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079025.com:2017-2018学年高中数学人教A版选修2-2练习:第1章 导数及其应用1.2.2 Word版含解析

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第一章 1.2 1.2.2

一、选择题

A 级 基础巩固

1.函数 y=(x+1)2(x-1)在 x=1 处的导数等于 导学号 84624138 ( D )

A.1

B.2

C.3

D.4

[解析] y′=[(x+1)2]′(x-1)+(x+1)2(x-1)′

=2(x+1)·(x-1)+(x+1)2=3x2+2x-1,

∴y′|x=1=4.

2.曲线 y=ln(x+2)在点 P(-1,0)处的切线方程是 导学号 84624139 ( A )

A.y=x+1 C.y=2x+1

B.y=-x+1 D.y=-2x+1

[解析] ∵y=ln(x+2),∴y′=x+1 2,

∴切线斜率 k=y′|x=-1=1, ∴切线方程为 y-0=1×(x+1),即 y=x+1. 3.设曲线 y=xn+1(n∈N*)在点(1,1)处的切线与 x 轴的交点的横坐标为 xn,则 x1·x2·…·xn

的值为 导学号 84624140 ( B )

A.1n

B.n+1 1

C.n+n 1

D.1

[解析] 对 y=xn+1(n∈N*)求导得 y′=(n+1)xn,令 x=1 得在点(1,1)处的切线的斜率 k

=n+1,在点(1,1)处的切线方程为 y-1=(n+1)(xn-1). 令 y=0,得 xn=n+n 1.

则 x1·x2·…·xn=12×23×34×…×n-n 1×n+n 1=n+1 1,故选 B.

4.(2016·泉州高二检测)若 f(x)=sinπ3-cosx,则 f ′(α)等于 导学号 84624141 ( A )

A.sinα

B.cosα

C.sinπ3+cosα

D.cos3π+sinα

[解析] ∵f(x)=sinπ3-cosx, ∴f ′(x)=sinx, ∴f ′(α)=sinα,故选 A. 5.设函数 f(x)=xm+ax 的导数为 f ′(x)=2x+1,则数列{f?1n?}(n∈N*)的前 n 项和是

导学号 84624142 ( A )

A.n+n 1

B.nn+ +21

C.n-n 1

D.n+n 1

[解析] ∵f(x)=xm+ax 的导数为 f ′(x)=2x+1,

∴m=2,a=1,∴f(x)=x2+x,

∴f(n)=n2+n=n(n+1),

∴数列{f?1n?}(n∈N*)的前 n 项和为:

Sn=1×1 2+2×1 3+3×1 4+…+n?n1+1?

=??1-12??+??12-13??+…+??1n-n+1 1??

=1-n+1 1=n+n 1,故选 A.

6.(2016·邯郸高二检测)已知二次函数 f(x)的图象如图所示,则其导函数 f ′(x)的图象大

致形状是 导学号 84624143 ( B )

[解析] 依题意可设 f(x)=ax2+c(a<0,且 c>0),于是 f ′(x)=2ax,显然 f ′(x)的图象 为直线,过原点,且斜率 2a<0,故选 B.
二、填空题 7.(2016·衡水中学高二检测)已知 f(x)=x3+ax-2b,如果 f(x)的图象在切点 P(1,-2) 处的切线与圆(x-2)2+(y+4)2=5 相切,那么 3a+2b=__-7__. 导学号 84624144 [解析] 由题意得 f(1)=-2?a-2b=-3,

又∵f ′(x)=3x2+a,∴f ′(1)=3+a,

∴f(x)的图象在点(1,-2)处的切线方程为 y+2=(3+a)(x-1),即(3+a)x-y-a-5=0,

∴|?3+a?×2+4-a-5|= ?3+a?2+1

5?a=-52,∴b=14,

∴3a+2b=-7. 8.(2016·全国卷Ⅲ文,16)已知 f(x)为偶函数,当 x≤0 时,f(x)=e-x-1-x,则曲线 y=f(x)

在点(1,2)处的切线方程是__y=2x__. 导学号 84624145 [解析] 当 x>0 时,-x<0,则 f(-x)=ex-1+x.又 f(x)为偶函数,所以 f(x)=f(-x)=eex+x,
所以当 x>0 时, f ′(x)=ex-1+1,则曲线 y=f(x)在点(1,2)处的切线的斜率为 f ′(1)=2,所以切线方程
为 y-2=2(x-1),即 y=2x. 三、解答题

9.求下列函数的导数: 导学号 84624146

(1)y=x(x2+1x+x13);(2)y=(

x+1)( 1 -1); x

(3)y=sin44x+cos44x;(4)y=11+ -

xx+11- +

x .
x

[解析] (1)∵y=x??x2+1x+x13??=x3+1+x12,

∴y′=3x2-x23.

(2)∵y=( x+1)?? 1x-1??=-x12+x-12,

∴y′=-12x-12-12x-32=-21 x??1+1x??.

(3)∵y=sin44x+cos44x

=??sin24x+cos24x??2-2sin24xcos24x

=1-12sin22x=1-12·1-2cosx=34+14cosx,

∴y′=-14sinx.

(4)∵y=11+ - xx+11- + xx=?1+1-xx?2+?1-1-xx?2

=21+-2xx=1-4 x-2,

∴y′=??1-4 x-2??′=-4?1?1--xx?2?′=?1-4 x?2.
10.已知函数 f(x)=axx2+-b6的图象在点 M(-1,f(-1))处的切线的方程为 x+2y+5=0,

求函数的解析式. 导学号 84624147

[解析] 由于(-1,f(-1))在切线上, ∴-1+2f(-1)+5=0, ∴f(-1)=-2.∵f′(x)=a?x2+?bx?2-+2bx??2ax-6?,

??-1a+-b6=-2,

???a?1+b??1++2b?- ?2 a-6?=-12,

解得 a=2,b=3(∵b+1≠0,∴b=-1 舍去). 故 f(x)=2xx2+-36.
B 级 素养提升 一、选择题 1.已知函数 f(x)的导函数为 f ′(x),且满足 f(x)=2xf ′(e)+lnx,则 f ′(e)=

导学号 84624148 ( C )

A.e-1

B.-1

C.-e-1

D.-e

[解析] ∵f(x)=2xf ′(e)+lnx,

∴f ′(x)=2f ′(e)+1x,

∴f ′(e)=2f ′(e)+1e,解得 f ′(e)=-1e,故选 C.

2.曲线 y=xsinx 在点??-π2,π2??处的切线与 x 轴、直线 x=π 所围成的三角形的面积为

导学号 84624149 ( A )

A.π22

B.π2

C.2π2

D.12(2+π)2

[解析] 曲线 y=xsinx 在点??-π2,π2??处的切线方程为 y=-x,所围成的三角形的顶点为

O(0,0),A(π,0),C(π,-π),∴三角形面积为π22. 二、填空题

3.(2016·太原高二检测)设函数 f(x)=cos( 3x+φ)(0<φ<π),若 f(x)+f ′(x)是奇函数,

则 φ=

π 6

. 导学号 84624150

[解析] f ′(x)=- 3sin( 3x+φ),

f(x)+f ′(x)=cos( 3x+φ)- 3sin( 3x+φ)

=2sin?? 3x+φ+56π??.

若 f(x)+f ′(x)为奇函数,则 f(0)+f ′(0)=0,

即 0=2sin??φ+56π??,∴φ+56π=kπ(k∈Z).

又∵φ∈(0,π),∴φ=π6.

4.(2015·陕西理,15)设曲线 y=ex 在点(0,1)处的切线与曲线 y=1x(x>0)上点 P 处的切线

垂直,则 P 的坐标为__(1,1)__. 导学号 846241514152

[解析] 设 f(x)=ex,则 f ′(x)=ex,所以 f ′(0)=1,因此曲线 f(x)=ex 在点(0,1)处的切 线方程为 y-1=1×(x-0),即 y=x+1;设 g(x)=1x(x>0),则 g′(x)=-x12,由题意可得 g′(xP) =-1,解得 xP=1,所以 P(1,1).故本题正确答案为(1,1).
三、解答题 5.偶函数 f(x)=ax4+bx3+cx2+dx+e 的图象过点 P(0,1),且在 x=1 处的切线方程为 y

=x-2,求 y=f(x)的解析式. 导学号 84624153

[解析] ∵f(x)的图象过点 P(0,1),∴e=1. 又∵f(x)为偶函数,∴f(-x)=f(x). 故 ax4+bx3+cx2+dx+e=ax4-bx3+cx2-dx+e. ∴b=0,d=0.∴f(x)=ax4+cx2+1. ∵函数 f(x)在 x=1 处的切线方程为 y=x-2, ∴切点为(1,-1).∴a+c+1=-1. ∵f ′(x)|x=1=4a+2c,∴4a+2c=1.∴a=52,c=-92. ∴函数 y=f(x)的解析式为 f(x)=52x4-92x2+1. 6.已知 f(x)=13x3+bx2+cx(b,c∈R),f ′(1)=0,x∈[-1,3]时,曲线 y=f(x)的切线斜

率的最小值为-1,求 b,c 的值. 导学号 84624154

[解析] f ′(x)=x2+2bx+c=(x+b)2+c-b2, 且 f ′(1)=1+2b+c=0.① (1)若-b≤-1,即 b≥1,则 f ′(x)在[-1,3]上是增函数,所以 f ′(x)min=f ′(-1)=- 1, 即 1-2b+c=-1.② 由①②解得 b=14,不满足 b≥1,故舍去. (2)若-1<-b<3,即-3<b<1,则 f ′(x)min=f ′(-b)=-1, 即 b2-2b2+c=-1.③ 由①③解得 b=-2,c=3 或 b=0,c=-1. (3)若-b≥3,即 b≤-3,则 f ′(x)在[-1,3]上是减函数, 所以 f ′(x)min=f ′(3)=-1, 即 9+6b+c=-1.④ 由①④解得 b=-94,不满足 b≤-3,故舍去. 综上可知,b=-2,c=3 或 b=0,c=-1.
C 级 能力拔高 (2016·德州模拟)设函数 f(x)=ax-bx,曲线 y=f(x)在点(2,f(2))处的切线方程为 7x-4y

-12=0. 导学号 84624156

(1)求 f(x)的解析式; (2)证明:曲线 y=f(x)上任一点处的切线与直线 x=0 和直线 y=x 所围成的三角形的面 积为定值,并求此定值. [解析] (1)f′(x)=a+xb2,又根据切线方程可知 x=2 时,y=12,y′=74,

?2a-2b=21
则有
??a+b4=74

,解???a=1 ??b=3

.

(2)设 P(x0,y0)为曲线上任一点,由 y′=1+x32知曲线在点 P(x0,y0)处的切线方程为 y -y0=(1+x320)(x-x0),
即 y-(x0-x30)=(1+x320)(x-x0).

令 x=0 得 y=-x60,从而得切线与直线 x=0 的交点坐标为(0,-x60). 令 y=x 得 y=x=2x0,从而得切线与直线 y=x 的交点坐标为(2x0,2x0). 所以点 P(x0,y0)处的切线与直线 x=0,y=x 所围成的三角形面积为12|-x60||2x0|=6. 故曲线 y=f(x)上任一点处的切线与直线 x=0,y=x 所围成的三角形的面积为定值,此 定值为 6.


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