导数
(选修2-2P18A7改编)曲线y=sin x
x在x=
π
2处的切线方程为()
A.y=0
B.y=2π
C.y=-
4
π2
x+
4
π
D.y=
4
π2
x
解析∵y′=x cos x-sin x
x2,∴y′|x=
π
2=-
4
π2
,
当x=π
2时,y=
2
π
,
∴切线方程为y-2
π
=-
4
π2⎝
⎛
⎭
⎪
⎫
x-
π
2
,即y=-
4
π2
x+
4
π
.
(2016·天津卷)已知函数f(x)=(2x+1)e x,f′(x)为f(x)的导函数,则f′(0)的值为________.
解析因为f(x)=(2x+1)e x,
所以f′(x)=2e x+(2x+1)e x=(2x+3)e x,
所以f′(0)=3e0=3.
(2017·西安月考)设曲线y=ax-ln(x+1)在点(0,0)处的切线方程为y=2x,则a=________.
解析y′=a-
1
x+1
,由题意得y′|x=0=2,即a-1=2,
所以a=3.
(2017·威海质检)已知函数f(x)=x ln x,若直线l过点(0,-1),并且与曲线y=f(x)相切,则直线l的方程为()
A.x+y-1=0
B.x-y-1=0
C.x+y+1=0
D.x-y+1=0
解析 ∵点(0,-1)不在曲线f (x )=x ln x 上, ∴设切点为(x 0,y 0).
又∵f ′(x )=1+ln x ,∴⎩⎪⎨⎪⎧y 0=x 0ln x 0,
y 0+1=(1+ln x 0)x 0,
解得x 0=1,y 0=0.
∴切点为(1,0),∴f ′(1)=1+ln 1=1. ∴直线l 的方程为y =x -1,即x -y -1=0.
(2015·全国Ⅱ卷)已知曲线y =x +ln x 在点(1,1)处的切线与曲线y =ax 2+(a +2)x +1相切,则a =________.
解析 法一 ∵y =x +ln x ,∴y ′=1+1
x ,y ′|x =1=2.
∴曲线y =x +ln x 在点(1,1)处的切线方程为y -1=2(x -1),即y =2x -1. ∵y =2x -1与曲线y =ax 2+(a +2)x +1相切, ∴a ≠0(当a =0时曲线变为y =2x +1与已知直线平行). 由⎩⎪⎨⎪⎧y =2x -1,y =ax 2
+(a +2)x +1消去y ,得ax 2+ax +2=0.
由Δ=a 2-8a =0,解得a =8. 法二 同法一得切线方程为y =2x -1.
设y =2x -1与曲线y =ax 2+(a +2)x +1相切于点(x 0,ax 20+(a +2)x 0+1).
∵y ′=2ax +(a +2),∴y ′|x =x 0=2ax 0+(a +2). 由⎩⎪⎨⎪⎧2ax 0+(a +2)=2,ax 20+(a +2)x 0+1=2x 0-1,解得⎩⎨⎧x 0=-12,a =8.
答案 8
(2017·西安质测)曲线f (x )=x 3-x +3在点P 处的切线平行于直线y =2x -1,则P
点的坐标为( ) A.(1,3)
B.(-1,3)
C.(1,3)和(-1,3)
D.(1,-3)
解析 f ′(x )=3x 2-1,令f ′(x )=2,则3x 2-1=2,解得x =1或x =-1,∴P (1,3)或(-1,3),经检验,点(1,3),(-1,3)均不在直线y =2x -1上,故选C. (2015·天津卷)已知函数f (x )=ax ln x ,x ∈(0,+∞),其中a 为实数,f ′(x )为f (x )的导函数,若f ′(1)=3,则a 的值为________.
解析 f ′(x )=a ⎝ ⎛
⎭⎪⎫ln x +x ·1x =a (1+ln x ),由于f ′(1)=a (1+ln 1)=a ,又f ′(1)=3,所以a =3.
(2016·全国Ⅲ卷)已知f (x )为偶函数,当x <0时,f (x )=ln(-x )+3x ,则曲线y =f (x )在点(1,-3)处的切线方程是________.
解析 设x >0,则-x <0,f (-x )=ln x -3x ,又f (x )为偶函数,f (x )=ln x -3x , f ′(x )=1
x -3,f ′(1)=-2,切线方程为y =-2x -1. 答案 2x +y +1=0
(2015·陕西卷)设曲线y =e x 在点(0,1)处的切线与曲线y =1
x (x >0)上点P 处的切线垂直,则P 的坐标为________.
解析 y ′=e x ,曲线y =e x 在点(0,1) 处的切线的斜率k 1=e 0=1,设P (m ,n ),y =1x (x >0)的导数为y ′=-1x 2(x >0),曲线y =1x (x >0)在点P 处的切线斜率k 2=-1m 2(m >0),因为两切线垂直,所以k 1k 2=-1,所以m =1,n =1,则点P 的坐标为(1,1). 答案 (1,1)
(2016·北京卷)设函数f (x )=x e a -x +bx ,曲线y =f (x )在点(2,f (2))处的切线方程为y =(e -1)x +4. (1)求a ,b 的值;
(2)求f (x )的单调区间.
解 (1)∵f (x )=x e a -x +bx ,∴f ′(x )=(1-x )e a -x +b .
由题意得⎩⎨⎧f (2)=2e +2,f ′(2)=e -1,即⎩⎨⎧2e a -2
+2b =2e +2,
-e a -2+b =e -1,
解得a =2,b =e.
(2)由(1)得f (x )=x e 2-x +e x ,
由f ′(x )=e 2-x (1-x +e x -1)及e 2-x >0知,f ′(x )与1-x +e x -1同号. 令g (x )=1-x +e x -1,则g ′(x )=-1+e x -1.
当x ∈(-∞,1)时,g ′(x )<0,g (x )在(-∞,1)上递减; 当x ∈(1,+∞)时,g ′(x )>0,g (x )在(1,+∞)上递增, ∴g (x )≥g (1)=1在R 上恒成立, ∴f ′(x )>0在R 上恒成立.
∴f (x )的单调递增区间为(-∞,+∞).
(2016·四川卷)已知a 为函数f (x )=x 3-12x 的极小值点,则a =( ) A.-4
B.-2
C.4
D.2
解析 f ′(x )=3x 2-12,∴x <-2时,f ′(x )>0,-2<x <2时,f ′(x )<0,x >2时, f ′(x )>0,∴x =2是f (x )的极小值点. 答案 D
(2016·全国Ⅲ卷)设函数f (x )=ln x -x +1.讨论f (x )的单调性; 解 依题意,f (x )的定义域为(0,+∞). f ′(x )=1
x -1,令f ′(x )=0,得x =1, ∴当0<x <1时,f ′(x )>0,f (x )单调递增. 当x >1时,f ′(x )<0,f (x )单调递减.
(2015·北京卷)设函数f (x )=x 2
2-k ln x ,k >0.求f (x )的单调区间和极值;
解 由f (x )=x 22-k ln x (k >0),得x >0且f ′(x )=x -k x =x 2
-k
x .由f ′(x )=0,解得x =k
(负值舍去).
f (x )与f ′(x )在区间(0,+∞)上的变化情况如下表:
所以,f f (x )在x =k 处取得极小值f (k )=k (1-ln k )
2.
(2017·西安调研)定积分⎠⎛01(2x +e x )d x 的值为( )
A.e +2
B.e +1
C.e
D.e -1
解析 ⎠⎛0
1(2x +e x )d x =(x 2+e x )⎪⎪⎪
1
0)=1+e 1-1=e.故选C.
(2015·全国Ⅱ卷)已知函数f (x )=ln x +a (1-x ).讨论f (x )的单调性; 解 f (x )的定义域为(0,+∞),f ′(x )=1
x -a .
若a ≤0,则f′(x )>0,所以f (x )在(0,+∞)上单调递增. 若a >0,则当x ∈⎝ ⎛
⎭⎪⎫0,1a 时,f ′(x )>0;
当x ∈⎝ ⎛⎭
⎪⎫
1a ,+∞时,f ′(x )<0,
所以f (x )在⎝ ⎛⎭⎪⎫0,1a 上单调递增,在⎝ ⎛⎭⎪⎫
1a ,+∞上单调递减.
综上,知当a ≤0时,f (x )在(0,+∞)上单调递增;
当a >0时,f (x )在⎝ ⎛⎭⎪⎫0,1a 上单调递增,在⎝ ⎛⎭
⎪⎫
1a ,+∞上单调递减.。