2021高考数学一轮复习课时作业13变化率与导数导数的计算理(含答案及解析)
高考数学一轮复习:
课时作业13 变化率与导数、导数的计算
[基础达标]
一、选择题
1.函数f (x )=(x +2a )(x -a )2
的导数为( ) A .2(x 2
-a 2
) B .2(x 2
+a 2
) C .3(x 2
-a 2
) D .3(x 2
+a 2
)
解析:∵f (x )=(x +2a )(x -a )2
=x 3
-3a 2
x +2a 3
, ∴f ′(x )=3(x 2
-a 2
). 答案:C
2.[2020·河南南阳月考]已知函数f (x )的导函数为f ′(x ),且满足f (x )=2xf ′(e)+ln x ,则f (e)=( )
A .e
B .-1
e
C .-1
D .-e
解析:由f (x )=2xf ′(e)+ln x ,得f ′(x )=2f ′(e)+1x ,则f ′(e)=2f ′(e)+1
e ,
所以f ′(e)=-1e ,故f (x )=-2
e
x +ln x ,所以f (e)=-1.故选C 项.
答案:C
3.[2020·山西太原模拟]已知函数f (x )=x ln x +a 的图象在点(1,f (1))处的切线经过原点,则实数a 的值为( )
A .1
B .0 C.1
e
D .-1 解析:∵f (x )=x ln x +a ,∴f ′(x )=ln x +1,∴f ′(1)=1,f (1)=a ,∴切线方程为y =x -1+a ,∴0=0-1+a ,解得a =1.故选A 项.
答案:A
4.[2020·河北示范性高中联考]已知函数f (x )是定义在R 上的奇函数,且当x <0时,
f (x )=
1-2ln -x
x
,则曲线y =f (x )在点(1,f (1))处的切线方程为( )
A .3x +y -4=0
B .3x +y +4=0
C .3x -y -2=0
D .3x -y -4=0
解析:若x >0,则-x <0,所以f (-x )=1-2ln x
-x .又函数f (x )是定义在R 上的奇函数,
所以f (x )=-f (-x )=1-2ln x x ,此时f ′(x )=2ln x -3
x
2
,f ′(1)=-3,f (1)=1,所以切线方程为y -1=-3(x -1),即3x +y -4=0.故选A 项.
答案:A
5.[2020·河南新乡模拟]若f (x )=a -2+a sin(2x )为奇函数,则曲线y =f (x )在x =0处的切线的斜率为( )
A .-2
B .-4
C .2
D .4
解析:∵f (x )是奇函数,∴a -2=0,得a =2,∴f (x )=2sin(2x ),f ′(x )=4cos(2x ),∴f ′(0)=4.∴曲线y =f (x )在x =0处的切线的斜率为4.故选D 项.
答案:D 二、填空题
6.[2019·全国卷Ⅰ]曲线y =3(x 2
+x )e x
在点(0,0)处的切线方程为________. 解析:∵y ′=3(x 2
+3x +1)e x
,
∴曲线在点(0,0)处的切线斜率k =y ′|x =0=3, ∴曲线在点(0,0)处的切线方程为y =3x . 答案:y =3x
7.[2020·天津十二重点中学联考]已知函数f (x )=(x 2
-a )ln x ,f ′(x )是函数f (x )的导函数,若f ′(1)=-2,则a 的值为________.
解析:∵f (x )=(x 2
-a )ln x (x >0),∴f ′(x )=2x ln x +x 2-a
x
,∴f ′(1)=1-a =-2,
得a =3.
答案:3
8.[2020·湖南湘东六校联考]已知曲线f (x )=e x
+x 2
,则曲线y =f (x )在(0,f (0))处的切线与坐标轴围成的图形的面积为________.
解析:由题意,得f ′(x )=e x
+2x ,所以f ′(0)=1.又f (0)=1,所以曲线y =f (x )在(0,f (0))处的切线方程为y -1=1×(x -0),即x -y +1=0,所以该切线与x ,y 轴的交点坐标分别为(-1,0),(0,1),所以该切线与坐标轴围成的图形的面积为12×1×1=1
2
.
答案:1
2
三、解答题
9.求下列函数的导数:
(1)y =(3x 3
-4x )(2x +1); (2)y =
x +cos x
x +sin x
;
(3)y =ln 2x +3x 2+1
.
解析:(1)解法一:因为y =(3x 3
-4x )(2x +1)=6x 4
+3x 3
-8x 2
-4x ,所以y ′=24x 3
+9x 2
-16x -4.
解法二:y ′=(3x 3
-4x )′(2x +1)+(3x 3
-4x )(2x +1)′=(9x 2
-4)(2x +1)+(3x 3
-4x )·2=24x 3
+9x 2
-16x -4.
(2)y ′=x +cos x ′x +sin x -x +cos x
x +sin x ′
x +sin x 2
=1-sin x x +sin x -x +cos x
1+cos x
x +sin x 2
=
-x cos x -x sin x +sin x -cos x -1
x +sin x 2
.
(3)y ′=
[ln
2x +3]′
x 2+1-ln 2x +3
x 2+1′
x 2+12
=2x +3′
2x +3
·
x 2+1-2x ln 2x +3x 2+1
2
=2
x 2+1-2x 2x +3ln 2x +3
2x +3x 2+12
.
10.已知函数f (x )=x 3
-4x 2
+5x -4. (1)求曲线f (x )在点(2,f (2))处的切线方程; (2)求经过点A (2,-2)的曲线f (x )的切线方程.
解析:(1)因为f ′(x )=3x 2
-8x +5,所以f ′(2)=1,又f (2)=-2, 所以曲线在点(2,f (2))处的切线方程为y +2=x -2, 即x -y -4=0.
(2)设曲线与经过点A (2,-2)的切线相切于点P (x 0,x 3
0-4x 2
0+5x 0-4),因为f ′(x 0)=3x 2
0-8x 0+5,
所以切线方程为y -(-2)=(3x 2
0-8x 0+5)(x -2), 又切线过点P (x 0,x 3
0-4x 2
0+5x 0-4),
所以x 3
0-4x 2
0+5x 0-2=(3x 2
0-8x 0+5)(x 0-2), 整理得(x 0-2)2
(x 0-1)=0,解得x 0=2或1,
所以经过A (2,-2)的曲线f (x )的切线方程为x -y -4=0或y +2=0.
[能力挑战]
11.[2020·江西南昌模拟]已知f (x )在R 上连续可导,f ′(x )为其导函数,且f (x )=e x +e -x -xf ′(1)·(e x -e -x
),则f ′(2)+f ′(-2)-f ′(0)f ′(1)=( )
A .4e 2
+4e -2
B .4e 2
-4e -2
C .0
D .4e 2
解析:函数f (-x )=e -x
+e x -(-x )f ′(1)·(e -x -e x
)=f (x ),即函数f (x )是偶函数,两边对x 求导数,得-f ′(-x )=f ′(x ).即f ′(-x )=-f ′(x ),则f ′(x )是R 上的奇函数,则f ′(0)=0,f ′(-2)=-f ′(2),即f ′(2)+f ′(-2)=0,则f ′(2)+f ′(-2)-f ′(0)f ′(1)=0.故选C 项.
答案:C
12.[2020·河北保定乐凯中学模拟]设函数f (x )=g (x )+x 2
,曲线y =g (x )在点(1,g (1))处的切线方程为y =2x +1,则曲线y =f (x )在点(1,f (1))处的切线的斜率为( )
A .2 B.14
C .4
D .-1
2
解析:因为曲线y =g (x )在点(1,g (1))处的切线方程为y =2x +1,所以g ′(1)=2.又
f ′(x )=
g ′(x )+2x ,故曲线y =f (x )在点(1,f (1))处的切线的斜率为f ′(1)=g ′(1)+2
=4.故选C 项.
答案:C
13.[2020·四川绵阳月考]过点A (2,1)作曲线f (x )=x 3
-3x 的切线,最多有( ) A .3条 B .2条 C .1条 D .0条
解析:设切点为P (x 0,x 3
0-3x 0).易知f ′(x 0)=3x 2
0-3,则切线方程为y -x 3
0+3x 0=(3x 2
-3)(x -x 0),代入(2,1)得,2x 3
0-6x 2
0+7=0.令y =2x 3
0-6x 2
0+7,则y ′=6x 2
0-12x 0.由y ′=0,得x 0=0或x 0=2,且当x 0=0时,y =7>0,x 0=2时,y =-1<0,所以方程2x 3
0-6x 2
0+7=0有3个解,则过点A (2,1)作曲线f (x )=x 3
-3x 的切线的条数是3条.故选A 项.
答案:A