第4课时 导数的运算法则
基础达标(水平一)
1.已知f(x)=x2f'(1),则f'(0)=( ).
A.0 B.1 C.2 D.3
【解析】因为f(x)=x2f'(1),所以f'(x)=2xf'(1),所以f'(0)=2f'(1)×0=0.
【答案】A
2.已知函数f(x)=ax3+3x2+2,若f'(-1)=4,则a的值是( ).
A.19/3 B.13/3 C.10/3 D.16/3
【解析】由f(x)=ax3+3x2+2,得f'(x)=3ax2+6x.
所以f'(-1)=3a-6=4,解得a=10/3.
【答案】C
3.若f(x)=ax2-bsin x,且f'(0)=1,f'(π/3)=1/2,则a+b等于( ).
A.1 B.0 C.-1 D.2
【解析】因为f'(x)=2ax-bcos x,
所以f'(0)=-b,f'(π/3)=2π/3a-bcosπ/3=2π/3a-1/2b,
所以{■(b="-" 1"," @2π/3 a"-" 1/2 b=1/2 "," )┤解得{■(a=0"," @b="-" 1"," )┤所以a+b=-1.
【答案】C
4.曲线y=xsin x在点("-" π/2 "," π/2)处的切线与x轴、直线x=π所围成的三角形的面积为( ).
A.π^2/2 B.π2 C.2π2 D.1/2(2+π)2
【解析】因为曲线y=xsin x在点("-" π/2 "," π/2)处的切线方程为y=-x,所以此切线与x轴、直线x=π所围成的三角形的面积为π^2/2.
【答案】A
5.若函数f(x)在(0,+∞)内可导,且f(ex)=x+ex,则f'(1)= .
【解析】∵f(ex)=x+ex=ln ex+ex,∴f(x)=ln x+x.
∴f'(x)=1/x+1,∴f'(1)=2.
【答案】2
6.若f(x)=x2-2x-4ln x,则f'(x)>0的解集为 .
【解析】由f(x)=x2-2x-4ln x,得函数定义域为(0,+∞),且f'(x)=2x-2-4/x=(2x^2 "-" 2x"-" 4)/x=2·(x^2 "-" x"-" 2)/x=2·("(" x+1")(" x"-" 2")" )/x>0,解得x>2,故f'(x)>0的解集为{x|x>2}.
【答案】{x|x>2}
7.设函数f(x)=x3+bx2+cx,若g(x)=f(x)-f'(x)是奇函数,求b+c的值.
【解析】∵函数f(x)=x3+bx2+cx,
∴f'(x)=3x2+2bx+c,
∴g(x)=f(x)-f'(x)=x3+(b-3)x2+(c-2b)x-c.
∵g(x)为奇函数,
∴b-3=0,-c=0,即 b=3,c=0,∴b+c=3.