高職高考數(shù)學復(fù)習第三章函數(shù)一、選擇題(每小題6分,共90分)1.已知函數(shù)f(x)=2x-1,則f(2)= (

) A.6 B.5 C.4 D.3第三章單元檢測【答案】D

【答案】B

【答案】B4.函數(shù)y=(x-3)2-1的頂點坐標是 (

) A.(3,1) B.(-3,1) C.(3,-1) D.(-3,-1)【答案】C

【答案】C6.函數(shù)f(x)=x2-6x+7的 (

) A.最小值為-2 B.最大值為-2 C.最小值為3 D.最大值為3【答案】A7.函數(shù)y=-2x+3在R上是 (

) A.增函數(shù) B.減函數(shù)

C.非單調(diào)函數(shù) D.不能判斷【答案】B8.已知函數(shù)f(x)是偶函數(shù),且f(1)=2,則[f(-1)]3= (

) A.-8 B.-2 C.2 D.8【答案】D9.下列函數(shù)是奇函數(shù)的是 (

)

A.f(x)=5x+3 B.f(x)=x2+1

C.f(x)=x3+x

D.f(x)=x3-2【答案】C10.已知函數(shù)f(x)為奇函數(shù),當x<0時,f(x)=2x+1,則f(3)= (

) A.-7 B.7 C.-5 D.5【答案】D11.已知函數(shù)f(x)是R上的減函數(shù),那么 (

)

A.f(5)>f(3)>f(1) B.f(5)>f(1)>f(3)

C.f(5)<f(3)<f(1) D.f(5)<f(1)<f(3)【答案】C12.若函數(shù)f(x)=ax2+bx+c滿足f(0)=3,對稱軸是直線x=-1,最小值為2,則該函數(shù)的表達式為 (

)

A.f(x)=x2-2x-3 B.f(x)=x2+2x-3

C.f(x)=x2-2x+3 D.f(x)=x2+2x+3【答案】D13.已知函數(shù)f(x)=4x2-mx+5在區(qū)間[-2,+∞)上是增函數(shù),在區(qū)間(-∞,-2]上是減函數(shù),則m= (

) A.16 B.-16 C.8 D.-8【答案】B14.若函數(shù)f(x)=x5+(m+2)x4+x3為奇函數(shù),則實數(shù)m的值為 (

) A.-1 B.1 C.-2 D.2【答案】C15.已知函數(shù)f(x)為偶函數(shù),當x>0時,f(x)=x2-ax,且f(-1)=2,則a=(

) A.-1 B.0 C.1 D.2【答案】A

【答案】117.若函數(shù)f(x)=x2+a,且f(1)=0,則a的值為

.

【答案】-118.已知函數(shù)f(x)在R上是減函數(shù),且f(t)<f(-1),則t的取值范圍為

.

【答案】(-1,+∞)19.已知函數(shù)f(x)是定義在R上的奇函數(shù),則f(0)=

.

【答案】020.函數(shù)y=2x2-5x-3的單調(diào)遞增區(qū)間是

.

22.(8分)已知函數(shù)f(x)=x2-2x-5.(1)求f(1),f[f(1)]的值;(2)若f(a)=10,求a的值.【解】(1)∵f(x)=x2-2x-5,

∴f(1)=12-2×1-5=-6,

f[f(1)]=f(-6)=(-6)2-2×(-6)-5=43.

(2)若f(a)=10,則a2-2a-5=10,即a2-2a-15=0.由(a-5)(a+3)=0,解得a=5或a=-3.23.(8分)若函數(shù)f(x)在R上單調(diào)遞減,且f(m2-2)>f(m),求m的取值范圍.【解】∵函數(shù)f(x)在R上單調(diào)遞減,且f(m2-2)>f(m),

∴m2-2<m,即m2-m-2<0.由(m-2)(m+1)<0,解得-1<m<2,

∴m的取值范圍為(-1,2).24.(8分)已知函數(shù)f(x)=x2+2.(1)寫出函數(shù)的頂點坐標;(2)判斷函數(shù)的奇偶性;(3)寫出函數(shù)的單調(diào)區(qū)間;(4)求函數(shù)在區(qū)間[-1,2]上的最大值和最小值.【解】(1)函數(shù)的頂點坐標為(0,2).

(2)函數(shù)為偶函數(shù).

(3)函數(shù)的單調(diào)遞增區(qū)間為[0,+∞),單調(diào)遞減區(qū)間為(-∞,0).

(4)∵x=0∈[-1,2],且當x=0時,y=2;當x=-1時,y=3;當x=2時,y=6,

∴當x=0時,