【名师一号】2014-2015学年高中数学人教版通用选修2-2双基限时练12]
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双基限时练(十二)1.下列各式中,正确的是( ) A .⎠⎛ab F ′(x )d x =F ′(b )-F ′(a )B.⎠⎛a b F ′(x)d x =F ′(a)-F ′(b)C .⎠⎛ab F ′(x )d x =F (b )-F (a ) D.⎠⎛ab F ′(x)d x =F(a)-F(b)答案 C2.∫π20( sin x -cos x)d x =( ) A .0 B .1 C .2D .π2解析 ∫π20(sin x -cos x)d x =∫π20sin x d x -∫π20cos x d x =(-cos x)⎪⎪⎪ π20-(sin x)⎪⎪⎪ π2=1-1=0. 答案 A3.若∫a 1(2x +1x )d x =3+ln 2(a>1),则a 的值为( ) A .6 B .4 C .3D .2解析 ∵⎠⎛1a (2x +1x )d x=(x 2+ln x)⎪⎪⎪ a 1=a 2+ln a -1, 又⎠⎛1a (2x +1x )d x =3+ln 2,∴a =2. 答案 D4.⎠⎛π-πcos x d x 等于( )A .2πB .πC .0D .1解析 ⎠⎛π-πcos x d x =sin x⎪⎪⎪ π-π=sinπ-sin (-π)=0. 答案 C5.设f(x)=⎩⎪⎨⎪⎧x 2(0≤x<1),2-x (1<x ≤2),则⎠⎛02f(x)d x 等于( )A .34 B .45 C .56D .不存在解析 ⎠⎛02f(x)d x =⎠⎛01x 2d x +⎠⎛12(2-x)d x=13+2-32=56. 答案 C6.由曲线y =x 2-1,直线x =0,x =2和x 轴围成的封闭图形的面积(如图阴影部分)是( )A .⎠⎛02(x 2-1)d xB .|⎠⎛02(x 2-1)d x |C.⎠⎛02|x 2-1|d xD .⎠⎛01(x 2-1)d x +⎠⎛12(x 2-1)d x答案 C7.若a =⎠⎛02x 2d x ,b =⎠⎛02x 3d x ,c =⎠⎛02 sin x d x ,则a ,b ,c 的大小关系是________.解析 ∵a =⎠⎛02x 2d x =13x 3⎪⎪⎪ 20=83,b =⎠⎛02x 3d x =14x4⎪⎪⎪ 20=4,⎠0∴b >a >c . 答案 b >a >c8.计算⎠⎛2-2( sin x +2)d x =________.解析 ⎠⎛2-2(sin x +2)d x =⎠⎛2-2sin x d x +⎠⎛2-22d x=(-cos x ) ⎪⎪⎪ 2-2+2x⎪⎪⎪ 2-2 =-cos2+cos(-2)+2×2-2×(-2) =8. 答案 89.设函数f (x )=ax 2+c (a ≠0),若0≤x 0≤1.且⎠⎛01f (x )d x =f (x 0),则x 0=________.解析 ∵⎠⎛01f (x )d x =⎠⎛01(ax 2+c )d x =⎝ ⎛⎭⎪⎫a 3x 3+cx ⎪⎪10=a3+c , 又⎠⎛1f (x )d x =f (x 0),∴ax 20+c =a 3+c .∵a ≠0,∴x 20=13,又0≤x 0≤1,∴x 0=33. 答案 3310.计算下列定积分:(1)⎠⎛14x -x 2x +x d x ;(2)⎠⎛02(2-|1-x |)d x ;(3)∫π2-π2(sin x -cos x )d x .解 (1)⎠⎛14x -x 2x +x d x =⎠⎛14(x +x )(x -x )x +x d x =⎠⎛14(x -x )d x =⎝ ⎛⎭⎪⎫23x 32-12x 2⎪⎪⎪41=⎝ ⎛⎭⎪⎫23×432-12×42-⎝ ⎛⎭⎪⎫23-12=163-8-23+12=-176.(2)∵y =2-|1-x |=⎩⎪⎨⎪⎧1+x ,0≤x ≤1,3-x ,1<x ≤2.∴⎠⎛02(2-|1-x |)d x =⎠⎛01(1+x )d x +⎠⎛12(3-x )d x =⎝ ⎛⎭⎪⎫x +12x 2⎪⎪⎪10+⎝⎛⎭⎪⎫3x -12x 2⎪⎪⎪21=32+4-52=3. (3)∫π2-π2(sin x -cos x )d x =(-cos x -sin x )⎪⎪⎪π2-π2=-1-1=-2.11.f (x )是一次函数,且⎠⎛01f (x )d x =5,⎠⎛01xf (x )d x =176,求f (x )的解析式.解 设f (x )=ax +b (a ≠0), 由⎠⎛01f (x )d x =5,⎠⎛01xf (x )d x =176, 得⎠⎛01(ax +b )d x =(12ax 2+bx )⎪⎪⎪10=12a +b , ⎠⎛01x (ax +b )d x =(13ax 3+12bx 2)⎪⎪⎪ 10=13a +12b ,∴⎩⎪⎨⎪⎧12a +b =5,13a +b 2=176,解得⎩⎪⎨⎪⎧a =4,b =3.∴f (x )=4x +3.12.求f (a )=⎠⎛01(6x 2+4ax +a 2)d x 的最小值.解 f (a )=⎠⎛01(6x 2+4ax +a 2)d x=⎠⎛016x 2d x +⎠⎛014ax d x +⎠⎛01a 2d x=2x 3 ⎪⎪⎪ 10+2ax 2⎪⎪⎪ 10+a 2x⎪⎪⎪ 10 =2+2a +a 2 =(a +1)2+1.∴当a =-1时,f (a )的最小值为1. 13.设F (x )=⎠⎛0x (t 2+2t -8)d t .(1)求F (x )的单调区间; (2)求F (x )在[1,3]上的最值.解 F (x )=⎠⎛0x (t 2+2t -8)d t =⎝ ⎛⎭⎪⎫13t 3+t 2-8t ⎪⎪⎪x=13x 3+x 2-8x ,定义域是(0,+∞).(1)F ′(x )=x 2+2x -8=(x +4)(x -2), ∵当x <-4或x >2时,F ′(x )>0; 当-4<x <2时,F ′(x )<0.又∵x >0,∴函数的增区间为(2,+∞),减区间为(0,2). (2)令F ′(x )=0,得x =2(x =-4舍去).又F (1)=-203,F (2)=-283,F (3)=-6, ∴F (x )在[1,3]上的最大值为-6,最小值是-283.。
双基限时练(十三)1.由曲线y =f (x )(f (x )≤0),x ∈[a ,b ],x =a ,x =b 和x 轴围成的曲边梯形的面积S 等于( )A.⎠⎛ab f(x)d x , B .-⎠⎛ab f(x)d xC .⎠⎛ab [f (x )-a ]d x D.⎠⎛ab [f(x)-b]d x答案 B2.如图,阴影部分的面积为()A .⎠⎛ab f (x )d xB.⎠⎛a b g(x)d xC .⎠⎛ab [f (x )-g (x )]d x D. ⎠⎛ab [f(x)+g(x)]d x解析 阴影部分的面积 S =⎠⎛ab f(x)d x +|⎠⎛ab g(x)d x|=⎠⎛a b f(x)d x -⎠⎛ab g(x)d x=⎠⎛ab [f(x)-g(x)]d x.答案 C3.曲线y =x 3与直线y =x 所围成图形的面积等于( ) A .⎠⎛1-1(x -x 3)d x B.⎠⎛1-1(x 3-x)d xC .2⎠⎛01(x -x 3)d x D .2⎠⎛0-1(x -x 3)d x解析 由⎩⎪⎨⎪⎧y =x 3,y =x ,得交点A(-1,-1),B(0,0),C(1,1),如下图所示.∴阴影部分的面积为S =2⎠⎛01(x -x 3)d x.答案 C4.曲线y =cos x(0≤x ≤32π)与坐标轴所围成的面积为( ) A .2 B .3 C .52 D .4解析 利用函数y =cos x 在0≤x ≤3π2的图知,所求面积为S =3∫π2cos x d x =3(sin x)⎪⎪⎪π20=3.答案 B5.如图阴影部分面积为( )A . 2 3B . 9-2 3C .323D .353解析 S =⎠⎛1-3(3-x 2-2x)d x=(3x -13x 3-x 2)⎪⎪⎪ 1-3 =53+9=323. 答案 C6.f(x)=⎩⎨⎧x +1 (-1≤x<0),cos x (0≤x ≤π2)的图象与x 轴所围成的封闭图形的面积为( )A .32 B . 1 C . 2D .12解析 根据定积分的几何意义结合图形可得所求封闭图形的面积为S =12×1×1+∫π20cos x d x =12+sin x⎪⎪⎪ π20=32.答案 A7.曲线y =1x 与直线y =x ,x =2所围成图形的面积为________. 解析 示意图如图所示,所求面积为S =⎠⎛12(x -1x )d x =(12x 2-ln x)⎪⎪⎪ 21=32-ln 2. 答案 32-ln 28.设函数f(x)=3x 2+c ,若⎠⎛01f(x)d x =5,则实数c 的值为________.解析 ∵⎠⎛01f(x)d x =⎠⎛01(3x 2+c)d x=(x 3+cx)⎪⎪⎪ 10=1+c =5, ∴c =4. 答案 49.设a>0,若曲线y =x 与直线x =a ,y =0所围成封闭图形的面积为a 2,则a =________.解析 依题意得,由y =x 与直线x =a ,y =0所围成封闭图形的面积S =⎠⎛0a x d x =23x 32| a 0=23a 32=a 2,∴a =49.答案 4910.求正弦曲线y = sin x ,x ∈[0,3π2]和直线x =3π2及x 轴所围成的平面图形的面积.解 如图,当x ∈[0,π]时,曲线y = sin x 位于x 轴上方,而当x ∈[π,3π2]时,曲线位于x 轴下方,因此所求面积应为两部分面积之和.∴S =⎠⎛0π sin x d x +|∫3π2π sin x d x |=⎠⎛0π sin x d x -∫3π2π sin x d x=-cos x⎪⎪⎪ π0+cos x⎪⎪⎪ 32ππ =2+1=3.11.如图,直线y =kx 分抛物线y =x -x 2与x 轴所围成图形为面积相等的两部分,求k 的值.解 抛物线y =x -x 2与x 轴两交点的横坐标为x 1=0,x 2=1,抛物线与x 轴所围成的面积S =⎠⎛01(x -x 2)d x =⎝ ⎛⎭⎪⎫x22-x 33⎪⎪⎪1=16. 抛物线y =x -x 2与直线y =kx 两交点的横坐标为0和1-k ,∴12S =∫1-k 0(x -x 2-kx)d x =⎝ ⎛⎭⎪⎫x 22-x 33-k 2x 2⎪⎪⎪1-k 0=16(1-k)3=112.∴(1-k)3=12,k =1-312=1-342.12.求曲线y =x 2和直线x =0,x =1,y =t 2,t ∈(0,1)所围成图形(如图阴影部分)的面积的最小值.解 由定积分与微积分基本定理得S =S 1+S 2=⎠⎛0t (t 2-x 2)d x +⎠⎛t1(x 2-t 2)d x =(t 2x -13x 3)⎪⎪⎪t0+⎝⎛⎭⎪⎫13x 3-t 2x ⎪⎪⎪1t=t 3-13t 3+13-t 2-13t 3+t 3=43t 3-t 2+13,t ∈(0,1).S ′=4t 2-2t =2t(2t -1).当0<t<12时,S ′<0;当12<t<2时S ′>0, ∴当t =12时,S 有最小值S min =14.。
双基限时练(十二)1.下列各式中,正确的是( )A .⎠⎜⎛a bF ′(x )d x =F ′(b )-F ′(a )B.⎠⎜⎛a bF ′(x)d x =F ′(a)-F ′(b)C .⎠⎜⎛a bF ′(x )d x =F (b )-F (a )D.⎠⎜⎛abF ′(x)d x =F(a)-F(b)答案 C2.∫π20( sin x -cos x)d x =( )A .0B .1C .2D .π2解析 ∫π20(sin x -cos x)d x=∫π20sin x d x -∫π20cos x d x=(-cos x)⎪⎪⎪ π20-(sin x)⎪⎪⎪ π20 =1-1=0. 答案 A3.若∫a 1(2x +1x)d x =3+ln 2(a>1),则a 的值为( )A .6B .4C .3D .2解析 ∵⎠⎜⎛1a(2x +1x )d x =(x 2+ln x)⎪⎪⎪ a 1=a 2+ln a -1,又⎠⎜⎛1a(2x +1x )d x =3+ln 2, ∴a =2. 答案 D4.⎠⎛π-πcos x d x 等于( )A .2πB .πC .0D .1解析 ⎠⎛π-πcos x d x =sin x⎪⎪⎪ π-π=sin π-sin (-π)=0.答案 C5.设f(x)=⎩⎪⎨⎪⎧x 2 (0≤x<1),2-x (1<x ≤2),则⎠⎜⎛2f(x)d x 等于( )A .34B .45C .56D .不存在解析 ⎠⎜⎛02f(x)d x =⎠⎜⎛01x 2d x +⎠⎜⎛12(2-x)d x =13x 3⎪⎪⎪ 10+(2x -12x 2)⎪⎪⎪ 21=13+2-32=56. 答案 C6.由曲线y =x 2-1,直线x =0,x =2和x 轴围成的封闭图形的面积(如图阴影部分)是( )A .⎠⎜⎛02(x 2-1)d x B .|⎠⎜⎛02(x 2-1)d x | C.⎠⎜⎛02|x 2-1|d x D .⎠⎜⎛01(x 2-1)d x +⎠⎜⎛12(x 2-1)d x 答案 C7.若a =⎠⎜⎛02x 2d x ,b =⎠⎜⎛02x 3d x ,c =⎠⎜⎛02 sin x d x ,则a ,b ,c 的大小关系是________.解析 ∵a =⎠⎜⎛2x 2d x =13x 3⎪⎪⎪ 20=83,b =⎠⎜⎛2x 3d x =14x 4⎪⎪⎪ 20=4,c =⎠⎜⎛2sin x d x =(-cos x )⎪⎪⎪ 20=-cos2+1<2.∴b >a >c . 答案 b >a >c8.计算⎠⎛2-2( sin x +2)d x =________.解析 ⎠⎛2-2(sin x +2)d x =⎠⎛2-2sin x d x +⎠⎛2-22d x=(-cos x )⎪⎪⎪ 2-2+2x⎪⎪⎪ 2-2=-cos2+cos(-2)+2×2-2×(-2) =8. 答案 89.设函数f (x )=ax 2+c (a ≠0),若0≤x 0≤1.且⎠⎜⎛01f (x )d x =f (x 0),则x 0=________. 解析 ∵⎠⎜⎛01f (x )d x =⎠⎜⎛01(ax 2+c )d x =⎝ ⎛⎭⎪⎫a 3x 3+cx ⎪⎪ 1=a3+c ,又⎠⎜⎛1f (x )d x =f (x 0),∴ax 20+c =a3+c . ∵a ≠0,∴x 20=13,又0≤x 0≤1,∴x 0=33.答案 3310.计算下列定积分:(1)⎠⎜⎛14x -x 2x +x d x ;(2)⎠⎜⎛02(2-|1-x |)d x ; (3)∫π2-π2(sin x -cos x )d x .解 (1)⎠⎜⎛14x -x 2x +x d x =⎠⎜⎛14(x +x )(x -x )x +x d x = ⎠⎜⎛14(x -x )d x =⎝ ⎛⎭⎪⎫23x 32-12x 2⎪⎪⎪41=⎝ ⎛⎭⎪⎫23×432-12×42-⎝ ⎛⎭⎪⎫23-12=163-8-23+12=-176.(2)∵y =2-|1-x |=⎩⎪⎨⎪⎧1+x ,0≤x ≤1,3-x ,1<x ≤2.∴⎠⎜⎛02(2-|1-x |)d x =⎠⎜⎛01(1+x )d x +⎠⎜⎛12(3-x )d x =⎝ ⎛⎭⎪⎫x +12x 2⎪⎪⎪1+⎝⎛⎭⎪⎫3x -12x 2⎪⎪⎪21=32+4-52=3. (3)∫π2-π2(sin x -cos x )d x =(-cos x -sin x )⎪⎪⎪π2-π2=-1-1=-2.11.f (x )是一次函数,且⎠⎜⎛01f (x )d x =5,⎠⎜⎛01xf (x )d x =176,求f (x )的解析式.解 设f (x )=ax +b (a ≠0),由⎠⎜⎛01f (x )d x =5,⎠⎜⎛01xf (x )d x =176, 得⎠⎜⎛01(ax +b )d x =(12ax 2+bx )⎪⎪⎪10=12a +b ,⎠⎜⎛01x (ax +b )d x =(13ax 3+12bx 2)⎪⎪⎪ 10=13a +12b ,∴⎩⎪⎨⎪⎧12a +b =5,13a +b 2=176,解得⎩⎪⎨⎪⎧a =4,b =3.∴f (x )=4x +3.12.求f (a )=⎠⎜⎛01(6x 2+4ax +a 2)d x 的最小值. 解 f (a )=⎠⎜⎛01(6x 2+4ax +a 2)d x =⎠⎜⎛016x 2d x +⎠⎜⎛014ax d x +⎠⎜⎛01a 2d x =2x 3⎪⎪⎪ 10+2ax2⎪⎪⎪ 10+a 2x⎪⎪⎪ 10=2+2a +a 2 =(a +1)2+1.∴当a =-1时,f (a )的最小值为1.13.设F (x )=⎠⎜⎛0x(t 2+2t -8)d t . (1)求F (x )的单调区间; (2)求F (x )在[1,3]上的最值.解 F (x )=⎠⎜⎛0x(t 2+2t -8)d t =⎝ ⎛⎭⎪⎫13t 3+t 2-8t ⎪⎪⎪x0=13x 3+x 2-8x ,定义域是(0,+∞).(1)F ′(x )=x 2+2x -8=(x +4)(x -2), ∵当x <-4或x >2时,F ′(x )>0; 当-4<x <2时,F ′(x )<0.又∵x >0,∴函数的增区间为(2,+∞),减区间为(0,2). (2)令F ′(x )=0,得x =2(x =-4舍去). 又F (1)=-203,F (2)=-283,F (3)=-6,∴F (x )在[1,3]上的最大值为-6,最小值是-283.。
双基限时练(三)1.设f ′(x 0)=0,则曲线y =f (x )在点(x 0,f (x 0))处的切线( ) A .不存在B .与x 轴垂直C .与x 轴平行D .与x 轴平行或重合 答案 D2.一木块沿某一斜面自由下滑,测得下滑的水平距离s 与时间t 之间的函数关系为s =18t 2,则当t =2时,此木块在水平方向的瞬时速度为( )A. 2B. 1C.12 D .14解析 s ′=lim Δt →0ΔsΔt =lim Δt →018(t +Δt )2-18t 2Δt=lim Δt →014tΔt +18(Δt )2Δt=lim Δt →0(14t +18Δt )=14t . ∴当t =2时,s ′=12. 答案 C3.若曲线y =h (x )在点P (a ,h (a ))处切线方程为2x +y +1=0,则( )A .h ′(a )<0B .h ′(a )>0C .h ′(a )=0D .h ′(a )的符号不定 解析 由2x +y +1=0,得h ′(a )=-2<0. ∴h ′(a )<0. 答案 A4.曲线y =9x 在点(3,3)处的切线方程的倾斜角α等于 ( ) A .45°B .60°C .135°D .120°解析 k =y ′=lim Δx →0ΔyΔx =lim Δx →09x +Δx -9x Δx=lim Δx →-9x (x +Δx )=-9x 2.∴当x =3时,tan α=-1.∴α=135°. 答案 C5.在曲线y =x 2上切线倾斜角为π4的点是( ) A .(0,0) B .(2,4) C .(14,116) D .(12,14)解析 y ′=lim Δx →0ΔyΔx =lim Δx →0(x +Δx )2-x 2Δx=lim Δx →02xΔx +(Δx )2Δx=lim Δx →0(2x +Δx )=2x . 令2x =tan π4=1,∴x =12,y =14. 故所求的点是(12,14). 答案 D6.已知曲线y =2x 2上一点A (2,8),则过点A 的切线的斜率为________.解析 k =f ′(2)=lim Δx →02(2+Δx )2-2×22Δx=lim Δx →08Δx +2(Δx )2Δx=lim Δx →0(8+2Δx )=8. 答案 87.若函数f (x )在x 0处的切线的斜率为k ,则极限lim Δx →0f (x 0-Δx )-f (x 0)Δx=________.解析 lim Δx →0f (x 0-Δx )-f (x 0)Δx=-lim Δx →f (x 0-Δx )-f (x 0)-Δx =-k .答案 -k8.已知函数f (x )在区间[0,3]上图象如图所示,记k 1=f ′(1),k 2=f ′(2),k 3=f ′(3),则k 1,k 2,k 3之间的大小关系为________.(请用“>”连接)解析 由f (x )的图象及导数的几何意义知,k 1>k 2>k 3. 答案 k 1>k 2>k 39.已知曲线y =2x 2上的点(1,2),求过该点且与过该点的切线垂直的直线方程.解 ∵f ′(1)=lim Δx →0f (1+Δx )-f (1)Δx=4,∴过点(1,2)的切线的斜率为4.设过点(1,2)且与过该点的切线垂直的直线的斜率为k ,则4k =-1,k =-14.∴所求的直线方程为y -2=-14(x -1), 即x +4y -9=0.10.已知曲线y =1t -x 上两点P (2,-1),Q ⎝ ⎛⎭⎪⎫-1,12.求:(1)曲线在点P 处、点Q 处的切线的斜率; (2)曲线在点P ,Q 处的切线方程.解 将P (2,-1)代入y =1t -x 得t =1,∴y =11-x.∴y ′=lim Δx →0f (x +Δx )-f (x )Δx =lim Δx →011-(x +Δx )-11-x Δx=lim Δx →01[1-(x +Δx )](1-x )=1(1-x )2.(1)曲线在点P 处的切线的斜率为y ′|x =2=1(1-2)2=1;曲线在点Q 处的切线的斜率为y ′|x =-1=1[1-(-1)]2=14.(2)曲线在点P 处的切线方程为 y -(-1)=x -2,即x -y -3=0. 曲线在点Q 处的切线方程为 y -12=14(x +1),即x -4y +3=0.11.已知点M (0,-1),F (0,1),过点M 的直线l 与曲线y =13x 3-4x +4在x =2处的切线平行.(1)求直线l 的方程;(2)求以点F 为焦点,l 为准线的抛物线C 的方程. 解 (1)∵f ′(2)=lim Δx →013(2+Δx )3-4(2+Δx )+4-⎝ ⎛⎭⎪⎫13×23-4×2+4Δx=0, ∴直线l 的斜率为0,其直线方程为y =-1.(2)∵抛物线以点F (0,1)为焦点,y =-1为准线,∴设抛物线的方程为x 2=2py ,则-p2=-1,p =2.故抛物线C 的方程为x 2=4y .12.已知曲线y =x 2+1,问是否存在实数a ,使得经过点(1,a )能作出该曲线的两条切线?若存在,求出实数a 的取值范围;若不存在,请说明理由.解 存在. 理由如下: ∵y =x 2+1,∴y ′=lim Δx →0ΔyΔx =lim Δx →0(x +Δx )2+1-(x 2+1)Δx= lim Δx →02xΔx +(Δx )2Δx=2x . 设切点坐标为(t ,t 2+1),∵y ′=2x ,∴切线的斜率为k =y ′|x =t =2t . 于是可得切线方程为y -(t 2+1)=2t (x -t ). 将(1,a )代入,得a -(t 2+1)=2t (1-t ), 即t 2-2t +a -1=0.∵切线有两条,∴方程有两个不同的解.故Δ=4-4(a -1)>0.∴a <2.故存在实数a ,使得经过点(1,a )能作出该曲线的两条切线,a 的取值范围是(-∞,2).。
【名师一号】2014-2015学年高中数学 第一章 导数及其应用单元综合测试 新人教版选修2-2(时间:120分钟,满分:150分)一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.设函数y =f (x )在(a ,b )上可导,则f (x )在(a ,b )上为增函数是f ′(x )>0的( ) A .必要不充分条件 B .充分不必要条件 C .充分必要条件 D .既不充分也不必要条件解析 y =f (x )在(a ,b )上f ′(x )>0⇒y =f (x )在(a ,b )上是增函数,反之,y =f (x )在(a ,b )上是增函数⇒f ′(x )≥0⇒f ′(x )>0.答案 A2.若曲线y =f (x )在点(x 0,f (x 0))处的切线方程是2x +y -1=0,则( ) A .f ′(x 0)>0 B .f ′(x 0)<0 C .f ′(x 0)=0D .f ′(x 0)不存在解析 曲线y =f (x )在点(x 0,f (x 0))处的切线的斜率为f ′(x 0)=-2<0. 答案 B3.曲线y =13x 3-2在点(-1,-53)处切线的倾斜角为( )A .30°B .45°C .135°D .150°解析 ∵y ′=x 2,k =tan α=y ′|x =-1=(-1)2=1, ∴α=45°. 答案 B4.曲线f (x )=x 3+x -2的一条切线平行于直线y =4x -1,则切点P 0的坐标为( ) A .(0,-1)或(1,0) B .(1,0)或(-1,-4) C .(-1,-4)或(0,-2)D .(1,0)或(2,8)解析 设P 0(x 0,y 0),则f ′(x 0)=3x 20+1=4, ∴x 20=1,∴x 0=1,或x 0=-1. ∴P 0的坐标为(1,0)或(-1,-4). 答案 B5.下列函数中,在(0,+∞)上为增函数的是( ) A .y =sin 2xB .y =x 3-xC .y =x e xD .y =-x +ln(1+x )解析 对于C ,有y ′=(x e x)′=e x+x e x=e x(x +1)>0. 答案 C6.下列积分值为2的是( ) A.⎠⎛05(2x -4)d xB .⎠⎛0πcos x d xC .⎠⎛131xd xD .⎠⎛0πsin x d x解析 ⎠⎛0πsin x d x =-cos x ⎪⎪⎪π0=-cos π+cos 0=2.答案 D7.函数f(x)在其定义域内可导,y =f(x)的图象如右图所示,则导函数y =f′(x)的图象为( )解析 由y =f(x)的图象知,有两个极值点,则y =f′(x)的图象与x 轴应有两个交点,又由增减性知,应选D 项.答案 D8.已知函数f(x)=x 3-3x 2-9x ,x∈(-2,2),则f(x)有( )A .极大值5,极小值为-27B .极大值5,极小值为-11C .极大值5,无极小值D .极小值-27,无极大值解析 f′(x)=3x 2-6x -9 =3(x +1)(x -3). 当x<-1时,f′(x)>0, 当-1<x<3时,f′(x)<0. ∴x=-1是f(x)的极大值点.且极大值为f(-1)=5,在(-2,2)内无极小值. 答案 C9.已知f(x)为三次函数,当x =1时f(x)有极大值4,当x =3时f(x)有极小值0,且函数f(x)过原点,则此函数是( )A .f(x)=x 3-2x 2+3xB .f(x)=x 3-6x 2+xC .f(x)=x 3+6x 2+9xD .f(x)=x 3-6x 2+9x解析 设f(x)=ax 3+bx 2+cx(a≠0),则f′(x)=3ax 2+2bx +c =3a(x -1)(x -3)=3ax 2-12ax +9a. 由题意得⎩⎪⎨⎪⎧=a +b +c =4,=27a +9b +3c =0,c =9a.解得a =1,b =-6,c =9. 所以f(x)=x 3-6x 2+9x. 答案 D10.由抛物线y =x 2-x ,直线x =-1及x 轴围成的图形的面积为( )A .23B .1C .43D .53解析 如图所示,阴影部分的面积为S 1=⎠⎛0-1(x 2-x)d x=(13x 3-12x 2)⎪⎪⎪ 0-1=56. S 2=⎪⎪⎪⎠⎛01(x 2-x)d x⎪⎪⎪ =-(13x 3-12x 2)⎪⎪⎪ 10=16, 故所求的面积为S =S 1+S 2=1. 答案 B11.函数f(x)=ax 3+bx 2+cx 在x =1a处有极值,则ac +2b 的值为( )A .-3B .0C .1D .3解析 f′(x)=3ax 2+2bx +c , 依题意知,3a×(1a )2+2b×1a +c =0,即3a +2ba +c =0, ∴2b+ac =-3. 答案 A12.设函数f(x)满足x 2f′(x)+2xf(x)=e x x ,f(2)=e 28,则x>0时, f(x)( )A .有极大值,无极小值B .有极小值,无极大值C .既有极大值又有极小值D .既无极大值也无极小值解析 由题意知,f′(x)=e xx3-x=e x -2x 2x3.令g(x)=e x -2x 2f(x),则g′(x)=e x-2x 2f′(x)-4xf(x)=e x-2[x 2f′(x)+2xf(x)]=e x-2e xx =e x ⎝ ⎛⎭⎪⎫1-2x .由g′(x)=0,得x =2.当x =2时,g(x)有极小值g(2)=e 2-2×22f(2)=e 2-8·e 28=0.∴g(x)≥0.当x>0时,f′(x)=x3≥0,故f(x)在(0,+∞)上单调递增,∴f(x)既无极大值也无极小值.答案 D二、填空题(本大题共4小题,每小题5分,共20分.把答案填在题中的横线上) 13.函数f(x)在R 上可导,且f ′(0)=2.∀x ,y ∈R ,若函数f (x +y )=f (x )f (y )成立,则f (0)=________.解析 令y =0,则有f (x )=f (x )f (0). ∵f ′(0)=2,∴f (x )不恒为0,∴f (0)=1. 答案 114.解析答案π2-1 15.若函数f(x)=13x 3-f′(1)·x 2+2x +5,则f′(2)=________.解析 ∵f′(x)=x 2-2f′(1)x+2, ∴f′(1)=1-2f′(1)+2. ∴f′(1)=1. ∴f′(x)=x 2-2x +2. ∴f′(2)=22-2×2+2=2. 答案 216.一物体以初速度v =9.8t +6.5米/秒的速度自由落下,且下落后第二个4 s 内经过的路程是________.解析 ⎠⎛48(9.8t +6.5)d t =(4.9t 2+6.5t)⎪⎪⎪84=4.9×64+6.5×8-4.9×16-6.5×4 =313.6+52-78.4-26 =261.2. 答案 261.2米三、解答题(本大题共6个小题,共70分.解答应写出文字说明、证明过程或演算步骤) 17.(10分)已知函数f(x)=13x 3-4x +m 在区间(-∞,+∞)上有极大值283.(1)求实数m 的值;(2)求函数f(x)在区间(-∞,+∞)的极小值. 解 f′(x)=x 2-4=(x +2)(x -2). 令f′(x)=0,得x =-2,或x =2. 故f(x)的增区间(-∞,-2)和(2,+∞), 减区间为(-2,2).(1)当x =-2,f(x)取得极大值, 故f(-2)=-83+8+m =283,∴m=4.(2)由(1)得f(x)=13x 3-4x +4,又当x =2时,f(x)有极小值f(2)=-43.18.(12分)用总长为14.8米的钢条制成一个长方体容器的框架,如果所制的容器的底面的长比宽多0.5米,那么高为多少时容器的容器最大?并求出它的最大容积.解 设容器底面宽为x m ,则长为(x +0.5)m ,高为(3.2-2x)m .由⎩⎪⎨⎪⎧3.2-2x>0,x>0,解得0<x<1.6,设容器的容积为y m 3,则有y =x(x +0.5)(3.2-2x)=-2x 3+2.2x 2+1.6x , y′=-6x 2+4.4x +1.6,令y′=0,即-6x 2+4.4x +1.6=0,解得x =1,或x =-415(舍去).∵在定义域(0,1.6)内只有一个点x =1使y′=0,且x =1是极大值点, ∴当x =1时,y 取得最大值为1.8. 此时容器的高为3.2-2=1.2 m .因此,容器高为1.2 m 时容器的容积最大,最大容积为1.8 m 3. 19.(12分)设函数f(x)=2x 3-3(a +1)x 2+6ax(a ∈R ). (1)当a =1时,求证:f (x )为R 上的单调递增函数; (2)当x ∈[1,3]时,若f (x )的最小值为4,求实数a 的值.解 (1)证明:当a =1时,f (x )=2x 3-6x 2+6x ,则f ′(x )=6x 2-12x +6=6(x -1)2≥0, ∴f (x )为R 上的单调增函数.(2)f ′(x )=6x 2-6(a +1)x +6a =6(x -1)(x -a ).①当a ≤1时,f (x )在区间[1,3]上是单调增函数,此时在[1,3]上的最小值为f (1)=3a -1,∴3a -1=4,∴a =53>1(舍去);②当1<a <3时,f (x )在(1,a )上是减函数,在区间(a,3)上是增函数,故在[1,3]上的最小值为f (a )=2a 3-3(a +1)a 2+6a 2=4.化简得(a +1)(a -2)2=0,∴a =-1<1(舍去),或a =2;③当a ≥3时,f (x )在区间(1,a )上是减函数,故f (3)为最小值, ∴54-27(a +1)+18a =4, 解得a =229<3(舍去).综上可知,a =2.20.(12分)设函数f (x )=a3x 3+bx 2+cx +d (a >0),且方程f ′(x )-9x =0的两根分别为1,4.(1)当a =3,且曲线y =f (x )过原点时,求f (x )的解析式; (2)若f (x )在(-∞,+∞)内无极值点,求a 的取值范围. 解 由f (x )=a3x 3+bx 2+cx +d ,得f ′(x )=ax 2+2bx +c ,∵f ′(x )-9x =ax 2+2bx +c -9x =0的两根分别为1,4,∴⎩⎪⎨⎪⎧a +2b +c -9=0,16a +8b +c -36=0,(*)(1)当a =3时,由(*)得⎩⎪⎨⎪⎧2b +c -6=0,8b +c +12=0,解得b =-3,c =12.又∵曲线y =f (x )过原点,∴d =0. 故f (x )=x 3-3x 2+12x .(2)由于a >0,所以“f (x )=a3x 3+bx 2+cx +d 在(-∞,+∞)内无极值点”,等价于“f ′(x )=ax 2+2bx +c ≥0在(-∞,+∞)内恒成立”.由(*)式得2b =9-5a ,c =4a . 又Δ=(2b )2-4ac =9(a -1)(a -9),解⎩⎪⎨⎪⎧a >0,Δ=a -a -,得a ∈[1,9],即a 的取值范围是[1,9].21.(12分)已知函数f (x )=ax 3+bx 2的图象经过点M (1,4),曲线在点M 处的切线恰好与直线x +9y =0垂直.(1)求实数a ,b 的值;(2)若函数f (x )在区间[m ,m +1]上单调递增,求m 的取值范围. 解 (1)∵f (x )=ax 3+bx 2的图象经过点M (1,4), ∴a +b =4.①又f ′(x )=3ax 2+2bx ,则f ′(1)=3a +2b ,由条件f ′(1)(-19)=-1,得3a +2b =9.②由①,②解得a =1,b =3.(2)f (x )=x 3+3x 2,f ′(x )=3x 2+6x , 令f ′(x )=3x 2+6x ≥0,得x ≥0,或x ≤-2, 若函数f (x )在区间[m ,m +1]上单调递增,则 [m ,m +1]⊆(-∞,-2]∪[0,+∞), ∴m ≥0,或m +1≤-2,即m ≥0,或m ≤-3, ∴m 的取值范围是(-∞,-3]∪[0,+∞). 22.(12分)已知函数f (x )=(x +1)ln x -x +1. (1)若xf ′(x )≤x 2+ax +1,求a 的取值范围; (2)证明:(x -1)f (x )≥0. 解 (1)f ′(x )=x +1x +ln x -1=ln x +1x,xf ′(x )=x ln x +1,题设xf ′(x )≤x 2+ax +1等价于ln x -x ≤a . 令g (x )=ln x -x ,则g ′(x )=1x-1.当0<x <1时,g ′(x )>0; 当x ≥1时,g ′(x )≤0,x =1是g (x )的最大值点, g (x )≤g (1)=-1.综上,a 的取值范围是[-1,+∞). (2)由(1)知,g (x )≤g (1)=-1, 即g (x )+1≤0,即ln x -x +1≤0, 当0<x <1时,f (x )=(x +1)ln x -x +1=x ln x +(ln x -x +1)≤0; 当x ≥1时,f (x )=ln x +(x ln x -x +1)=ln x +x (ln x +1x-1)=ln x -x (ln 1x -1x+1)≥0.所以(x -1)f (x )≥0.。
双基限时练(十二)
1.下列各式中,正确的是( ) A .⎠⎛a
b F ′(x )d x =F ′(b )-F ′(a )
B.⎠⎛a b F ′(x)d x =F ′(a)-F ′(b)
C .⎠⎛a
b F ′(x )d x =F (b )-F (a ) D.⎠⎛a
b F ′(x)d x =F(a)-F(b)
答案 C
2.∫π
20( sin x -cos x)d x =( ) A .0 B .1 C .2 D .π
2
解析 ∫π
20(sin x -cos x)d x =∫π20sin x d x -∫π
20cos x d x =(-cos x) ⎪⎪⎪ π20-(sin x)
⎪⎪⎪ π20 =1-1=0. 答案 A
3.若∫a 1(2x +1
x )d x =3+ln 2(a>1),则a 的值为( ) A .6 B .4 C .3 D .2 解析 ∵⎠
⎛1
a (2x +1
x )d x
=(x 2
+ln x)
⎪
⎪⎪ a 1=a 2+ln a -1, 又⎠⎛1
a (2x +1
x )d x =3+ln 2,
∴a =2. 答案 D
4.⎠
⎛π-πcos x d x 等于( )
A .2π
B .π
C .0
D .1
解析 ⎠
⎛π-πcos x d x =sin x
⎪
⎪⎪ π-π=sinπ-sin (-π)=0. 答案 C
5.设f(x)=⎩
⎪⎨⎪⎧
x 2
(0≤x<1),
2-x (1<x ≤2),则⎠⎛02f(x)d x 等于( )
A .34
B .4
5 C .5
6 D .不存在
解析 ⎠⎛0
2f(x)d x =⎠⎛0
1x 2d x +⎠⎛1
2(2-x)d x
=13x 3 ⎪⎪⎪ 10+(2x -12x 2)
⎪⎪⎪ 2
1 =13+2-32=5
6. 答案 C
6.由曲线y =x 2-1,直线x =0,x =2和x 轴围成的封闭图形的面积(如图阴影部分)是( )
A . ⎠⎛0
2(x 2-1)d x
B .|⎠⎛0
2(x 2-1)d x |
C.⎠⎛0
2|x 2-1|d x
D .⎠⎛0
1(x 2-1)d x +⎠⎛1
2(x 2-1)d x
答案 C
7.若a =⎠⎛0
2x 2d x ,b =⎠⎛0
2x 3d x ,c =⎠⎛0
2 sin x d x ,则a ,b ,c 的大小关
系是________.
解析 ∵a =⎠⎛0
2x 2
d x =13x 3
⎪⎪⎪ 20=8
3, b =⎠⎛0
2x 3
d x =14x
4
⎪⎪⎪ 2
0=4, c =⎠⎛0
2 sin x d x =(-cos x )
⎪
⎪⎪
20=-cos2+1<2. ∴b >a >c . 答案 b >a >c
8.计算⎠
⎛2-2( sin x +2)d x =________.
解析 ⎠⎛2-2(sin x +2)d x =⎠⎛2-2sin x d x +⎠
⎛2-22d x
=(-cos x ) ⎪
⎪⎪ 2-2+2x
⎪
⎪⎪ 2
-2 =-cos2+cos(-2)+2×2-2×(-2) =8. 答案 8
9.设函数f (x )=ax 2+c (a ≠0),若0≤x 0≤1.且
⎠⎛0
1
f (x )d x =f (x 0),则x 0=________.
解析 ∵⎠⎛01
f (x )d x =⎠⎛0
1
(ax 2
+c )d x =⎝ ⎛⎭⎪⎫a 3x 3+cx ⎪⎪
10=a 3+c , 又⎠⎛
1
f (x )d x =f (x 0),∴ax 2
0+c =a 3
+c .
∵a ≠0,∴x 2
0=13, 又0≤x 0≤1,∴x 0=33. 答案 3
3
10.计算下列定积分:
(1)⎠
⎛1
4x -x 2x +x d x ;
(2)⎠⎛0
2(2-|1-x |)d x ;
(3)∫π2-π
2(sin x -cos x )d x .
解 (1)⎠⎛1
4x -x 2x +x d x =⎠
⎛1
4
(x +x )(x -x )x +x d x =
⎠⎛1
4
(x -x )d x =⎝ ⎛⎭⎪⎫23x 32-12x 2⎪⎪
⎪
4
1=
⎝ ⎛⎭⎪⎫23×432-12×42-⎝ ⎛⎭⎪⎫23-12=163
-8-23+12=-176. (2)∵y =2-|1-x |=⎩
⎪⎨⎪⎧
1+x ,0≤x ≤1,3-x ,1<x ≤2.
∴⎠⎛0
2(2-|1-x |)d x =⎠⎛0
1(1+x )d x +⎠⎛1
2(3-x )d x =⎝
⎛
⎭⎪⎫x +12x 2⎪
⎪⎪
10+
⎝
⎛
⎭⎪⎫3x -12x 2⎪⎪
⎪
2
1=32
+4-52
=3. (3)∫π2-π
2(sin x -cos x )d x =(-cos x -sin x )⎪
⎪
⎪
π2-π
2=-1-1=-2.
11.f (x )是一次函数,且⎠⎛0
1f (x )d x =5,⎠⎛0
1xf (x )d x =17
6,求f (x )的解析
式.
解 设f (x )=ax +b (a ≠0), 由⎠⎛0
1
f (x )d x =5,⎠
⎛0
1xf (x )d x =17
6, 得⎠
⎛01(ax +b )d x =(12ax 2
+bx )⎪⎪
⎪
1
0=12
a +
b , ⎠
⎛0
1
x (ax +b )d x =(13ax 3+12bx 2)
⎪⎪⎪ 10=13a +1
2b , ∴⎩⎪⎨⎪⎧
12a +b =5,
13a +b 2=176,解得⎩
⎪⎨⎪⎧
a =4,
b =3.
∴f (x )=4x +3.
12.求f (a )=⎠⎛0
1(6x 2+4ax +a 2)d x 的最小值.
解 f (a )=⎠⎛0
1(6x 2+4ax +a 2)d x
=⎠⎛0
16x 2d x +⎠⎛0
14ax d x +⎠⎛0
1a 2d x
=2x
3
⎪
⎪⎪ 10+2ax 2
⎪
⎪⎪ 10+a 2x
⎪
⎪⎪ 10 =2+2a +a 2 =(a +1)2+1.
∴当a =-1时,f (a )的最小值为1. 13.设F (x )=⎠⎛0
x (t 2+2t -8)d t .
(1)求F (x )的单调区间; (2)求F (x )在[1,3]上的最值.
解 F (x )=⎠⎛0
x (t 2
+2t -8)d t =⎝ ⎛⎭⎪⎫13t 3+t 2-8t ⎪⎪
⎪
x
=13
x 3+x 2-8x ,定义域是(0,+∞).
(1)F ′(x )=x 2+2x -8=(x +4)(x -2), ∵当x <-4或x >2时,F ′(x )>0; 当-4<x <2时,F ′(x )<0.
又∵x >0,∴函数的增区间为(2,+∞),减区间为(0,2). (2)令F ′(x )=0,得x =2(x =-4舍去). 又F (1)=-203,F (2)=-28
3,F (3)=-6, ∴F (x )在[1,3]上的最大值为-6,最小值是-28
3.。