福州市高中毕业班数学质量检测答案

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福建福州市2023-2024高一上学期期末质量检测数学试卷及答案

2023-2024学年第一学期福州市四校教学联盟1月期末学业联考高一数学试卷考试范围：必修一命题教师：审核教师：考试时间：1月3日完卷时间：120分钟满分：150分一、单项选择题：本大题共8小题，每小题5分，满分40分。

在每小题所给出的四个选项中，只有一个选项是符合题意的。

1．集合A={x∣−2<x≤2},B={−2,−1,0,1},则A∩B=A．{−1,1,2}B．{−2,−1,0,1}C．{−1,0,1}D．{−2,−1,0,1,2}2．若a>b>0，c>d，则下列结论正确的是3．函数y=−|ln(x−1)|的图象大致是A．B．C．D．4．命题p：α是第二象限角或第三象限角，命题q：cosα<0，则p是q的A．充要条件B．必要不充分条件C．充分不必要条件D．既不充分也不必要条件A．110%B．120%C．130%D．140%7．命题“对∀x∈[1,2]，ax2−x+a>0”为真命题的一个充分不必要条件可以是8．已知f(x)=ax2−1是定义在R上的函数，若对于任意−3≤x1<x2≤−1，都有f(x1)−f(x2)<2，则实数x1−x2a的取值范围是二、多项选择题：本大题共4小题，每小题5分，满分20分。

在每小题所给出的四个选项中，有多个选项是符合题意的。

9．下列大小关系正确的是A．20.3<20.4B．30.2<40.2C．log23<log48D．log23>log32 10．设正实数x，y满足x+y=2，则下列说法正确的是A．当k>1，有1个零点B．当k>1时，有3个零点C．当k<0时，有9个零点D．当k=−4时，有7个零点三、填空题：本大题共4小题，每小题5分，满分20分。

13．已知扇形的圆心角是2rad，其周长为6cm，则扇形的面积为cm2．四、解答题：本大题共6小题，满分70分。

除第17小题10分以外，每小题12分。

福建省福州市2024届高三上学期第一次质量检测数学试题含答案

福建省福州市2024届高三上学期第一次质量检测数学试题含答案本文为福建省福州市2024届高三上学期第一次质量检测数学试题及答案的解析。

为了方便阅读，我们将试题分为多个小题，并提供相应的解答。

第一大题：选择题1. 若函数 f(x) 的定义域为实数集 R，且关于 x 的方程 f(x) = a 有且仅有一个实根，则实数 a 的取值范围是？解答：由于函数 f(x) 只有一个实根，所以必然存在一个实数 a，使得 f(a) = 0。

因此，实数 a 在实数集 R 中的取值范围为全体实数。

2. 已知函数 f(x) = ax^2 + bx + c 在区间 [2, 4] 上的最大值为 5，最小值为 1，且在区间 (2, 4) 上函数的图象单调递减，则函数 f(x) 是递增函数的区间为？解答：根据题意，函数 f(x) 在区间 (2, 4) 上为递减函数。

由于 f(x) 是一个二次函数，其图象开口方向由二次项系数 a 的正负决定。

由于函数在区间 [2, 4] 上的最大值为 5，最小值为 1，而图象单调递减，则函数必然在区间 (2, 4) 上为递减函数。

因此，函数 f(x) 是递增函数的区间为空集。

...第二大题：解答题1. 给定一个等差数列 {an}，已知 a1 = 2，d = 3。

求该等差数列的通项公式。

解答：设等差数列的通项公式为 an = a1 + (n-1)d。

代入已知条件 a1 = 2，d = 3，得到 an = 2 + (n-1)3。

因此，该等差数列的通项公式为 an = 3n - 1。

2. 已知函数 f(x) 的导函数为 f'(x) = 2x + 1，且 f(1) = 3。

求函数 f(x) 在点 x = 2 处的函数值。

解答：根据题意，函数 f(x) 的导函数为 f'(x) = 2x + 1。

对导函数进行积分，得到 f(x) = x^2 + x + C，其中 C 为常数。

由于 f(1) = 3，代入得到 1^2 + 1 + C = 3，解得 C = 1。

【教师版】福建省福州市普通高中2023届高三毕业班质量检测(二检)数学试题

2023年福州市普通高中毕业班质量检测数学试卷一､单项选择题：本题共8小题，每小题5分，共40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{}12A x x =-≤，{|2x B x =，则A B =（）A ．112x x ⎧⎫-⎨⎬⎩⎭≤≤B．{|1x x -≤ C ．12x x ⎧⎫⎨⎬⎩⎭≤ D ．{|3}x x ≤1．【答案】D【解析】因为{}12{|13}A x x x x =-=-≤≤≤，1{|22x B x x x ⎧⎫==⎨⎬⎩⎭≤≤，所以{|3}x A B x = ≤．故选：D2．已知(1i)24i z +=-，则z = （）A ．2 BC ．4D ．102．【答案】B【解析】因为(1i)24i z +=-，所以24i1i z -=+，所以24i 1i z -===+． 3．若二项式2213nx x ⎛⎫+ ⎪⎝⎭展开式中存在常数项，则正整数n 可以是（）A ．3B ．5C ．6D ．73．【答案】C【解析】二项式2213n x x ⎛⎫+ ⎪⎝⎭展开式的通项为224121C (3)3C rr n r n r r n r r n n T x xx ---+⎛⎫== ⎪⎝⎭，令240n r -=，解得：2n r =，又因为0r n ≤≤且r 为整数，所以n 为2的倍数，所以6n =，故选：C ．4．为培养学生“爱读书、读好书、普读书”的良好习惯，某校创建了人文社科类、文学类、自然科学类三个读书社团．甲、乙两位同学各自参加其中一个社团，每位同学参加各个社团的可能性相同，则这两位同学恰好参加同一个社团的概率为（） A ．13B ．12C ．23D ．344．【答案】A【解析】甲、乙两位同学各自随机选择一个社团，共有9种情况，这两位同学恰好参加同一个社团，共有3种情况，故这两位同学恰好参加同一个社团的概率3193P ==．故选：A 5．已知2b a = ，若a 与b 的夹角为120︒，则2a b - 在b上的投影向量为（）A ．3b -B ．32b -C ．12b - D ．3b5．【答案】B【解析】2223(2)22cos1202a b b a b b a b b b -⋅=⋅-=⋅⋅︒-=- ，2a b ∴- 在b 上的投影向量为23(2)322b a b b b b b b b b b--⋅⋅=⋅=-．故选：B 6．已知221:(2)(3)4O x y -+-= ，1O 关于直线210ax y ++=对称的圆记为2O ，点E ，F 分别为1O ，2O 上的动点，EF 长度的最小值为4，则=a（）A ．32-或56B ．56-或32C ．32-或56- D ．56或326．【答案】D【解析】由题易知两圆不可能相交或相切，则如图，当EF 过两圆圆心且与对称轴垂直又接近于对称轴时， EF 长度最小，此时圆心1O 到对称轴的距离为4，4，22(27)16(4)a a +=+，解得32a =或56a =．故选：D7．已知三棱锥-P ABC 的四个顶点都在球O的球面上，PA PB PC AB ====2π3ACB ∠=，则球O 的体积为（） A ．3πB ．27π8C ．9π2D ．9π7．【答案】C 【解析】由PA PB PC ==可知点P 在底面ABC 内的射影为ABC △的外心，所以是圆锥模型，设ABC △外接圆半径为r，则2sinABr ACB===∠，所以r =2h ==，设外接球半径为R ，则222()R h R r =-+，即22(2)2R R =-+，解得32R =，所以球O 的体积为344279πππ3382V R ==⨯=．故选：C ． 8．已知函数()f x ，()g x 的定义域均为R ，(1)f x +是奇函数，且(1)()2f x g x -+=，()(3)2f x g x +-=，则（）A ．()f x 为奇函数B ．()g x 为奇函数C ．201()40k f k ==∑D ．201()40k g k ==∑8．【答案】D【解析】因为()(3)2f x g x +-=，所以(3)()2f x g x ++=，又(1)()2f x g x -+=，则有(3)(1)f x f x +=-，因为(1)f x +是奇函数，所以(1)(1)x x f f =--+，可得(3)(1)f x f x +=-+，即有(2)()f x f x +=-与(4)(2)f x f x +=-+，即(4)()f x f x +=，所以()f x 是周期为4的周期函数，故()g x 也是周期为4的周期函数．因为()(2)f x f x --=+，所以()()f x f x -=，所以()f x 为偶函数．故A 错误；由(1)f x +是奇函数，则(1)0f =，所以(3)0f =，又(2)(4)(2)(0)0f f f f +=+=，所以201()5[(1)(2)(3)(4)]0k f k f f f f -=+++=∑，所以C 选项错误；由(1)0f =得(0)2g =，所以B 选项错误；因为(2)2(5)2(1)2g f f =-=-=，(1)(3)[2(4)][2(6)]4[(4)(2)]4g g f f f f +=-+-=-+=，所以(0)(1)(2)(3)8g g g g +++=，所以201()5[(0)(1)(2)(3)]40k g k g g g g ==+++=∑，所以D 正确．二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分．9．已知函数π()2sin 23f x x ⎛⎫=+ ⎪⎝⎭，则（）A ．()f x 在区间π,02⎡⎤-⎢⎥⎣⎦单调递增 B ．()f x 在区间[0,π]有两个零点 C ．直线π12x =是曲线()y f x =的对称轴 D ．直线2π43y x =+是曲线()y f x =的切线 9．【答案】BCD【解析】对于A ，当π,02x ⎡⎤∈-⎢⎥⎣⎦，则22,333πππx ⎡⎤+∈-⎢⎥⎣⎦，因为sin y x =在2ππ,33⎡⎤-⎢⎥⎣⎦先减后增，所以π,02⎡⎤-⎢⎥⎣⎦不是()f x 的增区间，故错误；对于B ，当[0,π]x ∈时，π72,333x ππ⎡⎤+∈⎢⎥⎣⎦，令π23x π+=或2π，解得3x π=或5π6，所以()f x 在区间[0,π]有两个零点，故正确；对于C ，令π12x =，则ππsin 2sin 132x ⎛⎫+== ⎪⎝⎭，所以直线π12x =是曲线()y f x =的对称轴，故正确；对于D ，由π()2sin 23f x x ⎛⎫=+ ⎪⎝⎭可得π()4cos 23f x x ⎛⎫'=+ ⎪⎝⎭，令π()4cos 243f x x ⎛⎫'=+= ⎪⎝⎭，即π22π,3x k k +=∈Z ，解得ππ,6x k k =-+∈Z ，当0k =时，π6x =-，代入()f x 可得π06f ⎛⎫-= ⎪⎝⎭，所以()f x 在点π,06⎛⎫- ⎪⎝⎭处的切线为π46y x ⎛⎫=+ ⎪⎝⎭，化简得2π43y x =+，故D 正确．故选：BCD10．已知曲线222:1424x y C m +=-，则（）A ．若m >，则C 是椭圆B ．若m <<C 是双曲线 C ．当C 是椭圆时，若m 越大，则C 越接近于圆D ．当C 是双曲线时，若m 越小，则C 的张口越大 10．【答案】BD【解析】对于A ，2m =满足m >，代入曲线C 中，得22144x y +=，即224x y +=，表示以(0,0)为圆心，半径为2的圆，故A 错误；对于B ，当m <<24240m --<≤，所以24(24)0m -<，故C 是双曲线，故B 正确；对于C ，当C 是椭圆时，2240m ->，且2244m -≠，解得22m >且24m ≠，即m >且2m ≠，所以当m 越接近2时，椭圆C 越接近于圆．所以C 错误．对于D ，当曲线222:1424x y C m +=-为双曲线时，2240m -<，整理成2221442x y m-=-，则2a =，b =，当m 越小，则b =b y x a=的斜率越大，图象的张口就越大，故D 正确．故选：BD ．11．在棱长为2的正方体1111ABCD A B C D -中，E ，F 分别为棱BC ，1CC 的中点，P 为线段EF 上的动点，则（）A ．线段DP 长度的最小值为2B ．三棱锥1D A AP -的体积为定值C ．平面AEF 截正方体所得截面为梯形D ．直线DP 与1AA 所成角的大小可能为π311．【答案】BC【解析】对于A ：DE DF == ，EF =，∴当点P 为EF 的中点时，此时DP 长度的最小，最=，A 错；对于B ：//EF 平面11AA D D ，∴点P 到平面11AA D D 的距离等于E 到平面11AA D D 的距离，故三棱锥1D A AP -的体积等于三棱锥1P A AD -的体积也等于三棱锥1E A AD -的体积，为114222323⨯⨯⨯⨯=，B 正确；对于C ：如图平面AEF 截正方体所得截面为四边形AEFD ，因为11////EF BC AD ，且111122EF BC AD ==，所以四边形AEFD 为梯形，C 正确；对于D ：过点P 作PN BC ⊥于点N ，连接DN ，则PN ⊥平面ABCD ，所以1//PN AA ，则直线DP 与AA 1所成角的大小为DPN ∠或其补角，令EN a =，则PN a =，1CN a =-，DN =，tan DNDPN PN ∴∠===令225211415,0155t a a a a ⎛⎫=-+=-+ ⎪⎝⎭≤≤，01a ≤≤，当0a =时，1DP AA ⊥，当01a <≤，11a ≥，而22521141555t a a a ⎛⎫=-+=-+ ⎪⎝⎭在[1,)+∞上单调递增，所以21451455t ⎛⎫-+= ⎪⎝⎭≥，即tan 2DPN ∠>≥故直线DP 与1AA 所成角的大小不可能为π3，D 错．故选：BC12．若x ，y 满足223x xy y ++=，则（）A ．2x y +≤B ．21x y +-≥C ．228x y xy +-≤D ．221x y xy +-≥12．【答案】AD【解析】由223x xy y ++=，可得2213324x y y ⎛⎫++= ⎪⎝⎭，令1,2,x y y θθ⎧+=⎪⎪=可得sin 2sin x y θθθ⎧=-⎪⎨=⎪⎩，对于A 、B：22sin 2sin [x y θθθθ+=-+=∈-，故A 正确，B 错误；对于C 、D：22323sin )2sin x y xy xy θθθ+-=-=--⋅23sin 4sin 32(1cos 2)θθθθθ=-+=-+-1542cos 254sin 2[1,9]26πθθθ⎫⎛⎫=-+=-+∈⎪ ⎪⎪⎝⎭⎝⎭，故C 错误，D 正确．故选：AD ．三､填空题：本大题共4小题，每小题5分，共20分．把答案填在题中的横线上．13．若3cos 5α=-，α是第三象限角，则tan 2α=___________． 13．【答案】247-【解析】因为3cos 5α=-，且α是第三象限角，所以4sin 5α==-，则sin 4tan cos 3ααα==，所以282tan 243tan 2161tan 719ααα===---，故答案为：247-． 14．利率变化是影响某金融产品价格的重要因素经分析师分析，最近利率下调的概率为60%，利率不变的概率为40%．根据经验，在利率下调的情况下该金融产品价格上涨的概率为80%，在利率不变的情况下该金融产品价格上涨的概率为40%．则该金融产品价格上涨的概率为__________． 14．【答案】64%##1625##0.64 【解析】由题意可知金融产品价格上涨的概率为：60%80%40%40%064.⨯+⨯=，故答案为：0.64 15．已知曲线32()362f x x x x =-++在点P 处的切线与在点Q 处的切线平行，若点P 的纵坐标为1，则点Q 的纵坐标为__________． 15．【答案】11 【解析】2()366f x x x '=-+，函数()f x '的对称轴为1x =，(1)6f =，所以函数()f x 关于点(1,6)对称，因为函数()f x 在点P 处的切线与在点Q 处的切线平行，所以点P 和点Q 关于点(1,6)对称，即12p Q y y +=，而1P y =，11Q y ∴=．16．已知椭圆22:1126x y C +=，直线l 与C 在第二象限交于A ，B 两点（A 在B 的左下方），与x 轴，y 轴分别交于点M ，N ，且::1:2:3MA AB BN =，则l 的方程为____________________． 16．【答案】40x y -+=【解析】如图，由条件得B 点为线段MN 中点，设B 点坐标为00(,)x y ，得0(2,0)M x 、0(0,2)N y ，由:1:2MA AB =得A 坐标为005,33x y A ⎛⎫ ⎪⎝⎭，将A B ､坐标分别代入221126x y +=中，得220022001,126251,12969x y x y ⎧+=⎪⎪⎨⎪+=⎪⨯⨯⎩解得0022x y =-⎧⎨=⎩，则M 、N 坐标分别为()4,0-、(0,4)，故直线方程为144x y+=-，即40x y -+=，所以直线l 的方程为40x y -+=．解法二：设(,0)M m ，(0,)N n ，则0m <，0n >，由::1:2:3MA AB BN =，可知5,66m n A ⎛⎫ ⎪⎝⎭，,22m n B ⎛⎫ ⎪⎝⎭，则AB 中点2,33m n P ⎛⎫ ⎪⎝⎭，由22MN OP b k k a ⋅=-，得221222n n n m m m -⋅=-=-，22m n ∴=，m n =-，1MNk ∴=，此时5,66m m A ⎛⎫- ⎪⎝⎭，将点A 坐标代入椭圆方程，解得：4m =-，故102,33A ⎛⎫- ⎪⎝⎭，所以直线l 的方程为21033y x -=+，即40x y -+=．四､解答题：本大题共6小题，共70分．解答应写出文字说明､证明过程或演算步骤．17．记ABC △的内角A ，B ，C 的对边分别为a ，b ，c ．已知2222b a c -=．（1）求tan tan BA的值：（2）求C 的最大值．17．【答案】（1）tan 3tan B A =-；（2）π6．【解析】（1）由余弦定理可得2222cos b c a ac B =+-，代入2222b a c -=，得到2222(2cos )2c a ac B a c +--=，················································1分化简得22cos 0c ac B +=，即2cos 0c a B +=．·······························································2分由正弦定理可得sin 2sin cos 0C A B +=，即sin()2sin cos 0A B A B ++=，····························3分展开得sin cos cos sin 2sin cos 0A B A B A B ++=，即3sin cos cos sin A B A B =-，······················4分所以tan 3tan BA=-．······································································································5分（2）由2222b a c -=得2222b a c =-，故222cos 2a b c C ab+-=···········································6分 222222b a a b ab-+-=···································································································7分2233444a b a b ab b a +==+=≥·········································································8分当且仅当223b a =，即b =时等号成立．··································································9分因为(0,π)C ∈，所以π6C ≤，所以C 的最大值为π6．·····················································10分18．如图，在四棱锥P ABCD -中，底面ABCD 是梯形，//AB CD ，AD CD ⊥，24CD AB ==，PAD △ 是正三角形，E 是棱PC 的中点．（1）证明：//BE 平面PAD ；（2）若AD =，平面PAD ⊥平面ABCD ，求直线AB 与平面PBC 所成角的正弦值．18．【答案】（1）证明见解析；（2．【解析】（1）取PD 中点F ，连接AF ，EF ．//EF CD ，12EF CD =，//AB CD ，12AB CD =， //EF AB ∴，EF AB =，··························································································1分∴四边形ABEF 为平行四边形，//BE AF ∴，·······························································3分又BE ⊄ 平面PAD ，AF ⊂平面PAD ，//BE ∴平面PAD ．········································4分（2）取AD 中点O ，BC 中点G ，连接PO ，OG ，可得PO AD ⊥，//OG CD ．平面PAD ⊥平面ABCD ，平面PAD 平面ABCD AD =，PO AD ⊥，PO ⊂平面PAD ， PO ∴⊥平面ABCD ．·······························································································5分 AD CD ⊥ ，//OG CD ，OG OA ∴⊥．····································································6分以O 为原点，以AD ，OG ，OP 所在直线为x 轴､y 轴､z 轴，建立如图所示空间直角坐标系．7分因为2AB =，4CD =，AD =，PAD △是等边三角形，所以PD AD ==12OA AD ==，3PO ==，所以A，2,0)B，(4,0)C ，(0,0,3)P ．则(0,2,0)AB =，(2,0)BC =-，(2,3)BP =-．·············································9分设平面PBC 的法向量为(,,)n x y z = ，由0BC n ⋅= ，0BP n ⋅=，可得20230y y z ⎧-+=⎪⎨-+=⎪⎩，令1x =，可得y =z =，从而n =是平面PBC 的一个法向量．······························································10分则cos ,AB n AB n AB n ⋅〈〉===⋅····································································11分所以直线AB 与平面PBC．·························································12分 19．欧拉函数*()()n n ϕ∈N 的函数值等于所有不超过正整数n ，且与n 互质的正整数的个数，例如：(1)1ϕ=，(4)1ϕ=．（1）求2(3)ϕ，3(3)ϕ；（2）令1(3)2n n a ϕ=，求数列3log n n a a ⎧⎫⎨⎬⎩⎭的前n 项和． 19．【答案】（1）2(3)6ϕ=；3(3)18ϕ=；（2）1321443n n -+-⨯．【解析】（1）不超过9，且与其互质的数即为[1,9]中排除掉3，6，9剩下的正整数，则22(3)336ϕ=-=；································································································2分不超过27，且与其互质的数即为[1,27]中排除掉3，6，9，12，15，18，21，24，27剩下的正整数，则33(3)3918ϕ=-=．······························································································4分（2）*[32,3]()k k k -∈N 表示任意相邻的三个正整数，其中与3n 互质的为32k -与31k -两个，故分别取11,2,,3n k -= 可得[1,3]n 中与3n 互质的正整数个数为123n -⨯，所以1(3)23n n ϕ-=⨯，11(3)32n n n a ϕ-==，···································································7分所以31log 13n n n a n a --=．·································································································8分设数列3log n n a a ⎧⎫⎨⎬⎩⎭的前n 项和为n T ．1221122103333n n n n n T ----=+++++ ，23111221033333n n n n n T ---∴=+++++ ，···················9分两式相减得：11211111111113331133333313n n n n n n n T ---⨯--⎛⎫-=+++-=- ⎪⎝⎭- ····························10分 11111212233223n n nn n --+=--=-⨯⨯，··············································································11分则131213212223443n n n n n T -++⎛⎫=-=- ⎪⨯⨯⎝⎭．··········································································12分 20．脂肪含量（单位：%）指的是脂肪重量占人体总重量的比例．某运动生理学家在对某项健身活动参与人群的脂肪含量调查中，采用样本量比例分配的分层随机抽样，如果不知道样本数据，只知道抽取了男性120位，其平均数和方差分别为14和6，抽取了女性90位，其平均数和方差分别为21和17．（1）试由这些数据计算出总样本的均值与方差，并对该项健身活动的全体参与者的脂肪含量的均值与方差作出估计．（结果保留整数）（2）假设全体参与者的脂肪含量为随机变量X ，且2(17,)X N σ～，其中2σ近似为（1）中计算的总样本方差．现从全体参与者中随机抽取3位，求3位参与者的脂肪含量均小于12.2%的概率．附：若随机变量X 服从正态分布2(,)N μσ，则()0.6827P X μσμσ-+≈≤≤，(22)0.9545P X μσμσ-+≈≤≤4.7≈4.8≈，30.158650.004≈．20．【答案】（1）总样本的均值为17，方差为23；据此估计该项健身活动全体参与者的脂肪含量的总体均值为17，方差为23；（2）0.004【解析】（1）把男性样本记为12120,,,x x x ，其平均数记为x ，方差记为2x s ；把女性样本记为1290,,,y y y ，其平均数记为y ，方差记为2y s ．则14x =，26x s =；21y =，217y s =．记总样本数据的平均数为z ，方差为2s ．由14x =，21y =，根据按比例分配的分层随机抽样总样本平均数与各层样本平均数的关系，可得总样本平均数为120901209012090z x y =+++·····························································2分12014902117210⨯+⨯==，··························································································3分根据方差的定义，总样本方差为1209022211121()(0i i i i s x z y z ==⎡⎤=-+-⎢⎥⎣⎦∑∑·································4分 1209022111(()210i i i i x x x z y x y z ==⎡⎤=-+-+-+-⎢⎥⎣⎦∑∑ 由1201201112)0(0iii i x x x x ==-=-=∑∑可得120120112()2()(0iii i x x x z x z x x ==--=--=∑∑同理，9090112(2((0iii i y y y z y z y y ==--=--=∑∑，·····················································5分因此，1201209090222221111()(1)(()210i i i i i i s x x x z y y y z ====⎡⎤=-+-+-+-⎢⎥⎣⎦∑∑∑∑ {}22221120(90(210x y s x z s y z ⎡⎤⎡⎤=+-++-⎣⎦⎣⎦·····························································6分所以{}2221120[6(1417)]90[17(2117)]23210s =⨯+-+⨯+-≈，所以总样本的均值为17，方差为23，并据此估计该项健身活动全体参与者的脂肪含量的总体均值为17，方差为23．······················7分（2）由（1）知223 =，所以(17,23)X N ~4.8≈，所以(12.221.8)(17 4.817 4.8)0.6827P X P X =-+≈≤≤≤≤，··································8分1(12.2)(10.6827)0.158652P X <≈⨯-=，··································································9分因此(3,0.15865)X B ～，····························································································10分所以333(3)C 0.158650.004P X ==⨯≈．······································································11分所以3位参与者的脂肪含量均小于12.2%的概率为0.004．················································12分 21．已知抛物线2:2(0)E y px p =>，过点(2,0)-的两条直线1l ，2l 分别交E 于A 、B 两点和C 、D 两点．当1l 的斜率为23时，AB =．（1）求E 的标准方程：（2）设G 为直线AD 与BC 的交点，证明：点G 必在定直线上． 21．【答案】（1）24y x =；（2）证明见解析，定直线为2x =．【解析】（1）当1l 的斜率为23时，得1l 方程为2(2)3y x =+，·············································1分由222(2)3y pxy x ⎧=⎪⎨=+⎪⎩，消元得2340y py p -+=，2(3)440p p ∆=--⨯>，123y y p +=，124y y p =；······························································································································2分由弦长公式得AB ===，········3分2=，解得2p =或29p =-（舍去），2p =满足0∆>，从而E 的标准方程为24y x =．····················································································4分（2）法一：因为直线1l ，2l 分别交E 于A 、B 两点和C ，D 两点，所以直线斜率存在．设直线AB 的方程为1()2y k x =+，设211,4y A y ⎛⎫ ⎪⎝⎭，222,4y B y ⎛⎫⎪⎝⎭，由12(2)4y k x y x=+⎧⎨=⎩，消去x 得211480k y y k -+=，···························································5分则128y y =．············································································································6分设直线CD 的方程为2()2y k x =+，233,4y C y ⎛⎫⎪⎝⎭，244,4y D y ⎛⎫⎪⎝⎭，同理22(2)4y k x y x=+⎧⎨=⎩，消去x 得222480k y y k -+=，可得348y y =．··································7分直线AD 方程为241112241444y y y y y x y y ⎛⎫--=- ⎪⎝⎭-，即1441414y y y x y y y y =+++，化简得14144()0x y y y y y -++=，···············································································8分同理，直线BC 方程为23234()0x y y y y y -++=，·························································9分因为(2,0)-在抛物线的对称轴上，由抛物线的对称性可知，交点G 必在垂直于x 轴的直线上，所以只需证G 的横坐标为定值即可．由141423234()04()0x y y y y y x y y y y y -++=⎧⎨-++=⎩，消去y ，因为直线AD 与BC 相交，所以2314y y y y +≠+，解得2314142323144[()(())]()y y y y y y y y x y y y y +-+=+-+···········································································10分 1232341241343241231423142314231488888[()()]4[()()2]4[()()]4[()()]y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y +--+--+-+===+-++-++-+=．11分所以点G 的横坐标为2，即直线AD 与BC 的交点G 在定直线2x =上．······························12分法二：设直线AB 方程为2x my =-，由224x my y x=-⎧⎨=⎩，消去x 得2480y my -+=，··············5分设211,4y A y ⎛⎫ ⎪⎝⎭，222,4y B y ⎛⎫⎪⎝⎭，则128y y =．···································································6分设直线CD 的方程为2x ny =-，233,4y C y ⎛⎫⎪⎝⎭，244,4y D y ⎛⎫⎪⎝⎭，同理可得348y y =．················7分直线AD 方程为241112241444y y y y y x y y ⎛⎫--=- ⎪⎝⎭-，即1441414y y y x y y y y =+++，化简得14144()0x y y y y y -++=，···············································································8分同理，直线BC 方程为23234()0x y y y y y -++=．·························································9分因为(2,0)-在抛物线的对称轴上，由抛物线的对称性可知，交点G 必在垂直于x 轴的直线上，所以只需证G 的横坐标为定值即可．由141423234()04()0x y y y y y x y y y y y -++=⎧⎨-++=⎩，消去y ，因为直线AD 与BC 相交，所以2314y y y y +≠+，解得2314142323144[()(())]()y y y y y y y y x y y y y +-+=+-+···········································································10分1232341241343241231423142314231488888[()()]4[()()2]4[()()]4[()()]y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y y +--+--+-+===+-++-++-+=．11分。

福州市高中毕业班文科数学质量检查

福州市高中毕业班文科数学质量检查2021年福州市高中毕业班质量反省数学〔文科〕试卷参考答案及评分规范一、选择题〔本大题共12小题，每题5分.〕 1．A 2．B 3．D 4．B 5．D 6．C 7．C8．B9．A10．C11．A12．D二、填空题〔本大题共4小题，每题4分，共16分．〕 13．i14．1415．116．1()3n三、解答题〔本大题共6小题，共74分．〕 17．解：〔Ⅰ〕由得112n n a a +=+，即112n n a a +-=． ········································ 1分 ∴ 数列{}n a 是以12为首项，以12d =为公差的等差数列． ································· 2分 ∵ 1(1),n a a n d =+- ················································································· 3分 ∴ 11(1)222n na n =+-=〔*n N ∈〕． ·························································· 6分〔Ⅱ〕由〔Ⅰ〕得141(1)22nb n n n n ==++⋅， ················································· 7分 ∴ 114()1n b n n =-+． ············································································ 9分 ∴ 111114[(1)()()]2231n T n n =-+-++-+14(1)1n =-+41n n =+． ···················· 12分 18．解：〔Ⅰ〕事情A 包括的基身手情为：{,}a b 、{,}a c 、{,}a x 、{,}a y 、{,}b c 、{,}b x 、{,}b y 、{,}c x 、{,}c y ，{,}x y ，共10个． ······································································ 6分注：⑴ 漏写1个情形扣2分，扣完6分为止；多写情形一概扣3分．〔Ⅱ〕方法一：记〝至少有1扇门被班长关闭〞为事情B ．∵ 事情B 包括的基身手情有{,}a x 、{,}a y 、{,}b x 、{,}b y 、{,}c x 、{,}c y ，{,}x y ，共7个． ···················································································································· 9分∴ 7()10P B =． ······················································································ 12分方法二：事情〝2个门都没被班长关闭〞包括的基身手情有{,}a b 、{,}a c 、{,}b c ，共3个． ·································································· 8分∴ 2个门都没被班长关闭的概率1310P =， ···················································· 10分 ∴ 至少有1个门被班长关闭的概率23711010P =-=．······································ 12分 19．方法一：由sin()04x π-≠，得4x k ππ-≠〔k ∈Z 〕，即4x k ππ≠+〔k ∈Z 〕，∴ 函数()f x 定义域为{|,}4x x k k ππ≠+∈Z ． ··············································· 2分 ∵cos 2(),sin()4x f x x π=-22cos sin ()cos sin )cos sin 4x x f x x x x x x π-∴==+=+-， ······································ 5分注：以上的5分全部在第Ⅱ小题计分.〔Ⅰ〕()sin()121243f ππππ=+===································ 8分〔Ⅱ〕令322(242Z)k x k k πππππ+<+<+∈， ·········································· 10分得522(44Z),k x k k ππππ+<<+∈ ··························································· 11分 ∴ 函数()f x 的单调递减区间为5(2,2)44k k ππππ++(Z)k ∈． ····················· 12分注：先生假定未求函数的定义域且将单调递减区间求成闭区间，只扣2分．方法二：由sin()04x π-≠，得4x k ππ-≠〔k ∈Z 〕，即4x k ππ≠+〔k ∈Z 〕，∴ 函数()f x 定义域为{|,}4x x k k ππ≠+∈Z ． ············································· 2分 ∵cos 2(),sin()4x f x x π=-sin 2()2sin()cos()444())4sin()sin()44x x x f x x x x ππππππ---∴==---， ························ 5分〔Ⅰ〕()cos())121246f ππππ=--= ··························· 8分〔Ⅱ〕令22()4k x k k Z ππππ<-<+∈， ·················································· 10分得522(44Z)k x k k ππππ+<<+∈， ························································· 11分 ∴ 函数()f x 的单调递减区间为5(2,2)44k k ππππ++(Z)k ∈． ····················· 12分方法三：〔Ⅰ〕∵cos(2)cos126ππ⨯==，1sin()sin 41262πππ-==， ∴2()1122f π=． ········································································ 3分下同方法一、二．20．解：〔Ⅰ〕依题意，点P 坐标为(,0)a ． ················································ 1分 ∵ 6OP OQ ⋅=，点Q 坐标为(3,3)，∴ 3306a +⨯=，解得2a =． ··································································· 3分∴ 椭圆C 的方程为22149x y +=． ································································ 4分〔Ⅱ〕过点Q (3,3)且斜率为32的直线AB 方程为33(3)2y x -=-，即3230x y --=． ···················································································· 5分方法一：设点A 、B 的坐标区分为11(,)x y 、22(,)x y ，由221,493230,x y x y ⎧+=⎪⎨⎪--=⎩消去x 并整理得，2812270y y +-=． ·································· 6分 ∴ 1212327,28y y y y +=-=-， ··································································· 7分 ∴ 2212121295463()()4444y y y y y y -=+-=+=， ∴12||y y -=． ················································································ 9分 ∵ 直线AB 与x 轴的交点为(1,0)M ， ∴AOB ∆的面积AOB OMA OMBS S S ∆∆∆=+121211||(||||)1||22OM y y y y =⋅+=⨯⨯-． ·············· 12分方法二：设点A 、B 的坐标区分为11(,)x y 、22(,)x y ，由221,493230,x y x y ⎧+=⎪⎨⎪--=⎩消去y 并整理得22230x x --=， ············ 6分 ∴12,x x ==， ········································· 7分 ∴12|||AB x x =-= ·· 9分 ∵ 点O 到直线AB的距离d ===， ····································· 10分 ∴AOB ∆的面积1122AOB S AB d ∆=⋅⋅==． ·························· 12分方法三：设点A 、B 的坐标区分为11(,)x y 、22(,)x y ，由221,493230,x y x y ⎧+=⎪⎨⎪--=⎩消去y 并整理得22230x x --=， ············ 6分 ∴12,x x ==， ········································· 8分 ∵ 直线AB 与y 轴的交点为3(0,)2M -，∴AOB ∆的面积AOB OMA OMB S S S ∆∆∆=+12113||(||||)222OM x x =⋅+=⨯⨯=12分方法四：设点A 、B 的坐标区分为11(,)x y 、22(,)x y ，由221,493230,x y x y ⎧+=⎪⎨⎪--=⎩消去y 并整理得22230x x --=， ········································ 6分 ∴ 121231,2x x x x +=⋅=-， ······································································· 7分∴12||AB x x -==·············································································································· 9分∵ 点O 到直线AB的距离d ===， ··································· 10分 ∴AOB ∆的面积1122AOB S AB d ∆=⋅⋅==． ·························· 12分 21．〔Ⅰ〕证明：在菱形ABCD 中，∵ BD AC ⊥，∴ BD AO ⊥. ························································································· 1分 ∵ EF AC ⊥，∴PO EF ⊥,∵ 平面PEF ⊥平面ABFED ，平面PEF平面ABFED EF =，且PO ⊂平面PEF ，∴ PO ⊥平面ABFED , ·········································································· 2分 ∵ BD ⊂平面ABFED ，∴ PO BD ⊥. ························································································· 3分 ∵ AOPO O =，所以BD ⊥平面POA .······················································ 4分〔Ⅱ〕连结OB ，设AO BD H =.由〔Ⅰ〕知，AC BD ⊥． ∵ 60DAB ∠=︒，4BC =，∴ 2BH =，CH = ········································································· 5分设OH x =〔0x <<由〔Ⅰ〕知，PO ⊥平面ABFED ,故POB ∆为直角三角形． ∴ 222222()PB OB PO BH OH PO =+=++，∴222224)2162(10PB x x x x =++=-+=+． ······················ 7分事先x =PB 取得最小值，此时O 为CH 中点． ········································· 8分 ∴ 14CEF BCD S S ∆∆=， ················································································· 9分 ∴ 3344BCD ABD BFED S S S ∆∆==梯形， ································································ 10分 ∴ 1211,33ABD BFED V S PO V S PO ∆=⋅=⋅梯形． ················································· 11分 ∴1243ABD BFED V S V S ∆==梯形．∴ 当PB 取得最小值时，12:V V 的值为4:3． ················································ 12分22．解：〔Ⅰ〕22(1)(1)()2x x f x x x x+-'=-+=-〔0x >〕， ····························· 1分由()0,0f x x '>⎧⎨>⎩得，01x <<；由()0,0f x x '<⎧⎨>⎩得，1x >. ∴ ()f x 在(0,1)上为增函数，在(1,)+∞上为减函数． ······································· 3分 ∴ 函数()f x 的最大值为(1)1f =-． ···························································· 4分〔Ⅱ〕∵ ()a g x x x =+， ∴ 2()1a g x x'=-．〔ⅰ〕由〔Ⅰ〕知，1x =是函数()f x 的极值点，又∵ 函数()f x 与()a g x x x =+有相反极值点， ∴ 1x =是函数()g x 的极值点，∴ (1)10g a '=-=，解得1a =． ································································· 7分经检验，事先1a =，函数()g x 取到极小值，契合题意． ···································· 8分〔ⅱ〕∵ 211()2f e e =--，(1)1f =-，(3)92ln3f =-+， ∵ 2192ln321e -+<--<-，即 1(3)()(1)f f f e<<， ∴ 11[,3]x e ∀∈，1min 1max ()(3)92ln3,()(1)1f x f f x f ==-+==-． ···················· 9分由〔ⅰ〕知1()g x x x =+，∴21()1g x x '=-. 事先1[,1)x e ∈，()0g x '<；事先(1,3]x ∈，()0g x '>．故()g x 在1[,1)e 为减函数，在(1,3]上为增函数．∵ 11110(),(1)2,(3)333g e g g e e =+==+=，而 11023e e <+<, 1(1)()(3),g g g e ∴<<∴ 21[,]x e e ∀∈，2min 2max 10()(1)2,()(3)3g x g g x g ====． ······························ 10分 ① 当10k ->，即1k >时，关于121,[,]x x e e ∀∈，不等式12()()11f xg x k -≤-恒成立12max 1[()()]k f x g x ⇔-≥-12max [()()]1k f x g x ⇔≥-+12()()(1)(1)123f x g x f g -≤-=--=-，∴ 312k ≥-+=-，又∵ 1k >，∴ 1k >． ····························································································· 12分 ② 当10k -<，即1k <时，关于121,[,]x x e e ∀∈，不等式12()()11f x g x k -≤-12min 1[()()]k f x g x ⇔-≤-12min [()()]1k f x g x ⇔≤-+．∵ 121037()()(3)(3)92ln32ln333f x g x f g -≥-=-+-=-+， ∴ 342ln33k ≤-+．又∵1k <，∴ 342ln33k ≤-+．综上，所求的实数k 的取值范围为34(,2ln3](1,)3-∞-++∞． ··························· 14分。

福建省福州市2024届高三上学期第一次质量检测数学试题及参考答案

2023~2024学年福州市高三年级第一次质量检测数学试题（完卷时间120分钟；满分150分）第Ⅰ卷一、选择题：本题共8小题，每小题5分，共40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．已知复数z 满足11i z=−，则在复平面内，z 对应的点在（）A ．第一象限B ．第二象限C ．第三象限D ．第四象限2．已知集合{}21A x x=<，{}0B x x =>，则A B = （）A ．（0，1）B ．()0,+∞C ．()1,−+∞D ．(),−∞+∞3．已知点()0,2P x 在抛物线C ：24y x =上，则P 到C 的准线的距离为（） A ．4B ．3C ．2D ．14．“二十四节气”是中国古代劳动人民伟大的智慧结晶，其划分如图所示．小明打算在网上搜集一些与二十四节气有关的古诗．他准备在春季的6个节气与夏季的6个节气中共选出3个节气，若春季的节气和夏季的节气各至少选出1个，则小明选取节气的不同情况的种数是（）A ．90B ．180C ．270D ．3605．一个正四棱台形油槽可以装煤油3190000cm ，其上、下底面边长分别为60cm 和40cm ，则该油槽的深度为（） A ．75cm 4B ．25cmC ．50cmD ．75cm6．一个袋子中有大小和质地相同的4个球，其中有2个红球，2个黄球，每次从中随机摸出1个球，摸出的球不再放回，则第二次摸到黄球的条件下，第一次摸到红球的概率为（） A ．13B ．12C ．23D ．347．已知1ea =，ln b =，ln c =，则（） A ．a b c >> B ．b c a >> C ．a c b >>D ．c a b >>8．若定义在R 上的函数()()sin cos 0f x x x ωωω=+>的图象在区间[]0,π上恰有5条对称轴，则ω的取值范围为（） A ．1721,44B ．1725,44C ．1725,44D ．3341,44二、选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求，全部选对的得5分，部分选对的得2分，有选错的得0分．9．某市抽查一周空气质量指数变化情况，得到一组数据：80，76，73，82，86，75，81．以下关于这组数据判断正确的有（） A ．极差为13B ．中位数为82C ．平均数为79D ．方差为12410．已知圆M ：221x y +=，直线l ：(1y k x =−，则（）A ．l 恒过定点)1−B ．若l 平分圆周M ，则k =C ．当k =l 与圆M 相切D ．当k <<时，l 与圆M 相交11．已知函数()332f x x ax =−+有两个极值点．则（） A ．()f x 的图象关于点()0,2对称 B ．()f x 的极值之和为-4C ．a ∃∈R ，使得()f x 有三个零点D ．当01a <<时，()f x 只有一个零点12．已知正四棱柱1111ABCD A B C D −的底面边长为2，球O 与正四棱柱的上、下底面及侧棱都相切，P 为平面1CDD 上一点，且直线BP 与球O 相切，则（） A ．球O 的表面积为4π B ．直线1BD 与BP 夹角等于45°C ．该正四棱柱的侧面积为D ．侧面11ABB A 与球面的交线长为2π第Ⅱ卷注意事项：用0.5毫米黑色签字笔在答题卡上书写作答．在试题卷上作答，答案无效．三、填空题；本大题共4小题，每小题5分，共20分．13．已知向量()1,2a = ，()1,2b λλ=+−，若a b ⊥ ，则实数λ的值为__________．14．将圆周16等分，设每份圆弧所对的圆心角为θ，则sin cos θθ的值为__________．15．已知定义城为R 的函数()f x 同时具有下列三个性质，则()f x =__________．（写出一个满足条件的函数即可） ①()()()f x y f x f y +=+；②()f x ′是偶函数；③当0x y +>时，()()0f x f y +<．16．已知双曲线C ：()222210,0x y a b a b−=>>的左焦点为F ，两条渐近线分别为1l ，2l ．点A 在1l 上，点B在2l 上，且点A 位于第一象限，原点O 与B 关于直线AF 对称、若2AF b =，则C 的离心率为__________．四、解答题：本大题共6小题，共70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10分）已知等比数列{}n a 的前n 项和为n S ，且12n n a S +=+．（1）求{}n a 的通项公式；（2）若221log n n b a −=，求数列{}n b 的前n 项和n T ． 18．（本小题满分12分）记ABC △的内角A ，B ，C 所对的边分别为a ，b ，c ，已知b =，6B π=．（1）若2c =，求a ；（2）求ABC △面积的最大值． 19．（本小题满分12分）国际上常采用身体质量指数（Body Mass Index ，缩写BMI ）来衡量人体肥瘦程度，其计算公式是()()22kg BMI m=体重单位:身高单位:．为了解某公司员工的身体肥瘦情况，研究人员从该公司员工体检数据中，采用比例分配的分层随机抽样方法抽取了50名男员工、30名女员工的身高和体重数据．计算得到他们的BMI（1）若该公司男员工有1500名，则该公司共有多少名员工？（2）以频率估计概率，分别从该公司男、女员工中各随机抽取2名员工，求抽到的员工中至少有一名是肥胖的概率．20．（本小题满分12分）如图，在底面为菱形的四棱锥M ABCD −中，2AD BD MB ===，MA MD ==（1）求证：平面MAD ⊥平面ABCD ；（2）已知2MN NB =，求直线BN 与平面ACN 所成角的正弦值．21．（本小题满分12分）已知椭圆E ：22143x y +=的右焦点为F ，左、右顶点分别为A ，B ．点C 在E 上，()4,P P y ，()4,Q Q y 分别为直线AC ，BC 上的点．（1）求P Q y y ⋅的值；（2）设直线BP 与E 的另一个交点为D ，求证：直线CD 经过F ．22．（本小题满分12分）已知函数()ln f x x a =−，记曲线()y f x =在点()()11,x f x 处的切线为l ，l 在x 轴上的截距为()220x x >．（1）当1e x =，1a =时，求切线方程；（2）证明：12e e a a x x −≥−．答案一、单项选择题：本题共8小题，每小题5分，共40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．【考查意图】本小题以复数为载体，主要考查复数的基本运算、几何意义等基础知识；考查运算求解能力、推理论证能力；考查数学运算、逻辑推理等数学核心素养，体现基础性．【答案】A ．【解析】由11i z =−得11i1i 2z+==−，应选A ． 2．【考查意图】本小题以不等式为载体，主要考查集合运算等基础知识；考查运算求解能力、推理论证能力；考查数学运算、逻辑推理等核心素养，体现基础性．【答案】C ．【解析】{}11A x x =−<<，{}0B x x =>，故()1,A B =−+∞ ，应选C ． 3．【考查意图】本小题以抛物线为载体，主要考查抛物线的图象和性质、直线与抛物线的位置关系等基础知识；考查运算求解能力、推理论证能力；考查数形结合思想、化归与转化思想；考查直观想象、逻辑推理、数学运算等核心素养，体现基础性．【答案】C ．【解析】抛物线24y x =的准线为1x =−，由P C ∈得01x =，故P 到准线的距离为2，应选C ．4．【考查意图】本小题以二十四节气为载体，主要考查排列与组合等基础知识；考查运算求解能力、推理论证能力和应用意识；考查数学运算、逻辑推理等核心素养，体现基础性和应用性．【答案】B ．【解析】根据题意可知，小明可以选取1春2夏或2春1夏．其中1春2夏的不同情况有：1266C C 90⋅=种；2春1夏的不同情况有：2166C C 90⋅=种，所以小明选取节气的不同情况有：9090180+=种．应选B ．5．【考查意图】本小题以正四棱台形油槽为载体，主要考查空间几何体的体积等基础知识；考查空间想象能力、推理论证能力、运算求解能力；考查数形结合思想、化归与转化思想；考查直观想象、逻辑推理、数学运算等核心素养，体现基础性和应用性．【答案】D ．【解析】设正四棱台的高，即深度为cm h ，依题意，得()22190000604060403h=++×，解得75h =，应选D ．6．【考查意图】本小题主要考查条件概率、全概率公式等基础知识；考查推理论证能力、运算求解能力与创新意识；考查化归与转化思想；考查数学建模、逻辑推理、数据分析等核心素养，体现综合性、应用性与创新性．【答案】C ．【解析】解法一：记第i 次摸到红球为事件i A ，摸到黄球为事件()1,2i B i =，则()()()()()21211211211123232P B P A P B A P B P B B =+=×+×=，()()()12121221433P A B P A P B A ==×=，故()()()1212223P A B P A B P B ==．应选C ．解法二：记第i 次摸到红球为事件i A ，摸到黄球为事件()1,2i B i =．由抽签的公平性可知()22142P B ==，又()12221433P A B ×==×，所以()()()1212223P A B P A B P B ==．应选C ． 7．【考查意图】本小题以数的大小比较为载体，主要考查函数与导数等基础知识；考查运算求解能力、推理论证能力、应用意识；考查数学建模、数学运算、逻辑推理等核心素养，体现基础性、应用性和综合性．【答案】A ．【解答】解法一：1ln e e e a ==，ln 2ln 4ln 24b ==，ln 55c ，令()ln x f x x =，()21ln xf x x−′=，当e x ≥时，()0f x ′≤，故()f x 在区间[)e,+∞上单调递减，所以a b c >>．==>，所以ln ln >b c >．在同一坐标系中作出函数()2xf x =，()2g x x =的图象，如图所示，由图可知，()()e e f g <，即e22e <，所以e22e 2e 2e <，即11e22e <，所以111ln 2ln e 2e e<=，即b a <．（令()ln x f x x=，()21ln xf x x −′=，当0e x <<时，()0f x ′>，故()f x 在区间()0,e 上单调递增，所以1ln e ln 2e e 2a b ==>=．）综上，a b c >>．应选A ．8．【考查意图】本小题以三角函数为载体，考查三角函数的图象与性质、三角恒等变换等基础知识；考查抽象概括能力、推理论证能力、应用意识；考查数形结合思想；考查直观想象、逻辑推理、数学运算等核心素养，体现基础性和综合性．【答案】A ．【解析】由已知，()4f x x πω=+，令42x k ππωπ+=+，k ∈Z ，得()414k x πω+=，k ∈Z ，依题意知，有5个整数k 满足()4104k ππω+≤≤，即0414k ω≤+≤，所以0,1,2,3,4k =，则4414451ω×+≤<×+，故172144ω≤<，应选A ．二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分．9．【考查意图】本小题主要考查极差、中位数、平均数、方差等基础知识；考查推理论证能力、运算求解能力；考查化归与转化思想；考查数据分析等核心素养，体现基础性．【答案】AC ．10．【考查意图】本小题以直线与圆为载体，考查直线的方程、圆的方程、直线与圆的位置关系等基础知识；考查运算求解能力；考查直观想象、逻辑推理等核心素养；体现基础性和综合性．【答案】BC ．【解析】依题意，l恒过定点()1−，选项A 错误；若l 平分圆周M ，则l 经过圆M 的圆心()0,0，代入直线方程得k =B 正确；圆心()0,0O 到l的距离dk =1d r ==，l 与圆M 相切，选项C 正确；若l 与圆M相交，则1d <，即)2211k −<+，即0k <<，故选项D 错误．综上，应选BC ．11．【考查意图】本小题以三次函数为载体，主要考查函数与导数等基础知识；考查运算求解能力、推理论证能力、应用意识；考查数学建模、数学运算、逻辑推理等核心素养，体现基础性、应用性和综合性．【答案】ACD ．【解答】()f x 的图象可由奇函数()33g x x ax =−的图象向上平移2个单位长度得到，故()f x 的图象关于点()0,2对称，选项A 正确．设()f x 的极值点分别为()1212,x x x x <，则由对称性可知120x x +=，故()()12224f x f x +=×=，即()f x 的极值之和为4，选项B 错误．依题意，方程()2330f x x a ′=−=有两异根，则0a >，1x =2x =，()f x在区间(−∞上单调递增，在区间(上单调递减，在区间)+∞单调递增．由图象可知，当()()120f x f x >>时，()f x 的图象与x 轴有3个交点，即()f x 有3个零点，选项C 正确．当01a <<时，(32210f=−+=−>，此时()f x 只有一个零点，选项D 正确．综上，应选ACD ．12．【考查意图】本小题以正四棱柱为载体，主要考查球、直线与平面的位置关系等基础知识；考查空间想象能力、推理论证能力、运算求解能力；考查化归与转化思想；考查直观想象、逻辑推理等核心素养，体现基础性、应用性和综合性．【答案】BCD ．【解答】如图，设球O 与下底面相切于点1O ，则1OO ⊥平面ABCD ，连接1O A ，则1OAO ∠为直线OA 与平面ABCD 所成的角．因为球O与正四棱柱的侧棱相切，所以其半径11R OO O A===，所以428S ππ=⋅=表，四棱柱的侧面积为()24××，故选项A 错误，C 正确．依题意，1BB ，BP 均为球O 的切线，1BD 经过球心O ，所以111B BD PBD ∠=∠，又111B D BB =，所以11145PBD B BD ∠=∠=°，选项B 正确．对于选项D ，棱1AA 的中点F ，即球O 与棱1AA 的切点应为交线上的点，故交线应为过F 的圆．截面圆的圆心即为矩形11ABB A 的中心E ，在Rt OEF △中，OFR ==，112OEBC ==，所以截面圆半径1r EF ==，周长为2π，该选项正确．综上，应选BCD ．三、填空题：本大题共4小题，每小题5分，共20分．13．【考查意图】本小题以平面向量为载体，主要考查平面向量的基本运算等基础知识；考查运算求解能力、推理论证能力；考查数学运算、逻辑推理、直观想象等核心素养，体现基础性．【答案】5．【解析】由a b ⊥得()()1220λλ++−=，解得5λ=． 14．【考查意图】本小题以圆的等分为载体，考查三角恒等变换等基础知识；考查推理论证能力，抽象概括能力；考查逻辑推理等核心素养；体现基础性与应用性．．【解析】依题意，得8πθ=，所以11sin cos sin 2sin 224πθθθ===． 15．【考查意图】本小题以函数的性质为载体，考查函数的奇偶性、函数与导数等基础知识；考查推理论证能力；考查逻辑推理等核心素养；体现基础性、综合性与应用性．【答案】x −（答案不唯一，()0kx k <均可）．16．【考查意图】本小题以双曲线为载体，主要考查双曲线的离心率、双曲线的图象和性质、直线与双曲线的位置关系等基础知识；考查运算求解能力、推理论证能力；考查数形结合思想、化归与转化思想；考查直观想象、逻辑推理、数学运算等核心素养，体现基础性和综合性．【答案】2．【解答】依题意，1l 的方程为by x a=，2AF l ⊥，设垂足为P ，则FP b =．因为22AFb FP ==，所以点F ，A 关于直线2l 对称，FOP AOP ∠=∠，又1l ，2l 关于y 轴对称，所以1l 的倾斜角为1180603×°=°，故tan 60ba=°=，所以离心率2e =．四、解答题：本大题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤．17．【命题意图】本小题主要考查等差数列、等比数列、递推数列及数列求和等基础知识，考查运算求解能力、逻辑推理能力和创新能力等，考查化归与转化思想、分类与整合思想、函数与方程思想、特殊与一般思想等，考查逻辑推理、数学运算等核心素养，体现基础性和综合性．满分10分．【解答】（1）解法一：由12n n a S +=+得21322,2,a S aS =+ =+设等比数列{}n a 的公比为q ，所以()()12112,12,a q a q q −=−−= 解得12,2,a q == 或12,a q =− = （舍去）．所以2n n a =．（2）212212log log 221n n n b a n −−===−，故11b =，()()12121122n n b b n n n −−=−−−−=≥ ，所以{}n b 是首项为1，公差为2的等差数列，所以()()1212122n n n b b n n T n ++−===．解法二：（1）因为12n n a S +=+，① 所以当2n ≥时，12n n a S −=+，② ①－②得12n n a a +=，所以等比数列{}n a 的公比12n na qa +==．由①式得212a a =+，得12a =，所以2n n a =．（2）12n n T b b b =++⋅⋅⋅+2123221log log log n a a a −++⋅⋅⋅+()21321log n a a a −⋅⋅⋅= ()13212log 2n ++⋅⋅⋅+−=()12122log 2n n+− =2n =．18．【命题意图】本小题主要考查正弦定理、余弦定理及三角恒等变换等基础知识，考查逻辑推理能力、运算求解能力等，考查化归与转化思想、函数与方程思想、数形结合思想等，考查数学运算、逻辑推理等核心素养，体现基础性和综合性．满分12分．【解答】解法一：（1）因为b =，2c =，6B π=，根据余弦定理得2222cos b a c ac B =+−，所以22224cos6a a π=+−，即220a −+=，解得1a =．（2）根据余弦定理，得2222cos a b c ac B =+−，所以(222222cos 226a c ac a c ac ac π=+−=+−≥−=，（当且仅当1a c ==时取等号），即(22ac ≤=+，所以ABC △面积(1111sin sin 2222644ABC S ac B ac ac π===≤×△，即ABC △．解法二：（1）因为b =，2c =且6B π=，根据正弦定理，得sin sin b c B C=，2sin C=，即sin C =，因为c b >，所以C B >，所以566C ππ<<，所以4C π=或34C π=，当4C π=时，()1sin sin sin 642A B C ππ =+=+= ，根据正弦定理，得sin sin a bA B=，所以sin 1sin b Aa B ==+；当34C π=时，()31sin sin sin 642A B C ππ =+=+=×= ，根据正弦定理，得sin sin a bA B=，所以sin 1sin b A a B ==；综上，1a =．（2）略，同解法一．解法三：（1）因为b =，2c =且6B π=，根据正弦定理，得sin sin b cB C=， 2sin C=，即sin C =，因为c b >，所以C B >，所以566C ππ<<，所以4C π=或34C π=，当4C π=时，()76412A B C πππππ =−+=−+= ，根据正弦定理，得sin sin a b A B=，所以sin sin cos cos sin sin 343434b A a B ππππππ ==+=+ ；sin cos cos sin 13434ππππ ++；当34C π=时，()36412A B C πππππ =−+=−+= ，根据正弦定理，得sin sin a b A B =，所以sin sin cos cos sin sin 343434b A a B ππππππ ==−=−sin cos cos sin 13434ππππ −− ；综上，1a =．（2）根据正弦定理，得sin sin sin ac b A C B ===，所以a A =，c C =，即(251sin sin 8sin sin 8sin cos 62aC A C A A A A A π ==−=+21cos 22sin 22sin 22sin 222A A A A A A −=−=+=++14sin 224sin 223A A A π +−+= 因为506A π<<，所以42333A πππ−<−<，所以当232A ππ−=，即512A π=时，sin 23A π −取得最大值为1，即ac最大值为4+，所以ABC △面积(1111sin sin 422644ABC S ac B ac ac π===≤×+△，即ABC △． 19．【命题意图】本小题主要考查分层抽样、独立事件的概率、互斥事件、对立事件的概率等基础知识；考查数学建模能力，运算求解能力，逻辑推理能力，创新能力以及阅读能力等；考查统计与概率思想、分类与整合思想等；考查数学抽象，数学建模和数学运算等核心素养；体现应用性和创新性．满分12分．【解】（1）设该公司共有x 名员工，依题意得1500505030x =+，解得2400x =，所以该公司共有2400名员工．（2）依题意，事件“抽到一名男员工不为肥胖”的概率为404505=，事件“抽到一名女员工不为肥胖”的概率为2793010=，由事件的独立性，得抽到的两个男员工都不存在肥胖的概率为44165525×=，抽到的两个女员工都不存在肥胖的概率为99811010100×=，设事件M 为“抽到的员工中至少有一名是肥胖”，则事件M 为“抽到的员工都不存在肥胖”，所以()811632410025625P M =×=，所以()3243011625625P M =−=，所以抽到的员工中至少有一名是肥胖的概率为301625． 20．【命题意图】本小题主要考查直线与直线、直线与平面、平面与平面的位置关系，直线与平面所成角等基础知识；考查空间想象能力，逻辑推理能力，运算求解能力等；考查化归与转化思想，数形结合思想，函数与方程思想等；考查直观想象，逻辑推理，数学运算等核心素养；体现基础性和综合性．满分12分．【解答】（1）取AD 的中点为O ，连结OM ，OB ，因为四边形ABCD 是为菱形，且2AD BD ==，所以ABD △为正三角形，所以BO AD ⊥，且BO =．因为MAMD ==，所以MO AD ⊥，所以1MO =，又因为2MB =，所以222MO BO MB +=，所以MO BO ⊥，因为AD BO O ∩=，AD ⊂平面ABCD ，BO ⊂平面ABCD所以MO ⊥平面ABCD ，又因为MO ⊂平面MAD ，所以平面MAD ⊥平面ABCD ．（2）由（1）知，OA ，OB ，OM 两两垂直，故以O 为坐标原点，分别以OA ，OB ，OM 为x ，y ，z 轴的正方向建立如图所示的空间直角坐标系O xyz −．则()1,0,0A，()B，()C −，()0,0,1M，13N ，所以()3,CA =，12,3CN =，()2,0,0CB = ，设平面ACN 的法向量为(),,n x y z = ，则0,0,n CA n CN ⋅= ⋅=即30,120,3x x y z = +=取1x =，则()3n − ．因为()0,BM = ，则cos ,BM n BM n BM n⋅=== ，所以直线BN 与平面ACN21．【命题意图】本小题主要考查椭圆的标准方程及简单几何性质，直线与圆、椭圆的位置关系，平面向量等基础知识；考查运算求解能力，逻辑推理能力，直观想象能力和创新能力等；考查数形结合思想，函数与方程思想，化归与转化思想等；考查直观想象，逻辑推理，数学运算等核心素养；体现基础性，综合性与创新性．满分12分．【解答】（1）依题意，()2,0A −，()2,0B ．设()11,C x y ，则2211143x y +=，直线AC 方程为()1122y y x x ++，令4x =得1162P y y x =+，直线BC 方程为()1122y yx x −−，令4x =得1122Q y y x =−，所以2121124P Q y y y x =− 2121123144x x ×− =− 9=−，即P Q y y ⋅的值为9−．（2）设()22,D x y ，()4,P t ，则直线AP 方程为()26t y x =+，直线BP 的方程为()22t y x =−，由()222,63412t y x x y =+ +=得()222227441080t x t x t +++−=，所以2124108227t x t −−=+，即21254227t x t −=+，故()112182627t t y x t=+=+．由()222,23412t y x x y =− +=得()2222344120t x t x t +−+−=，所以22241223t x t −=+，即222263t x t −=+，故()2226223t t y x t −=−=+．所以()()122111x y x y −−−2222222736918273327t t t t t t t t−−−=⋅−⋅++++ ()()()222262733270327t t t t t −−+−=++，又()1,0F ，所以向量()111,FC x y =− ，与()221,FD x y =− 共线，所以直线CD 经过F ．解法二：（1）依题意，()2,0A −，()2,0B ．设()11,C x y ，则2211143x y +=，所以111122AC BC y y k k x x ⋅=⋅+− 21214y x =− 21213144x x −−= 34=−．即B 344242Q P AP Q y y k k −=⋅=⋅+−，故P Q y y 的值为9−．（2）设()11,C x y ，()22,D x y ，()4,P t ．要证直线CD 经过()1,0F ，只需证向量()111,FC x y =− ，与()221,FD x y =− 共线，即证()()122111x y x y −=−．（*）因为()2222112014343x y −+==+，所以111123246P AC y x y k x y −==−⋅=+，同理可得222223242P BD y x y k x y +==−⋅=−，所以()()21122123AC BD x y k k x y −==+，即1221123620x y x y y y −++=，① 同理可得1221123260x y x y y y −+++=，②①－②得12211244440x y x y y y −+−=，即()()122111x y x y −=−．所以（*）式成立，命题得证．22．【命题意图】本小题主要考查导数，函数的单调性、零点、不等式等基础知识；考查逻辑推理能力，直观想象能力，运算求解能力和创新能力等；考查函数与方程思想，化归与转化思想，分类与整合思想等；考查逻辑推理，直观想象，数学运算等核心素养；体现基础性、综合性和创新性．满分12分．【解答】（1）()1f x x′=，当1e x =，1a =时，()1ln e 10f x =−=，即切点为()e,0，所以所求切线斜率()1e ek f ′=，所以所求的切线方程为()1e e y x =−，即11ey x =−．（2）由于()11ln f x x a =−，所以切线l 的方程为()()1111ln y x a x x x −−=−．令0y =，得()()1111ln x a x x x −−=−，解得()2111ln x x x x a =−−．（*）由20x >，得11e a x +<．构造函数()()ln g x x x x a =−−，所以()ln g x a x ′=−，所以当0e a x <<时，()0g x ′>，()x 单调递增；当e a x >时，()0g x ′<，()g x 单调递减．故()()max e e a a g x g ==．所以2e a x ≤．若1e a x ≤，由（*）式知12x x ≤，所以12e a x x ≤≤，故12e e a a x x −≥−．若1e a x >，则()()()121212e e e e 2e a a a a a x x x x x x −−−=−−−=+−，所以()12111e e 2ln 2e a a a x x x x x a −−−=−−−．构造函数()()()12ln 2e e e aa a x x x x a x ϕ+=−−−<<，所以()()1ln 0x a x ϕ′=+−>，故()x ϕ在区间()1e ,e a a +上单调递增，所以()()e 0a x ϕϕ>=，所以()1112ln 2e 0a x x x a −−−>，即所以12e e 0a a x x −−−>，即12e e a a x x −>−．综上，不等式成立12e e a a x x −≥−成立（当且仅当1e a x =时取等号）．。

高中数学试题-2024~2025学年福州市高三年级第一次质量检测数学答案

2024-2025学年福州市高三年级第一次质量检测数学答案一、单项选择题：本题共8小题，每小题5分，共40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。

有多项符合题目要求。

全部选对的得6分，部分选对的得部分分，有选错的得0分。

15. （13分）已知数列{}n a 满足12a =，132n n a a +=+．（1）证明：数列{}1n a +是等比数列；（2）求{}n a 的前n 项和n S ．【解法一】（1）证明：因为132n n a a +=+，且12a =，所以10n a +≠， ··················································································· 1分所以1132111n n n n a a a a ++++=++ ········································································ 3分 3(1)31n n a a +==+， ···································································· 5分又113a +=，所以数列{}1n a +是以3为首项，3为公比的等比数列． ······························· 6分（2）由（1）得13n n a +=，所以31n n a =−， ············································· 8分所以()()()2313131n n S =−+−++−()233333n n =++++− ···························································· 10分13313n n +−=−− ············································································ 12分 133.2n n +−=− ············································································ 13分【解法二】（1）证明：因为132n n a a +=+，所以()113331n n n a a a ++=+=+， ····························································· 2分因为12a =，所以1130a +=≠，所以10n a +≠， ········································ 4分所以1131n n a a ++=+, 所以数列{}1n a +是以3为首项，3为公比的等比数列． ······························· 6分（2）略，同解法一． 16. （15分）已知ABC △的内角,,A B C 的对边分别为,,a b c，且2cos cos cos a C C B =⋅．（1）求角C ；（2）若4a =，b =D 为AB 中点，求CD 的长．【解法一】（1）因为2cos cos cos a C C B =+⋅，由正弦定理，得2sin cos cos sin A C B C B C =·············································· 2分()B C =+ ·································································· 4分()πA −A =，······································································ 6分因为0πA <<，则sin 0A ≠，所以cos C =， ·········································· 7分由于0πC <<，则π6C =； ···································································· 8分（2）因为D 为AB 中点，故()12CD CA CB =+， ······································ 10分22111πcos 4426CA CB CA CB =++ ············································ 13分 1113164442=⨯+⨯+ 314=，················································································· 14分所以CD ． ······································································ 15分【解法二】（1）因为2coscos cos a CC B =⋅，由余弦定理，得2222222cos 22a b c a c b a C ab ac+−+−=··························· 2分， ···························································· 4分所以cos C =， ················································································ 6分由于0πC <<，则π6C =； ···································································· 8分（2）由（1）知，π6ACB ∠=，在ABC △中，由余弦定理，得2222cos c a b ab ACB =+−∠··································································· 10分 22424=+−⨯ 7=， ··························································································· 11分故c =， ······················································································· 12分因为D 为AB 中点，所以cos cos 0ADC BDC ∠+∠=，故222222022AD CD AC BD CD BC AD CD BD CD +−+−+=⨯⨯⨯⨯， ·········································· 13分22222240CD CD +−+−=，故CD ． ··········································································· 15分【解法三】（1）略，同解法一或解法二；（2）由（1）知，π6ACB ∠=，在ABC △中，由余弦定理，得2222cos c a b ab ACB =+−∠··································································· 10分22424=+−⨯ 7=， ··························································································· 11分故c =， ······················································································· 12分所以222cos 2b c a A bc+−=2224+−==， ············································································· 13分在ACD △中，由余弦定理，得2222cos CD AC AD AC AD A =+−⋅222⎛=+− ⎝⎭⎝314=， ······················································································· 14分故CD ． ··········································································· 15分 17. （15分）如图，在四棱锥S ABCD −中，BC ⊥平面SAB ，∥AD BC ，1SA BC ==，SB =，o 45SBA ∠=．（1）求证：SA ⊥平面ABCD ；（2）若12AD =，求平面SCD 与平面SAB 的夹角的余弦值．【解法一】（1）在△SAB 中，因为1SA =，o 45SBA ∠=，SB =，由正弦定理，得sin sin SA SBSBA SAB=∠∠， ········································································· 1分所以1sin 45︒， ······································································ 2分所以sin 1SAB ∠=，因为0180SAB ︒<∠<︒，所以90SAB ∠=︒，所以SA AB ⊥． ··················································································· 4分因为BC ⊥平面SAB ，SA ⊂平面SAB ，所以BC SA ⊥， ··················································································· 5分又BCAB B =，所以SA ⊥平面ABCD ； ········································································· 6分（2）解：由（1）知SA ⊥平面ABCD ，又,⊂AB AD 平面ABCD ，所以SA AB ⊥，SA AD ⊥，因为BC ⊥平面SAB ， ··········································································· 7分 ⊂AB 平面SAB ，所以BC AB ⊥，因为∥AD BC ，所以AD AB ⊥，所以,,SA AD AB 两两垂直． ··································································· 8分以点A 为原点，分别以AD ，AB ，AS 所在直线为x 轴，y 轴，z 轴建立如图所示的空间直角坐标系， ················································································ 9分则1(1,1,0),,0,0,2(0,0,1),D S C ⎛⎫ ⎪⎝⎭所以()1,1,1SC =−，1,0,12SD ⎛⎫=− ⎪⎝⎭，设平面SCD 的法向量为1(,,)x y z =n ，则11,,SC SD ⎧⊥⎪⎨⊥⎪⎩n n 即110,10,2SC x y z SD x z ⎧⋅=+−=⎪⎨⋅=−=⎪⎩n n 取2x =，则()12,1,1=−n ， ·································································· 11分显然平面SAB 的一个法向量()21,0,0=n ， ················································ 12分所以cos ⋅=⋅121212n n n ,n n n ····································································· 13分==········································································· 14分所以平面SCD 与平面SAB ． ··································· 15分【解法二】（1）证明：设AB x =，在△SAB 中，因为1SA =，o 45SBA ∠=，SB =，由余弦定理，得2222cos SA SB AB SB S AB BA =∠+−⋅， · (1)分所以212co 5s 4x =+−︒， (2)分所以221x +−=，所以2210x x −+=，解得1x =． ························································································ 3分所以2222SA AB SB +==，所以SA AB ⊥． ················································ 4分因为BC ⊥平面SAB ，SA ⊂平面SAB ，所以BC SA ⊥， ··················································································· 5分又BCAB B =，所以SA ⊥平面ABCD ； ········································································· 6分（2）略，同解法一．【解法三】（1）设AB x =，在△SAB 中，因为1SA =，o 45SBA ∠=，SB =，由余弦定理，得2222cos SA SB AB SB S AB BA =∠+−⋅， (1)分所以212co 5s 4x =+−︒， ································································ 2分所以221x+−=，所以2210x x−+=，解得1x=． ························································································3分所以2222SA AB SB+==，所以SA AB⊥．················································4分因为BC⊥平面SAB，BC⊂平面ABCD，所以平面ABCD⊥平面SAB；·································································5分又平面ABCD平面SAB AB=，SA AB⊥，SA⊂平面SAB，所以SA⊥平面ABCD；·········································································6分（2）由（1）知SA⊥平面ABCD，过B作BM SA，则BM⊥平面ABCD，又,AB BC⊂平面ABCD，所以BM AB⊥，BM BC⊥，因为BC⊥平面SAB，···········································································7分又⊂AB平面SAB，所以BC AB⊥，所以,,BM BA BC两两垂直．··································································8分以点B为原点，分别以BA，BC，BM所在直线为x轴，y轴，z轴建立如图所示的空间直角坐标系， ················································································9分则1(0,1,0),1,(,0,21,0,1),CS D⎛⎫⎪⎝⎭所以()1,1,1SC=−−，11,,02CD⎛⎫=−⎪⎝⎭，设平面SCD的法向量为1(,,)x y z=n，则11,,SCCD⎧⊥⎪⎨⊥⎪⎩nn即110,10,2SC x y zCD x y⎧⋅=−+−=⎪⎨⋅=−=⎪⎩nn取2y=，则()11,2,1=n， ···································································· 11分显然平面SAB的一个法向量()20,1,0=n， ··············································· 12分所以cos⋅=⋅121212n nn,nn n····································································· 13分=。

5月福州市高中毕业班数学综合质检试卷文科附答案

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2024福建省福州市高三下学期4月末质量检测数学试题及答案

（在此卷上答题无效）2023～2024 学年福州市高三年级4月份质量检测数学试题（完卷时间 120 分钟；满分 150 分）友情提示：请将所有答案填写到答题卡上！请不要错位、越界答题！一、单项选择题：本题共8小题，每小题5分，共40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101M xx= +…，则M =R ð A. {}1x x <−B. {}1x x −…C. {}1x x >−D. {}1x x −…2. 设,a b ∈R ，则“0ab <”是“0a ba b”的 A ．充分不必要条件B ．必要不充分条件C ．充要条件D ．既不充分也不必要条件3. 等轴双曲线经过点(3,1)−，则其焦点到渐近线的距离为A ．B ．2C ．4D6. 54(1)(12)x x −+的展开式中2x 的系数为A ．14−B ．6−C ．34D ．747. 数列{}n a 共有5项，前三项成等差数列，且公差为d ，后三项成等比数列，且公比为q .若第2项等于2，第1项与第4项的和等于10，第3项与第5项的和等于30，则d q −= A ．1 B ．2 C ．3 D ．48. 四棱锥E ABCD −的顶点均在球O 的球面上，底面ABCD 为矩形，平面BEC ⊥平面ABCD ，BC =，1CD CE ==，2BE =，则O 到平面ADE 的距离为A ．13B ．14C D二、多项选择题：本题共3小题，每小题6分，共18分.在每小题给出的四个选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9. 在一次射击比赛中，甲、乙两名选手的射击环数如下表，则下列说法正确的是甲乙87 90 96 91 86 90 86 92 87 95A ．甲选手射击环数的极差大于乙选手射击环数的极差B ．甲选手射击环数的平均数等于乙选手射击环数的平均数C ．甲选手射击环数的方差大于乙选手射击环数的方差D ．甲选手射击环数的第75百分位数大于乙选手射击环数的第75百分位数 10. 已知函数()()sin 2f x x ϕ=+满足ππ33f x f x+=−，且π()(π)2f f >，则A ．1sin 2ϕ=B ．1sin 2ϕ=−C ．()y f x =的图象关于点13(π,0)12对称D ．()f x 在区间π(,π)2单调递减11. 已知函数()(e e )e e x x x x f x ax −−=+−+恰有三个零点123,,x x x ，且123x x x <<，则 A ．1230x x x ++= B ．实数a 的取值范围为(0,1] C ．110ax +>D ．31ax a +>三、填空题：本大题共3小题，每小题5分，共15分.12. 若向量(3,4)=−a 在向量b (2,1)=−上的投影向量为λb ，则λ等于__________.13. 倾斜角为π3的直线经过抛物线C ：212y x =的焦点F ，且与C 交于A ，B 两点，Q 为线段AB 的中点，P 为C 上一点，则PF PQ +的最小值为__________.14. 如图，六面体111ABCDA C D的一个面ABCD 是边长为2的正方形，111,,AA CC DD 均垂直于平面ABCD ，且11AA =，12CC =，则该六面体的体积等于__________，表面积等于_________.1A四、解答题：本大题共5小题，共77 分.解答应写出文字说明、证明过程或演算步骤.15. （13分）已知数列{}n a 满足12a =，12n n a a n −=+（2n …）. （1）求数列{}n a 的通项公式；（2）记数列1{}na 的前n 项和为n S ，证明：1n S <.16. （15分）甲企业生产线上生产的零件尺寸的误差X 服从正态分布2(0,0.2)N ，规定(0.2,0.2)X ∈−的零件为优等品，(0.6,0.6)X ∈−的零件为合格品．（1）从该生产线上随机抽取100个零件，估计抽到合格品但非优等品的个数（精确到整数）；（2）乙企业拟向甲企业购买这批零件，先对该批零件进行质量抽检，检测的方案是：从这批零件中任取2个作检测，若这2个零件都是优等品，则通过检测；若这2个零件中恰有1个为优等品，1个为合格品但非优等品，则再从这批零件中任取1个作检测，若为优等品，则通过检测；其余情况都不通过检测.求这批零件通过检测时，检测了2个零件的概率（精确到0.01）.（附：若随机变量2~(,)N ξµσ，则()0.6827P µσξµσ−<<+=，(22)0.9545P µσξµσ−<<+=，(33)0.9973P µσξµσ−<<+=）17. （15分）如图，以正方形ABCD 的边AB 所在直线为旋转轴，其余三边旋转120°形成的面围成一个几何体ADF BCE −．设P 是»CE 上的一点，G ，H 分别为线段AP ，EF 的中点.（1）证明：GH ∥平面BCE ；（2）若BP AE ⊥，求平面BPD 与平面BPA 夹角的余弦值.C18. （17分）点P 是椭圆E ：22221(0)x y a b a b+=>>上（左、右端点除外）的一个动点，1(,0)F c −，2(,0)F c 分别是E 的左、右焦点．（1）设点P 到直线2:a l x c =的距离为d ，证明2||PF d为定值，并求出这个定值；（2）12PF F △的重心与内心（内切圆的圆心）分别为G ，I ，已知直线IG 垂直于x 轴．（ⅰ）求椭圆E 的离心率；（ⅱ）若椭圆E 的长轴长为6，求12PF F △被直线IG 分成两个部分的图形面积之比的取值范围．19. （17分）记集合{}(),000()()|,()(),,()()f x x D L l x kx b x x D f x l x x D f x l x ∈==+∈∀∈∃∈=R 且…，集合{}(),000()()|,()(),,()()f x x D T l x kx b x x D f x l x x D f x l x ∈==+∈∀∈∃∈=R 且….若(),()f x x D l x L ∈∈，则称直线()y l x =为函数()f x 在D 上的“最佳上界线”；若(),()f x x D l x T ∈∈，则称直线()y l x =为函数()f x 在D 上的“最佳下界线”.（1）已知函数2()f x x x =−+，0()1l x kx =+.若0(),()f x x l x L ∈∈R ，求k 的值；（2）已知()e 1x g x =+.（ⅰ）证明：直线()y l x =是曲线()y g x =的一条切线的充要条件是直线()y l x =是函数()g x 在R 上的“最佳下界线”；（ⅱ）若()ln(1)h x x =−，直接写出集合(),(1,)(),h x x g x x L T ∈+∞∈R I 中元素的个数（无需证明）.2023～2024 学年福州市高三年级4月份质量检测参考答案与评分细则一、选择题：本大题考查基础知识和基本运算．每小题5分，满分40分.1．D 2．C 3．A 4．C 5．D 6．B 7．B 8．A二、选择题：本大题考查基础知识和基本运算．每小题6分，满分18分．全部选对的得6分，部分选对的得部分分，有选错的得0分．9．ABC 10．BC 11．ACD三、填空题：本大题考查基础知识和基本运算．每小题5分，满分15分．12．2− 13．8 14．6，22四、解答题：本大题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15. 【考查意图】本小题主要考查递推数列与数列求和等基础知识，考查运算求解能力、推理论证能力等；考查分类与整合、化归与转化等思想方法；考查数学运算、逻辑推理等核心素养；体现基础性和综合性.满分13分.解：（1）因为12,2n n a a n n −=+…，所以12n n a a n −−=，···································1分当2n …时，112211()()()n n n n n a a a a a a a a −−−=−+−++−+L ，所以22242n a n n =+−+++L ，·························································3分所以(22),22n n n a n +=…，所以2,2n a n n n =+…，··································4分又因为12a =，···············································································5分所以2*,n a n n n =+∈N .······································································6分（2）由（1）可知2*(1),n a n n n n n =+=+∈N ，·············································7分所以()111111n a n n n n ==−++，····························································9分所以11111223(1)(1)n S n n n n =++++××−+L 1111111122311n n n n =−+−++−+−−+L ，·····································11分所以111n S n =−+，·········································································12分又因为1n …，所以1n S <.·································································································13分16.【考查意图】本小题主要考查正态分布、全概率公式、条件概率等基础知识，考查数学建模能力、逻辑思维能力和运算求解能力等，考查分类与整合思想、概率与统计思想等，考查数学建模、数据分析、数学运算等核心素养，体现基础性、综合性和应用性．满分15分. 解：（1）依题意得，0,0.2µσ==，···························································1分所以零件为合格品的概率为(0.60.6)(33)0.9973P X P X µσµσ−<<=−<<+=， ···································································································2分零件为优等品的概率为(0.20.2)()0.6827P X P X µσµσ−<<=−<<+=，·····3分所以零件为合格品但非优等品的概率为0.99730.68270.3146P =−=，···········5分所以从该生产线上随机抽取100个零件，估计抽到合格品但非优等品的个数为1000.314631×≈.·············································································6分（2）设从这批零件中任取2个作检测，2个零件中有2个优等品为事件A ，恰有1个优等品，1个为合格品但非优等品为事件B ，从这批零件中任取1个检测是优等品为事件C ，这批产品通过检测为事件D ，····························································8分则D A BC =+，且A 与BC 互斥，·······················································9分所以()()()P D P A P BC =+·································································10分()()(|)P A P B P C B =+·························································11分221220.68270.68270.31460.6827C C =×+×××21.62920.6827=×，····························································12分所以这批零件通过检测时，检测了2个零件的概率为 ()(|)()P AD P A D P D =···········································································13分 220.68271.62920.6827=× 11.6292= 0.61≈.············································································· 15分答：这批零件通过检测时，检测了2个零件的概率约为0.61.17.【考查意图】本小题主要考查直线与平面平行的判定定理、直线与平面垂直的判定与性质定理、平面与平面的夹角、空间向量、三角函数的概念等基础知识，考査直观想象能力、逻辑推理能力、运算求解能力等,考查数形结合思想、化归与转化思想等，考査直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性.满分15分.解法一：（1）在正方形ABEF 中，连接AH 并延长，交BE 的延长线于点K ，连接PK .···································································································2分因为,G H 分别为线段,AP EF 中点，所以HF HE =，所以Rt AFH △≌Rt KEH △，所以AH KH =，·····························4分所以GH PK ∥.································5分又因为,GH BCE PK BCE ⊄⊂面面，所以GH BCE ∥面.···········································································7分（2）依题意得，AB BCE ⊥面，又因为BP BCE ⊂面，所以AB BP ⊥.又因为BP AE ⊥,AB AE A =I ,,AB AE ABEF ⊂面，所以BP ABEF ⊥面，········································································8分又BE ABEF ⊂面，所以BP BE ⊥，·····················································9分解法二：（1）证明：取BP 的中点Q ，连接,GQ EQ . ·····1分因为,G H 分别为线段,AP EF 的中点，所以GQ AB ∥，12GQ AB =，····························2分又因为,AB EF AB EF =∥，所以,GQ HE GQ HE =∥，·································3分所以四边形GQEH 是平行四边形，······················································4分所以GH QE ∥，··············································································5分又因为,GH BCE QE BCE ⊄⊂面面，所以GH BCE ∥面.············································································7分（2）同解法一.····················································································15分所以GH BCE ∥面.············································································7分（2）同解法一.····················································································15分18.【考查意图】本小题主要考查圆、椭圆的标准方程及简单几何性质，直线与椭圆的位置关系等基础知识，考査直观想象能力、逻辑推理能力、运算求解能力等,考查数形结合思想、化归与转化思想、分类与整合思想等，考査直观想象、逻辑推理、数学运算等核心素养，体现基础性、综合性与创新性.满分17分.解法一：（1）依题意，222b c a +=．··························································1分设00(,)P x y ，则2200221x y a b +=，0ax a −<<，所以2||PF =，所以20||||cPF x a a==−，············································3分又a c >，所以0c a x a >，20a x c >，所以20||c PF a x a =−，20a d x c =−所以0220||c a x PF c a a d a x c−==−，即2||PF d 为定值，且这个定值为ca．··················4分（2）（ⅰ）依题意，00(,)33x yG ，设直线IG 与x 轴交于点C ，因为IG ⊥x 轴，所以0(,0)3xC ，·······················5分所以001202||||()()333x x FC F C c c x −=+−−=，··········································6分因为△PF 1F 2的内切圆与x 轴切于点C ，所以121202||||||||3PF PF F C F C x −=−=，·················································7分又因为12||||2PF PF a +=，解得02||3x PF a =−，··········································8分由（1）得20||cPF a x a =−，所以003x ca x a a −=−，所以椭圆E 的离心率13c e a ==．·························10分（ⅱ）由26a =，得3a =，又13c a =，所以1c =，2228b a c =−=，所以椭圆E 的方程为22198x y +=．······················································11分根据椭圆对称性，不妨设点P 在第一象限或y 轴正半轴上，即003x <…，00y <…，又1(1,0)F −，2(1,0)F ，所以直线PF 1的方程为00(1)1yy x x =++，设直线IG 与PF 1交于点D ，因为03D x x =，所以000(3)3(1)D y x y x +=+，△F 1CD 的面积1S 与△PF 1F 2的面积S 之比为00200100(3)1(1)233(1)(3)118(1)22x y x x x S S x y ++×++==+××，················································13分令2(3)()18(1)x f x x +=+（03x <…），则2(3)(1)(1))(18x x x f x −′++=，当[0,1)x ∈，()0f x ′<，当(1,3)x ∈，()0f x ′>，所以函数()f x 在[0,1)单调递减，在(1,3)单调递增.又因为1(0)2f =，4(1)9f =，1(3)2f =，所以()f x 的值域是41[,]92，所以14192S S ……，··········································································15分所以11415S S S −……，·······································································16分根据对称性，△PF 1F 2被直线IG 分成两个部分的图形面积之比的取值范围是45[,54．········17分解法二：（1）同解法一···········································································4分（2）（ⅰ）依题意，00(,33x yG ，设直线IG 与x 轴交于点C ，因为IG ⊥x 轴，所以0(,0)3xC ，·······················5分所以001202||||()()333x x FC F C c c x −=+−−=，··········································6分因为△PF 1F 2的内切圆与x 轴切于点C ，所以121202||||||||3PF PF F C F C x −=−=，·················································7分又因为12||||2PF PF a +=，得0102||,3||.3x PF a x PF a=+ =− ···········································8分所以00,3,3x a x a =+=−两式平方后取差，得00443cx ax =对任意0x 成立，所以椭圆E 的离心率13c e a ==．························································10分（ⅱ）同解法一···················································································17分解法三：（1）同解法一···········································································4分（2）（ⅰ）依题意，00(,33x y G ，因为IG ⊥x 轴，设点I 坐标为0(,)3xt .··········5分可求直线1PF 方程为00()yy x c x c=++，则点I 到直线1PF 的||t =，·································6分即()222200000()()()3x y c t x c t y x c +−+=++ ，化简得22000002()()()033x xy t t c x c y c +++−+=，①同理，由点I 到直线2PF 的距离等于||t ，可得22000002()()()033x xy t t c x c y c +−−−−=，②············································7分将式①−式②，得00084233t cx y cx ⋅=⋅，则04y t =.·····································8分将04yt =代入式①，得2200001()()()016233y xx c x c c +++−+=，化简得220022198x y c c +=，得229c a =，所以椭圆E 的离心率13c e a ==．························································10分（ⅱ）同解法一···················································································17分19.【考查意图】本小题主要考查集合、导数、不等式等基础知识，考查逻辑推理能力、直观想象能力、运算求解能力和创新能力等，考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想等，考查数学抽象、逻辑推理、直观想象、数学运算等核心素养，体现基础性、综合性与创新性.满分17分. 解：（1）依题意，因为0(),()f x x l x L ∈∈R ，所以2,1x x x kx ∀∈−++R …，且0x R ，20001x x kx −+=+，····················1分令2()(1)1x x k x φ=−+−−，()214k ∆=−−，则()0x φ…，且0()0x φ= ，所以0,0,∆ ∆ ……所以0∆= ，···································································3分即()2140k −−=，解得3k =或1−.··············································································4分（2）（ⅰ）先证必要性.若直线()y l x =是曲线()y g x =的切线，设切点为()00,e 1x x +，因为()e x g x ′=，所以切线方程为()000e 1e ()x x y x x −+=−，即()000e (1)e 1x x l x x x =+−+（*）.························································5分一方面，()()00g x l x =，所以000,()()x g x l x ∃∈=R ，································6分。

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参考答案-、选择题 1 . C 2 . A 3 . C 4 . B 5 . B 6 . A 8 . C 9 . B 10 . A11 . B 12 . A 二、填空题4 •主？ 15 .(理)82 ;(文) 916~3.①②③三、解答题 17.解：T A 、 B > <2是厶ABC 勺三个内角•••A+B+C =Tt•・y = tB Ccos — 2 B C2si n2^~Ccos —2A tg iBC 2(s in cos 2 2 o B C 2cos cos2B . cos sin 2 2tg|+ Atg 2因此，任意交换两个角的位置,18.解：(I )设 X 1 Vx 2 tg2y 的值不变.10则 f(x 1)—f(x2) 2X1 +1)(2 x12x1)(12aL•「X 1Vx 2 , •2为2 又••• a V0,「.1 —2.•.当 a <0时， f (x •••当 a <0时，y = f(n)设P ( x , y )是函数X 1a X i X 22 x 1)—f(x )是R 上的增函数 y =g (x )的图象上的一点,X2V (x 2)v 0 即 fV f ( x 2) 分(x 1) 4P(x , y )关于直线x =1成对称的点为P'(x ',y')x x' 1 则T1y y' 由已知有 • • y = 2y 2-x—1y'=2 —1(x )=2 2-x(川)•••(x)=0=2V 0 x a••• 2X12Xa = 0△= (— 1)2—4a=1—4a>01 1 4a2x■ 1 4a分•••当aVO 时，1V ..1 4a •••2 x = 1一上没有实数解.2 由 2 x = 11 4a 2解得： x =log 2 1_J__4a2•••当aVO 时，f ( x )= O 的解为 x =log -_—_4a 分219. (I)证：取AC 中点D,连ED ••• E 是AB 的中点1• ED // — B'C . 22••• E ' C 丄 AC•DE 丄AC 2 分又'/△ABC 是底角等于30°的等腰三角形 • ED 丄AC 3 分BD A DE = D 「.AC 丄面 BDE• AC 丄BE ，即AC 丄BA' 4 分ED =BD =A D •tg3 0 =2在厶BDE 中，由余弦定理得：2八+ BD —2 ED •2EB = EDBD cos4=2+4—2•△ BDE 是等腰直角三角形， DE 丄BE ED 是异面直线 AC 与BA 的距离为运 8 分（川）连A' D ■/EB =EA '=ED = ,2• A ' D 丄 BD 又AC 丄面BEDA'D 面BED • A ' D ± AC • A ' D 丄面 ABC M A ' D =2V B'— ABC =A ' D=丄】(BD AC) A'D3 2 =8 .3 11 分3V B '—BEC=V C —BEB'=2V CABB' 2V B 'ABC（H ）解：由（I ）小题知/EDB 是二面角B '— AC — B 的一个平面角1)20.解：（I ）2000年底除了危旧住房和新型住房外的其它形式的住房面积为 a -( - l )a -a 1 分4 312每年拆除危旧住房面积为 1 1 a ——a 10 330a a1=(1 4 20%)5 a 12a 一5 a 2= (1 20%)2a 4 12a5 a 3= (1 20%)3a4 12一般地依题意，得： -a ------- ------a3 10 31 2a — a3 301 a 3 a 4分3301 1 1 10 nan5a -(1 20%)—412 a , 亠，、n5a-(1 20%)412 30 a n =a(1 10)(n ),a由- (1 20%) n 5a 4a412an43得— 1.2na 41211) (n n Ig 1.2 Ig 43 Ig 3 Ig 43 Ig 3 n14.372lg 2 Ig 3 1 n 15取 15 105 答：至少再经过5年才能使该地区的居民住房总面积翻两番 .12分21.解：（I ）：a n?2n 1 n当n >2时，a a 3 3a n 1n 1 a 2(n 1} 1 ②由①一②得弐n•'•a n =(a 2—1) n 当n = 1时，由已知得a --a n^( a 1)n(n) S n =( a 2—1) I ① 2na[a 2(n °n 1] a 2(a 2 1)2n — 2・a21=a2n —2・a [1 + 2 (n >2)—1 (n €N) a2+3(4分a 2)2 + 42) 3+-+(n — 1)2 n-2(a ) +n(a2) "J2 2• •a S n =(a — 1)①—②得：(1 — (a 2) T a >0且a 工12— 1^0 n= — [1 + a1 a 2n1 a 2aa 2)S n2、 2 / 2、 )+3(a)2a +2(a2(a —1) [1++ -+(n —1)(2、 2 / 2、 a )+…+(a )2) —1n —1/ 2、 n+ n(a ):—nna 2n2+(a2) 2+ -+(a2）—i(a 2)S n n im(a 2n1)n2na2n0(a 2 1(a 21)（川）（理）本题即a n a n 1 2 a n 2n n 1 n 2• a n an 1a n2 - 212分nn 1 n 2“ 2 八 2n 2“ 2 2n 2 八 2n 2 (a 1)a (a 1)a2(a 1)a (a 2 2n1)a2(1 2 a 2a 4)2n a 2 2(a1)(1 2a 2)(1 a 2) 2 2 2 2n 2(a 1) (1 2a )a 0(a 0) a n a n 1 2 n n 1 a n 2n 214 22.解：（理）（I ）解：设椭圆 C： (x 1)22 笃 1(a b b 2 0)•/2C=2,A • ••右准线方程为设M (x , y ）、连接PB 则| P 2 C =1 x = a 十 P ( x 0, 2 A | +| 0） •(| PA | 2• (2 a )—2・2 y o = ±(+ | PB | y PB | 2 -1) I 2 )-2| PA |・| PB |=4 I =4 AB 2由x a y y 0 消去a ,得y =±（ x — 2)T0v| y 0 |<1,2•・0< a — 1<1, 1< 2 a < 2••2< x <3即M 点的轨迹方程是 y = ±( x —2) (2< x < 3 = 7分 (H)解：设 ZABQ = a , a €(- 4 ，2），则 8 分 | AB |=2,| P A | =| B Q | = 2co s a| PQ | = | AB | —2 | BQ | co s a = =2- -4 c 2 os a• •周长 L =(2—4c :os 2 、 a ) + 4c os a + 2 = — 4(co 10s(a 2 1)当cos a =［，即a =—时，周长L 取最大值5. 一 3 112 此时 | BQ |=1, I AQ |= 3 2 a =| BQ | + | AQ |=1 b 2 2 a 2 2 仝 2 •••所求椭圆C 的方程为(x 1)2 2y— 1 14 分 2 2 一 3 2（文）（I ）解：设 D 为AB 中点，连接 DQ DR PB当丨PQ |= 1时，△ DPQ 为正三角形，边长为 1.从而 △DPA 、A DQE 也是边长为1的正三角形.•••I PA |=1,| PB |= ,3 /•2 a =| PA | + | PB |=1+3 312 2V3• -a=, a2 2又••• c = 1a 2 1y o(a 2 1)消去 a ,得 y =±( x —2) '^0<| y o |<1 ••0< a • •2< x <3即M 点的轨迹方程是 y =±( x — 2) (2< x <3 )14分异—c 2=异—1=2•椭圆C 的方程〔为(x 1) 2 12_y_ 1 6 分2、 3321)2 22(n )解：设椭圆C: (x2；2 1(ab0)abV2 C =2,. C =1•右准线方程为 x 2=a+ 1 8 分又设M ( x , y ), P ( x o , y o )•••| PA | 2+| PB | 221 AB |•(| PA | 2+ | PB | 2)—2| PA |•|即(2a 2)— 2・ 2| y o |=4•- b2.•.y o = ±(a2—1)11 分PB=421<1,1< a <2。