课件习题详解第一章答案

解得x2（２）x433x431x7即x13x15x15或x15x4或x(4)x24x24或x24x2或x — —ｘ ｘ1y 5x2令5x20x25x5,5 2依题有x+30xx0x3且x0即x[33yarcsin1x11x1x2, 4y

arcsinx则有 0且1x1,两式取交集得x[-x x 5根据题意5根据题意得4

{x1x[1,2）U(2,6令x23x20x1或x2x(,,1)U(27依题意得2x0且0x3x1令0x211x1定义域[-a

x[a,1(0,]时，定义域为21fxsinxfxsinxsinxfxfx 12依题意得0x-a1且0x+a112fxfx1又xR

x

xlog210

fxfx ax

ax3fxax12

2

,fx

1ax2

21ax

fxf又定义域为R奇函x4fxa1,fxx

1

1

fax

1又xR5当x0时x0,fx1x,fx1x1当x0时x0fx1x,fx1x1当x0时-x=0,fxfx综上fxfx且xR原函数是6由已知得fxx2x1而fxx2x1两式对应法则不1是

ysin2x1cos2x,T

2是

ysinx1sin2x,设Usinx,Vsin2x则T2

T23是4

Tycosx2T2 x3,0x依题意在[0,2)内可得fxx3的图像，由此可得fxx232xx43,4x依题意把x4x1.5x4x0f4e41sin1sin

1,f1.53,f442115,f0y u,u1sin2ycos2x3,ycosu,u3yeu,usinv,vx

v,v2x4y4arcsin1x3,y4arcsinu,uv3,v156ylnsin2x,ylnu,uv2,vsin (1x2fx1x2fxfx1x2,ffx1[f(x)]2x42x2f f(1x2)

fx

11x2

2x21设tx1则x21t22

fxx2 2设tsinx则cosx12t22

fx22x2,x

0,x

2当x0时，gxx20ggxgx23当x0时gxx20fgxfx2当x0时gx0fgxf00,综上有fgx 依题意有fgxx1

x当a2时,ylg2sinx是复合函数此时ylguu2vvsin由2sinx0xR,即定义域为当a2时,ylg2sinx,而2sinx此时ylg2sinx不是复合函数

4x2,0x2

y

x 2x2,2x4y当0x2时y04y当2x3时y(2,4],此时xy2y综上可得反函数y

4xx2,x0,x2,x(2,2当x1时yx解得：xy，y当1x4时yx2,解得：x y,y2当x4y=2x解得：xlogyy[162综上可得反函数y

x,xx,x22

17.(1)Qd

由1002.5p5p50

p

Q（2）6元，则有1002.5p5p650p32,Q 18.8C(89681828平均成本为CC(Q)112 一、填空题x24 x2或x

自测 由 x2

x

，得函数的定义（-，- 设tx1xt1f(tt1)2t11t23t 有f ) x (xf(ex)ex,1ex

xex,0xex,0x1sin21sin21sin2(logx)asin21cos2ysinx ，而cos2x的周期为，sin21cos2y1lg(x2)解得x10y12y10x1f[(x

f[(x)]1

sin所以(xarcsin(1x2f(sinx3cos2x312sin2x22sin2xf(t22tf(cosx)22cos2aa 由f[f(x)] 2x3 a 2ax2x

2ax6xa2x2a6)x2

2a6

a 22x2

22x 22x1 y

2x

122x122x1

22x

22xx1log(

g(x) 1x

21

2log2(111.C.由

x lnx0

xx

16x2 4x 2.B.两个函数定义域与对应关系均相同，所以是同一函数1.7,23x由3x

x[7,2)(2,

31，f[f(7)] 3149x2设t x

x

t af )bf(x) xaf(x)bf(x) xxfx

f(x

xaex 1

a当x0时f(xx此时gfxf2xx2xf(x)02当x0时fx1x01x2

此时g[fx1

,1xf(x)0当x0时f(x)1x x1，此时g[f(x)]f(x)1x,x g[fx

x2(1x)2

x1x由已知得

f(x),f(x)

1x,xf(x1)

x1x1

xxf(x21)

x1x21x21x21

xxxx1xsinx,xf(x)

x2,2xx22y 42