高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)一輪用書文科數(shù)學(xué)配北師版課時(shí)規(guī)范練14　導(dǎo)數(shù)的概念及運(yùn)算

課時(shí)規(guī)范練14導(dǎo)數(shù)的概念及運(yùn)算基礎(chǔ)鞏固組1.(2021山西臨汾一模)曲線f(x)=x2+2ex在點(diǎn)(0,f(0))處的切線方程為()A.x+2y+2=0 B.2x+y+2=0C.x-2y+2=0 D.2x-y+2=0答案:D解析:f(x)=x2+2ex的導(dǎo)數(shù)為f'(x)=2x+2ex,則在點(diǎn)(0,f(0))處的切線的斜率為f'(0)=2,且切點(diǎn)為(0,2),則切線的方程為y=2x+2,即2x-y+2=0.2.(2021江西宜春模擬)已知函數(shù)f(x)=x3-f'(1)x2+2的導(dǎo)數(shù)為f'(x),則f(x)的圖像在點(diǎn)(2,f(2))處的切線的斜率為()A.-8 B.8 C.12 D.16答案:B解析:因?yàn)閒'(x)=3x2-2f'(1)x,令x=1,得f'(1)=3-2f'(1),所以f'(1)=1,所以f'(x)=3x2-2x,f(x)的圖像在點(diǎn)(2,f(2))處的切線的斜率為f'(2)=8.3.已知f(x)=14x2+sin5π2+x,f'(x)為f(x)的導(dǎo)函數(shù),則f'答案:A解析:∵f(x)=14x2+sin5π2+x=14x2+cos∴f'(x)=12x-sinx∴函數(shù)f'(x)為奇函數(shù),排除B,D.又f'π2=π4-1<0,排除C4.(2021河南新鄉(xiāng)三模)已知函數(shù)f(x)=x4+ax,若limΔx→0f(A.36 B.12 C.4 D.2答案:C解析:根據(jù)題意,f(x)=x4+ax,則f'(x)=4x3+a,則f'(0)=a,若limΔx則limΔx→0f(2Δx)-f(-Δx)Δx5.(2021湖南岳陽模擬)函數(shù)f(x)的圖像如圖所示,則下列數(shù)值排序正確的是()A.0<f'(2)<f'(3)<f(3)-f(2)B.0<f'(3)<f(3)-f(2)<f'(2)C.0<f'(3)<f'(2)<f(3)-f(2)D.0<f(3)-f(2)<f'(2)<f'(3)答案:B解析:如圖所示,f'(2)是函數(shù)f(x)的圖像在x=2(即點(diǎn)A)處切線的斜率k1,f'(3)是函數(shù)f(x)的圖像在x=3(即點(diǎn)B)處切線的斜率k2,f(3)-f(2)3-2由圖像知,0<k2<kAB<k1,即0<f'(3)<f(3)-f(2)<f'(2).6.(2021河南洛陽二模)若曲線f(x)=lnx在點(diǎn)(1,0)的切線與曲線g(x)=12x2+mx+72也相切,則m=答案:-2或4解析:由f(x)=lnx的導(dǎo)數(shù)為f'(x)=1x,可得曲線f(x)=lnx在點(diǎn)(1,0)的切線斜率為1,切線的方程為y=x-1,聯(lián)立y=x-1,y=12x2+mx+72,可得x2+(2m-2)x+9=0,由切線與曲線g(x)=12x2+mx+72也相切,7.(2021河北邯鄲二模)寫出一個(gè)奇函數(shù)f(x),當(dāng)x>0時(shí),f(x)>0且其導(dǎo)數(shù)f'(x)<0,則f(x)=.

答案:1x(答案不解析:f(x)=1x為奇函數(shù),當(dāng)x>0時(shí),f(x)>0,且f'(x)=-1x2<8.已知函數(shù)f(x)=13x3-4x,x<0,-1答案:14或-解析:由題意得f'(x)=x2-4,x<0,所以0<a<1,1a2-9.(2021貴州貴陽高三期末)曲線f(x)=2x-ex與直線x-y+t=0相切,則t=.

答案:-1解析:∵f(x)=2x-ex,∴f'(x)=2-ex,切線x-y+t=0的斜率為k=1,設(shè)切點(diǎn)P(x0,y0),令f'(x0)=2-ex0=1,解得x0=0,代入f(x)=2x-ex得y0=2×0-e0=-1,∴切點(diǎn)坐標(biāo)為(0,-1),代入切線方程x-y+t=0中得到0+1+t=0,解得t=-10.(2021廣東廣州二模)已知函數(shù)f(x)=lnxx+a,且f'(1)=1,則a=,曲線y=f(x)在x=答案:0y-1e=解析:由f(x)=lnxx+a,則f'(x)=x+ax-lnx(x+a)2,因?yàn)閒'(1)=1,即11+a=1,解得a=0,所以f(x)=lnxx,f'(x)=1-lnxx2∴x0e2x0=e3,∴f(x0)=x0e2x0-e綜合提升組11.(2021黑龍江齊齊哈爾三模)已知函數(shù)f(x)=sinx和g(x)=cosx圖像的一個(gè)公共點(diǎn)為P(x0,y0),現(xiàn)給出以下結(jié)論:①f(x0)=g(x0);②f'(x0)=g'(x0);③f(x)和g(x)的圖像在點(diǎn)P處的切線的傾斜角互補(bǔ);④f(x)和g(x)的圖像在點(diǎn)P處的切線互相垂直.其中正確結(jié)論的序號(hào)是()A.①③ B.②④ C.②③ D.①④答案:A解析:對(duì)于①,因?yàn)閒(x0)=y0,g(x0)=y0,則f(x0)=g(x0),故①正確;對(duì)于②,因?yàn)閒(x)和g(x)在點(diǎn)P處的切線不平行且不重合,所以f'(x0)≠g'(x0),故②錯(cuò)誤;對(duì)于③,顯然f'(x0)+g'(x0)=0成立,故③正確;對(duì)于④,假設(shè)f(x)和g(x)的圖像在點(diǎn)P處的切線互相垂直,則有-cosx0sinx0=-1,即sin2x0=2,這與|sin2x0|≤1矛盾,故④錯(cuò)誤.12.(2021云南昆明一中模擬)函數(shù)f(x)=lnx圖像上一點(diǎn)P到直線y=2x的最短距離為()A.2 BC.(1+ln2)答案:C解析:設(shè)與直線y=2x平行且與曲線f(x)=lnx相切的直線的切點(diǎn)坐標(biāo)為(x0,lnx0),因?yàn)閒'(x)=1x,所以1x0=2,解得x0=12,則切點(diǎn)坐標(biāo)為12,-ln2,最短距離為點(diǎn)12,-ln2到直線y=2x的距離,即13.(2021廣西桂林模擬)設(shè)曲線y=lnx與y=(x+a)2有一條斜率為1的公切線,則a=()A.-1 B.-34 C.14答案:B解析:因?yàn)閥=lnx,所以y'=1又因?yàn)榍芯€的斜率為1,設(shè)切點(diǎn)為(x0,y0),所以1x0=1,解得x0=1,y0=0,所以切線方程為y=x-因?yàn)閥=(x+a)2,設(shè)切點(diǎn)(x,y),所以y'=2x+2a=1,解得x=12-a代入切線方程得y=-12-a,再將12-a,-12-a代入y=(x+a)2解得a=-314.(2021四川涼山三模)已知函數(shù)f(x)=ex-lnxx?1x+a,若直線y=0在點(diǎn)(b,f(b))處與曲線y=f(xA.1 B.0 C.-1 D.-1或1答案:C解析:由f(x)=ex-lnxx?1x+a可得f'(x)=ex-1-lnxx2+1x2=ex+lnxx2,因?yàn)橹本€y=0在點(diǎn)(b,f(b))處與曲線y=f(x所以eb·b=-lnbb=1b·ln1b,兩邊同時(shí)取以e為底的對(duì)數(shù),可得ln(eb即lneb+lnb=ln1b+lnln所以b+lnb=ln1b+lnln設(shè)g(x)=x+lnx,g'(x)=1+1x>0,函數(shù)在(0,+∞)上是遞增的所以b=ln1b,即b=-lnb,又因?yàn)閒(b)=所以f(b)=eb-lnbb解得a=-1.15.(2021云南紅河三模)丹麥數(shù)學(xué)家琴生在函數(shù)的凹凸性與不等式方面留下了很多寶貴的成果.定義:函數(shù)f(x)在(a,b)上的導(dǎo)函數(shù)為f'(x),f'(x)在(a,b)上的導(dǎo)函數(shù)為f″(x),若在(a,b)上f″(x)<0恒成立,則稱函數(shù)f(x)在(a,b)上的“嚴(yán)格凸函數(shù)”,稱區(qū)間(a,b)為函數(shù)f(x)的“嚴(yán)格凸區(qū)間”.則下列正確說法的序號(hào)為.

①函數(shù)f(x)=-x3+3x2+2在(1,+∞)上為“嚴(yán)格凸函數(shù)”;②函數(shù)f(x)=lnxx的“嚴(yán)格凸區(qū)間”為(0,e32);③函數(shù)f(x)=ex-m2x2在(1,4)為“嚴(yán)格凸函數(shù)”,則m答案:①②解析:f(x)=-x3+3x2+2的導(dǎo)函數(shù)f'(x)=-3x2+6x,f″(x)=-6x+6,故f″(x)<0在(1,+∞)上恒成立,所以函數(shù)f(x)=-x3+3x2+2在(1,+∞)上為“嚴(yán)格凸函數(shù)”,所以①正確;f(x)=lnxx的定義域?yàn)?0,+∞)且導(dǎo)函數(shù)f'(x)=1-lnxx2,f″(x)=2lnx-3x3,由f″(x)<0可得2lnx-3<0,解得x∈(0,e32),所以函數(shù)f(x)f(x)=ex-m2x2的導(dǎo)函數(shù)f'(x)=ex-mx,f″(x)=ex-m因?yàn)閒(x)在(1,4)為“嚴(yán)格凸函數(shù)”,故f″(x)<0在(1,4)上恒成立,所以ex-m<0在(1,4)上恒成立,即m>ex在(1,4)上恒成立,故m≥e4,所以③不正確.創(chuàng)新應(yīng)用組16.(2021浙江杭州二中模擬)函數(shù)f(x)=ax+sinx的圖像上存在兩條相互垂直的切線,則實(shí)數(shù)a的取值范圍是()A.{0,1} B.{0}C.[0,1) D.[1,+∞)答案:B解析:因?yàn)閒(x)=ax+sinx,所以f'(x)=a+cosx,因?yàn)楹瘮?shù)f(x)=ax+sinx的圖像上存在兩條相互垂直的切線,所以不妨設(shè)在x=x1和x=x2處的切線互相垂直,則(a+cosx1)·(a+cosx2)=-1,即a2+(cosx1+cosx2)a+cosx1cosx2+1=0,①因?yàn)閍的值一定存在,即方程①一定有解,所以Δ=(cosx1+cosx2)2-4(cosx1cosx2+1)≥0,即(cosx1-cosx2)2≥4,解得cosx1-cosx2≥2或cosx1-cosx2≤-2,又因?yàn)閨cosx|≤1,所以有cosx1=1,cosx2=-1或cosx1=-1,cosx2=1,Δ=0,所以方程①變?yōu)閍2=0,所以a=0.故選B.17.已知a-lnb=0,c-d=1,則(a-c)2+(b-d)2的最小值是