徐州市2017-2018学年度高三年级摸底考试高三摸底数学I试题定稿

绝密★启用前徐州市2017-2018学年度高三年级摸底考试数学I参考公式： 1．样本数据12,,,n x x x 的方差2211()n i i s x x n ==-∑，其中11n i i x x n ==∑；2．锥体的体积公式：13V Sh =，其中S 是锥体的底面面积，h 是高．一．填空题：本大题共14小题，每小题5分，共70分．请把答案填写在答题卡．．．相应位置．．．．． 1．设集合{1,2,3}A =，{2,4,6}B =，则A B = ▲ ．2．已知复数z 满足(1i)i z +=，其中i 为虚数单位，则复数z 的实部为▲ ． 3．函数1()2sin()34f x x π=+的周期为 ▲ ．4．已知一组数据：87,,90,89,93x 的平均数为90，则该组数据的方差为 ▲ ． 5．双曲线2213y x -=的离心率为 ▲ ． 6．从2个黄球，3个红球中随机取出两个球，则两球颜色 的正四棱锥的体积为 ▲ ．9．已知公差不为零的等差数列{}n a 的前n 项和为n S ，且26a =，若137,,a a a 成等比数列，则8S 的值为 ▲ ．（第7题）ABS ECD（第16题）10．如图，在半径为2的扇形AOB 中，90AOB ∠=，P 为AB 上的一点，若2OP OA ⋅=，则OP AB ⋅的值为 ▲ ．11．已知函数()e +1e x x f x -=-（e 为自然对数的底数），若2(21)42)(f x f x +->-，则实数x 的取值范围 为 ▲ ．12．已知实数,x y 满足223x y +=，||||x y ≠，则()()221422x y x y ++-的最小值为 ▲ ．13．已知点P 是圆22:4O x y +=上的动点，点(4,0)A ，若直线1y kx =+上总存在点Q ，使点Q 恰是线段AP 的中点， 则实数k 的取值范围为 ▲ ．14．已知函数32()2f x x x a =--，若存在(]0,x a ∈-∞，使0()0f x …，则实数a 的取值范 围为 ▲ ．二．解答题：本大题共6小题，共计90分．请在答题卡指定区域内作答．．．．．．．．．．，解答时应写出文字说明、证明过程或计算步骤． 15．(本小题满分14分)已知ABC △的内角A ，B ，C 所对的边分别为a ，b ，c ，且22cos a c b A +=．（1）求角B 的大小；（2）若b =，=4a c +，求ABC △的面积．16．（本小题满分14分）如图，在三棱锥S ABC -中，SA SC =，AB AC ⊥，D 为BC 的中点，E 为AC 上一点，且//DE 平面SAB ．求证：（1）直线//AB 平面SDE ；（2）平面ABC ⊥平面SDE ．O ABP （第10题）17．（本小题满分14分）如图，有一块半圆形空地，开发商计划建一个矩形游泳池ABCD 及其矩形附属设施EFGH ，并将剩余空地进行绿化，园林局要求绿化面积应最大化．其中半圆的圆心为O ，半径为R ，矩形的一边AB 在直径上，点C 、D 、G 、H 在圆周上，E 、F 在边CD 上，且3BOG π∠=，设BOC θ∠=．（1）记游泳池及其附属设施的占地面积为()f θ，求()f θ的表达式； （2）怎样设计才能符合园林局的要求？18．（本小题满分16分）如图，在平面直角坐标系xOy 中，椭圆2222:1(0)x y E a b a b+=>>的左顶点为(2,0)A -，离心率为12，过点A 的直线l 与椭圆E 交于另一点B ，点C 为y 轴上的一点.（1）求椭圆E 的标准方程；（2）若ABC △是以点C 为直角顶点的等腰直角三角形，求直线l 的方程.O BC（第18题）（第17题）OCDEFGH19．（本小题满分16分）已知数列{}n a 的前n 项和为n S ，满足21n n S a =-，*n ∈N ．数列{}n b 满足1(1)(1)n n nb n b n n +-+=+，*n ∈N ，且11b =． （1）求数列{}n a 和{}n b 的通项公式；（2）若n n c a =数列{}n c 的前n 项和为n T ，对任意的*n ∈N ，都有n n T nS a ≤-，求实数a 的取值范围；（3）是否存在正整数m ，n ，使1b ，m a ，n b （1n >）成等差数列，若存在，求出所有满足条件的m ，n ，若不存在，请说明理由．20．（本小题满分16分）已知函数()(1)e x f x ax =-(0a ≠,e 是自然对数的底数).（1）若函数()f x 在区间[]1,2上是单调减函数,求实数a 的取值范围；（2）求函数()f x 的极值；（3）设函数()f x 图象上任意一点处的切线为l ，求l 在x 轴上的截距的取值范围．徐州市2017-2018学年度高三年级摸底考试数学I 参考答案一．填空题：1． {2} 2． 12 3． 6 4．4 5． 2 6． 35 7． 4 8．9．88 10． 2-+ 11．(1,3)- 12．3513．4[,0]3-14．[1,0][2,)-+∞二．解答题： 15．（1）因为22cos a c b A +=，由正弦定理，得sin +2sin 2sin cos A C B A =． ···························································2分因为()C A B =π-+，所以()sin +2sin 2sin cos A A B B A +=．即sin +2sin cos 2cos sin 2sin cos A A B A B B A +=， 所以()sin 1+2cos 0A B ⋅=． ····························································································4分 因为sin 0A ≠，所以1cos 2B =-． ················································································6分 又因为0B π<<，所以23B π=．···················································································································7分（2）由余弦定理2222cos a c ac B b +-=及b =22+12a c ac +=，即()212a c ac +-=． ··································································································10分 又因为=4a c +， 所以4ac =， ···············································································································12分所以11=sin 422ABC S ac B =⨯=△·································································14分16. （1）因为//DE 平面SAB ，D E ⊂平面ABC ，平面SAB 平面ABC AB =，所以//DE AB ． ·················································3分 因为D E ⊂平面SDE ，AB ⊄平面SDE ， 所以//AB 平面SDE ． ···························································································6分（2）因为D 为BC 的中点，//DE AB ，所以E 为AC 的中点． 又因为SA SC =，所以SE AC ⊥， ············································································8分又AB AC ⊥，//DE AB ，所以DE AC ⊥． ···························································10分 ,DE SE ⊂平面SDE ，DE SE E =， 所以AC ⊥平面SDE ． ···························································································12分 因为AC ⊂平面ABC ，所以平面ABC ⊥平面SDE ． ····················································································14分 17．（1）由题意，2cos AB R θ=，sin BC R θ=，且HOG △ 为等边三角形，所以，HG R =，sin EH R θ=-， ·····························································2分 ()=ABCD EFGH f S S θ+2cos sin sin )R R R R θθθ=⋅+-2(2sin cos sin R θθθ=-，(0)3πθ∈，．·····························································6分 （2）要符合园林局的要求，只要()f θ最小， 由（1）知，22222()(2cos 2sin cos =(4cos cos 2)f R R θθθθθθ'=----）令()0f θ'=，即24cos cos 2=0θθ--，解得cos θ或cos θ（舍去），·························································10分令00cos 03πθθ⎛⎫∈ ⎪⎝⎭，， 当00θθ∈（，）时，()0,()f f θθ'<是单调减函数， 当03πθθ∈（，）时，()0,()f f θθ'>是单调增函数， 所以当0=θθ时，()f θ取得最小值.答：当θ满足cos θ时，符合园林局要求. ·····················································14分 18．（1）由题意可得： 212a e =⎧⎪⎨=⎪⎩，即212a c a =⎧⎪⎨=⎪⎩，从而有2223b a c =-=，所以椭圆E 的标准方程为：221x y +=．····································································4分191111当2n ≥时，21n n S a =-，-1-121n n S a =-，两式相减得12n n a a -=，从而数列{}n a 为首项1=1a ，公比=2q 的等比数列，从而数列{}n a 的通项公式为12n n a -=．由1(1)(1)n n nb n b n n +-+=+两边同除以(1)n n +，得111n nb b n n+-=+ 从而数列{}n b n 为首项11b =，公差1d =的等差数列，所以=n bn n，从而数列{}n b 的通项公式为2n b n =． ·····························································4分（2）由（1）得12n n c a n -==⋅， 于是221112232(1)22n n n T n n --=⨯+⨯+⨯++-⨯+⨯， 所以2312122232(1)22n n n T n n -=⨯+⨯+⨯++-⨯+⨯两式相减得211212222212nn nn n T n n ---=++++-⨯=-⨯-，所以-12+1n n T n =⋅（）， 由（1）得2121n n n S a =-=-，·················································································8分 因为对∀*n ∈N ，都有n n T nS a ≤-，即-12+1(21)n n n n a ⋅≤--（）恒成立， 所以21n a n ≤--恒成立，记21n n c n =--，所以min ()n a c ≤，············································································································10分 因为1+1(2(1)1)(21)n n n n c c n n +-=-+----210n =->， 从而数列{}n c 为递增数列，所以当=1n 时n c 取最小值1=0c ，于是0a ≤．···················································································································12分 （3）假设存在正整数m n ，（1n >），使1,,m n b a b 成等差数列，则1+=2n m b b a ， 即212m n += ，若n 为偶数，则21n +为奇数，而2m 为偶数，上式不成立.若n 为奇数，设21()n k k *=-∈N ，则22211+(21)4422m n k k k +=-=-+=， 于是212212m k k --+=，即212()12m k k --+=， 当1m =时，1k =，此时=21=1n k -与1n >矛盾；当2m …时，上式左边为奇数，右边为偶数，显然不成立. 综上所述，满足条件的实数对(,)m n 不存在．··························································16分20．（1）函数()f x 的导函数'()(1)e x f x ax a =-+，则'()0f x …在区间[]1,2上恒成立，且等号不恒成立，又e 0x>，所以10ax a -+…在区间[]1,2上恒成立， ·········································2分记()1g x ax a =-+，只需(1)0(2)0g g ⎧⎨⎩……， 即21010a a -⎧⎨-⎩…3?，解得13a …． ···················4分（2）由'()(1)e =0xf x ax a =-+，得1ax a-= ， ①当0a <时，有1(,),()0a x f x a -∈-∞'>；1(,),()0ax f x a -∈+∞'<， 所以函数()f x 在1(,)a x a -∈-∞单调递增，1(,)ax a-∈+∞单调递减， 所以函数()f x 在1ax a-=取得极大值1aa a e --⋅，没有极小值．②当0a >时，有1(,),()0a x f x a -∈-∞'<；1(,),()0ax f x a -∈+∞'>， 所以函数()f x 在1(,)a x a -∈-∞单调递减，1(,)ax a-∈+∞单调递增， 所以函数()f x 在1ax a-=取得极小值1aa a e --⋅，没有极大值．综上可知: 当0a <时，函数()f x 在1ax a-=取得极大值1aa a e --⋅，没有极小值；当0a >时，函数()f x 在1ax a-=取得极小值1aa a e --⋅，没有极大值．·········································································································································10分 （3）设切点为(,(1))tT t at e -，则曲线在点T 处的切线l 方程为(1)(1)()tty at e at a x t e --=-+-，当1a t a -=时，切线l 的方程为1=(1)e =e ata y at a ---⋅，其在x 轴上的截距不存在．当1a t a -≠时，令0y =，得切线l 在x 轴上的截距为11at x t at a -=--+(1)1at a a t at a -+-=--+11a t at a=-+-+1111t t a=-+-+ 1111211t a a t a=-++-+-+，··············································································12分 当110t a-+>时，1111211x t a a t a=-++-+-+112=a a +…， 当且仅当111=11t a t a-+-+，即1=t a 或1=2t a -时取等号；·····························14分当110t a-+<时，1111211x t a a t a=-++-+-+112=4a a -+-…， 当且仅当111=11t a t a-+-+，即1=t a 或1=2t a -时取等号.所以切线l 在x 轴上的截距范围是11,4,a a ⎛⎤⎡⎫-∞-+∞ ⎪⎥⎢⎝⎦⎣⎭.·····································16分。