1、【全程復(fù)習(xí)方略】2020版高考數(shù)學(xué) 2.12導(dǎo)數(shù)在實(shí)際問(wèn)題中的應(yīng)用及綜合應(yīng)用課時(shí)提升作業(yè) 理 北師大版一、選擇題1.(2020西安模擬)函數(shù)y=f(x)在定義域(-32,3)內(nèi)的圖像如圖所示,記y=f(x)的導(dǎo)函數(shù)為y=f(x),則不等式f(x)0的解集為()(A)-1,1243,83(B)-13,12,3)(C)(-32,121,2)(D)(-32,-1312,43)43,3)2.若對(duì)任意的x0,恒有l(wèi)nxpx-1(p0),則p的取值范圍是()(A)(0,1(B)(1,+)(C)(0,1)(D)1,+)3.(2020黃山模擬)在半徑為R的半球內(nèi)有一內(nèi)接圓柱,則這個(gè)圓柱的體積的最大值是()(A

2、)239R3(B)439R3(C)233R3(D)49R34.(2020宣城模擬)對(duì)于R上可導(dǎo)的任意函數(shù)f(x),若滿足(x-1)f(x)0,則必有()(A)f(0)+f(2)2f(1)5.(2020咸陽(yáng)模擬)函數(shù)y=2x3+1的圖像與函數(shù)y=3x2-b的圖像有三個(gè)不相同的交點(diǎn),則實(shí)數(shù)b的取值范圍是()(A)(-2,-1)(B)(-1,0)(C)(0,1)(D)(1,2)6.(2020沈陽(yáng)模擬)設(shè)f(x)是定義在R上的奇函數(shù),且f(2)=0,當(dāng)x0時(shí),有xf(x)-f(x)x20的解集是()(A)(-2,0)(2,+)(B)(-2,0)(0,2)(C)(-,-2)(2,+)(D)(-,-2)(

3、0,2)二、填空題7.已知函數(shù)f(x)=xsinx,xR,f(-4),f(43),f(-54)的大小關(guān)系為(用“0,且x1時(shí),f(x)lnxx-1+kx,求k的取值范圍.11.(2020合肥模擬)某唱片公司要發(fā)行一張名為春風(fēng)再美也比不上你的笑的唱片,包含新花好月圓荷塘月色等10首創(chuàng)新經(jīng)典歌曲.該公司計(jì)劃用x(百萬(wàn)元)請(qǐng)李子恒老師進(jìn)行創(chuàng)作,經(jīng)調(diào)研知:該唱片的總利潤(rùn)y(百萬(wàn)元)與(3-x)x2成正比的關(guān)系,當(dāng)x=2時(shí)y=32.又有x2(3-x)(0,t,其中t是常數(shù),且t(0,2.(1)設(shè)y=f(x),求其表達(dá)式及定義域(用t表示).(2)求總利潤(rùn)y的最大值及相應(yīng)的x的值.12.(2020淮北模擬

4、)已知函數(shù)f(x)=(a+1a)lnx+1x-x.(1)當(dāng)a1時(shí),討論f(x)在區(qū)間(0,1)上的單調(diào)性.(2)當(dāng)a0時(shí),求f(x)的極值.(3)當(dāng)a3時(shí),曲線y=f(x)上總存在不同兩點(diǎn)P(x1,f(x1),Q(x2,f(x2),使得曲線y=f(x)在P,Q兩點(diǎn)處的切線互相平行,證明:x1+x265.答案解析1.【解析】選B.由函數(shù)y=f(x)的圖像知,函數(shù)y=f(x)在-13,1,2,3)上是減少的,故f(x)0的解集為-13,12,3).2.【解析】選D.原不等式可化為lnx-px+10,令f(x)=lnx-px+1,故只需f(x)max0.由f(x)=1x-p,知f(x)在(0,1p)

5、上是增加的,在(1p,+)上是減少的.故f(x)max=f(1p)=-lnp,由-lnp0得p1.3.【解析】選A.設(shè)圓柱的高為h,則圓柱的底面半徑為R2-h2,圓柱的體積為V=(R2-h2)h=-h3+R2h(0hR),V=-3h2+R2=0,h=R3時(shí)V有最大值為V=239R3.4.【解析】選C.由(x-1)f(x)0,得x1時(shí),f(x)0;x1時(shí),f(x)0.因此,函數(shù)y=f(x)在(-,1上是減少的(或?yàn)槌?shù)函數(shù));在1,+)上是增加的(或?yàn)槌?shù)函數(shù)),所以f(0)f(1);f(2)f(1),故f(0)+f(2)2f(1).5.【解析】選B.由題意知方程2x3+1=3x2-b,即2x3

6、-3x2+1=-b有三個(gè)不相同的實(shí)數(shù)根,令f(x)=2x3-3x2+1,即函數(shù)y=f(x)=2x3-3x2+1與直線y=-b有三個(gè)交點(diǎn).由f(x)=6x2-6x=6x(x-1)知,函數(shù)y=f(x)在區(qū)間(-,0)上是增加的,在(0,1)上是減少的,在(1,+)上是增加的,故f(0)是函數(shù)的極大值,f(1)是函數(shù)的極小值,若函數(shù)y=f(x)=2x3-3x2+1與直線y=-b有三個(gè)交點(diǎn),則f(1)-bf(0),解得-1b0時(shí),有xf(x)-f(x)x20,則f(x)x0時(shí)是減少的,x2f(x)0,即為x3f(x)x0f(x)x0.f(2)=0,畫(huà)出y=f(x)x在x0時(shí)的示意圖,知0x2.同理,由

7、f(x)是奇函數(shù),則y=f(x)x是偶函數(shù),如圖,在x0,即為x3f(x)x0f(x)x0.f(-2)=0,x-2.綜上所述,不等式的解集是(-,-2)(0,2).7.【解析】f(x)=sinx+xcosx,當(dāng)x54,43時(shí),sinx0,cosx0,f(x)=sinx+xcosx0,則函數(shù)f(x)在x54,43上是減少的,f(43)f(4)f(54),又函數(shù)f(x)為偶函數(shù),f(43)f(-4)f(-54).答案:f(43)f(-4)0,即x(0,1時(shí),f(x)=ax3-3x+10可化為a3x2-1x3,設(shè)g(x)=3x2-1x3,則g(x)=3(1-2x)x4,所以g(x)在區(qū)間(0,12上

8、是增加的,在區(qū)間12,1上是減少的,因此g(x)max=g(12)=4,從而a4.當(dāng)x0f(1)+f(1)f(0),得到-4+2mmf(1)+f(1)f(2),得到-4+2m2+m由得到m6,即為所求.答案:m610.【解析】(1)由f(x)=alnxx+1+bx,得f(x)=ax+1x-lnx(x+1)2-bx2=ax+1-xlnxx(x+1)2-bx2,曲線y=f(x)在點(diǎn)(1,f(1)處的切線方程為x+2y-3=0,f(1)=b=1,f(1)=12a-b=-12,解得a=1,b=1.(2)由(1)知f(x)=lnxx+1+1x,所以f(x)-(lnxx-1+kx)=11-x22lnx+(

9、k-1)(x2-1)x.考慮函數(shù)h(x)=2lnx+(k-1)(x2-1)x(x0),則h(x)=(k-1)(x2+1)+2xx2.()設(shè)k0,由h(x)=k(x2+1)-(x-1)2x2知,當(dāng)x1時(shí),h(x)0,可得11-x2h(x)0;當(dāng)x(1,+)時(shí),h(x)0.從而當(dāng)x0,且x1時(shí),f(x)-(lnxx-1+kx)0,即f(x)lnxx-1+kx.()設(shè)0k0,故h(x)0.而h(1)=0,故當(dāng)x(1,11-k)時(shí),h(x)0,可得11-x2h(x)0,而h(1)=0,故當(dāng)x(1,+)時(shí),h(x)0,可得11-x2h(x)0,與題設(shè)矛盾.綜合得,k的取值范圍為(-,0.11.【解析】(

10、1)y=k(3-x)x2,當(dāng)x=2時(shí),y=32,k=8,y=f(x)=24x2-8x3.x2(3-x)(0,t,06t2t+1,即0t0,所以f(x)在(0,6t2t+1)上是增加的.ymax=f(6t2t+1)=864t2(2t+1)3.綜上所述,當(dāng)1t2,x=2時(shí),ymax=32;當(dāng)0t1,故01a10),當(dāng)a1時(shí),f(x)的減區(qū)間是(0,1a),(a,+),增區(qū)間是(1a,a).f(x)極小值=f(1a)=(a+1a)ln1a+a-1a=-(a+1a)lna+a-1a,f(x)極大值=f(a)=(a+1a)lna-a+1a,當(dāng)a=1時(shí),f(x)=-(x-1)2x20,f(x)無(wú)極值.當(dāng)0a2x1x2得:x1x24x1+x2,即a+1a=x1+x2