新人教版高中数学必修第一册课时跟踪检测(十二) 函数的表示法

新人教版高中数学必修第一册课时跟踪检测(十二)  函数的表示法
新人教版高中数学必修第一册课时跟踪检测(十二)  函数的表示法

课时跟踪检测(十二) 函数的表示法

A 级——学考合格性考试达标练

1.已知函数y =f (x )的对应关系如下表,函数y =g(x )的图象是如图的曲线ABC ,其中A(1,3),B(2,1),C(3,2),则f (g(2))的值为(

)

A .3

B .2

C .1

D .0

解析:选B 由函数g(x )的图象知,g(2)=1,则f (g(2))=f (1)=2. 2.如果f ????1x =x

1-x ,则当x ≠0,1时,f (x )等于( ) A .1x B .

1x -1

C .11-x

D .1x

-1

解析:选B 令1x =t ,则x =1

t ,代入f ????1x =x 1-x ,则有f (t )=1t

1-

1t =1t -1,∴f (x )=1x -1,故选B.

3.若f (x )是一次函数,2f (2)-3f (1)=5,2f (0)-f (-1)=1,则f (x )=( ) A .3x +2 B .3x -2 C .2x +3

D .2x -3

解析:选B 设f (x )=ax +b (a ≠0),由题设有

?????2(2a +b )-3(a +b )=5,2(0·a +b )-(-a +b )=1.解得?

????a =3,b =-2.所以选B . 4.设f (x )=2x +3,g(x )=f (x -2),则g(x )=( ) A .2x +1 D .2x -1 C .2x -3

D .2x +7

解析:选B ∵f (x )=2x +3,∴f (x -2)=2(x -2)+3=2x -1,即g(x )=2x -1,故选B . 5.若f (1-2x )=1-x 2x 2(x ≠0),那么f ????12等于( ) A .1 D .3 C .15

D .30

解析:选C 令1-2x =t , 则x =1-t 2

(t ≠1),

∴f (t )=4

(t -1)2-1(t ≠1),

即f (x )=

4

(x -1)2

-1(x ≠1),

∴f ????12=16-1=15.

6.已知函数f (x )=x -m

x ,且此函数图象过点(5,4),则实数m 的值为________.

解析:将点(5,4)代入f (x )=x -m

x ,得m =5.

答案:5

7.已知f (x )是一次函数,满足3f (x +1)=6x +4,则f (x )=________. 解析:设f (x )=ax +b (a ≠0), 则f (x +1)=a (x +1)+b =ax +a +b , 依题设,3ax +3a +3b =6x +4,

∴?

????3a =6,

3a +3b =4,∴?

????a =2,

b =-23

, 则f (x )=2x -23.

答案:2x -2

3

8.已知a ,b 为常数,若f (x )=x 2+4x +3,f (ax +b )=x 2+10x +24,则5a -b =________. 解析:由f (x )=x 2+4x +3,f (ax +b )=x 2+10x +24,得(ax +b )2+4(ax +b )+3=x 2+10x +24,即a 2x 2+(2ab +4a )x +b 2+4b +3=x 2

+10x +24,由系数相等得????

?a 2=1,2ab +4a =10,b 2+4b +3=24,解

得a =-1,b =-7或a =1,b =3,则5a -b =2.

答案:2

9.已知函数p =f (m )的图象如图所示.求:

(1)函数p =f (m )的定义域; (2)函数p =f (m )的值域;

(3)p 取何值时,只有唯一的m 值与之对应.

解:(1)观察函数p =f (m )的图象,可以看出图象上所有点的横坐标的取值范围是-3≤m ≤0或1≤m ≤4,由图知定义域为[-3,0]∪[1,4].

(2)由图知值域为[-2,2].

(3)由图知:p ∈(0,2]时,只有唯一的m 值与之对应.

10.已知f (x )为二次函数且f (0)=3,f (x +2)-f (x )=4x +2,求f (x )的解析式. 解:设f (x )=ax 2+bx +c(a ≠0),又f (0)=c =3,∴f (x )=ax 2+bx +3,∴f (x +2)-f (x )=a (x +2)2+b (x +2)+3-(ax 2+bx +3)=4ax +4a +2b =4x +2.∴?

??

??4a =4,4a +2b =2,解得

?

????a =1,b =-1,∴f (x )=x 2-x +3. B 级——面向全国卷高考高分练

1.函数y =f (x )(f (x )≠0)的图象与x =1的交点个数是( ) A .1 D .2 C .0或1

D .1或2

解析:选C 结合函数的定义可知,如果f :A →B 成立,则任意x ∈A ,则有唯一确定的B 与之对应,由于x =1不一定是定义域中的数,故x =1可能与函数y =f (x )没有交点,故函数f (x )的图象与直线x =1至多有一个交点.

2.若一次函数的图象经过点A(1,6)和B(2,8),则该函数的图象还可能经过的点的坐标为( )

A.????

12,5 B .????

14,4 C .(-1,3)

D .(-2,1)

解析:选A 设一次函数的解析式为y =k x +b (k ≠0),由该函数的图象经过点A(1,6)

和B(2,8),得?????k +b =6,2k +b =8,解得????

?k =2,b =4,

,所以此函数的解析式为y =2x +4,只有A 选项的坐标符合此函数的解析式.故选A.

3.已知函数f (x +1)=x 2-x +3,那么f (x -1)的表达式是( ) A .f (x -1)=x 2+5x -9 D .f (x -1)=x 2-x -3 C .f (x -1)=x 2-5x +9

D .f (x -1)=x 2-x +1

解析:选C f (x +1)=(x +1)2-3(x +1)+5,所以f (x )=x 2-3x +5,f (x -1)=(x -1)2

-3(x -1)+5=x 2-5x +9,故选C.

4.设f (x )=2x +a ,g(x )=1

4(x 2+3),且g(f (x ))=x 2-x +1,则a 的值为( )

A .1

D .-1

C .1或-1

D .1或-2

解析:选B 因为g(x )=14(x 2+3),所以g(f (x ))=14[(2x +a )2+3]=1

4(4x 2+4ax +a 2+3)

=x 2-x +1,求得a =-1.故选B.

5.已知函数f (2x +1)=3x +2,且f (a )=4,则a =________.

解析:因为f (2x +1)=32(2x +1)+12,所以f (a )=32a +12.又f (a )=4,所以32a +12=4,a =7

3.

答案:7

3

6.已知函数f (x )满足f (x )=2f ????

1x +3x ,则f (x )的解析式为________________. 解析:由题意知函数f (x )满足f (x )=2f ????1x +3x ,即f (x )-2f ????1x =3x ,用1x 代换上式中的x ,可得f ????1x -2f (x )=3x ,联立方程得???

f (x )-2f ????

1x =3x ,

f ????

1x -2f (x )=3x ,

解得f (x )=-x -2

x (x ≠0).

答案:f (x )=-x -2

x (x ≠0)

7.设二次函数f (x )满足f (x -2)=f (-x -2),且图象与y 轴交点的纵坐标为1,被x 轴截得的线段长为22,求f (x )的解析式.

解:法一:设f (x )=ax 2+bx +c(a ≠0). 由f (x -2)=f (-x -2)得4a -b =0;① 又因为|x 1-x 2|=b 2-4a c

|a |=22,

所以b 2-4a c =8a 2;② 又由已知得c =1.③

由①②③解得b =2,a =1

2,c =1,

所以f (x )=1

2

x 2+2x +1.

法二:因为y =f (x )的图象有对称轴x =-2, 又|x 1-x 2|=22,

所以y =f (x )的图象与x 轴的交点为(-2-2,0), (-2+2,0),故可设f (x )=a (x +2+2)(x +2-2). 因为f (0)=1,所以a =1

2

.

所以f(x)=1

2[(x+2)

2-2]=

1

2x

2+2x+1.

C级——拓展探索性题目应用练

某省两个相近重要城市之间人员交流频繁,为了缓解交通压力,特修一条专用铁路,用一列火车作为交通车,若该车每次拖4节车厢,一天能来回16次(来、回各算作一次),若每次拖7节车厢,则每天能来回10次.

(1)若每天来回的次数是车头每次拖挂车厢节数的一次函数,求此一次函数的解析式.

(2)在(1)的条件下,每节车厢能载乘客110人.问这列火车每天来回多少次才能使运营人数最多?并求出每天最多运营人数.

解:(1)设每天来回y次,每次拖x节车厢,则可设y=k x+b(k≠0).

由题意,得16=4k+b,10=7k+b,

解得k=-2,b=24,

所以y=-2x+24.

(2)设这列火车每天来回总共拖挂的车厢节数为S,则由(1)知S=xy,

所以S=x(-2x+24)=-2x2+24x=-2(x-6)2+72,

所以当x=6时,S max=72,此时y=12,

则每日最多运营的人数为110×72=7 920.

所以这列火车每天来回12次,才能使运营人数最多,每天最多运营人数为7 920.

新人教版必修一 Unit5 Period 1课时跟踪检测

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高中数学必修一幂函数及其性质

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人教语文必修四课时跟踪检测(十二) 苏武传+Word版含答案

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