廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練14 導(dǎo)數(shù)的概念及運(yùn)算 文

考點(diǎn)規(guī)范練14導(dǎo)數(shù)的概念及運(yùn)算一、基礎(chǔ)鞏固1.已知函數(shù)f(x)=3x+1,則limx0f(1-x)-f(1)x的值為()A.-13B.13C.23D.0答案A解析limx0f(1-x)-f(1)x=-limx0f(1-x)-f(1)-x=-f'(1)=-13×1-23=-13.2.已知曲線y=ln x的切線過原點(diǎn),則此切線的斜率為()A.eB.-eC.1eD.-1e答案C解析由題意可得y=lnx的定義域?yàn)?0,+),且y'=1x.設(shè)切點(diǎn)為(x0,lnx0),則切線方程為y-lnx0=1x0(x-x0).因?yàn)榍芯€過點(diǎn)(0,0),所以-lnx0=-1,解得x0=e,故此切線的斜率為1e.3.已知函數(shù)f(x)在R上滿足f(2-x)=2x2-7x+6,則曲線y=f(x)在點(diǎn)(1,f(1)處的切線方程是()A.y=2x-1B.y=xC.y=3x-2D.y=-2x+3答案C解析令x=1,得f(1)=1;令2-x=t,可得x=2-t,代入f(2-x)=2x2-7x+6得f(t)=2(2-t)2-7(2-t)+6,化簡(jiǎn)整理得f(t)=2t2-t,即f(x)=2x2-x,f'(x)=4x-1,f(1)=1,f'(1)=3,所求切線方程為y-1=3(x-1),即y=3x-2.4.已知y=f(x)是可導(dǎo)函數(shù),如圖,直線y=kx+2是曲線y=f(x)在x=3處的切線,令g(x)=xf(x),g'(x)是g(x)的導(dǎo)函數(shù),則g'(3)=()A.-1B.0C.2D.4答案B解析由題圖可知曲線y=f(x)在x=3處切線的斜率等于-13,故f'(3)=-13.g(x)=xf(x),g'(x)=f(x)+xf'(x),g'(3)=f(3)+3f'(3).又由題圖可知f(3)=1,g'(3)=1+3×-13=0.5.曲線f(x)=x3-x+3在點(diǎn)P處的切線平行于直線y=2x-1,則點(diǎn)P的坐標(biāo)為()A.(1,3)B.(-1,3)C.(1,3)和(-1,3)D.(1,-3)答案C解析f(x)=x3-x+3,f'(x)=3x2-1.設(shè)點(diǎn)P(x,y),則f'(x)=2,即3x2-1=2,解得x=1或x=-1,故P(1,3)或(-1,3).經(jīng)檢驗(yàn),點(diǎn)(1,3),(-1,3)均不在直線y=2x-1上,符合題意.故選C.6.已知直線y=kx+1與曲線y=x3+ax+b相切于點(diǎn)A(1,2),則ab等于()A.-8B.-6C.-1D.5答案A解析由題意得y=kx+1過點(diǎn)A(1,2),故2=k+1,即k=1.y'=3x2+a,且直線y=kx+1與曲線y=x3+ax+b相切于點(diǎn)A(1,2),k=3+a,即1=3+a,a=-2.將點(diǎn)A(1,2)代入曲線方程y=x3+ax+b,可解得b=3,即ab=(-2)3=-8.故選A.7.若函數(shù)y=f(x)的圖象上存在兩點(diǎn),使得函數(shù)的圖象在這兩點(diǎn)處的切線互相垂直,則稱y=f(x)具有T性質(zhì).下列函數(shù)中具有T性質(zhì)的是()A.y=sin xB.y=ln xC.y=exD.y=x3答案A解析設(shè)曲線上兩點(diǎn)P(x1,y1),Q(x2,y2),則由導(dǎo)數(shù)幾何意義可知,兩條切線的斜率分別為k1=f'(x1),k2=f'(x2).若函數(shù)具有T性質(zhì),則k1·k2=f'(x1)·f'(x2)=-1.A項(xiàng),f'(x)=cosx,顯然k1·k2=cosx1·cosx2=-1有無數(shù)組解,所以該函數(shù)具有性質(zhì)T;B項(xiàng),f'(x)=1x(x>0),顯然k1·k2=1x1·1x2=-1無解,故該函數(shù)不具有性質(zhì)T;C項(xiàng),f'(x)=ex>0,顯然k1·k2=ex1·ex2=-1無解,故該函數(shù)不具有性質(zhì)T;D項(xiàng),f'(x)=3x20,顯然k1·k2=3x12×3x22=-1無解,故該函數(shù)不具有性質(zhì)T.綜上,選A.8.若點(diǎn)P是曲線y=x2-ln x上任意一點(diǎn),則點(diǎn)P到直線y=x-2的距離的最小值為()A.1B.2C.22D.3答案B解析因?yàn)槎x域?yàn)?0,+),所以y'=2x-1x,令2x-1x=1,解得x=1,則曲線在點(diǎn)P(1,1)處的切線方程為x-y=0,所以兩平行線間的距離為d=22=2.故所求的最小值為2.9.(2018天津,文10)已知函數(shù)f(x)=exln x,f'(x)為f(x)的導(dǎo)函數(shù),則f'(1)的值為. 答案e解析f(x)=exlnx,f'(x)=exlnx+exx.f'(1)=eln1+e1=e.10.曲線y=log2x在點(diǎn)(1,0)處的切線與坐標(biāo)軸所圍三角形的面積等于. 答案12log2e解析y'=1xln2,k=1ln2,切線方程為y=1ln2(x-1),所圍三角形的面積為S=12×1×1ln2=12ln2=12log2e.11.(2018甘肅天水月考)設(shè)函數(shù)f(x)=g(x)+x2,曲線y=g(x)在點(diǎn)(1,g(1)處的切線方程為y=2x+1,則曲線y=f(x)在點(diǎn)(1,f(1)處切線的斜率為. 答案4解析由導(dǎo)數(shù)的幾何意義及條件,得g'(1)=2,函數(shù)f(x)=g(x)+x2,f'(x)=g'(x)+2x,f'(1)=g'(1)+2=4,曲線y=f(x)在點(diǎn)(1,f(1)處切線的斜率為4.12.若函數(shù)f(x)=12x2-ax+ln x存在垂直于y軸的切線,則實(shí)數(shù)a的取值范圍是. 答案2,+)解析f(x)=12x2-ax+lnx,f'(x)=x-a+1x.f(x)存在垂直于y軸的切線,f'(x)存在零點(diǎn),x+1x-a=0有解,a=x+1x2(x>0).二、能力提升13.若函數(shù)y=f(x),y=g(x)的導(dǎo)函數(shù)的圖象如圖所示,則y=f(x),y=g(x)的圖象可能是()答案D解析由y=f'(x)的圖象知y=f'(x)在(0,+)內(nèi)單調(diào)遞減,說明函數(shù)y=f(x)的切線的斜率在(0,+)內(nèi)也單調(diào)遞減,故可排除A,C.又由圖象知y=f'(x)與y=g'(x)的圖象在x=x0處相交,說明y=f(x)與y=g(x)的圖象在x=x0處的切線的斜率相同,故可排除B.故選D.14.若存在過點(diǎn)(1,0)的直線與曲線y=x3和y=ax2+154x-9都相切,則a等于()A.-1或-2564B.-1或214C.-74或-2564D.-74或7答案A解析因?yàn)閥=x3,所以y'=3x2.設(shè)過點(diǎn)(1,0)的直線與y=x3相切于點(diǎn)(x0,x03),則在該點(diǎn)處的切線斜率為k=3x02,所以切線方程為y-x03=3x02(x-x0),即y=3x02x-2x03.又點(diǎn)(1,0)在切線上,則x0=0或x0=32.當(dāng)x0=0時(shí),由y=0與y=ax2+154x-9相切,可得a=-2564;當(dāng)x0=32時(shí),由y=274x-274與y=ax2+154x-9相切,可得a=-1.15.(2018安徽六安模擬)給出定義:設(shè)f'(x)是函數(shù)y=f(x)的導(dǎo)函數(shù),f(x)是函數(shù)f'(x)的導(dǎo)函數(shù),若方程f(x)=0有實(shí)數(shù)解x0,則稱點(diǎn)(x0,f(x0)為函數(shù)y=f(x)的“拐點(diǎn)”.已知函數(shù)f(x)=3x+4sin x-cos x的“拐點(diǎn)”是M(x0,f(x0),則點(diǎn)M()A.在直線y=-3x上B.在直線y=3x上C.在直線y=-4x上D.在直線y=4x上答案B解析由題意,知f'(x)=3+4cosx+sinx,f(x)=-4sinx+cosx,由f(x0)=0,知-4sinx0+cosx0=0,即4sinx0-cosx0=0,所以f(x0)=3x0+4sinx0-cosx0=3x0,即點(diǎn)M(x0,3x0),顯然在直線y=3x上.故選B.16.已知f(x),g(x)分別是定義在R上的偶函數(shù)和奇函數(shù),且f(x)-g(x)=ex+x2+1,則函數(shù)h(x)=2f(x)-g(x)在點(diǎn)(0,h(0)處的切線方程是. 答案x-y+4=0解析f(x)-g(x)=ex+x2+1,且f(x)是偶函數(shù),g(x)是奇函數(shù),f(-x)-g(-x)=f(x)+g(x)=e-x+x2+1.f(x)=ex+e-x+2x2+22,g(x)=e-x-ex2.h(x)=2f(x)-g(x)=ex+e-x+2x2+2-e-x-ex2=32ex+12e-x+2x2+2.h'(x)=32ex-12e-x+4x,即h'(0)=32-12=1.又h(0)=4,切線方程為x-y+4=0.三、高考預(yù)測(cè)17.設(shè)曲線y=xex+x2在原點(diǎn)處的切線與直線x+ay+1=0垂直,則a=. 答案1解析由y=xex+x2得y'=ex+xex+2x,在原點(diǎn)處的切線的斜率k1=e0+0·e0+0=1,直線x+ay+1=0的斜率k2=-1a,由題意知k1k2=-1a×1=-1a=1.7