[套卷]四川省资阳市高中2014届高三第一次诊断性考试数学理试题

页1第四川省资阳市高中2014届高三第一次诊断性考试数学理试题本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）。

第Ⅰ卷1至2页，第Ⅱ卷3至4页。

考生作答时，须将答案答在答题卡上，在本试题卷、草稿纸上答题无效。

满分150分。

考试时间120分钟，考试结束后，将本试题卷和答题卡一并收回。

第Ⅰ卷 （选择题 共50分）注意事项：必须使用2B 铅笔在答题卡上将所选答案的标号涂黑。

一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1．已知集合A ＝{4，5，6，8}，B ={3，5，7，8}，则A ∩B ＝ （A ）{3，5} （B ）{6，8}（C ）{5，8}（D ）{3，4，5，6，7，8}2．已知向量a ＝(3, 4)，b ＝(1, 3)，则a －2b ＝ （A ）(1, 3) （B ）(1, －2) （C ）(2, 1)（D ）(2, －2)3．已知i 是虚数单位，a ，b ∈R ，且(i)i 2i a b +=-，则a ＋b ＝ （A ）1（B ）－1（C ）－2（D ）－34. 函数()lg(1)xf x x =-的定义域为（A ）(12)(2)+∞ ，， （B ）(01)(1)+∞ ，， （C ）(2)+∞，（D ）(1)+∞，5. 命题:p n ∀∈Z ，n ∈Q ，则 （A ）:p ⌝n ∀∉Z ，n ∉Q（B ）:p ⌝n ∀∈Z ，n ∉Q （C ）:p ⌝0n ∃∉Z ，0n ∈Q（D ）:p ⌝0n ∃∈Z ，0n ∉Q6. ABC ∆中，若222sin sin sin sin sin 0B C A B C +-+=，则A = （A ）23π（B ）56π （C ）3π（D ）6π 7. 若把函数sin y x ω=（0ω>）的图象向左平移3π个单位后与函数cos y x ω=的图象重合，则ω的值可能是 （A ）13（B ）12（C）32（D）238. 函数ln||y x x=的图象大致是（A）（B）（C）（D）9．已知函数()e ex xf x m-=-，若()f x'≥恒成立，则实数m的取值范围是（A）[0,)+∞（B）[2,)+∞（C）[3,)+∞（D）(,3]-∞10．如图，在边长为2的正六边形ABCDEF中，动圆Q的半径为1，圆心在线段CD（含端点）上运动，P是圆Q上及内部的动点，设向量AP mAB nAF=+（m，n为实数），则m n+的取值范围是（A）(1,2]（B）[5,6]（C）[2,5]（D）[3,5]页2第页 3第第Ⅱ卷 （非选择题 共100分）注意事项：必须使用0.5毫米黑色签字笔在答题卡上题目指示的答题区域内作答。

作图时可先用铅笔绘出，确认后再用0.5毫米黑色签字笔描清楚。

答在试题卷上无效。

二、填空题：本大题共5小题，每小题5分，共25分。

11．在平面直角坐标系中，角α的顶点与原点重合，始边与x 轴的非负半轴重合，若角α终边经过点(2,4)P ，则tan()4πα+=___________.12．若210a =，5log 10b =，则11a b+=______. 13．已知向量a ，b 的夹角为45 ，2||||a b ==， 且向量a 与λb a -垂直，则实数λ=________. 14．已知x ∈R ，根据右图所示的程序框图，则 不等式1()22f x x -+≥的解集是____________.15．在平面直角坐标系中，横坐标、纵坐标均为整数的点称为整点．如果函数()f x 的图象恰好通过k （*k ∈N ）个整点，则称()f x 为k 阶整点函数．给出下列函数：①()cos f x x =；②2()(1)f x x π=-；③21()()3x f x -=；④0.6()log (1)f x x =+；⑤1()1f x x =-.其中是1阶整点函数的序号有______________.（写出所有满足条件的函数的序号） 三、解答题：共6大题，共75分。

解答应写出文字说明，证明过程或演算步骤。

16.（本小题满分12分）在等比数列{}n a 中，11a =，且14a ，22a ，3a 成等差数列. （Ⅰ）求n a ；（Ⅱ）令2log n n b a =，求数列{}n b 的前n 项和n S .17.（本小题满分12分）设向量m (cos α=，1)，n (sin α=，2)，且m ∥n ，其中(0)2πα∈,.（Ⅰ）求sin α；（Ⅱ）若3sin()5αβ-=，(0)2πβ∈，，求cos β.页4第18.（本小题满分12分）设()f x 是定义在实数集R 上的奇函数，当0x >时，2()4f x x x =-+. （Ⅰ）求()f x 的解析式，并解不等式()f x x ≥；（Ⅱ）设1()2x g x m -=+，若对任意1[14]x ∈-，，总存在2[25]x ∈，，使12()()f x g x =，求实数m 的取值范围．19.（本小题满分12分）已知函数2()2sin()cos sin cos 3f x x x x x x π=-++（x ∈R ）.（Ⅰ）求()f x 在[0]π,内的单调递增区间；（Ⅱ）在ABC ∆中，B 为锐角，且()f B =，AC =D 是BC 边上一点，AB AD =，试求AD DC +的最大值．20.（本小题满分13分）如图，将一矩形花坛ABCD 扩建成一个更大的矩形花坛AMPN ，要求点B 在AM 上，点D 在AN 上，点C 在MN 上，3AB =米，2AD =米．（Ⅰ）要使扩建成的花坛面积大于272米，则AN 的长度应在什么范围内？（Ⅱ）当AN 的长度是多少米时，扩建成的花坛面积最小？并求出最小面积．21.（本小题满分14分）已知函数2()2ln f x x x ax =-+（a ∈R ）. （Ⅰ）当2a =时，求()f x 的图象在1x =处的切线方程；（Ⅱ）若函数()()g x f x ax m =-+在1[e]e,上有两个零点，求实数m 的取值范围；（Ⅲ）若函数()f x 的图象与x 轴有两个不同的交点12(0)(0)A x B x ，，，，且120x x <<， 求证：12()02x x f +'<（其中()f x '是()f x 的导函数）．资阳市高中2011级第一次诊断性考试数学参考答案及评分标准（理工类）一、选择题：CBDAD ，ACBCC.二、填空题：11.3-；12.1；；14.[04]，；15.①②④. 16.【解】（Ⅰ）设{}n a 的公比为q ，由14a ，22a ，3a 成等差数列，得13244a a a +=. 又11a =，则244q q +=，解得2q =. ∴12n n a -=（*N n ∈ ）.·········································· 6分页 5第（Ⅱ）12log 21n n b n -==-，∴11n n b b +-=，{}n b 是首项为0，公差为1的等差数列，它的前n 项和(1)2n n n S -=. ······································································································ 12分17.【解】（Ⅰ）∵m //n ，∴2cos sin αα=， ····································································· 2分又22sin cos 1αα+=，∴221sin sin 14αα+=，∴24sin 5α=，············································ 4分∵(0)2πα∈，，∴sin 0α>，故sin α=. ········································································· 6分（Ⅱ）∵(0)2πα∈，，(0)2πβ∈，，∴22ππαβ-<-<.∵3sin()5αβ-=，∴4cos()5αβ-=；sin α=cos α=． ·································· 9分cos βcos[()]ααβ=--cos cos()sin sin()ααβααβ=-+- ················································ 11分4355=····························································································· 12分 18.【解】（Ⅰ）当0x =时，()0f x =； ················································································· 1分当0x <时，有0x ->，由22()()[()4()]4f x f x x x x x =--=---+-=+. ····························· 3分∴()f x 的解析式为2240()40.x x x f x x x x ⎧-+≥⎪=⎨+<⎪⎩，，， ········································································· 4分当0x ≥时，()f x x ≥为24x x x -+≥，解得03x ≤≤；当0x <时，()f x x ≥为24x x x +≥，解得3x ≤-.故不等式()f x x ≥的解集是{|3x x ≤-或03}x ≤≤. ······································································ 6分 （Ⅱ）当10x -≤<时，22()4(2)4f x x x x =+=+-，知()[30)f x ∈-，；当04x ≤≤时， 22()4(2)4f x x x x =-+=--+，知()[04]f x ∈，，∴当1[14]x ∈-，时，1()[34]f x ∈-，. ··· 8分 ∵1()2x g x m -=+是R 上的增函数，∴当2[25]x ∈，时，2()[216]g x m m ∈++，， ··········· 9分 ∵对任意1[14]x ∈-，，总存在2[25]x ∈，使12()()f x g x =，∴[34][216]m m -⊆++，，， ···· 10分 则23,164,m m +≤-⎧⎨+≥⎩解得125m -≤≤-，故实数m 的取值范围是[125]--，. ······························ 12分 19.【解】（Ⅰ）21()2(sin )cos sin cos 2f x x x x x x x =+222sin cos sin )x x x x =--sin 2x x =2sin(2)3x π=-. ································· 2分由222232k x k πππππ-+≤-≤+，得51212k x k ππππ-+≤≤+（Z k ∈）. ··························· 3分 取0k =，得51212x ππ-≤≤，又[0]x π∈,，则5[0]12x π∈,； ················································ 4分取1k =，得11171212x ππ≤≤，又[0]x π∈,，则11[]12x ππ∈,.················································ 5分 ∴()f x 在[0]π，上的单调递增区间是5[0]12π,，11[]12ππ,. ················································· 6分（Ⅱ）由()f B =得sin(2)3B π-又02B π<<，则22333B πππ-<-<，从而233B ππ-=，∴3B π=. ············································································································ 8分由AB AD =知ABD ∆是正三角形，AB AD BD ==，∴AD DC BD DC BC +=+=， 在ABC ∆中，由正弦定理，得sin sin3BCBACπ=∠，即8sin BC BAC =∠.页6第∵D 是BC 边上一点，∴233BAC ππ<∠<sin 1BAC <∠≤，知8BC ≤.当26BAC C ππ∠==，时，AD CD +取得最大值8. ······························································· 12分【另】在ACD ∆中，由正弦定理，得sin sin()sin 33AD DC C C π==-8sin AD C =， 8sin()3CD C π=-，则8sin 8sin()3AD DC C C π+=+-18(sin sin )2C C C =-1sin )2C C =+8sin()3C π=+．∵23ADC π∠=，∴03C π<<，2333C πππ<+<， 当32C ππ+=，即6C π=时，AD DC +取得最大值8. ·························································· 12分20.【解】（Ⅰ）设AN x =（米），则2x >.∵DCN AMN ∆∆ ，∴DN DCAN AM=， 则23x x AM -=，32xAM x =-. ································································································· 2分 ∴花坛AMPN 的面积232x S AM AN x =⋅=-（2x >）. ·························································· 4分由27S >，得23272x x >-，则29180x x -+>，∴23x <<或6x >，故AN 的长度范围是23AN <<或6AN >（米）. ································································ 8分（Ⅱ）由2233[(2)4(2)4]22x x x S x x -+-+==--43[(2)4]242x x =-++≥-， ························ 12分 当且仅当422x x -=-，即4x =（米）时，等号成立.∴当AN 的长度是4米时，扩建成的花坛AMPN 的面积最小，最小值为242米. ············ 13分【另】（Ⅰ）设DN x =（米）(0x >)，则2AN x =+. ∵DCN ∆∽AMN ∆，∴DN DCAN AM=， 则32x x AM =+，63AM x=+. ································································································· 2分 ∴花坛AMPN 的面积612(3)(2)312S AM AN x x x x=⋅=++=++（0x >）. ······················ 4分由27S >，得1231227x x++>，则2540x x -+>，∴01x <<或4x >，故AN 的长度范围是23AN <<或6AN >（米）. ································································ 8分（Ⅱ）由612(3)(2)3121224S x x xx=++=++≥=， ················································· 12分当且仅当123x x=，即2x =（米）时，等号成立. ∴当AN 的长度是4米时，扩建成的花坛AMPN 的面积最小，最小值为242米. ············ 13分21.【解】（Ⅰ）当2a =时，2()2ln 2f x x x x =-+，2()22f x x x'=-+，切点坐标为(11)，，切线的斜率(1)2k f '==，则切线方程为12(1)y x -=-，即21y x =-. ····························· 2分（Ⅱ）2()2ln g x x x m =-+，则22(1)(1)()2x x g x x xx-+-'=-=，∵1[e]e x ∈，，故()0g x '=时，1x =.当11e x <<时，()0g x '>；当1e x <<时，()0g x '<.故()g x 在1x =处取得极大值(1)1g m =-. ·············································································· 4分又211()2e e g m =--，2(e)2e g m =+-，2211(e)()4e 0e e g g -=-+<，则1(e)()eg g <，页 7第∴()g x 在1[e]e，上的最小值是(e)g . ······················································································ 6分()g x 在1[e]e ，上有两个零点的条件是2(1)10,11()20,eeg m g m =->⎧⎪⎨=--≤⎪⎩解得2112e m <≤+， ∴实数m 的取值范围是21(12]e +，. ··························································································· 8分 （Ⅲ）∵()f x 的图象与x 轴交于两个不同的点12(0)(0)A x B x ，，，，∴方程22ln 0x x ax -+=的两个根为12x x ，，则211122222ln 0,2ln 0,x x ax x x ax ⎧-+=⎪⎨-+=⎪⎩两式相减得1212122(ln ln )()x x a x x x x -=+--.又2()2ln f x x x ax =-+，2()2f x x a x'=-+，则1212124()()2x x f x x a x x +'=-+++1212122(ln ln )4x x x x x x -=-+-. ······················································ 10分 下证1212122(ln ln )40x x x x x x --<+-（*），即证明2111222()ln 0x x x x x x -+<+，12x t x =， ∵120x x <<，∴01t <<，即证明2(1)()ln 01t u t t t -=+<+在01t <<上恒成立. ··················· 12分 ∵22222(1)2(1)114(1)()(1)(1)(1)t t t u t t t t t t t -+---'=+=-=+++，又01t <<，∴()0u t '>， ∴()u t 在(0,1)上是增函数，则()(1)0u t u <=，从而知2111222()ln 0x x x x x x -+<+，故（*）式＜0，即12()02x x f +'<成立. ····················································································· 14分。