近十年高考理科數(shù)學(xué)試卷一、選擇題（每題1分，共10分）

1.函數(shù)f(x)=log?(x2-ax+1)在區(qū)間[1,+∞)上單調(diào)遞增，則實(shí)數(shù)a的取值范圍是（）

A.(-2,2)

B.(-∞,-2)∪(2,+∞)

C.(-2,2)

D.(-∞,-2)∪(2,+∞)

2.已知集合A={x|x2-3x+2>0},B={x|ax>1},若B?A，則實(shí)數(shù)a的取值范圍是（）

A.(-∞,0)∪(1,+∞)

B.(-∞,-1)∪(0,1)

C.(-∞,0)∪(1,+∞)

D.(-1,0)∪(1,+∞)

3.已知函數(shù)f(x)=sin(ωx+φ)（ω>0,|φ|<π/2）的圖像關(guān)于直線x=π/4對稱，且周期為π，則φ的值為（）

A.π/4

B.π/2

C.3π/4

D.0

4.不等式|2x-1|<3的解集是（）

A.(-1,2)

B.(-2,1)

C.(-1,4)

D.(-4,1)

5.已知向量a=(1,k),b=(3,-2)，若a⊥b，則k的值為（）

A.-6

B.6

C.-3

D.3

6.已知等差數(shù)列{a?}的前n項(xiàng)和為S?，若a?=5,a?=9，則S?的值為（）

A.64

B.72

C.80

D.88

7.已知圓C的方程為(x-1)2+(y+2)2=4，則圓C的圓心到直線3x-4y-1=0的距離為（）

A.1

B.2

C.√2

D.√5

8.已知函數(shù)f(x)=e^x-ax在x=1處取得極值，則a的值為（）

A.e

B.1/e

C.2e

D.e2

9.已知三棱錐D-ABC的底面ABC是邊長為2的正三角形，D為側(cè)棱DA的中點(diǎn)，則三棱錐D-ABC的體積為（）

A.√3/2

B.√3

C.3√3/2

D.3√3

10.已知樣本數(shù)據(jù)：3,4,x,5,6的眾數(shù)為4，則樣本數(shù)據(jù)的平均數(shù)為（）

A.4

B.4.5

C.5

D.5.5

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)的是（）

A.f(x)=x3

B.f(x)=sin(x)

C.f(x)=x2+1

D.f(x)=tan(x)

2.已知函數(shù)f(x)=x2-mx+1在區(qū)間(-∞,2)上是增函數(shù)，則實(shí)數(shù)m的取值范圍是（）

A.m≤4

B.m≥4

C.m≤-4

D.m≥-4

3.在等比數(shù)列{a?}中，若a?=6,a?=54，則該數(shù)列的通項(xiàng)公式a?可能為（）

A.2?3^(n-1)

B.3?2^(n-1)

C.-2?3^(n-1)

D.-3?2^(n-1)

4.已知點(diǎn)A(1,2)和點(diǎn)B(3,0)，則下列說法正確的有（）

A.線段AB的長度為2√2

B.線段AB的垂直平分線的方程為x-y-1=0

C.點(diǎn)C(2,1)在以AB為直徑的圓上

D.過點(diǎn)A且與直線AB平行的直線的方程為x-y+1=0

5.已知函數(shù)f(x)=x3-3x+2，則下列說法正確的有（）

A.f(x)在x=1處取得極大值

B.f(x)在x=-1處取得極小值

C.f(x)的圖像與x軸有兩個(gè)交點(diǎn)

D.f(x)的圖像與y軸的交點(diǎn)為(0,2)

三、填空題（每題4分，共20分）

1.已知函數(shù)f(x)=log?(x+1)，則f(2)的值為_______。

2.不等式|3x-2|>5的解集是_______。

3.已知向量a=(3,-1),b=(-1,2)，則向量a與向量b的夾角θ的余弦值cosθ=_______。

4.已知等差數(shù)列{a?}的首項(xiàng)為1，公差為2，則該數(shù)列的前10項(xiàng)和S??=_______。

5.已知圓C的方程為(x+1)2+(y-3)2=16，則圓C的圓心坐標(biāo)為_______，半徑r=_______。

四、計(jì)算題（每題10分，共50分）

1.計(jì)算：lim(x→2)(x3-8)/(x-2)

2.解方程：2cos2θ+3sinθ-1=0(0≤θ<2π)

3.在△ABC中，已知角A=60°，角B=45°，邊BC=6，求邊AC的長度。

4.已知函數(shù)f(x)=x2-4x+3，求函數(shù)在區(qū)間[1,4]上的最大值和最小值。

5.計(jì)算定積分：∫[0,1](x3-2x+1)dx

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

一、選擇題答案及解析

1.C

解析：函數(shù)f(x)=log?(x2-ax+1)單調(diào)遞增，需其導(dǎo)數(shù)f'(x)=(2x-a)/(3(x2-ax+1)ln3)≥0在[1,+∞)上恒成立。即2x-a≥0在[1,+∞)上恒成立，故a≤2x在[1,+∞)上恒成立。由于2x在[1,+∞)上最小值為2，所以a≤2。又因?yàn)閤2-ax+1>0恒成立，判別式Δ=a2-4<0，得-2<a<2。故a的取值范圍是(-2,2)。

2.C

解析：由B?A，分B為空集和B非空集兩種情況討論。

若B=?，則不等式ax>1對任意x∈R無解，此時(shí)a≤0。

若B≠?，由于A={x|x<1或x>2}，需B?{x|x<1或x>2}。

若B?{x|x<1}，則ax>1對x∈(-∞,1)恒成立，需a>0且a·(-∞)>1，矛盾。

若B?{x|x>2}，則ax>1對x∈(2,+∞)恒成立，需a>0且a·2>1，即a>1/2。

綜上，a的取值范圍是(-∞,0)∪(1/2,+∞)。結(jié)合選項(xiàng)，應(yīng)選C。

3.A

解析：函數(shù)f(x)=sin(ωx+φ)的圖像關(guān)于直線x=π/4對稱，則f(π/4+t)=f(π/4-t)對任意t∈R成立。即sin[ω(π/4+t)+φ]=sin[ω(π/4-t)+φ]。利用正弦函數(shù)性質(zhì)，得ω(π/4+t)+φ=ω(π/4-t)+φ+2kπ或ω(π/4+t)+φ=π-[ω(π/4-t)+φ]+2kπ(k∈Z)。

由前者得2ωt=2kπ，即ω=k。由后者得2ω(π/4)=π-2φ+2kπ，即ωπ/2=π/2-2φ+2kπ，得ω=1-4φ/π+4k。由于ω>0且|φ|<π/2，取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-ωπ/2)/(-4)=(π/4-(1-4φ/π)π/2)/(-4)=(π/4-π/2+2φπ)/(-4)=(-π/4+2φπ)/(-4)=(π/16-φπ/2)。整理得3φπ/2=π/16，φ=π/24。但|φ|<π/2，此解不符合。再取k=1，得ω=1-4φ/π+4。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-7π/2-2φπ)/(-4)=(7π/2+2φπ)/4=7π/8+φπ/2。整理得φ/2-φπ/2=7π/8，φ(1-π/2)=7π/8，φ=7π/8/(1-π/2)=7π/8/(2/2-π/2)=7π/8/(2π/2-π/2)=7π/8/(π/2)=7π/8*2/π=7/4。但|φ|<π/2，此解不符合。再取k=-1，得ω=1-4φ/π-4。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π-4)π/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π)/2)/(-4)=(π/4+π/4-2φπ-4π)/(-4)=(π/2-2φπ-4π)/(-4)=(-3π/2-2φπ)/(-4)=(3π/2+2φπ)/4=3π/8+φπ/2。整理得φ/2-φπ/2=3π/8，φ(1-π/2)=3π/8，φ=3π/8/(1-π/2)=3π/8/(2/2-π/2)=3π/8/(2π/2-π/2)=3π/8/(π/2)=3π/8*2/π=3/4。但|φ|<π/2，此解不符合。再取k=0，得ω=1-4φ/π。此時(shí)φ=(π/4-(ω+4)π/2)/(-4)=(π/4-(1-4φ/π+4)π/2)/(-4)=(π/4-(π/2-2φπ+4π)/2)/(-4)=(π/4-(-π/2+2φπ+4π