重庆西南师大附中2009届高三第三次月考(数学理)

西南师大附中高2009级第三次月考数 学 试 题（理）2008年11月本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分．共150分，考试时间120分钟．第Ⅰ卷 （选择题 共50分）一、选择题：本大题共10小题，每题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.已知向量(82)(1)a b x == ，，，，若//a b，则x 的值为（ ） A ．14 B ．14- C ．4 D ．– 42.两圆32cos 3cos 42sin 3sin x x y y θθθθ=-+=⎧⎧⎨⎨=+=⎩⎩与的位置关系是（ ） A ．内切B ．外切C ．相离D ．内含3.“a = 3”是“直线210ax y --=与直线640x y c -+=平行”的（ ）条件A ．充要B ．充分而不必要C ．必要而不充分D ．既不充分也不必要4.已知直线l 1：122y x =+，l 2过点P （– 3，1），且l 1到l 2的角为45︒，则l 2的方程为（ ） A ．3100x y ++= B ．3100x y -+= C ．20x y ++= D ．30x y +=5.已知23()1x f x x +=-，函数()y g x =的图象与函数1(1)y f x -=+的图象关于直线y x =对称，则(11)g 等于（ ）A ．32B ．52C ．2711D ．2396.已知点P （x ，y ）是直线kx + y + 4 = 0（k > 0）上一动点，P A 、PB 是圆C ：2220x y y +-=的两条切线，A 、B 是切点，若四边形P ACB 的最小面积是2，则k 的值为（ ）A ．3B C ．D ．27.已知4a b ab +=，a 、b 均为正数，则使a b m +>恒成立的m 的取值范围是（ ）A ．m < 9B ．9m ≤C ．m < 8D ．8m ≤8.已知{}n a 为等差数列，若11101a a <-，且它的前n 项和n S 有最大值，那么当n S 取得最小正值时，n =（ ） A ．11B ．20C ．19D ．219.已知O 是△ABC 内一点，230OA OB OC ++=，则△AOC 与△BOC 的面积的比值为（ ） A ．32 B ．53C ．2D ．310.把数列{21n +}依次按第一个括号一个数，第二个括号两个数，第三个括号三个数，第四个括号四个数，第五个括号一个数，第六个括号两个数，…循环分别为（3），（5，7），（9，11，13），（15，17，19，21），（23），（25，27），（29，31，33），（35，37，39，41），（43） （45，47）…则第104个括号内各数之和为（ ）A ．2036B ．2048C ．2060D ．2072第II 卷 （共100分）二、填空题：本大题共6小题，每题4分，共24分． 11.若cos 2sin 4απα=⎛⎫- ⎪⎝⎭cos sin αα+的值为____________． 12.等差数列{}n a 的前n 项和为n S ，若31815618S S S =--=，，则18S =____________． 13.如图，某地一天从6时到14时的温度变化曲线近似满足函数sin()(02)y A x B ωϕϕπ=++≤<，则温度变化曲线的函数解析式为____________．14.已知函数2|1|(0)()1(0)x x f x x x -+≤⎧=⎨->⎩．那么不等式()0f x <的解集为_______________． 15.x 、y 满足约束条件：225040y x y x y ≥⎧⎪+-≥⎨⎪+-≤⎩，则|5|z x y =+-的最小值是______________．16对于函数f ( x ) = x | x | + px + q ，现给出四个命题：① q = 0时，f ( x )为奇函数② y = f ( x )的图象关于（0，q ）对称③ p = 0，q > 0时，方程f ( x ) = 0有且只有一个实数根 ④方程f ( x ) = 0至多有两个实数根其中正确命题的序号为_____________________．三、解答题：本题共6小题，共76分，解答应写出文字说明、证明过程或演算步骤． 17.(本小题满分12分)已知2(53c o s c os)(si a x x b x x fx a===+，，，，记函数． (1) 求函数f ( x )的最小正周期； (2) 当62x ππ≤≤时，求函数f ( x )的值域．18.(本小题满分12分) 已知2(1)()(0)2x p x pf x p x p+++=>+（1）若p > 1时，解关于x 的不等式()0f x ≥； （2）若()2f x >对24x ≤≤时恒成立，求p 的范围．19.(本小题满分12分) 数列{a n }中，a 1 = 1，当2n ≥时，其前n 项和满足21()2n n n S a S =-（1）求S n 的表达式；（2）设21n n Sb n =+， 数列{b n }的前n 项和为T n ，求T n ．20.(本小题满分14分) 已知圆C 经过P （4，– 2），Q （– 1，3）两点，且在y 轴上截得的线段长为，半径小于5． （1）求直线PQ 与圆C 的方程．（2）若直线l ∥PQ ，且l 与圆C 交于点A 、B ，90AOB ∠=︒，求直线l 的方程．21.(本小题满分12分) 已知函数1()|1|(0)f x x x=->， . （1）当0 < a < b ，且f ( a ) = f ( b )时，求证：ab >1；（2）若存在实数a ，b （a < b ），使得函数()y f x =的定义域为 [a ，b ]时，值域为[]ma mb ，(0m ≠)，求m 的取值范围．22.(本小题满分14分) 数轴上有一列点P 1，P 2，P 3，…，P n ，…，已知当2n ≥时，点P n 是把线段P n – 1 P n +1作n 等分的分点中最靠近P n+1的点，设线段P 1P 2，P 2P 3，…，P n P n + 1的长度分别为a 1，a 2，a 3，…，a n ，其中a 1 = 1． （1）写出a 2，a 3和a n （2n ≥，*n N ∈）的表达式；（2）证明a 1 + a 2 + a 3 +…+a n < 3（*n N ∈）；（3）设点M n ( n ，a n )（n > 2，*n N ∈），在这些点中是否存在两个点同时在函数2(0)(1)k y k x =>-的图像上，如果存在，请求出点的坐标；如果不存在，请说明理由．西南师大附中高2009级第三次月考数学试题参考答案（理）一、选择题：本大题共10小题，每题5分，共50分． （2008年11月）1．C2．B3．C4．B5．A6．D7．A8．C9．C10．D二、填空题：本大题共6小题，每题4分，共24分．11．12 12．36 13．310sin()2084y x ππ=++ 14．{|11}x x x <≠-且 15．1 16．①②③ 三、解答题：本题共6小题，共76分．17．解：(1) 2222()||cos 2cos sin 4cos f x a b b x x x x x =+=+++ ··························· 2分21cos2cos 5cos 12512xx x x x +=+++⨯+························ 4分 75sin(2)62x π=++ ·························································································· 6分 ∴ 22T ππ==． ································································································ 7分 (2) 当7262266x x πππππ≤≤≤+≤，即， ········································································· 8分1s i n (2)126x π-≤+≤有 ························································································· 10分 ∴ 17()[1]622x f x ππ≤≤时，函数的值域为，． ·············································· 12分18．解：(1) ()(1)()02x p x f x x p++=≥+··························································································· 1分 ①12{|1}2pp x p x x <<-≤≤->-时，解集为或 ·············································· 3分② p = 2时，解集为{|21}x x x ≥-≠-且 ···························································· 5分③ p > 2时，解集为{|1}2px p x x -≤<-≥-或 ·················································· 7分(2) 2(1)22x p x p x p+++>+2(1)42x p x p x p +++>+ ····················································································· 8分∴ 2(3)024x p x p x +-->≤≤对恒成立∴ 232(2)2411x x p x x x x ->=--+≤≤--对恒成立············································ 9分 ∵ 2()(2)[24]1g x x x =--+-在，上递减 ························································· 10分 ∴ max ()(2)2g x g == ··························································································· 11分 ∴ p > 2 ················································································································· 12分19．解：(1) 当2n ≥时，1n n n a S S -=-代入已知得211()()2n n n n S S S S -=-- ··························································································· 2分 化简得：111122n n n n S S S S --=- ··············································································· 3分两边同除以11112n n n n S S S S ---=得 ·········································································· 4分 ∴111(1)212(1)21n n n n S S =+-=+-=- ··························································· 6分 ∴ 121n S n =- ········································································································· 7分 (2) ∵ 1111121()2121(21)(21)22121n n S n b n n n n n n -====-++-+-+ ··························· 10分 ∴ 12n n T b b b =+++111111(1)2335212111(1)221n n n =-+-++--+=-+ 21nn =+ ······································································································· 12分 20．解：(1) PQ 为323(1)2014y x x y +-=⨯++-=--即 ····························································· 2分 C 在PQ 的中垂线32411()22y x ---=⨯-即y = x – 1上 ······································· 3分 设C （n ，n – 1），则2222||(1)(4)r CQ n n ==++- ··············································· 4分由题意，有222||r n =+ ················································································· 5分 ∴ 22122617n n n +=-+ ∴ n = 1或5，r 2 = 13或37（舍） ·························· 7分 ∴圆C 为22(1)13x y -+=························································································ 8分 解法二：设所求圆的方程为220x y Dx Ey F ++++=由已知得24220310448D E F D E F E F ⎧-+=-⎪--=⎨⎪-=⎩解得2108124D DE EF F =-=-⎧⎧⎪⎪==-⎨⎨⎪⎪=-=⎩⎩或 当2012D E F =-⎧⎪=⎨⎪=-⎩时，5r ；当1084D E F =-⎧⎪=-⎨⎪=⎩时，5r =（舍） ∴ 所求圆的方程为222120x y x +--= (2) 设l 为0x y m ++= ································································································ 9分由22(1)13x y m x y ++=⎧⎨-+=⎩，得222(22)120x m x m +-+-= ····································· 10分 设A （x 1，y 1），B （x 2，y 2），则212121212m x x m x x -+=-=， ························ 11分∵ 90AOB ∠=︒， ∴ 12120x x y y += ··························································· 12分∴ 1212()()0x x x m x m +++=∴ 2120m m +-= ∴ m = 3或 – 4（均满足0∆>） ∴ l 为3040x y x y ++=+-=或 ···································································· 14分21．解：(1) ∵ x > 0，∴111()1101x xf x x x⎧-≥⎪⎪=⎨⎪-<<⎪⎩，，∴ f ( x )在（0，1）上为减函数，在(1)+∞，上是增函数． 由0 < a < b ，且f ( a ) = f ( b )，可得 0 < a <1 < b 和1111a b -=-．即112a b+=． ·············································· 3分∴2ab = a + b> ························································································ 4分1>，即ab >1． ··························································································· 5分(2) 若存在实数a ，b （a < b ），使得函数y = f ( x )的定义域为[a ，b ]时，值域为[ma ，mb ]．则a > 0，m > 0．① 当(01)a b ∈，，时，由于()f x 在（0，1）上是减函数，故11,11.mb ama b⎧-=⎪⎪⎨⎪-=⎪⎩．此时得a = b 不符合题意，所以a ，b 不存在． ·········································································· 7分 ② 当(01)a ∈，或[1)b ∈+∞，时，由（II ）知0在值域内，值域不可能是[ma ，mb ]，所以a ，b 不存在． ····································································································· 9分故只有[1)a b ∈+∞，，．∵1()1f x x=-在[1)+∞，上是增函数， ∴(),().f a ma f b mb =⎧⎨=⎩ 即 11,11.ma amb b⎧-=⎪⎪⎨⎪-=⎪⎩ ······································································· 10分a ，b 是方程210mx x -+=的两个根．即关于x 的方程210mx x -+=有两个大于1的实根．设这两个根为x 1，x 2．则121x x m +=，121x x m= ．∴12120,(1)(1)0,(1)(1)0.x x x x ∆>⎧⎪-+->⎨⎪-->⎩ 即 140,120.m m ->⎧⎪⎨->⎪⎩ 解得 104m <<．故m 的取值范围是104m <<． ······················································································ 12分 另解：设2()1g x mx x =-+，则01401(1)000411122m g m m m m⎧⎧⎪⎪∆>->⎪⎪>⇒><<⎨⎨⎪⎪⎪⎪><⎩⎩解得 22．解：(1) 由已知11(1)n n n n P P n P P -+=-，令n = 2，P 1P 2 = P 2P 3，所以a 2 = 1， ······································································· 1分令n = 3，P 2P 3 = 2P 3P 4，所以312a =， ···································································· 2分同理，111n n a a n -=-．所以1211111111(2)112122(1)!n n n a a a n n n n n n n --=====≥------ ······· 5分 (2) 因为21111(2)(1)!1234(1)22222n n n n -=≤=≥-⨯⨯⨯⨯⨯-所以123221111111111!2!(1)!222n n a a a a n -++++=++++≤+++++-1211()1213()3(2)1212n n n ---=+=-<≥-．而n = 1时，易知a 1 = 1 < 3成立，所以*1233()n a a a a n N ++++<∈ ········· 10分(3) 假设有两个点A （p ，a p ），B （q ，a q ）*(22)p q p q N p q ≠∈>>，、，且，，都在函数2(1)ky x =-即22(1)(1)p q k ka a p q ==--，．所以22(1)(1)(1)!(1)!p q k k p q --==--，．消去k 得22(1)(1)(1)!(1)!p q p q --=--，……① 以下考查数列{b n }，2!n n b n =的增减情况，22221(1)(1)31!(1)!(1)!(1)!n n n n n n n n b b n n n n -----+-=-==----， 当n > 2时，n 2 – 3n + 1 > 0，所以对于函数{b n }有b 2 > b 3 > b 4 > … > b n > … 所以①式不能成立，所以，不可能有两个点同时在函数2(1)ky x =-图像上． ································ 14分。