinx+2x﹣2﹣a〔x2+2cosxh′〔x=ex〔cosx﹣sinx+2x﹣2+ex〔﹣sinx﹣cosx+2﹣a〔2x﹣2sinx=2〔x﹣sinx〔ex﹣a=2〔x﹣sinx〔ex﹣elna．令u〔x=x﹣sinx,則u′〔x=1﹣cosx≥0,∴函數(shù)u〔x在R上單調(diào)遞增．∵u〔0=0,∴x＞0時(shí),u〔x＞0；x＜0時(shí),u〔x＜0．〔1a≤0時(shí),ex﹣a＞0,∴x＞0時(shí),h′〔x＞0,函數(shù)h〔x在〔0,+∞單調(diào)遞增；x＜0時(shí),h′〔x＜0,函數(shù)h〔x在〔﹣∞,0單調(diào)遞減．∴x=0時(shí),函數(shù)h〔x取得極小值,h〔0=﹣1﹣2a．〔2a＞0時(shí),令h′〔x=2〔x﹣sinx〔ex﹣elna=0．解得x1=lna,x2=0．①0＜a＜1時(shí),x∈〔﹣∞,lna時(shí),ex﹣elna＜0,h′〔x＞0,函數(shù)h〔x單調(diào)遞增；x∈〔lna,0時(shí),ex﹣elna＞0,h′〔x＜0,函數(shù)h〔x單調(diào)遞減；x∈〔0,+∞時(shí),ex﹣elna＞0,h′〔x＞0,函數(shù)h〔x單調(diào)遞增．∴當(dāng)x=0時(shí),函數(shù)h〔x取得極小值,h〔0=﹣2a﹣1．當(dāng)x=lna時(shí),函數(shù)h〔x取得極大值,h〔lna=﹣a[ln2a﹣2lna+sin〔lna+cos〔lna+2]．②當(dāng)a=1時(shí),lna=0,x∈R時(shí),h′〔x≥0,∴函數(shù)h〔x在R上單調(diào)遞增．③1＜a時(shí),lna＞0,x∈〔﹣∞,0時(shí),ex﹣elna＜0,h′〔x＞0,函數(shù)h〔x單調(diào)遞增；x∈〔0,lna時(shí),ex﹣elna＜0,h′〔x＜0,函數(shù)h〔x單調(diào)遞減；x∈〔lna,+∞時(shí),ex﹣elna＞0,h′〔x＞0,函數(shù)h〔x單調(diào)遞增．∴當(dāng)x=0時(shí),函數(shù)h〔x取得極大值,h〔0=﹣2a﹣1．當(dāng)x=lna時(shí),函數(shù)h〔x取得極小值,h〔lna=﹣a[ln2a﹣2lna+sin〔lna+cos〔lna+2]．綜上所述：a≤0時(shí),函數(shù)h〔x在〔0,+∞單調(diào)遞增；x＜0時(shí),函數(shù)h〔x在〔﹣∞,0單調(diào)遞減．x=0時(shí),函數(shù)h〔x取得極小值,h〔0=﹣1﹣2a．0＜a＜1時(shí),函數(shù)h〔x在x∈〔﹣∞,lna是單調(diào)遞增；函數(shù)h〔x在x∈〔lna,0上單調(diào)遞減．當(dāng)x=0時(shí),函數(shù)h〔x取得極小值,h〔0=﹣2a﹣1．當(dāng)x=lna時(shí),函數(shù)h〔x取得極大值,h〔lna=﹣a[ln2a﹣2lna+sin〔lna+cos〔lna+2]．當(dāng)a=1時(shí),lna=0,函數(shù)h〔x在R上單調(diào)遞增．a(chǎn)＞1時(shí),函數(shù)h〔x在〔﹣∞,0,〔lna,+∞上單調(diào)遞增；函數(shù)h〔x在〔0,lna上單調(diào)遞減．當(dāng)x=0時(shí),函數(shù)h〔x取得極大值,h〔0=﹣2a﹣1．當(dāng)x=lna時(shí),函數(shù)h〔x取得極小值,h〔lna=﹣a[ln2a﹣2lna+sin〔lna+cos〔lna+2]．8．〔2017?北京已知函數(shù)f〔x=excosx﹣x．〔1求曲線y=f〔x在點(diǎn)〔0,f〔0處的切線方程；〔2求函數(shù)f〔x在區(qū)間[0,]上的最大值和最小值．[解答]解：〔1函數(shù)f〔x=excosx﹣x的導(dǎo)數(shù)為f′〔x=ex〔cosx﹣sinx﹣1,可得曲線y=f〔x在點(diǎn)〔0,f〔0處的切線斜率為k=e0〔cos0﹣sin0﹣1=0,切點(diǎn)為〔0,e0cos0﹣0,即為〔0,1,曲線y=f〔x在點(diǎn)〔0,f〔0處的切線方程為y=1；〔2函數(shù)f〔x=excosx﹣x的導(dǎo)數(shù)為f′〔x=ex〔cosx﹣sinx﹣1,令g〔x=ex〔cosx﹣sinx﹣1,則g〔x的導(dǎo)數(shù)為g′〔x=ex〔cosx﹣sinx﹣sinx﹣cosx=﹣2ex?sinx,當(dāng)x∈[0,],可得g′〔x=﹣2ex?sinx≤0,即有g(shù)〔x在[0,]遞減,可得g〔x≤g〔0=0,則f〔x在[0,]遞減,即有函數(shù)f〔x在區(qū)間[0,]上的最大值為f〔0=e0cos0﹣0=1；最小值為f〔=ecos﹣=﹣．9．〔2017?天津設(shè)a∈Z,已知定義在R上的函數(shù)f〔x=2x4+3x3﹣3x2﹣6x+a在區(qū)間〔1,2內(nèi)有一個(gè)零點(diǎn)x0,g〔x為f〔x的導(dǎo)函數(shù)．〔Ⅰ求g〔x的單調(diào)區(qū)間；〔Ⅱ設(shè)m∈[1,x0∪〔x0,2],函數(shù)h〔x=g〔x〔m﹣x0﹣f〔m,求證：h〔mh〔x0＜0；〔Ⅲ求證：存在大于0的常數(shù)A,使得對(duì)于任意的正整數(shù)p,q,且∈[1,x0∪〔x0,2],滿足|﹣x0|≥．[解答]〔Ⅰ解：由f〔x=2x4+3x3﹣3x2﹣6x+a,可得g〔x=f′〔x=8x3+9x2﹣6x﹣6,進(jìn)而可得g′〔x=24x2+18x﹣6．令g′〔x=0,解得x=﹣1,或x=．當(dāng)x變化時(shí),g′〔x,g〔x的變化情況如下表：x〔﹣∞,﹣1〔﹣1,〔,+∞g′〔x+﹣+g〔x↗↘↗所以,g〔x的單調(diào)遞增區(qū)間是〔﹣∞,﹣1,〔,+∞,單調(diào)遞減區(qū)間是〔﹣1,．〔Ⅱ證明：由h〔x=g〔x〔m﹣x0﹣f〔m,得h〔m=g〔m〔m﹣x0﹣f〔m,h〔x0=g〔x0〔m﹣x0﹣f〔m．令函數(shù)H1〔x=g〔x〔x﹣x0﹣f〔x,則H′1〔x=g′〔x〔x﹣x0．由〔Ⅰ知,當(dāng)x∈[1,2]時(shí),g′〔x＞0,故當(dāng)x∈[1,x0時(shí),H′1〔x＜0,H1〔x單調(diào)遞減；當(dāng)x∈〔x0,2]時(shí),H′1〔x＞0,H1〔x單調(diào)遞增．因此,當(dāng)x∈[1,x0∪〔x0,2]時(shí),H1〔x＞H1〔x0=﹣f〔x0=0,可得H1〔m＞0即h〔m＞0,令函數(shù)H2〔x=g〔x0〔x﹣x0﹣f〔x,則H′2〔x=g′〔x0﹣g〔x．由〔Ⅰ知,g〔x在[1,2]上單調(diào)遞增,故當(dāng)x∈[1,x0時(shí),H′2〔x＞0,H2〔x單調(diào)遞增；當(dāng)x∈〔x0,2]時(shí),H′2〔x＜0,H2〔x單調(diào)遞減．因此,當(dāng)x∈[1,x0∪〔x0,2]時(shí),H2〔x＞H2〔x0=0,可得得H2〔m＜0即h〔x0＜0,．所以,h〔mh〔x0＜0．〔Ⅲ對(duì)于任意的正整數(shù)p,q,且,令m=,函數(shù)h〔x=g〔x〔m﹣x0﹣f〔m．由〔Ⅱ知,當(dāng)m∈[1,x0時(shí),h〔x在區(qū)間〔m,x0內(nèi)有零點(diǎn)；當(dāng)m∈〔x0,2]時(shí),h〔x在區(qū)間〔x0,m內(nèi)有零點(diǎn)．所以h〔x在〔1,2內(nèi)至少有一個(gè)零點(diǎn),不妨設(shè)為x1,則h〔x1=g〔x1〔﹣x0﹣f〔=0．由〔Ⅰ知g〔x在[1,2]上單調(diào)遞增,故0＜g〔1＜g〔x1＜g〔2,于是|﹣x0|=≥=．因?yàn)楫?dāng)x∈[1,2]時(shí),g〔x＞0,故f〔x在[1,2]上單調(diào)遞增,所以f〔x在區(qū)間[1,2]上除x0外沒有其他的零點(diǎn),而≠x0,故f〔≠0．又因?yàn)閜,q,a均為整數(shù),所以|2p4+3p3q﹣3p2q2﹣6pq3+aq4|是正整數(shù),從而|2p4+3p3q﹣3p2q2﹣6pq3+aq4|≥1．所以|﹣x0|≥．所以,只要取A=g〔2,就有|﹣x0|≥．10．〔2017?XX已知函數(shù)f〔x=x3﹣ax2,a∈R,〔1當(dāng)a=2時(shí),求曲線y=f〔x在點(diǎn)〔3,f〔3處的切線方程；〔2設(shè)函數(shù)g〔x=f〔x+〔x﹣acosx﹣sinx,討論g〔x的單調(diào)性并判斷有無極值,有極值時(shí)求出極值．[解答]解：〔1當(dāng)a=2時(shí),f〔x=x3﹣x2,∴f′〔x=x2﹣2x,∴k=f′〔3=9﹣6=3,f〔3=×27﹣9=0,∴曲線y=f〔x在點(diǎn)〔3,f〔3處的切線方程y=3〔x﹣3,即3x﹣y﹣9=0〔2函數(shù)g〔x=f〔x+〔x﹣acosx﹣sinx=x3﹣ax2+〔x﹣acosx﹣sinx,∴g′〔x=〔x﹣a〔x﹣sinx,令g′〔x=0,解得x=a,或x=0,①若a＞0時(shí),當(dāng)x＜0時(shí),g′〔x＞0恒成立,故g〔x在〔﹣∞,0上單調(diào)遞增,當(dāng)x＞a時(shí),g′〔x＞0恒成立,故g〔x在〔a,+∞上單調(diào)遞增,當(dāng)0＜x＜a時(shí),g′〔x＜0恒成立,故g〔x在〔0,a上單調(diào)遞減,∴當(dāng)x=a時(shí),函數(shù)有極小值,極小值為g〔a=﹣a3﹣sina當(dāng)x=0時(shí),有極大值,極大值為g〔0=﹣a,②若a＜0時(shí),當(dāng)x＞0時(shí),g′〔x＞0恒成立,故g〔x在〔﹣∞,0上單調(diào)遞增,當(dāng)x＜a時(shí),g′〔x＞0恒成立,故g〔x在〔﹣∞,a上單調(diào)遞增,當(dāng)a＜x＜0時(shí),g′〔x＜0恒成立,故g〔x在〔a,0上單調(diào)遞減,∴當(dāng)x=a時(shí),函數(shù)有極大值,極大值為g〔a=﹣a3﹣sina當(dāng)x=0時(shí),有極小值,極小值為g〔0=﹣a③當(dāng)a=0時(shí),g′〔x=x〔x+sinx,當(dāng)x＞0時(shí),g′〔x＞0恒成立,故g〔x在〔0,+∞上單調(diào)遞增,當(dāng)x＜0時(shí),g′〔x＞0恒成立,故g〔x在〔﹣∞,0上單調(diào)遞增,∴g〔x在R上單調(diào)遞增,無極值．11．〔2017?天津設(shè)a,b∈R,|a|≤1．已知函數(shù)f〔x=x3﹣6x2﹣3a〔a﹣4x+b,g〔x=exf〔x．〔Ⅰ求f〔x的單調(diào)區(qū)間；〔Ⅱ已知函數(shù)y=g〔x和y=ex的圖象在公共點(diǎn)〔x0,y0處有相同的切線,〔i求證：f〔x在x=x0處的導(dǎo)數(shù)等于0；〔ii若關(guān)于x的不等式g〔x≤ex在區(qū)間[x0﹣1,x0+1]上恒成立,求b的取值范圍．[解答]〔Ⅰ解：由f〔x=x3﹣6x2﹣3a〔a﹣4x+b,可得f'〔x=3x2﹣12x﹣3a〔a﹣4=3〔x﹣a〔x﹣〔4﹣a,令f'〔x=0,解得x=a,或x=4﹣a．由|a|≤1,得a＜4﹣a．當(dāng)x變化時(shí),f'〔x,f〔x的變化情況如下表：x〔﹣∞,