外研社杯全国英语阅读大赛.

外研社杯全国英语阅读大赛.
外研社杯全国英语阅读大赛.

2016年浙江省高考数学试卷(文科)

一、选择题

1.(5分)已知全集U={1,2,3,4,5,6},集合P={1,3,5},Q={1,2,4},则(?U P)∪Q=()

A.{1}B.{3,5}C.{1,2,4,6}D.{1,2,3,4,5}

2.(5分)已知互相垂直的平面α,β交于直线l,若直线m,n满足m∥α,n⊥β,则()

A.m∥l B.m∥n C.n⊥l D.m⊥n

3.(5分)函数y=sinx2的图象是()

A.B.C.

D.

4.(5分)若平面区域,夹在两条斜率为1的平行直线之间,则这两

条平行直线间的距离的最小值是()

A.B.C.D.

5.(5分)已知a,b>0且a≠1,b≠1,若log a b>1,则()

A.(a﹣1)(b﹣1)<0 B.(a﹣1)(a﹣b)>0 C.(b﹣1)(b﹣a)<0 D.(b ﹣1)(b﹣a)>0

6.(5分)已知函数f(x)=x2+bx,则“b<0”是“f(f(x))的最小值与f(x)的最小值相等”的()

A.充分不必要条件 B.必要不充分条件

C.充分必要条件D.既不充分也不必要条件

7.(5分)已知函数f(x)满足:f(x)≥|x|且f(x)≥2x,x∈R.()A.若f(a)≤|b|,则a≤b B.若f(a)≤2b,则a≤b

C.若f(a)≥|b|,则a≥b D.若f(a)≥2b,则a≥b

8.(5分)如图,点列{A n}、{B n}分别在某锐角的两边上,且|A n A n+1|=|A n+1A n+2|,A n≠A n+1,n∈N*,|B n B n+1|=|B n+1B n+2|,B n≠B n+1,n∈N*,(P≠Q表示点P与Q不重合)若d n=|A n B n|,S n为△A n B n B n+1的面积,则()

A.{S n}是等差数列B.{S n2}是等差数列

C.{d n}是等差数列 D.{d n2}是等差数列

二、填空题

9.(6分)某几何体的三视图如图所示(单位:cm),则该几何体的表面积是cm2,体积是cm3.

10.(6分)已知a∈R,方程a2x2+(a+2)y2+4x+8y+5a=0表示圆,则圆心坐标是,半径是.

11.(6分)已知2cos2x+sin2x=Asin(ωx+φ)+b(A>0),则A=,b=.12.(6分)设函数f(x)=x3+3x2+1,已知a≠0,且f(x)﹣f(a)=(x﹣b)(x ﹣a)2,x∈R,则实数a=,b=.

13.(4分)设双曲线x2﹣=1的左、右焦点分别为F1、F2,若点P在双曲线上,

且△F1PF2为锐角三角形,则|PF1|+|PF2|的取值范围是.

14.(4分)如图,已知平面四边形ABCD,AB=BC=3,CD=1,AD=,∠ADC=90°,沿直线AC将△ACD翻折成△ACD′,直线AC与BD′所成角的余弦的最大值是.

15.(4分)已知平面向量,,||=1,||=2,=1,若为平面单位向量,则||+||的最大值是.

三、解答题

16.(14分)在△ABC中,内角A,B,C所对的边分别为a,b,c,已知b+c=2acosB.(1)证明:A=2B;

(2)若cosB=,求cosC的值.

17.(15分)设数列{a n}的前n项和为S n,已知S2=4,a n+1=2S n+1,n∈N*.(Ⅰ)求通项公式a n;

(Ⅱ)求数列{|a n﹣n﹣2|}的前n项和.

18.(15分)如图,在三棱台ABC﹣DEF中,平面BCFE⊥平面ABC,∠ACB=90°,BE=EF=FC=1,BC=2,AC=3.

(Ⅰ)求证:BF⊥平面ACFD;

(Ⅱ)求直线BD与平面ACFD所成角的余弦值.

19.(15分)如图,设抛物线y2=2px(p>0)的焦点为F,抛物线上的点A到y 轴的距离等于|AF|﹣1,

(Ⅰ)求p的值;

(Ⅱ)若直线AF交抛物线于另一点B,过B与x轴平行的直线和过F与AB垂直的直线交于点N,AN与x轴交于点M,求M的横坐标的取值范围.

20.(15分)设函数f(x)=x3+,x∈[0,1],证明:

(Ⅰ)f(x)≥1﹣x+x2

(Ⅱ)<f(x)≤.

2016年浙江省高考数学试卷(文科)

参考答案与试题解析

一、选择题

1.(5分)已知全集U={1,2,3,4,5,6},集合P={1,3,5},Q={1,2,4},则(?U P)∪Q=()

A.{1}B.{3,5}C.{1,2,4,6}D.{1,2,3,4,5}

【分析】先求出?U P,再得出(?U P)∪Q.

【解答】解:?U P={2,4,6},

(?U P)∪Q={2,4,6}∪{1,2,4}={1,2,4,6}.

故选:C.

【点评】本题考查了集合的运算,属于基础题.

2.(5分)已知互相垂直的平面α,β交于直线l,若直线m,n满足m∥α,n⊥β,则()

A.m∥l B.m∥n C.n⊥l D.m⊥n

【分析】由已知条件推导出l?β,再由n⊥β,推导出n⊥l.

【解答】解:∵互相垂直的平面α,β交于直线l,直线m,n满足m∥α,

∴m∥β或m?β或m与β相交,l?β,

∵n⊥β,

∴n⊥l.

故选:C.

【点评】本题考查两直线关系的判断,是基础题,解题时要认真审题,注意空间思维能力的培养.

3.(5分)函数y=sinx2的图象是()

A.B.C.

D.

【分析】根据函数奇偶性的性质,以及函数零点的个数进行判断排除即可.【解答】解:∵sin(﹣x)2=sinx2,

∴函数y=sinx2是偶函数,即函数的图象关于y轴对称,排除A,C;

由y=sinx2=0,

则x2=kπ,k≥0,

则x=±,k≥0,

故函数有无穷多个零点,排除B,

故选:D.

【点评】本题主要考查函数图象的识别和判断,根据函数奇偶性和函数零点的性质是解决本题的关键.比较基础.

4.(5分)若平面区域,夹在两条斜率为1的平行直线之间,则这两

条平行直线间的距离的最小值是()

A.B.C.D.

【分析】作出平面区域,找出距离最近的平行线的位置,求出直线方程,再计算距离.

【解答】解:作出平面区域如图所示:

∴当直线y=x+b分别经过A,B时,平行线间的距离相等.

联立方程组,解得A(2,1),

联立方程组,解得B(1,2).

两条平行线分别为y=x﹣1,y=x+1,即x﹣y﹣1=0,x﹣y+1=0.

∴平行线间的距离为d==,

故选:B.

【点评】本题考查了平面区域的作法,距离公式的应用,属于基础题.

5.(5分)已知a,b>0且a≠1,b≠1,若log a b>1,则()

A.(a﹣1)(b﹣1)<0 B.(a﹣1)(a﹣b)>0 C.(b﹣1)(b﹣a)<0 D.(b ﹣1)(b﹣a)>0

【分析】根据对数的运算性质,结合a>1或0<a<1进行判断即可.

【解答】解:若a>1,则由log a b>1得log a b>log a a,即b>a>1,此时b﹣a>0,b>1,即(b﹣1)(b﹣a)>0,

若0<a<1,则由log a b>1得log a b>log a a,即b<a<1,此时b﹣a<0,b<1,即(b﹣1)(b﹣a)>0,

综上(b﹣1)(b﹣a)>0,

故选:D.

【点评】本题主要考查不等式的应用,根据对数函数的性质,利用分类讨论的数学思想是解决本题的关键.比较基础.

6.(5分)已知函数f(x)=x2+bx,则“b<0”是“f(f(x))的最小值与f(x)的最小值相等”的()

A.充分不必要条件 B.必要不充分条件

C.充分必要条件D.既不充分也不必要条件

【分析】求出f(x)的最小值及极小值点,分别把“b<0”和“f(f(x))的最小值与f(x)的最小值相等”当做条件,看能否推出另一结论即可判断.

【解答】解:f(x)的对称轴为x=﹣,f min(x)=﹣.

(1)若b<0,则﹣>﹣,∴当f(x)=﹣时,f(f(x))取得最小值f(﹣)=﹣,

即f(f(x))的最小值与f(x)的最小值相等.

∴“b<0”是“f(f(x))的最小值与f(x)的最小值相等”的充分条件.

(2)设f(x)=t,则f(f(x))=f(t),

∴f(t)在(﹣,﹣)上单调递减,在(﹣,+∞)上单调递增,

若f(f(x))=f(t)的最小值与f(x)的最小值相等,

则﹣≤﹣,解得b≤0或b≥2.

∴“b<0”不是“f(f(x))的最小值与f(x)的最小值相等”的必要条件.

故选:A.

【点评】本题考查了二次函数的性质,简易逻辑关系的推导,属于基础题.

7.(5分)已知函数f(x)满足:f(x)≥|x|且f(x)≥2x,x∈R.()A.若f(a)≤|b|,则a≤b B.若f(a)≤2b,则a≤b

C.若f(a)≥|b|,则a≥b D.若f(a)≥2b,则a≥b

【分析】根据不等式的性质,分别进行递推判断即可.

【解答】解:A.若f(a)≤|b|,则由条件f(x)≥|x|得f(a)≥|a|,

即|a|≤|b|,则a≤b不一定成立,故A错误,

B.若f(a)≤2b,

则由条件知f(x)≥2x,

即f(a)≥2a,则2a≤f(a)≤2b,

则a≤b,故B正确,

C.若f(a)≥|b|,则由条件f(x)≥|x|得f(a)≥|a|,则|a|≥|b|不一定成立,故C错误,

D.若f(a)≥2b,则由条件f(x)≥2x,得f(a)≥2a,则2a≥2b,不一定成立,即a≥b不一定成立,故D错误,

故选:B.

【点评】本题主要考查不等式的判断和证明,根据条件,结合不等式的性质是解决本题的关键.综合性较强,有一定的难度.

8.(5分)如图,点列{A n}、{B n}分别在某锐角的两边上,且|A n A n+1|=|A n+1A n+2|,A n≠A n+1,n∈N*,|B n B n+1|=|B n+1B n+2|,B n≠B n+1,n∈N*,(P≠Q表示点P与Q不重合)若d n=|A n B n|,S n为△A n B n B n+1的面积,则()

A.{S n}是等差数列B.{S n2}是等差数列

C.{d n}是等差数列 D.{d n2}是等差数列

【分析】设锐角的顶点为O,再设|OA1|=a,|OB1|=c,|A n A n+1|=|A n+1A n+2|=b,|B n B n+1|=|B n+1B n+2|=d,由于a,c不确定,判断C,D不正确,设△A n B n B n+1的底边B n B n+1上的高为h n,运用三角形相似知识,h n+h n+2=2h n+1,由S n=d?h n,可得S n+S n+2=2S n+1,进而得到数列{S n}为等差数列.

【解答】解:设锐角的顶点为O,|OA1|=a,|OB1|=c,

|A n A n+1|=|A n+1A n+2|=b,|B n B n+1|=|B n+1B n+2|=d,

由于a,c不确定,则{d n}不一定是等差数列,

{d n2}不一定是等差数列,

设△A n B n B n+1的底边B n B n+1上的高为h n,

由三角形的相似可得==,

==,

两式相加可得,==2,

即有h n+h n+2=2h n+1,

由S n=d?h n,可得S n+S n+2=2S n+1,

﹣S n+1=S n+1﹣S n,

即为S n

+2

则数列{S n}为等差数列.

另解:可设△A1B1B2,△A2B2B3,…,A n B n B n+1为直角三角形,

且A1B1,A2B2,…,A n B n为直角边,

即有h n+h n+2=2h n+1,

由S n=d?h n,可得S n+S n+2=2S n+1,

即为S n

﹣S n+1=S n+1﹣S n,

+2

则数列{S n}为等差数列.

故选:A.

【点评】本题考查等差数列的判断,注意运用三角形的相似和等差数列的性质,考查化简整理的推理能力,属于中档题.

二、填空题

9.(6分)某几何体的三视图如图所示(单位:cm),则该几何体的表面积是80 cm2,体积是40cm3.

【分析】根据几何体的三视图,得出该几何体下部为长方体,上部为正方体的组合体,结合图中数据求出它的表面积和体积即可.

【解答】解:根据几何体的三视图,得;

该几何体是下部为长方体,其长和宽都为4,高为2,

表面积为2×4×4+2×42=64cm2,体积为2×42=32cm3;

上部为正方体,其棱长为2,

表面积是6×22=24 cm2,体积为23=8cm3;

所以几何体的表面积为64+24﹣2×22=80cm2,

体积为32+8=40cm3.

故答案为:80;40.

【点评】本题考查了由三视图求几何体的表面积与体积的应用问题,也考查了空间想象和计算能力,是基础题.

10.(6分)已知a∈R,方程a2x2+(a+2)y2+4x+8y+5a=0表示圆,则圆心坐标是(﹣2,﹣4),半径是5.

【分析】由已知可得a2=a+2≠0,解得a=﹣1或a=2,把a=﹣1代入原方程,配方求得圆心坐标和半径,把a=2代入原方程,由D2+E2﹣4F<0说明方程不表示圆,则答案可求.

【解答】解:∵方程a2x2+(a+2)y2+4x+8y+5a=0表示圆,

∴a2=a+2≠0,解得a=﹣1或a=2.

当a=﹣1时,方程化为x2+y2+4x+8y﹣5=0,

配方得(x+2)2+(y+4)2=25,所得圆的圆心坐标为(﹣2,﹣4),半径为5;当a=2时,方程化为,

此时,方程不表示圆,

故答案为:(﹣2,﹣4),5.

【点评】本题考查圆的一般方程,考查圆的一般方程化标准方程,是基础题.

11.(6分)已知2cos2x+sin2x=Asin(ωx+φ)+b(A>0),则A=,b=1.【分析】根据二倍角的余弦公式、两角和的正弦函数化简左边,即可得到答案.【解答】解:∵2cos2x+sin2x=1+cos2x+sin2x

=1+(cos2x+sin2x)

=sin(2x+)+1,

∴A=,b=1,

故答案为:;1.

【点评】本题考查了二倍角的余弦公式、两角和的正弦函数的应用,熟练掌握公式是解题的关键.

12.(6分)设函数f(x)=x3+3x2+1,已知a≠0,且f(x)﹣f(a)=(x﹣b)(x ﹣a)2,x∈R,则实数a=﹣2,b=1.

【分析】根据函数解析式化简f(x)﹣f(a),再化简(x﹣b)(x﹣a)2,根据等式两边对应项的系数相等列出方程组,求出a、b的值.

【解答】解:∵f(x)=x3+3x2+1,

∴f(x)﹣f(a)=x3+3x2+1﹣(a3+3a2+1)

=x3+3x2﹣(a3+3a2)

∵(x﹣b)(x﹣a)2=(x﹣b)(x2﹣2ax+a2)=x3﹣(2a+b)x2+(a2+2ab)x﹣a2b,且f(x)﹣f(a)=(x﹣b)(x﹣a)2,

∴,解得或(舍去),

故答案为:﹣2;1.

【点评】本题考查函数与方程的应用,考查化简能力和方程思想,属于中档题.

13.(4分)设双曲线x2﹣=1的左、右焦点分别为F1、F2,若点P在双曲线上,且△F1PF2为锐角三角形,则|PF1|+|PF2|的取值范围是.

【分析】由题意画出图形,以P在双曲线右支为例,求出∠PF2F1和∠F1PF2为直角时|PF1|+|PF2|的值,可得△F1PF2为锐角三角形时|PF1|+|PF2|的取值范围.【解答】解:如图,

由双曲线x2﹣=1,得a2=1,b2=3,

∴.

不妨以P在双曲线右支为例,当PF2⊥x轴时,

把x=2代入x2﹣=1,得y=±3,即|PF2|=3,

此时|PF1|=|PF2|+2=5,则|PF1|+|PF2|=8;

由PF1⊥PF2,得,

又|PF1|﹣|PF2|=2,①

两边平方得:,

∴|PF1||PF2|=6,②

联立①②解得:,

此时|PF1|+|PF2|=.

∴使△F1PF2为锐角三角形的|PF1|+|PF2|的取值范围是().

故答案为:().

【点评】本题考查双曲线的简单性质,考查双曲线定义的应用,考查数学转化思

想方法,是中档题.

14.(4分)如图,已知平面四边形ABCD,AB=BC=3,CD=1,AD=,∠ADC=90°,沿直线AC将△ACD翻折成△ACD′,直线AC与BD′所成角的余弦的最大值是.

【分析】如图所示,取AC的中点O,AB=BC=3,可得BO⊥AC,在Rt△ACD′中,AC=.作D′E⊥AC,垂足为E,D′E=.CO=,CE==,EO=CO﹣CE=.过点B作BF∥AC,作FE∥BO交BF于点F,则EF⊥AC.连接D′F.∠FBD′为直线AC与BD′所成的角.则四边形BOEF为矩形,BF=EO=.EF=BO=.则∠FED′为二面角D′﹣CA﹣B的平面角,设为θ.利用余弦定理求出D′F2的最小值即可得出.

【解答】解:如图所示,取AC的中点O,∵AB=BC=3,∴BO⊥AC,

在Rt△ACD′中,=.

作D′E⊥AC,垂足为E,D′E==.

CO=,CE===,

∴EO=CO﹣CE=.

过点B作BF∥AC,作FE∥BO交BF于点F,则EF⊥AC.连接D′F.∠FBD′为直线AC与BD′所成的角.

则四边形BOEF为矩形,∴BF=EO=.

EF=BO==.

则∠FED′为二面角D′﹣CA﹣B的平面角,设为θ.

则D′F2=+﹣2×cosθ=﹣5cosθ≥,cosθ=1时取等号.

∴D′B的最小值==2.

∴直线AC与BD′所成角的余弦的最大值===.

也可以考虑利用向量法求解.

故答案为:.

【点评】本题考查了空间位置关系、空间角,考查了空间想象能力、推理能力与计算能力,属于难题.

15.(4分)已知平面向量,,||=1,||=2,=1,若为平面单位向量,则||+||的最大值是.

【分析】由题意可知,||+||为在上的投影的绝对值与在上投影的绝对值的和,由此可知,当与共线时,||+||取得最大值,即.【解答】解:||+||=,

其几何意义为在上的投影的绝对值与在上投影的绝对值的和,

当与共线时,取得最大值.

∴=.

故答案为:.

【点评】本题考查平面向量的数量积运算,考查向量在向量方向上的投影的概念,

考查学生正确理解问题的能力,是中档题.

三、解答题

16.(14分)在△ABC中,内角A,B,C所对的边分别为a,b,c,已知b+c=2acosB.(1)证明:A=2B;

(2)若cosB=,求cosC的值.

【分析】(1)由b+c=2acosB,利用正弦定理可得:sinB+sinC=2sinAcosB,而sinC=sin (A+B)=sinAcosB+cosAsinB,代入化简可得:sinB=sin(A﹣B),由A,B∈(0,π),可得0<A﹣B<π,即可证明.

(II)cosB=,可得sinB=.cosA=cos2B=2cos2B﹣1,sinA=.利用cosC=﹣cos(A+B)=﹣cosAcosB+sinAsinB即可得出.

【解答】(1)证明:∵b+c=2acosB,

∴sinB+sinC=2sinAcosB,

∵sinC=sin(A+B)=sinAcosB+cosAsinB,

∴sinB=sinAcosB﹣cosAsinB=sin(A﹣B),由A,B∈(0,π),

∴0<A﹣B<π,∴B=A﹣B,或B=π﹣(A﹣B),化为A=2B,或A=π(舍去).

∴A=2B.

(II)解:cosB=,∴sinB==.

cosA=cos2B=2cos2B﹣1=,sinA==.

∴cosC=﹣cos(A+B)=﹣cosAcosB+sinAsinB=+×=.

【点评】本题考查了正弦定理、和差公式、倍角公式、同角三角函数基本关系式、诱导公式,考查了推理能力与计算能力,属于中档题.

17.(15分)设数列{a n}的前n项和为S n,已知S2=4,a n+1=2S n+1,n∈N*.(Ⅰ)求通项公式a n;

(Ⅱ)求数列{|a n﹣n﹣2|}的前n项和.

【分析】(Ⅰ)根据条件建立方程组关系,求出首项,利用数列的递推关系证明数列{a n}是公比q=3的等比数列,即可求通项公式a n;

(Ⅱ)讨论n的取值,利用分组法将数列转化为等比数列和等差数列即可求数列{|a n﹣n﹣2|}的前n项和.

【解答】解:(Ⅰ)∵S2=4,a n+1=2S n+1,n∈N*.

∴a1+a2=4,a2=2S1+1=2a1+1,

解得a1=1,a2=3,

=2S n+1,a n=2S n﹣1+1,

当n≥2时,a n

+1

﹣a n=2(S n﹣S n﹣1)=2a n,

两式相减得a n

+1

即a n

=3a n,当n=1时,a1=1,a2=3,

+1

=3a n,

满足a n

+1

∴=3,则数列{a n}是公比q=3的等比数列,

则通项公式a n=3n﹣1.

(Ⅱ)a n﹣n﹣2=3n﹣1﹣n﹣2,

设b n=|a n﹣n﹣2|=|3n﹣1﹣n﹣2|,

则b1=|30﹣1﹣2|=2,b2=|3﹣2﹣2|=1,

当n≥3时,3n﹣1﹣n﹣2>0,

则b n=|a n﹣n﹣2|=3n﹣1﹣n﹣2,

此时数列{|a n﹣n﹣2|}的前n项和T n=3+﹣=,

则T n==.

【点评】本题主要考查递推数列的应用以及数列求和的计算,根据条件建立方程组以及利用方程组法证明列{a n}是等比数列是解决本题的关键.求出过程中使用了转化法和分组法进行数列求和.

18.(15分)如图,在三棱台ABC﹣DEF中,平面BCFE⊥平面ABC,∠ACB=90°,

BE=EF=FC=1,BC=2,AC=3.

(Ⅰ)求证:BF⊥平面ACFD;

(Ⅱ)求直线BD与平面ACFD所成角的余弦值.

【分析】(Ⅰ)根据三棱台的定义,可知分别延长AD,BE,CF,会交于一点,并设该点为K,并且可以由平面BCFE⊥平面ABC及∠ACB=90°可以得出AC⊥平面BCK,进而得出BF⊥AC.而根据条件可以判断出点E,F分别为边BK,CK的中点,从而得出△BCK为等边三角形,进而得出BF⊥CK,从而根据线面垂直的判定定理即可得出BF⊥平面ACFD;

(Ⅱ)由BF⊥平面ACFD便可得出∠BDF为直线BD和平面ACFD所成的角,根据条件可以求出BF=,DF=,从而在Rt△BDF中可以求出BD的值,从而得出cos∠BDF的值,即得出直线BD和平面ACFD所成角的余弦值.

【解答】解:(Ⅰ)证明:延长AD,BE,CF相交于一点K,如图所示:∵平面BCFE⊥平面ABC,且AC⊥BC;

∴AC⊥平面BCK,BF?平面BCK;

∴BF⊥AC;

又EF∥BC,BE=EF=FC=1,BC=2;

∴△BCK为等边三角形,且F为CK的中点;

∴BF⊥CK,且AC∩CK=C;

∴BF⊥平面ACFD;

(Ⅱ)∵BF⊥平面ACFD;

∴∠BDF是直线BD和平面ACFD所成的角;

∵F为CK中点,且DF∥AC;

∴DF为△ACK的中位线,且AC=3;

∴;

又;

∴在Rt△BFD中,,cos;

即直线BD和平面ACFD所成角的余弦值为

【点评】考查三角形中位线的性质,等边三角形的中线也是高线,面面垂直的性质定理,以及线面垂直的判定定理,线面角的定义及求法,直角三角形边的关系,三角函数的定义.

19.(15分)如图,设抛物线y2=2px(p>0)的焦点为F,抛物线上的点A到y 轴的距离等于|AF|﹣1,

(Ⅰ)求p的值;

(Ⅱ)若直线AF交抛物线于另一点B,过B与x轴平行的直线和过F与AB垂直的直线交于点N,AN与x轴交于点M,求M的横坐标的取值范围.

【分析】(Ⅰ)利用抛物线的性质和已知条件求出抛物线方程,进一步求得p值;(Ⅱ)设出直线AF的方程,与抛物线联立,求出B的坐标,求出直线AB,FN 的斜率,从而求出直线BN的方程,根据A、M、N三点共线,可求出M的横坐标的表达式,从而求出m的取值范围.

【解答】解:(Ⅰ)由题意可得,抛物线上点A到焦点F的距离等于A到直线x=﹣1的距离,

由抛物线定义得,,即p=2;

(Ⅱ)由(Ⅰ)得,抛物线方程为y2=4x,F(1,0),可设(t2,2t),t≠0,t≠±1,

∵AF不垂直y轴,

∴设直线AF:x=sy+1(s≠0),

联立,得y2﹣4sy﹣4=0.

y1y2=﹣4,

∴B(),

又直线AB的斜率为,故直线FN的斜率为,

从而得FN:,直线BN:y=﹣,

则N(),

设M(m,0),由A、M、N三点共线,得,

于是m==,得m<0或m>2.

经检验,m<0或m>2满足题意.

∴点M的横坐标的取值范围为(﹣∞,0)∪(2,+∞).

【点评】本题考查抛物线的简单性质,考查直线与圆锥曲线位置关系的应用,考查数学转化思想方法,属中档题.

20.(15分)设函数f(x)=x3+,x∈[0,1],证明:

(Ⅰ)f(x)≥1﹣x+x2

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