高三数学板块考点专练-——函数的图象与性质

1

板块考点专练(二) 函数的图象与性质

命题点1 函数的概念及其表示 1.导学号94950236(2014·江西卷)已知函数f (x )=5|x |,g (x )=ax 2-x (a ∈R ),若f [g (1)]=1,则a =( )

A .1

B .2

C .3

D .-1

解析:选A 因为f [g (1)]=1,且f (x )=5|x |,所以g (1)=0,即a ·12-1=0,解得a =1. 2.导学号94950237(2012·安徽卷)下列函数中,不满足f (2x )=2f (x )的是( )

A .f (x )=|x |

B .f (x )=x -|x |

C .f (x )=x +1

D .f (x )=-x

解析:选C 对于选项A ,f (2x )=|2x |=2|x |=2f (x );对于选项B ,f (x )=x -|x |=⎩⎪⎨⎪⎧

0(x ≥0)2x (x <0),当x ≥0时,

f (2x )=0=2f (x ),当x <0时,f (2x )=4x =2·2x =2f (x ),恒有f (2x )=2f (x );对于选项D ,f (2x )=-2x =2(-x )=

2f (x );对于选项C ,f (2x )=2x +1=2f (x )-1. 3.导学号94950238(2014·浙江卷)设函数f (x )=⎩⎪⎨⎪⎧

x 2+x ,x <0,-x 2,x ≥0,若f (f (a ))≤2,则实数a 的取值范围是__________________.

解析:f (x )的图象如图,由图象知.满足f (f (a ))≤2时,得f (a )≥-2,而满足f (a )≥-2

高三数学板块考点专练-——函数的图象与性质

时,a ≤ 2.

答案:(-∞,2]

命题点2 函数的基本性质 1.导学号94950239(2014·湖南卷)下列函数中,既是偶函数又在区间(-∞,0)上单调递增的是( )

A .f (x )=1x

2 B .f (x )=x 2+1 C .f (x )=x

3 D .f (x )=2-

x

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