（新課標(biāo)）高三數(shù)學(xué)一輪復(fù)習(xí) 大題沖關(guān)集訓(xùn)（一）理.docVIP

大題沖關(guān)集訓(xùn)(一)1.(2014高考安徽卷)設(shè)函數(shù)f(x)=1+(1+a)x-x2-x3,其中a0.(1)討論f(x)在其定義域上的單調(diào)性;(2)當(dāng)x0,1時(shí),求f(x)取得最大值和最小值時(shí)的x的值.解:(1)f(x)的定義域?yàn)?-,+),f(x)=1+a-2x-3x2.令f(x)=0,得x1=-1-4+3a3,x2=-1+4+3a3,x1x2.所以f(x)=-3(x-x1)(x-x2).當(dāng)xx2時(shí),f(x)0;當(dāng)x1x0.故f(x)在(-,-1-4+3a3)和(-1+4+3a3,+)內(nèi)單調(diào)遞減,在(-1-4+3a3,-1+4+3a3)內(nèi)單調(diào)遞增.(2)因?yàn)閍0,所以x10.當(dāng)a4時(shí),x21.由(1)知,f(x)在0,1上單調(diào)遞增.所以f(x)在x=0和x=1處分別取得最小值和最大值.當(dāng)0a4時(shí),x21.由(1)知,f(x)在0,x2上單調(diào)遞增,在x2,1上單調(diào)遞減.所以f(x)在x=x2=-1+4+3a3處取得最大值.又f(0)=1,f(1)=a,所以當(dāng)0a1時(shí),f(x)在x=1處取得最小值;當(dāng)a=1時(shí),f(x)在x=0處和x=1處同時(shí)取得最小值;當(dāng)1a0時(shí),f(x)單調(diào)增區(qū)間為(0,1c),f(x)單調(diào)減區(qū)間為(1c,+).(2)f(x)x2恒成立,即ln x-cxx2恒成立,clnxx-x,當(dāng)x(0,+)時(shí)恒成立.設(shè)g(x)=lnxx-x,g(x)=1-lnx-x2x2,g(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減.g(x)max=g(1)=-1,c-1.即c的取值范圍為(-1,+).3.(2014涼州一診)已知函數(shù)f(x)=(ax-2)ex在x=1處取得極值.(1)求a的值;(2)求函數(shù)f(x)在m,m+1上的最小值;(3)求證:對(duì)任意x1,x20,2,都有|f(x1)-f(x2)|e.(1)解:f(x)=aex+(ax-2)ex=(ax+a-2)ex.由已知得f(1)=0,即(2a-2)e=0,解得a=1.當(dāng)a=1時(shí),在x=1處函數(shù)f(x)=(x-2)ex取得極小值,所以a=1.(2)解:由(1)知f(x)=(x-2)ex,f(x)=ex+(x-2)ex=(x-1)ex.所以函數(shù)f(x)在(-,1)上單調(diào)遞減,在(1,+)上單調(diào)遞增.當(dāng)m1時(shí),f(x)在m,m+1上單調(diào)遞增,f(x)min=f(m)=(m-2)em.當(dāng)0m1時(shí),m1m+1,f(x)在m,1上單調(diào)遞減,在1,m+1上單調(diào)遞增,f(x)min=f(1)=-e.當(dāng)m0時(shí),m+11,f(x)在m,m+1上單調(diào)遞減,f(x)min=f(m+1)=(m-1)em+1.綜上,f(x)在m,m+1上的最小值f(x)min=(m-2)em,m1,-e,0m1,(m-1)em+1,m0.(3)證明:由(1)知f(x)=(x-2)ex,f(x)=(x-1)ex.令f(x)=0得x=1.因?yàn)閒(0)=-2,f(1)=-e,f(2)=0,所以當(dāng)x0,2時(shí),f(x)max=0,f(x)min=-e,所以,對(duì)任意x1,x20,2,都有|f(x1)-f(x2)|f(x)max-f(x)min=e.4.(2014臨沂市質(zhì)檢)已知函數(shù)f(x)=ln x.(1)若直線y=x+m與函數(shù)f(x)的圖象相切,求實(shí)數(shù)m的值;(2)證明曲線y=f(x)與曲線y=x-1x有唯一的公共點(diǎn);(3)設(shè)0ab,比較f(b)-f(a)2與b-ab+a的大小,并說明理由.(1)解:f(x)=1x,設(shè)切點(diǎn)為(x0,y0),則k=1x0=1,x0=1,y0=ln x0=ln 1=0,代入y=x+m,得m=-1.(2)證明:令h(x)=f(x)-(x-1x)=ln x-x+1x,則h(x)=1x-1-1x2=-x2+x-1x2=-(x-12)2-34x20,h(x)在(0,+)上單調(diào)遞減.又h(1)=ln 1-1+1=0,x=1是函數(shù)h(x)唯一的零點(diǎn),故點(diǎn)(1,0)是兩曲線唯一的公共點(diǎn).(3)解:lnb-lna2-b-ab+a=12ln ba-ba-1ba+1,0a1.構(gòu)造函數(shù) (x)=12ln x-x-1x+1(x1),則(x)=12x-x+1-(x-1)(x+1)2=12x-2(x+1)2=(x-1)22x(x+1)20, (x)在(1,+)上單調(diào)遞增,又當(dāng)x=1時(shí), (1)=0,x1時(shí), (x)0,即12ln xx-1x+1,則有12ln baba-1ba+1成立,即lnb-lna2b-ab+a.即f(b)-f(a)2b-ab+a.5.(2015湖北省八市聯(lián)考)定義在r上的函數(shù)g(x)及二次函數(shù)h(x)滿足g(x)+2g(-x)=ex+2ex-9,h(-2)=h(0)=1且h(-3)=-2.(1)求g(x)和h(x)的解析式;(2)對(duì)于x1,x2-1,1,均有h(x1)+ax1+5g(x2)-x2g(x2)成立,求a的取值范圍;(3)設(shè)f(x)=g(x),x0,h(x),x0,討論方程ff(x)=2的解的個(gè)數(shù)情況.解:(1)g(x)+2g(-x)=ex+2ex-9,g(-x)+2g(x)=e-x+2e-x-9,即g(-x)+2g(x)=2ex+1ex-9, 由聯(lián)立解得g(x)=ex-3.h(x)是二次函數(shù),且h(-2)=h(0)=1,可設(shè)h(x)=ax(x+2)+1,由h(-3)=-2,解得a=-1.h(x)=-x(x+2)+1=-x2-2x+1.g(x)=ex-3,h(x)=-x2-2x+1.(2)設(shè)(x)=h(x)+ax+5=-x2+(a-2)x+6,f(x)=ex-3-x(ex-3)=(1-x)ex+3x-3,依題意知,當(dāng)-1x1時(shí),(x)minf(x)max.f(x)=-ex+(1-x)ex+3=-xex+3在-1,1上單調(diào)遞減,f(x)min=f(1)=3-e0,f(x)在-1,1上單調(diào)遞增,f(x)max=f(1)=0,(-1)=7-a0,(1)=a+30,解得-3a7,實(shí)數(shù)a的取值范圍為-3,7.(3)f(x)的圖象如圖所示.令t=f(x),則f(t)=2.t1=-1,t2=ln 5,f(x)=-1有兩個(gè)解,f(x)=ln 5有3個(gè)解.ff(x)=2有5個(gè)解.6.已知函數(shù)f(x)=ax-1-ln x(ar).(1)討論函數(shù)f(x)的單調(diào)性;(2)若函數(shù)f(x)在x=1處取得極值,不等式f(x)bx-2對(duì)x(0,+)恒成立,求實(shí)數(shù)b的取值范圍;(3)當(dāng)xye-1時(shí),證明不等式exln(1+y)eyln(1+x).(1)解:函數(shù)的定義域是(0,+),且f(x)=a-1x=ax-1x.當(dāng)a0時(shí),ax-10,從而f(x)0時(shí),若0x1a,則ax-10,從而f(x)eyln(1+x)成立,只需證ex+1ln(x+1)ey+1ln(y+1).構(gòu)造函數(shù)h(x)=e