今年高考1卷數(shù)學(xué)試卷一、選擇題（每題1分，共10分）

1.函數(shù)f(x)=log?(x-1)的定義域是？

A.(-∞,1)

B.[1,+∞)

C.(1,+∞)

D.(-1,+∞)

2.已知集合A={x|x2-3x+2=0},B={x|ax=1},若A∩B={2},則a的值為？

A.1/2

B.1

C.2

D.1/4

3.不等式|2x-1|<3的解集是？

A.(-1,2)

B.(-2,1)

C.(-1,1)

D.(-2,2)

4.若f(x)是奇函數(shù),且f(1)=2,則f(-1)的值為？

A.-2

B.1

C.0

D.2

5.函數(shù)f(x)=sin(x)+cos(x)的最小正周期是？

A.2π

B.π

C.π/2

D.4π

6.已知等差數(shù)列{a?}中,a?=3,d=2,則a?的值為？

A.7

B.9

C.11

D.13

7.直線y=2x+1與直線y=-x+4的交點(diǎn)坐標(biāo)是？

A.(1,3)

B.(3,1)

C.(-1,-1)

D.(-3,-1)

8.圓x2+y2-4x+6y-3=0的圓心坐標(biāo)是？

A.(2,-3)

B.(-2,3)

C.(2,3)

D.(-2,-3)

9.已知三角形ABC中,∠A=60°,∠B=45°,BC=2,則AB的值為？

A.√2

B.2√2

C.2

D.√3

10.函數(shù)f(x)=e?的導(dǎo)數(shù)是？

A.e?

B.x?

C.lnx

D.1

二、多項(xiàng)選擇題（每題4分，共20分）

1.下列函數(shù)中，在其定義域內(nèi)是奇函數(shù)的有？

A.f(x)=x3

B.f(x)=sin(x)

C.f(x)=x2+1

D.f(x)=|x|

2.若函數(shù)f(x)=ax2+bx+c的圖像開口向上，且頂點(diǎn)在x軸上，則下列說法正確的有？

A.a>0

B.b2-4ac=0

C.c<0

D.f(x)在x軸上存在唯一零點(diǎn)

3.下列不等式解集為R的有？

A.x2+1>0

B.|x|+1>0

C.x2-2x+1>0

D.sin(x)+1≥0

4.已知等比數(shù)列{b?}中,b?=1,q=2,則下列說法正確的有？

A.b?=16

B.b?=2??1

C.數(shù)列的前n項(xiàng)和Sn=2?-1

D.數(shù)列{b?}是遞增數(shù)列

5.下列命題中，正確的有？

A.相似三角形的對應(yīng)角相等

B.勾股定理適用于任意三角形

C.圓的半徑是通過圓心且垂直于弦的線段

D.一個(gè)角為60°的等腰三角形是等邊三角形

三、填空題（每題4分，共20分）

1.函數(shù)f(x)=√(x-1)的定義域是[1,+∞)。

2.已知f(x)=x2-mx+1,若f(1)=3,則實(shí)數(shù)m的值為-1。

3.不等式組{x>0;x-1<2}的解集是(0,3)。

4.已知點(diǎn)A(1,2)和點(diǎn)B(3,0),則線段AB的長度是√8。

5.在等差數(shù)列{a?}中,若a?=5,a?=9,則該數(shù)列的公差d是2。

四、計(jì)算題（每題10分，共50分）

1.解方程：2x2-7x+3=0

2.計(jì)算極限：lim(x→2)(x2-4)/(x-2)

3.求函數(shù)f(x)=sin(x)+cos(x)在區(qū)間[0,π/2]上的最大值和最小值。

4.已知等比數(shù)列{a?}中,a?=3,q=-2,求該數(shù)列的前5項(xiàng)和S?。

5.計(jì)算不定積分：∫(x3-2x+1)dx

本專業(yè)課理論基礎(chǔ)試卷答案及知識(shí)點(diǎn)總結(jié)如下

一、選擇題答案及解析

1.答案：C

解析：函數(shù)f(x)=log?(x-1)有意義需滿足x-1>0，解得x>1，故定義域?yàn)?1,+∞)。

2.答案：C

解析：由x2-3x+2=0得A={1,2}。因A∩B={2}，則2∈B，代入ax=1得a=1/2，但需驗(yàn)證x=1時(shí)是否滿足ax=1，1/2×1=1/2≠1，故a=1/2時(shí)B={2}，滿足條件。

3.答案：A

解析：|2x-1|<3等價(jià)于-3<2x-1<3，解得-2<2x<4，即-1<x<2。

4.答案：A

解析：由奇函數(shù)定義f(-x)=-f(x)，故f(-1)=-f(1)=-2。

5.答案：A

解析：f(x)=sin(x)+cos(x)=√2sin(x+π/4)，其最小正周期為2π。

6.答案：D

解析：a?=a?+4d=3+4×2=11。

7.答案：A

解析：聯(lián)立方程組{y=2x+1;y=-x+4}，代入得2x+1=-x+4，解得x=1，代入y=2x+1得y=3，故交點(diǎn)為(1,3)。

8.答案：C

解析：圓方程化為標(biāo)準(zhǔn)式：(x-2)2+(y+3)2=16，圓心為(2,-3)。

9.答案：C

解析：由正弦定理：AB/sinC=BC/sinA，sinC=sin(180°-60°-45°)=sin75°=(√6+√2)/4，AB=BC×sinA/sinC=2×sin60°/(√6+√2)/4=2×√3/(√6+√2)=2√3(√6-√2)/(6-2)=√3(√6-√2)=2。

10.答案：A

解析：f'(x)=d/dx(e?)=e?。

二、多項(xiàng)選擇題答案及解析

1.答案：A,B

解析：f(x)=x3是奇函數(shù)(滿足f(-x)=-f(x))；f(x)=sin(x)是奇函數(shù)(滿足f(-x)=-sin(x)=-f(x))；f(x)=x2+1是偶函數(shù)(滿足f(-x)=x2+1=f(x))；f(x)=|x|是偶函數(shù)(滿足f(-x)=|-x|=|x|=f(x))。

2.答案：A,B,D

解析：開口向上需a>0；頂點(diǎn)在x軸上意味著判別式b2-4ac=0且a≠0(題目已隱含)；頂點(diǎn)在x軸上即函數(shù)有唯一零點(diǎn)，故f(x)在x軸上存在唯一零點(diǎn)。c的符號(hào)不確定，例如f(x)=x2-4x+4=(x-2)2，a=1>0,b=-4,c=4,b2-4ac=0，但c=4>0。

3.答案：A,B

解析：x2+1>0對任意實(shí)數(shù)x恒成立；|x|+1>0對任意實(shí)數(shù)x恒成立；x2-2x+1=(x-1)2≥0，解集為R；sin(x)+1≥0即sin(x)≥-1，解集為R。

4.答案：A,B,C

解析：b?=b?q3=1×23=8；b?=b?q??1=1×2??1=2??1；S?=a?(1-q?)/(1-q)=3(1-2?)/(1-(-2))=3(1-32)/3=-31。數(shù)列{b?}是遞增數(shù)列需q>1，此處q=-2，故是遞減數(shù)列。

5.答案：A,D

解析：相似三角形的定義要求對應(yīng)角相等，對應(yīng)邊成比例。勾股定理只適用于直角三角形。圓的半徑是連接圓心與圓上任意一點(diǎn)的線段。一個(gè)角為60°的等腰三角形，若頂角為60°，則三邊相等，為等邊三角形；若底角為60°，則兩腰相等，也為等腰三角形，但不是等邊三角形。題目表述為“一個(gè)角為60°的等腰三角形”，通常指頂角為60°的情況，故為等邊三角形。

三、填空題答案及解析

1.答案：[1,+∞)

解析：見選擇題第1題解析。

2.答案：-1

解析：f(1)=12-m×1+1=3，即1-m+1=3，解得m=-1。

3.答案：(0,3)

解析：見選擇題第3題解析。

4.答案：√8

解析：|AB|=√[(3-1)2+(0-2)2]=√[22+(-2)2]=√(4+4)=√8。

5.答案：2

解析：a?=a?+2d，9=5+2d，解得d=2。

四、計(jì)算題答案及解析

1.解方程：2x2-7x+3=0

解：(2x-1)(x-3)=0

得2x-1=0或x-3=0

x=1/2或x=3

答案：x=1/2或x=3

2.計(jì)算極限：lim(x→2)(x2-4)/(x-2)

解：原式=lim(x→2)[(x-2)(x+2)]/(x-2)

=lim(x→2)(x+2)(x≠2時(shí)，可約分)

=2+2

=4

答案：4

3.求函數(shù)f(x)=sin(x)+cos(x)在區(qū)間[0,π/2]上的最大值和最小值。

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